Some noob questions

[Mr D]

***WARNING***

Extreme noob question alert!

I've only been messing with electronics for about a week and i'm really struggling to get my head around the basics.

I bought myself a Fluke 115 multimeter as a starting point.

So, i grabbed a selectable voltage wall-wart i had lying around and started to measure it.

If i set it to 3V, i measure around 2 amps of current.

If i change the voltage to 12V, i still measure around 2 amps of current.

Sooooo, my first (of many) noob questions is: if 3V and 12V both push 2 amps of current down the wire, why would you need any more then 3 volts?

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And one other quick, forum related question: is there any way to subscribe to a topic, so that i get an email when there's a reply? I can't see this option anywhere when i post or reply.

[Ice-Tea]

Most electric stuff has a certain working voltage. A fan, per example, would like 12V and will not work properly at 3V. On the other hand, an IC may have been designed for 3V3 and will let out the magic smoke at 12V.

Note also that by measuring the output terminals of your wallwart with your DMM set to measure current you are effectively short-circuiting the terminals.

[Mr D]

Thanks, but that doesn't answer the question in my noob-brain.

The current is what does work in your target device, right?

If the current is the same, whether i set it to 3 or 12V, why is 12V necessary?

About your other point: why is short-circuiting the wall-wart a problem?

[Ice-Tea]

Why does a car run on diesel but not on gasoline? Because that was what it was designed for.

But to answer your question: one of the things you're not taking into account is power. 2A@3V gives you 6W but 24W at 12V.

Short-circuiting *anything* is a problem. In essence, you are using an-almost-zero-ohm resistor as a load (aka short circuit). Electronic sources that are well designed will provide their maximum current (in this case 2A) and the output voltage will drop to almost zero (check ohms law and you'll understand why). Less well designed sources will blow up, overheat, be in hickup mode, catch fire, blow the fuse of your DMM...

[Old Printer]

From another noob with a bit more time. Current needs to be flowing, doing something. To measure current (amps) you need to have a circuit (complete) where the current can flow and do work, like light a bulb or turn a motor. All you can really measure like you are doing is voltage (potential).

[Terry01]

YouTube and Google are your friend with all these things buddy.

I'm quite new myself, well around 18 months or there about into my electronics hobby. Compared to some of the guys and girls here I am well new!
I found watching a few different YouTube videos on the same subject I just got 1 way of teaching or the other.
You'll probably be the same yourself.   

[Mr D]

Thanks!

I see the pattern there:

2x3=6
2X12=24

But apart from the pattern, i don't understand why this should be the case.

An amp is (as far as i understand it and loosely speaking) the number of electrons flowing down the wire per second.

So why should 2 amps give you more "power" when being pushed by 12 volts instead of 3?

And just a side question which i guess is related: 1 amp is equal to 1 coulomb, right? And a coulomb is a measure of electric charge travelling down the wire.

So why bother talking about amps, why not just talk about coulombs? What's the difference?

[Mr D]

To try to answer my own question, is it the case that with the 12V supply, if a resistance is introduced, the PS will just push a bit harder to maintain the 2 amps, whereas with the 3V supply, it will run out of steam earlier at a certain level of resistance?

[bsudbrink]

The "water model" is a good starting point.  Think of a garden hose.  If you leave the end of the hose unrestricted, you can have a large volume of water flow, but it will only "jump" a meter or so from the end of the hose.  Now consider a child's "squirt gun" toy.  It can only move a relatively small volume of water, but it can move it close to 10 meters.  The "water hose" case is high amperage, low voltage.  The "squirt gun" case is high voltage, low amperage.

[Mr D]

Thanks, that helps a bit.

But doesn't the analogy break down in the sense that in the squirt gun, the water travels much faster down the thin pipe?

Whereas with electron flow the speed is the same? Or is it not?

Also, does the analogy break down in the sense that in the water example, the diameter of the pipe is also key, but in the metal wire, this i less important?

I'm not trying to be argumentative, just trying trying to understand. I've heard a few different variations of the water analogy, but it only gets me so far!

And how would you describe "power" in the water analogy?

[Terry01]

I think you've got it the paddy way round buddy. Volts=power current=flow of volts.
Current is the flow of volts and volts is the potential energy between 2 points.
Think about filling a sink with a hose(current) the water(volts) then filling a paddling pool with the same hose(current). You'd crank the tap up more when filling the pool compared to the sink but the hose/"current" stays the same but the water(volts) is more powerful.

It'll click soon 

[bsudbrink]

Absolutely, the analogy is very rough.  But it is a starting point.  To go to a real world example, you have probably seen a Van De Graaff generator.  It can make sparks that "jump" through several centimeters of air.  If you have ever "taken a hit" from one, you know it hurts but generally it won't kill you.  They let kids play with them in science museums (with supervision).  On the other hand, most vacuum tube (valve) devices have voltages and currents inside them that can easily kill but they are constructed inside of grounded metal cases with small clearances and no sparks fly.  The Van De Graaff generator is very high voltage and almost no amperage.  The valve device is much lower voltage but much higher current.

[Mr D]

@Terry: i've read quite a few times that voltage absolutely doesn't flow. But you're talking a flow of volts.

[PA4TIM]

You forget the resistance of the load.

There are voltage sources and current sources. If you feed a lamp with 1A from a current source the voltage will be as high as needed to "push" one amp through the resistance of the lamp. If the lamp is 10 ohm the voltage will be 10V. If the lamp is 20 ohm the voltage will be 20V. The source runs out of steam (voltage) if the resistance becomes so high it needs more volt as the source can deliver the current drops. Just Ohms law.

If you use a voltage source and you set the voltage at 50V, the current through the 10 ohm lamp will be 5A and through the 20 ohm amp it will be 2,5A

The wall wart thing is a voltage source. The load (the lamp, a resistor, motor, circuit) will draw the amount of current it needs and as the source it can deliver things are fine. 'If the load draws more current as the voltage source can deliver the voltage drops.  Again Ohms law.

And then there is power. The current times the voltage.

Don't get to attached to the flowing electrons story. 

[rstofer]

Also, does the analogy break down in the sense that in the water example, the diameter of the pipe is also key, but in the metal wire, this i less important?

If I want to push a lot of current through a wire, it will need to be larger than if I want to just push a few milliamps.  Wire has a property of Ohms per 1000 feet (usually).  So, I divide the resistance by 1000 to get Ohms per foot and maybe again by 12 to get Ohms per inch.

Now that I know the resistance, I can calculate the voltage drop for any current I want.  There are standards like 1000 amps per square inch of cross section for substation bus bar but that's a side issue.  Voltage dropped in wire is usually a waste of voltage and creates heat (voltage drop times current given in watts).  I don't want heating (usually) so I use a bigger conductor if this gets out of hand.

I'm not trying to be argumentative, just trying trying to understand. I've heard a few different variations of the water analogy, but it only gets me so far!

And how would you describe "power" in the water analogy?

If we take power as volts (pressure) times amps (gallons per minute) I suppose we can find some units that will ultimately make sense and it will be "Fluid Power" measured in HP after applying a constant.

http://web.applied.com/assets/attachments/779D4407-D2AE-6FAA-7DA1CEDE2268977B.pdf

3rd table entry...

Where the water analogy comes unglued is with inductance.  It sort of works ok for capacitance but, basically, we only use the analogy for voltage, resistance and amps.  It is just a crutch to give people an intuitive sense of what is happening.  It is talked about for the first half hour of the first lecture on DC circuits and never mentioned again.

[PA4TIM]

I think you've got it the paddy way round buddy. Volts=power current=flow of volts.
Current is the flow of volts and volts is the potential energy between 2 points.
Think about filling a sink with a hose(current) the water(volts) then filling a paddling pool with the same hose(current). You'd crank the tap up more when filling the pool compared to the sink but the hose/"current" stays the same but the water(volts) is more powerful.

It'll click soon 

Current is the stuff that makes the lamp burn. It needs a difference in voltage between two points and something between those points to flow through. If the voltage is high enough and the (resistance not to high) current can flow.

[Mr D]

If you use a voltage source and you set the voltage at 50V, the current through the 10 ohm lamp will be 5A and through the 20 ohm amp it will be 2,5A

I don't understand this. The more power-hungry the lamp, the less amps it needs? Seems like the wrong way round?!

Or is it the case that the amps are what's left over after work has been extracted by the resistance?

[rstofer]

In the usual workings of electronics, current is a result, not a cause.  Yes, there are current sources but they are rare unless working with op amps built from discrete components.  I can't recall ever playing with them but I do know they exist.

Usually, I have a voltage source (battery or power supply) and I have a load (as simple as a resistor or as complex as a PC) and the result of operating the load on the source results in a certain current flow.  The current is the result, not the cause.

It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will.  When you get to Kirchhoff's Laws, you will be writing equations in terms of current flow or voltage drop.  This is where it will all make sense.  In fact, it isn't even possible to analyze a simple op amp circuit without writing Kirchhoff's Current Law at the inputs (explicitly or just winging it).  It's simple to write (in the case of linear feedback, less so if the device is used as an integrator) but key to understanding the closed loop gain is calculated.

Get the idea that water pressure is kind of like voltage, pipe friction resistance is kind of like electrical resistance and water flow rate is like current and then put the analogy away.  It was just created to give an intuitive feeling toward the 3 fundamental units.

[PA4TIM]

Current sources are not very rare, but voltage sources are much more common.  Some loads need a current source, like a led, because they will draw more current then they can handle. Other parts will need a voltage. The voltage divided by the resistance will be the current draw. And the voltage times the current gives the power the load dissipates and that should be not more as the load can handle or the smoke escapes.

[rstofer]

If you use a voltage source and you set the voltage at 50V, the current through the 10 ohm lamp will be 5A and through the 20 ohm amp it will be 2,5A

I don't understand this. The more power-hungry the lamp, the less amps it needs? Seems like the wrong way round?!

Or is it the case that the amps are what's left over after work has been extracted by the resistance?

No, this is bass ackwards!

Current is a result in most cases.  A lamp will have a rated voltage and a rated power.  Let's say 100W at 120V...  So from P=E^2 / R (E squared over R), we can calculate that the hot resistance is 144 Ohms.  We can calculate the hot current from P = I^2 * R or I = sqrt(P / R) or, in this case, 0.8333 Amps.  We can also use P = I * E to get watts so 100 = 0.8333 * 120 (which it does).  All of the manipulations of Ohm's Law should be consistent,  P, I, E and R are all tied together.

So, what is really happening?  We apply a voltage to a lamp.  The internal resistance (assume hot) connected to a certain voltage will produce a certain current flow.  That current flow times the voltage will result in a certain amount of power being dissipated.  Mostly, it will come off as heat in the case of incandescent lamps.

What I have avoided saying is that the incandescent lamp changes resistance as the filament heats  up.  The value is much lower when cold.  You can just grab a 100W lamp and measure the cold resistance if you are interested.  I suspect it will be quite low and nowhere near 144 Ohms.  But I don't want to go there...

If you really want to chase the topic down a rathole, calculate the instantaneous current when you apply 120V to that cold filament resistance.  There are a lot of reasons why the current may not go quite that high but it does explain why lamps burn out when they are switched on.

[PA4TIM]

I don't understand this. The more power-hungry the lamp, the less amps it needs? Seems like the wrong way round?!

10 ohm draws more current as 20 ohm when the voltage is the same. 20 ohm is less powerhungry as 10 ohm. 1 ohm is even more power hungy.
10V / 10 ohm = 1A , this is 10V x 1A = 10W
10V / 20 ohm = 0,5A, this is 10V x 0,5A = 5W, so less power

It will become even more fun if you get to AC and paracitics 

[Mr D]

My brain hurts.

Just one more before i go to bed:

I don't get it. You're talking about a load drawing the current. The greater the load, the more current.

But at the same time the circuit has resistance.

So it's pulling and pushing back at the same time??

[rstofer]

In the usual workings of electronics, current is a result, not a cause.  Yes, there are current sources but they are rare unless working with op amps built from discrete components.  I can't recall ever playing with them but I do know they exist.

Usually, I have a voltage source (battery or power supply) and I have a load (as simple as a resistor or as complex as a PC) and the result of operating the load on the source results in a certain current flow.  The current is the result, not the cause.

It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will.  When you get to Kirchhoff's Laws, you will be writing equations in terms of current flow or voltage drop.  This is where it will all make sense.  In fact, it isn't even possible to analyze a simple op amp circuit without writing Kirchhoff's Current Law at the inputs (explicitly or just winging it).  It's simple to write (in the case of linear feedback, less so if the device is used as an integrator) but key to understanding how the closed loop gain is calculated.

Get the idea that water pressure is kind of like voltage, pipe friction resistance is kind of like electrical resistance and water flow rate is like current and then put the analogy away.  It was just created to give an intuitive feeling toward the 3 fundamental units.

[bsudbrink]

Nothing "pulls".  Everything "pushes".  Except maybe inductors but don't try to understand those yet.

[PA4TIM]

A lower resistance is a higher load.
In Dutch to make it more clear for the TS: "Iets belast een voeding meer, de belasting is hoger omdat de weerstand lager is. Er kan dus meer stroom lopen, dat maakt de boel warmer en vormt zo een zwaardere beslasting.

https://www.youtube.com/channel/UCOausWDNRDJikQ11gSLj7nA , this man is a Very good teacher

[rstofer]

Dave has a playlist of Fundamentals videos.  w2aew has some great videos on circuits.  Both of these sources are somewhat beyond Ohms Law but they are both excellent resources.

Digilent has an entire curriculum:
https://learn.digilentinc.com/classroom/realanalog/

MIT has OpenCourseware
https://www.google.com/search?q=mit+opencourseware+electronics+video+lectures

Khan Academy has an Electrical Engineering track but they are best known for their math lectures:
https://www.khanacademy.org/welcome?learn=1

I can't begin to list all the excellent resources, there are tutorials all over the Internet.

[Mr D]

In Dutch to make it more clear for the TS: "Iets belast een voeding meer, de belasting is hoger omdat de weerstand lager is. Er kan dus meer stroom lopen, dat maakt de boel warmer en vormt zo een zwaardere beslasting.

Haha, thanks, i can read it but i'm actually English!

Thanks all, now to bed. Sorry for coming across as a dumb-ass, i have zero background in this stuff.

I can't promise not to resurrect this one tomorrow!

[rstofer]

My brain hurts.

Just one more before i go to bed:

I don't get it. You're talking about a load drawing the current. The greater the load, the more current.

But at the same time the circuit has resistance.

So it's pulling and pushing back at the same time??

Load isn't a fundamental unit of electricity.  Volts, amps and resistance are fundamental units.

Intuitively, we think of a heavy load as drawing more current than a light load and for equal voltages, we get to the idea that a heavy load draws more current and therefore must have less resistance because, for a fixed voltage, current and resistance have an inverse relationship.  E = I / R   Hold E constant and to increase current, you need to decrease resistance.  Increase resistance and current must decrease.

We could also consider load in terms of power and this would allow for different voltages as well as different resistances and therefore different current.  We don't need to go there.

Ohm's Law does it all!
https://www.electronics-tutorials.ws/dccircuits/dcp_2.html

[rstofer]

Haha, thanks, i can read it but i'm actually English!

Thanks all, now to bed. Sorry for coming across as a dumb-ass, i have zero background in this stuff.

I can't promise not to resurrect this one tomorrow!

Going to bed at midnight?  That's prime time for doing homework as I recall!
Don't worry about resurrecting the topic.  It is the most fundamental discussion you will ever have on the topic and it needs to be nailed down or you simply can't move forward.

[Jwillis]

The water analogy is correct if explained  correctly .
Think of a tank of water with a pipe attached to the bottom. The amount of water in the tank is the "charge". The pressure of the water is called "volts " .The more pressure the higher the voltage. The flow or rate that the water moves  is called "Amps" or Current. The size of the pipe is called resistance. The bigger the pipe the less resistance. Now voltage and current are dependent on each other. No pressure on the pipe no flow or current. Put a plug in the pipe Even with increased pressure there is no flow.

Now lets consider your circuit is a paddle wheel at the output of the pipe.The paddle wheel can only handle so much flow before the water pours over the sides. This is the flow rating or current rating .no mater how much water is dumped into the paddle wheel it will only except so much.Lets say 1 amp . So your paddle wheel requires a certain amount of pressure and flow for it to move but also has a maximum pressure and flow before it breaks.
You set your pressure an flow at a required amount to make the paddle wheel move. Just like 3 volts and  1 amp.Every thing works fine.
Now lets say the flow remains the same but you increase the pressure .12 volts at 1 amp and like dirt in front of a pressure washer your paddle wheel  promptly explodes and flies apart .

Just like a circuit only excepts a certain amount of flow or amps .If you increase the pressure or voltage beyond its rated requirements it will burn out .A circuit .when working correctly ,will only take so many amps no matter what the voltage is. This is also why voltage is the dangerous part of electricity. Very high voltage with very small amps will kill where as very high current and very small volts will do little.

Now lets consider the size of pipe .Just like wire ,the bigger the pipe or wire the less resistance there is.

Hope this helps

[Mr D]

Thanks, i appreciate you taking the time, it helps.

One thing i don't understand though:

Imagine you were designing a complicated plumbing system in a building, moving water around the building through various pipes, valves and tanks at various pressures.

Say an alien landed and asked you to explain the physics of the system to him.

You wouldn't start to talk about analogies with electricity, volts, current etc, right?

So, when talking about electricity, why isn't it possible to explain it in terms of what it actually physically is?

Why is it always necessary to explain via analogy?

This is not meant as a criticism of your attempt to explain it. I just don't understand why it's never explained without resorting to analogy!

   

[agehall]

Why is it always necessary to explain via analogy?

It is not.

HOWEVER, going down the physics route to explain quickly becomes very complicated and if you understand that, you probably would not be asking the question in the first place. So the easiest way to get a good understanding of what is happening, it is usually easier to use an analogy to something most people already know.

[Mr D]

The flow or rate that the water moves  is called "Amps" or Current.

Ok, but what is being measured here? The physical speed that the current flows? Because water can move at different speeds. Can charge also move a different speeds?

Or is it the amount of electric charge per second that passes a certain point?

Because in the water analogy, you could have the same volume of water moving past a fixed point, whether it's a wide, deep, slow moving stream, or a 1 meter diameter pipeline at extremely high pressure.

[agehall]

Now you are getting into things way beyond basics.

Current is charge moving over time. Wikipedia does a good job of describing what it is. https://en.wikipedia.org/wiki/Electric_current

[Mr D]

Going down the physics route to explain quickly becomes very complicated and if you understand that, you probably would not be asking the question in the first place. So the easiest way to get a good understanding of what is happening, it is usually easier to use an analogy to something most people already know.

OK, but when talking about flow of water, you're not really explaining it at a fundamental level, in terms of molecules and atoms, you're also generalizing.

So what is it about electricity that is so fundamentally less intuitive, that makes it so hard to understand, one needs to resort to an analogy of water?

[Ice-Tea]

So what is it about electricity that is so fundamentally less intuitive, that makes it so hard to understand, one needs to resort to an analogy of water?

Have you seen water? I would assume so. Have you actually seen electricity? I would think not.

So, there

[PA4TIM]

It is possible to explain it on a fundamental level but that makes it very complex unless you are an ace in Physics.
Now we only talk about DC voltage and current.
The water model represents the functional model. Water as current is more or less usable to explain the effect.

In real life electrons are travelling from minus to plus, besides that they travel very slow while the energy is moving very fast. Then there are electric and magnetic fields, static or dynamic. Water won't induce a water flow in a pipe next to it. A current flow can do that. Water will flow in the pipe, DC current does, AC, too but AC can also move on the surface of the wire and to make it even more complex, move outside the wire and that brings us to transmission lines and things like wave behavior that leads us to quantum mechanics. There is no way to understand that straight away and the amount of things you need to know is huge. There are only a hand full of people on this forum who really know the deepest fundamentals.

I had the same problem when I started. I started with radio and every time I got my head around something "that opened a door" it turned out there were even more closed doors after that. And then is the problem, what path to follow. There is not one path, there are many running parallel and they are forking out constantly too. You need a very high drive to follow a lot of paths parallel

Now you talk about a simple restive load and that is already hard, but in real there is not just resistance, there is "complex resistance" called impedance and split out in reactance and resistance. You need to know this. But to know this you need to know a lot about inductance and capacitance. A real resistor also has inductance (and capacitance to other parts of the circuit) A trace on a pcb is a resistance for DC but has transmission line behavior for AC. Without that knowledge you can understand a lot of things practically, the rest is called voodoo by many and they can do a lot without knowing that part. Knowing that makes life more easy.
And that also brings in a lot of math

You are now struggling to learn maybe 0,001% of the fundamentals. By far the easy part and a lot of people do not understand more then this and still can do a lot.   

[hamster_nz]

Maybe a car analogy would help (not very likely, but I'll try  ).

Think of a car as being an electron (all identical cars, because all elections are exactly identical), and motorways are wires. You pick a reference point across the motorway and count each car as it passes, and how fast it is traveling.

In this analogy, current is the count of cars crossing that line in a standard unit of time. (this is a good analogy, as one amp is 6.242×10^18 electrons passing in one second).

Voltage could be thought of as the average speed of cars (this part isn't a very good analogy, as voltage is more a 'difference in pressure' between two points).

But what is clear in this analogy is that the two things (count of cars & average speed of cars) can be independent of each other - you can have a few cars moving slowly (low voltage and low current), or a few cars moving very fast (high voltage and high current), or you could lots of cars moving slowly (high current, low voltage), or lots of cars moving fast (high current high voltage). It all depends on the local conditions at the time - there might be heavy fog with snow and ice, or it might be a clear day.

So when you measure 2A on your plugpack, that is measuring 12,484,000,000,000,000,000 electrons entering the meter every second (and the same number exiting too).

The confusing bit is the voltage, and it is not just because of my broken analogy. The plugpack is engineered such that it will at most only let a given number of electrons flow per second, regardless of the voltage setting. So should the plugpack is set to 12V and more than 2A wants to flow, then the electronics in the plugpack reduces the output voltage (the energy per electron) until only 2A will flow.

[PA4TIM]

http://wiki.c2.com/?SpeedOfElectrons
Electrons moving through a cable is a big simplification on its own.  There is a drift speed and a thermal velocity. The electrons move slow but the charge moves fast.
From the link:

For example, for a copper wire of radius 1 mm carrying a steady current of 10 Amps, the drift velocity is only about 0.024 cm/sec!

[alanb]

Mr D. It sounds from your original question that you are connecting your multimeter directly to the power supply whilst in the current mode. This is something that you should not do. In current mode you should connect the meter in series between the power supply and the load. With the meter connected directly to the power source in current mode the current is only limited by the maximum current capability of the supply and this can blow the fuse in the meter or worse. In your case the maximum current from the supply appears to be 2 amps thats why you get the same result on both voltage ranges.

[rstofer]

All of these complications with the analogy is why it is used for the first half hour of the first class and then put away, never to be discussed again.

All that is hoped for is an understanding that volts pushes current through resistance.  The analogy should stop there because all the other electrical effects can't be described with the analogy.  I had forgotten about mutual inductance, mentioned above.  That clearly doesn't happen with water in a hose.

It is sufficient to understand that higher pressure in a hose will result in more water flow.  Add to that the fact that more water will flow through a big pipe than a small pipe for the same pressure and you have just run out of the analogy.  We're done, it's time to move to Ohm's Law and start playing with DC circuits.

OTOH, it isn't necessary to understand the physics behind electron flow.  At least not until AC circuits and probably not until RF or semiconductor physics.  Ohm's Law and Kirchhoff's Laws are sufficient for most anything a hobbyist is likely to run across unless they are into amateur radio.

The analogy is useful in that most people have played with a garden hose but beyond the trivial pressure, flow rate, pipe resistance, the analogy isn't helpful.  Even flow resistance isn't a linear kind of thing.  There are boundary conditions, laminar and turbulent flow and other issues that are not appropriate for consideration in the analogy.  That's why there is the entire study of fluid dynamics.  I wonder if they use the electrical analogy for the first five minutes of the discussion.

[asmfan]

It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will. 

For all practical purposes this statement is a useful simplification, and holds true when working with ohmic materials.
This simplification does not work when working with non-ohmic materials such as when experimenting with static electricity.

Usually you see Ohm's law written in the following form:

V = I x R

However, in this form it is tempting to define voltage in terms of resistance and current. It is important to realize that R is the resistance of an ohmic material and is independent of V in Ohm's law. In fact, Ohm's law does not say anything about voltage; rather, it defines resistance in terms of it and cannot be applied to other areas of physics such as static electricity, because there is no current flow. In other words, you don't define voltage in terms of current and resistance; you define resistance in terms of voltage and current. That's not to say that you can't apply Ohm's law to predict what voltage must exist across the a known resistance, given a measured current. In fact, this is done all the time in circuit analysis.

- Paul Scherz and Simon Monk. "Practical Electronics for Inventors" Fourth Edition pg 24

[rstofer]

It doesn't matter much because Ohm's Law relates all 3 variables and we can swap them back and forth at will. 

For all practical purposes this statement is a useful simplification, and holds true when working with ohmic materials.
This simplification does not work when working with non-ohmic materials such as when experimenting with static electricity.

And perfectly adequate for the purposes of a newbie thread.  There is no point in expanding the discussion at this point in the learning process.  Close enough...

[Mr D]

Many thanks to all for being exceedingly patient with me, i'm gonna take my time and read through this thread a few times.

I want to ask one more favour. Could someone tell me the values of a few resistors i should buy that i can use to connect to my Fluke and power source so i can experiment with calculating ohms law? I guess the resistance should be in a range that it'll be clear on my multimeter, but not so low that i risk damaging my power supply or Fluke.

And does anyone have some links to a good lab power supply that will be useful for testing Ohm's law? Doesn't have to be very cheap, but i also don't want to spend more than necessary. There's so many available, i have no idea what to get.

[Rick Law]

The flow or rate that the water moves  is called "Amps" or Current.

Ok, but what is being measured here? The physical speed that the current flows? Because water can move at different speeds. Can charge also move a different speeds?

Or is it the amount of electric charge per second that passes a certain point?

Because in the water analogy, you could have the same volume of water moving past a fixed point, whether it's a wide, deep, slow moving stream, or a 1 meter diameter pipeline at extremely high pressure.

Mr. D,

We need to back up a bit here.  Your wall-wart has the potential to supply energy to whatever you plug the power output into.

That potential to provide energy is measured in Voltage and Current it can supply to the load. Voltage is measured in V (volts), and Current is measured in A (Amps or Amperage).  Current is generally denoted as I.  When your output is merely your DMM, the DMM itself is a load.  This is actually unsafe, but lets ignore that safety issue for the time being and just understand what went on.

* When the wall-wart is set to 3 Volt.  The 2A reading means the DMM is taking up 2Amps when the DMM itself is the load. 2A*3V=6Watt, it is using up 6 watts of power.  The DMM is a heater putting out 6 Watts of power somewhere inside the DMM.
* When the wall-wart is set to 12 Volt.  The 2A reading means the DMM is again taking up 2Amps when the DMM itself is the load. 2A*12V=24Watt, it is using up 24 watts of power.  This is a lot of power, when applied for enough time, it may melt the whole darn thing.

Even at 6 Watt, it is a lot of heat to dissipate.  Your DMM can cook itself.  Since your DMM so far survived, lets continue to see what is going on.

Case 1:
(Your DMM is set to measure current)

Imagine you have a FAN connected to that output with the DMM in-line (in series) like this:
  power+  to  DMM (red)
  DMM (black)  to  FAN+
  FAN-  to  power-

Now your DMM is measuring the current flowing through the DMM itself.  Since the DMM is in series with the FAN, the current is the same as the current flowing through the FAN.  (Just as a garden hose, the amount of water going through the hose is the same as the amount of water going through the nozzle.)

Case 2:

(First make sure your DMM is set to measure voltage)
On the other hand, if you have the FAN and the DMM connected in parallel and then to power like this:
 power+  to  DMM(red) and DMM(red) is also connected to FAN+
 power-  to  DMM(black) and DMM(black) is also connected to FAN-

Now, your DMM is measuring the voltage being supplied by the power.  Consider voltage (aka potential) is the pressure (how hard it is pushing).  The higher the voltage (the harder it pushes), the more current would flow assuming the darn thing didn't burst.  The amount of energy being delivered would be V*I (ie:  Wattage = voltage times current, recall, current is denoted as I).  With your DMM, it seems pushing at 3V or 12V results in the same current (2Amps), so it was likely designed limit - just as the diameter of a garden hose limits the amount of water that can flow.

Why your initial test was unsafe:
If you understand what is described thus far, I can explain why your initial connection (with DMM measuring current directly connected to the output) is unsafe.  Without the FAN in between, nothing is there to use the power.  So, the DMM is taking the full load.  Good that your DMM limits that to just 2A - it is still not good for long, but it survives for the short duration of heating.

Case 1 and 2 wrap up:
Now if you want to try that yourself, find a FAN that operates in 12V and 3V - or use an incandescence light bulb such as a bulb for your car.  Make sure it is incandescence bulb.  (LED bulbs designed to function at specific voltage.  It would not do well in our voltage-changing experiment.)   When you connect the bulb at 12V, you should see it lights up - you can measure the current (when in series as depicted in case1 above), or you can measure the voltage (when in parallel as depicted in case2 above), now you can evaluate the Wattage the bulb is putting out.

Incandescence auto bulbs (as resisters) is a good way for your experimentation until you can buy some real resisters.  Remember, resisters (bulb) will heat up.  A 24watt 12volt bulb will dissipate the heat properly as long as you don't over-volt the darn thing by applying more volt than it can take.

Depending on where you are, mail order may be the best way to get some real resisters.  Tayda is good place to get electronic parts.  Very few real resisters would take 24watts (as your initial connection).
https://www.taydaelectronics.com/

If you understand the post thus far, you can dig into Ohms law where you can begin to figure our how V=I*R works.


Case 3:

If you got this far, an interesting thing to see would be to remove the light bulb with your DMM (set to voltage) connected to the power alone.  This is the NO-LOAD voltage put out by your wall-wart.  Depending on design, this should be higher than when the FAN was also connected.

Footnote:  when in series, the DMM actually caused a very small voltage drop, but that is too deep a lesson for now.

[rstofer]

Many thanks to all for being exceedingly patient with me, i'm gonna take my time and read through this thread a few times.

I want to ask one more favour. Could someone tell me the values of a few resistors i should buy that i can use to connect to my Fluke and power source so i can experiment with calculating ohms law? I guess the resistance should be in a range that it'll be clear on my multimeter, but not so low that i risk damaging my power supply or Fluke.

I bought an assortment kit from Jameco many years ago.  Maybe something similar is available to you?
https://www.jameco.com/z/00081832-540-Piece-1-4-Watt-5-Carbon-Film-Resistor-Component-Kit_81832.html

100, 220, 330, 470, 1k, 2.2k, 4.7k, 10k, 22k, 47k, 100k all 5% ought to cover it.  Notice how the values are just 10x multiples of the original sequence?  That's a property of common resistors

https://ecee.colorado.edu/~mcclurel/resistorsandcaps.pdf

And does anyone have some links to a good lab power supply that will be useful for testing Ohm's law? Doesn't have to be very cheap, but i also don't want to spend more than necessary. There's so many available, i have no idea what to get.

I didn't have a lab type supply for decades.  I used batteries or wall warts and occasionally a dedicated open frame power supply.  I finally broke down and bought the Rigol DP832.  It is not inexpensive and there are many cheaper alternatives but it works really well and I'm glad I bought it.

A 3 output supply is handy when working with dual rail op amps where you need +15V, -15V and maybe 5V for some kind of supporting logic.

Having a supply with adjustable current limiting saves on parts.  It does no good to have an analog knob with no display of the limit value, you can have it set too high and not even know it.  A true lab supply has an indication of the current limit value.  Very important feature.

You can search the forum for 'power supply' and find MANY threads.  A lot of people want to build their own and there are a number of users recommending it.  Maybe so...  I buy power supplies, the only one I built was an old Heathkit back around '70.  There are some power supplies on eBay if that's a help.  Some are older HP supplies and these are probably quite good.  There are also some Chinese versions of dubious quality.  So, prices are from cheap to expensive with features to match.

Of course, I have built the project level supplies with LM7805s and things like that.  They were specific to the project, not general purpose.

[rstofer]

Why your initial test was unsafe:
If you understand what is described thus far, I can explain why your initial connection (with DMM measuring current directly connected to the output) is unsafe.  Without the FAN in between, nothing is there to use the power.  So, the DMM is taking the full load.  Good that your DMM limits that to just 2A - it is still not good for long, but it survives for the short duration of heating.

Well, we know the 2A number but we really don't know the voltage at the DMM.  The power supply has some internal resistance (known later in your studies as the Thevenin Equivalent Resistance) so some of the voltage was dropped internally and some of it was dropped in the meter.  We don't know the resistance of the current measuring circuit in the DMM but it's going to be LOW.  Say it's 0.1 Ohm...  If so, the meter was only dissipating 0.4W (I squared times R).  The rest of the voltage was dropped internally within the power supply and there are a number of ways that can occur and not all of them result in excess heating.

Why do we care about this?  Well, Thevenin is going to play an important part in circuit analysis but more important, it's going to lead us to the fact that we need 2 DMMs (minimum).  It does no good to measure the current through a device unless we can simultaneously measure the voltage across it.  The power supply may be the limiting factor, not the device.

This is all a lot of fun.  Ohm's Law, Kirchhoff's Laws, Thevenin's Theorem - these are the tools of circuit analysis.  You go nowhere without understanding them.

[Rick Law]

Why your initial test was unsafe:
If you understand what is described thus far, I can explain why your initial connection (with DMM measuring current directly connected to the output) is unsafe.  Without the FAN in between, nothing is there to use the power.  So, the DMM is taking the full load.  Good that your DMM limits that to just 2A - it is still not good for long, but it survives for the short duration of heating.

Well, we know the 2A number but we really don't know the voltage at the DMM.  The power supply has some internal resistance...
...
...
This is all a lot of fun.  Ohm's Law, Kirchhoff's Laws, Thevenin's Theorem - these are the tools of circuit analysis.  You go nowhere without understanding them.

Don't scare the guy...  Let him in gently.  I think he has to figure out a lot more before he gets there.

I merely want to let him know he can't stick the probes into just anywhere when it is set to current...  Like sticking it into the AC outlet.

Now we can relabel the DMM settings as Crispy and Extra Crispy.

[rstofer]

Don't scare the guy...  Let him in gently.  I think he has to figure out a lot more before he gets there.

I merely want to let him know he can't stick the probes into just anywhere when it is set to current...  Like sticking it into the AC outlet.

Now we can relabel the DMM settings as Crispy and Extra Crispy.

One thing we all know: If you try to measure volts with the probes in the current position, the fuses may blow.  Or the meter may just do an imitation of a hand grenade.

The user just has to pay careful attention to where they put probes and what they expect to have happen.  LOOK at the meter and verify the test lead connections BEFORE sticking a probe somewhere.  Some meters (like my Fluke 189) will flash a warning if the probe position isn't correct for the range selected.  That's a nice feature

There was a time before meters had fuses and CAT ratings are kind of a new thing as well.  In spite of that, many of us are still here.

NOTE:  The EEVblog Brymen BM235 warns when the test leads are in the wrong position.  So does the new EEVblog 121GW.  These are professional meters.

The Aneng 8008 does NOT warn of incorrect test lead connections.  Apparently, it is expendable.  As is the user...  I like the meter and use if in preference to my other meters but I do need to be aware of the limitations.  Since I work on low energy projects, it's not much of a risk.

[Brumby]

Firstly, let me address the water analogy.

It is a very good analogy when taken as a guide to help your thinking, but - like all analogies - it will break down if you try and take it too far.  This is what you are doing when you start talking about the speed of the water flow.  You will get caught up with details that simply are not part of the analogy and you will get totally confused.

It would be better (not perfect) to think of the quantity of water flowing - not the speed.


Secondly, you absolutely have the right idea of wanting to understand Ohm's Law!!  Once you do, the water analogy will make a lot more sense and you will probably see where a few of its limitations arise - but you will also be less likely to need it.


Third - but perhaps the most important at this juncture - DO NOT put your meter into current mode and connect it directly across a power source unless you really have to and already know what to expect.  It can be very dangerous and it is not a habit that you want to develop.  To understand why, you really need to know Ohm's Law and apply it in the real world.

[Brumby]

The first thing to understand is that an 'ideal' multimeter measuring current will have zero resistance across its terminals which means it will allow as much current to pass as the circuit it is in can provide.  Such an uncontrolled flow can be a source of danger.  In a real multimeter, however, this resistance will not actually be zero but it will be as close to zero as the designer can get - so it is always wise to think of your multimeter (in current mode) as being a simple piece of wire between between one probe and the other.

Now imagine what would happen if you stuck a piece of wire into the active and neutral of a mains power point that was switched on!!!  Some fireworks, a blown fuse or both - as well as a risk of injury to yourself.  This is the sort of problem that can happen when you don't use measuring equipment properly.

In the case of your wall wart, this is exactly the sort of thing you were doing - but you were fortunate in that things were on a smaller scale and the meter was designed to be able to carry that current.  I would also guarantee you were overloading the wall wart and if you had left it connected it would have heated up very quickly and very likely have been destroyed.

To understand why you kept getting 2A, I will offer you the following very simplified diagram.  This represents your wall wart - but only in a very basic sense.  It is missing a lot of detail that you would find in a real device, but it shows why you would get the same current reading.  To make the maths simpler, I will use the 'ideal' current measuring meter that has zero ohms between the probes.



The story goes like this...
As shown, there are four sections of the transformer that generate 3V each - but the wire that is used in each section has a resistance of 1.5 ohms.
 * When you select 3V, there is only 1 section used and it has a resistance of 1.5 Ω.  Using Ohm's Law V=IR, you get a current of 2A.
 * When you select 6V, there are now two sections used and while this gives you 6V, the resistance across the two sections is 3 Ω.  Again, using Ohm's Law, you get a current of 2A.
 * When you select 9V, there are now three sections used and while this gives you 9V, the resistance across the three sections is 4.5 Ω.  Once more using Ohm's Law, you get a current of 2A.
 * When you select 12V, there are now four sections used and while this gives you 12V, the resistance across the four sections is 6 Ω.  Yet again using Ohm's Law, you get a current of 2A.

This resistance is buried inside the wall wart and something you can't really do anything about it, other than include it in your calculations.  This sort of internal resistance has a name: "Equivalent Series Resistance" or ESR and it is found in every electrical and electronic device - unless it happens to be a superconductor.

It is this ESR and how it is distributed through the wall wart that gives rise to the "constant 2A" phenomenon you observed.


For those wanting to challenge the above, I would like to add that there are several ways of providing a multi-voltage wall wart - and the above is only showing one of those in a very basic conceptual form, purely for the purposes of education, not as a construction project.

[ArthurDent]

The wall-wart that you have should have a rating label on it that will tell you what the voltage and safe current rating of the supply is. It sounds like this is a supply that no matter what voltage selection you choose, the short circuit current is limited to 2 amps. You are lucky that it is limited because you never should try to measure the output of an unknown supply by setting a DMM to current and measuring directly across the supply output. A lot of wall-warts would be destroyed by what amounts to a direct short across the output.

Imagine what would happen if you tried this by setting the DMM to current and plugging the leads into a wall outlet. I believe your mains are 230 VAC and with enough current to run appliances. If you plugged the DMM leads into a wall outlet it would blow the DMM’s internal 11A fuse, trip a breaker, or create an arc and burn the probe ends; or some combination of the above. 

As to the confusion about why 2 amps isn’t adequate for everything, what does the work is power, not current or voltage alone. A hair dryer has far different power requirements than a LED nightlight but both may be designed to operate from the same mains voltage.

An automotive bulb may be designed to operate at 12V/1A and an incandescent bulb for your house may be designed to operate at 230V/1A. Both bulbs are designed to load the proper supply at 1A but if you try to run the 230V/1A bulb on 12V it will have too high a resistance at 12V to even glow and will draw very little current while the 12V/1A auto bulb will brightly flash once and be destroyed on 230V because its resistance is far too low for the mains voltage and it will try to draw a very large current and burn out.

To measure current a DMM is generally placed in series with a load like a 12V/1A auto bulb to measure the actual current the bulb draws. A DMM set to current has a very low resistance between the leads so it doesn’t have a noticeable effect on the voltage being delivered to the load. To accomplish this, internally a DMM measures the voltage drop across a short heavy calibrated wire called a ‘shunt’ and the voltage drop across this shunt is displayed as current, and the total series resistance of the DMM leads, the shunt, and fuse is probably a fraction of an ohm. If you are measuring a 1A load, the voltage drop across the DMM is probably a couple of tenths of a volt. For instance IxR=E (the voltage drop across the DMM) might be 1Ax.2ohm=0.2V drop so if you have a 12V supply connected to a 1A load with the DMM reading current placed in series with the load, the load will see 12V-0.2V or 11.8V and the DMM will display about 1A or whatever the actual current at 11.8V is.

[Jwillis]

With the analogy, flow is defined as an amount over time. Speed is not a factor in the "flow" of electrons .We'll use yet another analogy to explain .Think of a Newtons cradle and that best represents how electrons move down a wire.Although in reality its much more chaotic with many other factors . Analogies are used to to best explain a process in the simplest terms.Its much easier to picture in the mind something that may be familiar.
It's not an attempt to insult your intelligence in any way.Just a simple way to explain the relationship between voltage , current and resistance .

[KL27x]

The water analogy isn't sorta kinda a half-ass analogy. It is pretty close to perfect.

A high pressure washer and your kitchen faucet might produce the same volume of water per time, but the water coming out of the high pressure washer has more velocity. Thus, it has more kinetic energy. Thus, it can do more work.

In the water analogy, it is most common to refer to the height of the water as a measure of its potential energy. But if you were to pour it to "ground," that turns into kinetic energy. In absence of "resistance," 1 gallon of water poured from 100 feet high would be moving faster by the time it reaches the ground than 1 gallon of water poured from 3 feet.

So, this is why hydroelectric generators are positioned at the bottom of tall water falls, not just stuck in the middle of a big, slow moving river that passes 1000x the volume of water per unit time.

Your adjustable PSU (most likely) has a linear regulator. All of the water it can pour starts out higher than 12V. When you set it under this level, the regulator absorbs the excess "velocity," and turns that energy into heat. But when you pass it through your 0.1ohm current meter DMM, the actual voltage you would record across your DMM leads (if you put another DMM on there on voltage setting) would be close to zero (maybe 1V or 2V tops)*, no matter what voltage you set it to. Almost all the energy is just turned into heat in your PSU. There are other sources of resistance in the PSU that are higher in sum than the 0.1R shunt in your DMM which are busy turning this energy into heat.

*ALL real PSU have internal resistance called impedance in this specific instance. 12V PSU with lower impedance (stronger PSU, essentially) will get that reading higher than one with higher output impedance. This is not necessarily proportional to the max current output, but there's usually a correlation. This stronger PSU would make your DMM do more of the job of turning this energy to heat.

[Mr D]

Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?

[rstofer]

Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?

1 Ohm is a very small number when dealing with electronics.  It is a very high number when dealing with busbar inside a utility substation.  All numbers are relative.

When I think in terms of electronics, I think 220 Ohms for a bright LED from 5V, 330 Ohms if I want it a little dimmer and 470 Ohms as about as dim as I want to go (RED LED).  I might use 330 Ohms from 3.3V if I want a particularly dim surface mount LED - something that might be used for diagnostics.  Later on you will learn to calculate the resistor value for a particular LED current which will result in a specific amount of luminous intensity.  Later on...

I tend to think in terms of 1k, 2.2k, 4.7k and 10k for most transistor projects and 1 Megohm (plus 1 ufd) for op amp integrators (time constant = 1 second, this will come up later...).

The problem with high resistance circuits is that they are easily swayed by noise.  The problem with low resistance circuits it that they tend to use more power.

Here is a calculator that should answer any specific questions you may have:

https://chemandy.com/calculators/round-wire-resistance-calculator.htm

There are MANY other calculators on the Internet.  Some of them are more appropriate for industrial wiring.

[james_s]

Why is it always necessary to explain via analogy?

This is not meant as a criticism of your attempt to explain it. I just don't understand why it's never explained without resorting to analogy!

Well the water analogy works well for many people because you can see, touch and feel flowing water. Ok I suppose you can feel electricity too if the voltage is high enough, but that's not really a recommended way to learn about it!

[Rick Law]

Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?

1 ohm is very little from the scale of resisters you can buy.  You can get (commonly) resisters as low as 0.01 ohm (and may be lower still, but less common) all the way up to multiple giga ohms in one resistor.

But 1 ohm is a lot if you don't want it - for example, on some of my cheaper switches, the contact is about 0.2 ohm to 0.5 ohm!  That is, if I put the switch to the ON position and I measure the resistance at the solder legs - it is 0.2 ohms!
 
Your DMM's probes has wires and so it has resistance as well.  When you put it in resistance mode measuring ohms, and you short the the tip of your two probes, you are measuring the resistance of your probe's wire and the contact resistance between the probes, sockets, etc.  You would probably read 0.25 ohm range - may be less, may be more.

As to your question about 10 cm sheet of copper  - can't answer that because resistance will depend on how thick.  Take a look at this "wire gauge table" - it will show you how many ohms (or milli-ohms) per unit length of copper wire, as well as the cross sectional area for a specific gauge.

https://en.wikipedia.org/wiki/American_wire_gauge

[rstofer]

Suppose I had a 480V source and a load current of 1200 Amps (typical for a small unit substation).  Now, assume I wanted to limit the busbar voltage drop to 1% or 4.8 Volts.

R = V / I or 4.8V / 1200A = 0.004 Ohms or 4 milliOhms.  That is 1/250 of 1 Ohm - a pretty small number.  But it is typical of busbar for a unit substation.  The bolted connections have to be done just right!

This is in the world of electrical, not electronics but it does point out that 1 Ohm is not the lower limit of resistance, 0 Ohms is, although it can't be achieved without involving a superconductor.

It's important to remember that Ohm's Law is a Law, not a suggestion.  You can't treat it like a speed limit!

And, yes, I am aware that there are boundary conditions on the Law but they are not relevant to this discussions.  Google tells all about it but it is still irrelevant at this level.

[james_s]

You don't need to be working on a substation for 1 Ohm to be substantial. I recently built a controller for the motor in a cordless electric lawnmower that can draw up to 40A from a 24V battery pack. If the mosfet had an on-resistance of 1 Ohm it would be trying to drop 40V which is more than the battery can provide. The result is the mosfet would be dissipating a huge amount of power and the motor would never be able to draw anywhere near 40A with only a 24V battery.

Even if it is only 0.1 Ohm it would drop 4V and dissipate 160 Watts which is far too much. 

[Mr D]

Many thanks to all who are taking the time, i'm reading through all replies carefully and i believe it's having some positive effect!

How's this?:

Say i have a 3V battery in a simple circuit with only a switch (admittedly not a very sensible circuit, but for argument's sake).

Is it fair to say?: With the switch open, i'll measure around 3 volts across that open switch because (as there's no current flowing), that is the potential charge that could flow if i were foolish enough to close the switch?

So that's why if i remove the battery from the circuit and only measure across the terminals of the battery i'll measure 3 volts (ish), whereas if i short-circuit the terminals (not good, i know) and measure, i'll measure almost 0V as the maximum current possible is flowing at that moment so the potential current is almost zero?

Am i on the right track?

[Old Printer]

I believe the reason your meter would read zero is because the dead short removes any difference between the prob tips, there is nothing to measure. Educated theory may offer another answer, but this is how I see it.

[rstofer]

Am i on the right track?

More or less...  The short circuit of the switch will have a resistance of essentially (but not really) 0 Ohms.  There will be no voltage drop across the switch because when you multiply by zero, the answer is 0.  This isn't a special case, it follows from E = I * R.  R = 0 so regardless of I, E will be 0.   Ohm's LAW

Better, and safer, experiment:  Connect the battery positive lead to your DMM red lead with the meter set to read milliamps.  Connect the meter black lead to 2 each 1k resistors in series.  You should read 3V / 2k = 1.5 mA.  Now short one of the resistors.  You should read 3V / 1k = 3 mA.

You can expand the experiment into a voltage divider circuit.  You will probably want to measure voltage rather than current.  Take out the meter, set it back to volts and reconnect the series resistors across the power supply.  Clip the black lead to the wire headed to the negative battery terminal.  You should read 1/2 the battery voltage at the center of the two series resistors. 

Now change the  top resistor to 100 Ohms.  The reading should be 0.9 * the battery voltage.  The calculation is easy:  Battery voltage divided by total series resistance (to get the current through the loop) multiplied by the bottom resistor value (to get E = I * R for the bottom resistor and the loop current).  Play with this until you really have a handle on the simple circuit.  Swap the location of the 1k and 100 Ohm.  The result would be 0.1 * the battery voltage.

The formula for the simple two resistor voltage divider is  V = (E / (R1 + R2)) * R2 when R2 is the bottom resistor.  The E / (R1 + R2) calculates the loop current I and then V = I * R2, the bottom resistor.

I like millamps, I don't like amps.  Actually shorting a battery, particularly the more exotic batteries, will produce a LOT of current and it only takes a few seconds for the battery to catch fire.  Then you have to run down the hall to the bathroom and flood the flaming mess with water.  Just sayin'...

[Brumby]

So that's why if i remove the battery from the circuit and only measure across the terminals of the battery i'll measure 3 volts (ish), whereas if i short-circuit the terminals (not good, i know) and measure, i'll measure almost 0V as the maximum current possible is flowing at that moment so the potential current is almost zero?

Am i on the right track?

I'll say "Yes", with one qualification.  The phrase "potential current" isn't correct.  I would use the word "voltage" instead.  Everything else reads like you are understanding correctly.

To understand where the voltage of the battery went, just look back at the internal resistance of the power source I mentioned earlier - ESR.  The battery chemistry not only provides the voltage, but presents a certain amount of resistance as well - so the "equivalent circuit" of a battery is a voltage source and a resistor in series.  As a result, when you short out a battery, the voltage generated within the battery is lost through the resistance of the battery.

[james_s]

I like millamps, I don't like amps.  Actually shorting a battery, particularly the more exotic batteries, will produce a LOT of current and it only takes a few seconds for the battery to catch fire.  Then you have to run down the hall to the bathroom and flood the flaming mess with water.  Just sayin'...

Years ago I learned rather quickly that putting a 9V battery in your pocket with keys and loose change is a bad idea.

[Mr D]

Thanks, it's getting ever so slightly clearer.

But now i've come up against the following stumbling block:

Say we have a simple circuit which is a 10v battery and a resistor.

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Let's make the resistor even worse. What's the worst resistor imaginable? No resistor!!!

Now if we measure either "side" of this non-resistor, we certainly won't see 10v on our MM.

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

[james_s]

What do you mean "worse" resistor?

As you decrease the value of the resistor, it will draw more current from the battery which means it will dissipate more power. This breaks down when the resistance becomes so low that the source can't supply as much current as it is trying to draw and the voltage will sag below 10V. If you could have a 0.0000 Ohm resistor across it the voltage would sag down to 0 Volts but in reality you can never have quite zero.

[rstofer]

Thanks, it's getting ever so slightly clearer.

But now i've come up against the following stumbling block:

Say we have a simple circuit which is a 10v battery and a resistor.

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Let's make the resistor even worse. What's the worst resistor imaginable? No resistor!!!

Now if we measure either "side" of this non-resistor, we certainly won't see 10v on our MM.

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would NOT suggest doing it to the car battery.

We know from Ohm's Law that the internal resistance will drop a voltage E = I * R.  It really is that simple!  The thing is, R is quite low so for a given E, I can be quite high.  Like the car battery - it has a very low internal resistance.

This is the Thevenin Equivalent Resistance I mentioned way back near the top and several others have mentioned.

When you put a light load (1A (10 Ohm 10V) or 2A (5 Ohm 10V) on a source with sufficiently low internal resistance, you will measure essentially the full source voltage.  When you put a very low resistance across the source, the internal resistance dominates and the majority of the voltage is dropped inside the device.  Take that internal voltage drop V and square it.  Then divide by the size of the resistor to get the power dissipated in Watts (E squared over R).  Notice how a resistor less than 1 Ohm drives the power up pretty quick.

That's also why I brought up the voltage divider experiment above.  I was using reasonable resistors such that internal heating of the source wasn't likely to occur.  But the top resistor does exactly the same thing when you slide it inside the battery and call it the Thevenin Equivalent Resistance.  The internal resistance forms a voltage divider with a very low value for the top resistor.

Later on you will find that maximum power transfer to the load occurs when the load resistance is equal to the internal resistance.  This is a consequence of the voltage divider equation given above.

If the external load resistance is higher than the internal resistance, the majority of the voltage drop is outside the battery.  Think 1 Ohm internal resistance and 10k load resistance.  Essentially all of the voltage appears across the load resistance.  Now turn it around:  Think of the 1 Ohm internal resistance and a 0.1 Ohm load resistance.  In this case only 10% (approx) of the voltage drop is at the load and the internal resistance is dropping 90% (approx).  And you can bet the battery is heating up.

You can't get around the necessity of wrapping your head around Ohm's Law.  It explains everything but you need to play with it to get an intuitive feel.  What happens to I when you vary E for constant R.  What happens to I when you vary R with constant E?  What happens to E when you vary R with constant I?  Sit down with a scratch pad and lay down that voltage divider and see how the number actually work.  Same with the voltage divider using the battery internal resistance (assume 1 Ohm just for a number) and various external resistors or even an external 2 resistor divider.

[james_s]

As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would suggest doing it to the car battery.

I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying.

[rstofer]

As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would suggest doing it to the car battery.

I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying.

Fixed!  Thanks...

[ArthurDent]

"I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying."

Years ago I owned a MG1100, a inexpensive little car that had a metal ratchet-type device to hold the hood (bonnet) up when you opened it. I was driving on a rough dirt road once and the car just died and smoke was billowing from under the hood. What had happened is the small bolts holding the upper end of the metal device had vibrated out and it had dropped directly across the battery terminals, which had no insulating covers. The sub-ohm resistance of the device forced enough current through the internal cell connectors of the battery so that one melted and the battery was now an open circuit. It got hot but no boom.

Oh, the car was a standard so pushing it and popping the clutch turned the engine over fast enough so residual magnetism in the actual generator allowed the car to start and I got home.

[Brumby]

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

To be very clear - the relationship NEVER breaks down.

You just need to catch up with your understanding - and don't fret ... you ARE getting there.

The answer to your confusion is the same as I (and others) have already mentioned - Equivalent Series Resistance.  You really need to get that concept drilled into your thinking - and then the whole mystery will be solved (well, at least this corner of the world of electronics.)

UNLESS you have a power source that is superconductive, there will ALWAYS be some amount of Equivalent Series Resistance.  For car batteries, this is very low - but it is not zero.  For alkaline batteries it is significantly higher - even though it is still small on the scale of resistance.  Your average 18650 lithium batteries are somewhere in between.

Whenever you look at a power source, you must ALWAYS consider there is a resistor of some value hidden inside it that you can't get around.

With this in mind, you have made an error in the following:

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Assuming we have a meter with a 10M input impedance, we can ignore the effect it has on the power source.  (Yes, even this will have an effect on the power source, but it will be very, very small.)  Let's say we measure the voltage (with no other resistors) and you get a reading of 10V.

Now, when you connect you 10 ohm resistor, you will NOT get a reading of 10V.  It will be slightly less.  This is because the current flowing through the circuit goes through TWO resistances.  The obvious one is the 10 ohm resistor you've connected across the power source.  The second one is the hidden internal resistance of the power source - the ESR.  Each of these will drop a number of volts according to Ohm's Law - and the only one you can measure directly is that of the external 10 ohm resistor you used.  You have to calculate the voltage drop caused by the ESR.

When you replace the 10 ohm resistor with a 5 ohm resistor, the current will increase.  You are still facing a circuit with 2 resistances - the external 5 ohm one and the ESR within the power source.  Note that the ESR is the same as it was for the 10 ohm resistor - so with increased current, Ohm's Law tells us that the voltage drop because of the ESR will be greater.  This means the voltage you measure across your 5 ohm resistor will be even lower than it was with the 10 ohm resistor.


If you are any good with algebra, you can perform the above as a real experiment and you should be able to calculate the ESR of the power source.  If you can do that, you will have mastered this fundamental!

[Brumby]

If you are any good with algebra, you can perform the above as a real experiment and you should be able to calculate the ESR of the power source.

Just be careful about the power rating of the resistors.  To do the above as a real experiment, the 10 ohm resistor will need to be able to handle around up to 10W - and the 5 ohm resistor, up to 40W.  They are going to heat up quite a bit.  If the ESR is very low, then these figures will be pretty close to what you will encounter, however as the ESR rises, the lower the power rating for these resistors will need to be...

BUT

be very aware that the ESR will also cause the power source to heat up - and that could provide some spectacular and dangerous outcomes if left to run for more than a couple of seconds.


PS.  When choosing a power source for the above experiment, don't use a lab power supply.  A lab power supply will have regulated output and possibly current limiting that will thwart any attempt at calculation.
 Use a battery.

[Mr D]

Thanks, i'm starting to get an intuitive feeling for this (although it's still foggy )

But i'm still struggling with some aspects of voltage.

Say we have a 9v battery. We measure for voltage across the terminals and measure 9v(ish).

Now, we touch only the + probe of the MM against the + of the battery while leaving the  MM - probe dangling. We measure 0v as expected.

Now, we lay a 10cm piece of copper wire on the table.

With the +MM still attached to the battery +, we touch the - probe to the end of the copper wire. We measure 0v as expected.

Now we attach the end of copper wire to the - battery terminal. Now if we touch the end of the copper wire with the MM -, which still having the + attached to the battery +, we measure 9v.

Soooooo, my question is: how does the MM "know" what is at the end of the copper wire? How does it "know" that in the last step, i attached the negative battery terminal to the copper wire?

As far as i can see it, there are only two possibilities:

Either a) some information must have travelled along the copper wire from - battery terminal to the MM - probe and into the MM.

Or b) some information must have travelled from the battery + terminal, into the MM+ probe, through the MM, out of the MM - probe, along the copper wire, into the - battery terminal, through the battery, out of the battery + terminal, into the MM+ probe, and into the MM.

Is it one or the other? Or a mixture of both? Or have i (yet again) fundamentally failed to understand something?

[ArthurDent]

"Soooooo, my question is: how does the MM "know" what is at the end of the copper wire? How does it "know" that in the last step, i attached the negative battery terminal to the copper wire?"

Using "know" is not how you should visualize it, the circuit is just following the laws of physics. An exact analogy is a lighting circuit in your house. If you turn a switch on (like connecting a lead to the copper wire you have on the table), current can now flow in the completed circuit and the light controlled by the switch will go on. Turn the switch off to open the circuit and the light goes off.

If you have a table lamp plugged into a wall outlet, the lamp can be on but when you unplug the lamp you break the circuit (remove the lead from the copper wire you have on the table), and the table lamp goes off. This is more obvious because you can actually see the physical separation between between the plug and the wall outlet. The only "know" part is your knowingly completing or opening a circuit to control the flow of current.

Don't overthink this. 

[Brumby]

You really are overthinking things.

With your circuit setup, just before you make the final connection, both the +ve and -ve probes of the DMM are at 9V with respect to the -ve connection of the battery (I use this reference point as it is pretty much the standard.)  Because there is no difference in potential between the meter probes, it will display 0V ... as expected.

The piece of wire touching the -ve probe of your DMM is also at 9V with respect to the -ve connection of the battery.

Then, when you connect the wire (at 9V) to the -ve of the battery (at 0V), you changed the equilibrium.  The circuit is now closed and current can flow.  How much current will depend on the resistance of the circuit and the voltage of all the points around the circuit will change.  Ohm's Law will now be able to put numbers to each of those.


To be honest, I would suggest you ease up a bit in trying to "understand" things to such a level of detail.  It is only going to confuse you and frustrate your efforts in learning electronics.

Understanding how a bipolar transistor actually works at the fundamental physics level is a classic example.  You don't really need to know in order to design, build and fault-find circuits.  All you really need is to understand the datasheet.

Be kind to yourself and:

Don't overthink this. 

[Mr D]

Ok, so when you put a MM across the terminals of a battery to measure voltage, there is in fact some current flowing all the way through the MM to the - terminal of the battery, right?

So the MM is basically just a resistor, and as we know the resistance (or, the MM knows it's own resistance in voltage-measure mode) and the MM can measure the current, the MM can, thanks to Ohm's law, tells us the voltage?

So a MM measures current in both voltage-measure and current-measure mode, the only difference being that in the voltage measure mode, the MM has a very high resistance?

 

[Brumby]

Correct.

Some people might argue that the DMM is measuring voltage - but it's sort of an indirect process and it takes some current flow to work.

[rstofer]

Back up to the days before the DMM and consider the analog Volt-Ohm-Milliammeter (VOM).  In every range, whether measuring resistance, current or voltage, the meter needle was displaced by current flow through the windings of the meter movement.  Always current flow!

When measuring current, the meter resistance was placed in parallel with a shunt resistor (a really low value resistor) such that only a percentage of the measured current was flowing through the meter.  The vast majority of the current was flowing through the shunt resistor.  But the current flow through the meter deflected the needle.

When measuring voltage, the meter resistance was placed in series with a resistor (a much higher value) such that the full scale maximum voltage would produce just enough current to fully deflect the meter.  Side issue:  This conversion factor was given on the meter face as Ohms/Volt Full Scale and a given scale, say 10V, would have a value like 10,000 Ohms/Volt Full Scale - the meter, when on the 10V scale, imposed a 100k Ohm load (10V scale * 10,000 Ohms/Volt) on the circuit.  For high impedance (vacuum tube) circuits, this could produce an error in the measurement.  Cheaper meters had values around 5k Ohms/Volt Full Scale.  DMMs are usually around 10M Ohm no matter the voltage scale.

When measuring resistance, there is a battery to produce the voltage for the meter.  It really measures current (the ONLY thing the analog meter movement really measures) based on the known battery voltage and the meter was scaled to read Ohms.

There is still a known relationship between the meter full scale current and full scale voltage and it is exactly the meter resistance.  When you buy an analog meter (VOM or panel meter) these values are given in the datasheet.

The same kind of thing happens in a DMM except that current isn't necessary to cause a magnetic field to move the needle.  In the case of the DMM, it measures voltage.  In current mode, it measures the voltage drop across the low value shunt resistor.  In resistance mode, it uses batteries to drive the current through the circuit and, one way or another, measures the voltage produced by that current flowing through a known resistor.  In voltage mode, is measures the voltage across a known internal resistor voltage divider.  The actual analog to digital converter measures voltage.

The DMM is always measuring voltage, the VOM is always measuring current.  But it doesn't matter!  Ohm's Law takes care of the details.  So while I say the DMM is always measuring voltage, current is involved somewhere in the circuit (causing a voltage drop across a known measuring resistor) and Ohm's Law is used to understand the relationship.

[Mr D]

Correct.

Some people might argue that the DMM is measuring voltage - but it's sort of an indirect process and it takes some current flow to work.

Great, this has been a stumbling block for me! Now it's clear. So actually when we say we measure voltage, what we actually mean is we measure current across a known resistance, and infer the voltage.

And when we measure current with an DMM, it's actually a two step process. First (or maybe second?) it does the voltage measurement, then, as it has the voltage and resistance, it can tell you the current? So in current measure mode, the DMM has a parallel measurement path?

Wait, nope, i still don't understand. Is it possible to measure the current without knowing BOTH the voltage and resistance?

Or put another way: to measure either, voltage, current or resistance, you ALWAYS need to know the other two? Or is one ever enough?

[PA4TIM]

Buy the Art of electronics, this is one of the best books about electronics. It goes very deep. If I'm correct it was written for university use. Buy the third edittion. The difference with the 1980's second edition is huge. (I have second and it still is my bible)

A DMM has an input resistance. The bad ones a few Mohm, the very good ones over 10 Gohm. So there is always a small current flow through your meter (and so influencing the circuit  ) . But there are electrometers and that is a different beast. They measures the electric field.

A voltage exist without any current flow, all it needs is one point that is more positive charged as an other point. If you connect those points a current will flow.

Measuring a voltage is not so easy as many people think. You disturb the circuit and for some things the meter is the wrong type. A nice example: If you connect a diode between the battery and a DMM you will measure the voltage of the battery minus the voltdrop of the diode. That is f.i. if you use a normal 10Mohm (or less) . If you measure it with a 10G meter or electrometer  the current is to small to bring the diode in forward conduction and you measure the electric field. 
Here I show some things that can go wrong while measuring a voltage. (i have also a video over pittfalls for small current measurements.)
https://youtu.be/bH5n5iMIqIw

[rstofer]

Wait, nope, i still don't understand. Is it possible to measure the current without knowing BOTH the voltage and resistance?

Or put another way: to measure either, voltage, current or resistance, you ALWAYS need to know the other two? Or is one ever enough?

There are 3 variables for Ohm's Law (E, I, R) and you need to know two to calculate the 3rd.

[rstofer]

This might be a worthwhile program but it costs a few bucks (I just signed up, it cost me $15):

https://www.eevblog.com/forum/beginners/beginners-that-are-interested-in-learning-electronics-from-the-ground-up-)/

This program is MASSIVE!  There are 134 lectures totaling 86+ hours!

The author has published quite a few books in addition to this program.  The books are available on Alibris.Com - just search by Author.

Price seems to be flexible.  The landing page says $20 but when I checked out it was $15, probably because I have bought other programs.  So, maybe it costs $20 to a newcomer.  It's a BARGAIN!  I have smoked parts that cost more than that!

[Mr D]

@PA4TIM, nice, will check out your videos!

@rstofer, thanks, but that link you posted doesn't go anywhere, could you try again?

[rstofer]

@rstofer, thanks, but that link you posted doesn't go anywhere, could you try again?

It's another thread in the Beginner's Forum and I'll be darned if I can link it.  So, here is the Udemy link
https://www.udemy.com/crash-course-electronics-and-pcb-design/

I don't know why I can't link the thread but the title is correct and it's just a few items down this page.

[drussell]

I don't know why I can't link the thread but the title is correct and it's just a few items down this page.

You just missed the last two characters )/ before the /url tag.

https://www.eevblog.com/forum/beginners/beginners-that-are-interested-in-learning-electronics-from-the-ground-up-)/

[rstofer]

Yup!  That works!

[ArthurDent]

“If you wish to converse with me, define your terms.”  ― Voltaire

Mr D -“So actually when we say we measure voltage, what we actually mean is we measure current across a known resistance, and infer the voltage.”

That isn’t the way to describe it at all.

If you have a 9 volt battery and you put a 10 ohm, or a 10,000 ohm, or a 10 megohm resistor in series with the positive terminal, the voltage at the unconnected end of whichever resistor you choose to negative is still exactly 9 volts. This is because there is no current flow and no voltage drop until you connect a load between the unconnected end of the resistor and the negative terminal. Until you have a completed circuit this voltage is only potential energy and is doing absolutely no work. You need a complete circuit or there is no current flow. The ‘work’ could be lighting a lamp, running a motor, or deflecting a meter needle.

You don’t “…measure current across a known resistance, and infer the voltage”, you measure the VOLTAGE drop ACROSS a known resistance and convert that to current. You measure the current flowing THROUGH a known resistance and convert that to VOLTAGE. You could measure the voltage across a known resistor and display a voltage but that would NOT be the voltage between the known resistor and the negative terminal, which is what you really want.

From the 9V positive terminal if you put a 9K resistor in series with a 1K resistor to the negative battery terminal you have a total resistance across the 9V battery of 10K and because the 1K resistor is 1/10 of the total, the potential between the junction of the 9K/1K and negative is 1/10th of the 9V or 0.9V.  If you connect a voltmeter across the 1K resistor with a 10Mohm input resistance almost all of the current will still flow through the 1K resistor and about 1/10,000th will flow through the meter so the reading is basically unaffected. Now if you use a meter with a 1K input resistance, the current from the 9K resistor will split with half going through the 1K resistor and half going through the 1K meter. The equivalent circuit is now 9K in series with 500 ohms (2 parallel 1K) and the meter reading will now be way low. 

[Mr D]

Thanks for the clarification.

But is it not indeed the case that when we say we measure voltage or resistance with a DMM, what in fact is happening is that the DMM is measuring current, then telling us (via Ohm's law) what the voltage or resistance is?

[rstofer]

Thanks for the clarification.

But is it not indeed the case that when we say we measure voltage or resistance with a DMM, what in fact is happening is that the DMM is measuring current, then telling us (via Ohm's law) what the voltage or resistance is?

In the case of the DMM, it is actually measuring the voltage drop across a resistor.  It is also true that the voltage drop is proportional to the current flow through the loop.  You can look  at it either way but the internals of a DMM are fundamentally voltage measuring.  The internals of a V-O-M are current measuring.  But it doesn't matter because we know R_internal and we can move back and forth between I & E with ease.

Here is a simplified diagram of a V-O-M.  Remember that the needle moves in response to current through the winding resistance.

https://www.allaboutcircuits.com/textbook/direct-current/chpt-8/multimeters/

I couldn't find a simplified diagram of a DMM but you could just substitute a winding resistor where the meter is shown and then measure the voltage across that resistor.  It would be acceptable for discussion if not entirely correct.

The DMM measures voltage, the V-O-M measures current.  They both get the same answer.

Back to the water analogy:  I was reminded today that the underlying units for Volts is Joules / Coulomb.  How much energy in Joules does it take to move so many Coulombs of charge.  An electric potential of 1 Volt will use 1 Joule of energy to move 1 Coulomb of charge.  And that's why I did EE and not Physics!

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html#c1

That is the reason we use the water analogy - Volts is just hard to describe when we get down to physical units.  We just rename it EMF (Electro Motive Force) or Volts (as a fundamental unit) and move on! Quickly!

[ArthurDent]

Measuring resistance is a variation of what I stated above: “You measure the current flowing THROUGH a known resistance and convert that to VOLTAGE”; only to measure resistance you pass a known current through an unknown resistor and measure the voltage and convert that to resistance. If ohm’s law only has 3 terms and you are trying to solve for one, the other two must be known.

R=E/I

[Brumby]

Wait, nope, i still don't understand. Is it possible to measure the current without knowing BOTH the voltage and resistance?

Well, yes it is - but not with a DMM.  You can do this using an analogue meter movement.  If the basic sensitivity of the meter is 50uA, then you can push through anything from zero to 50uA and it will give you the current.  You don't need to know the resistance or the voltage.  Of course, since the coil of the meter will have some resistance, there will be a voltage drop across it, but you don't need to know either of these to measure your current.

However, if you want to use this meter movement to measure currents higher than 50uA, then you WILL need to know voltages and resistances in order to build the appropriate shunt circuit.  Ohm's Law all the way!!

Or put another way: to measure either, voltage, current or resistance, you ALWAYS need to know the other two? Or is one ever enough?

With resistance, you have a passive quantity and you need to "excite" the component and determine its response in order to calculate that quantity ... So, yes, voltage and current are both needed.  Always.

With current, you can - in some cases - measure it directly, but with any sort of conditioning network, Ohm's Law is a must.  In the case of a DMM, you absolutely need it.

Voltages are an interesting case and while there are some techniques that can be used to measure a voltage without using Ohm's Law, they are not the sort of thing you would see very often, if at all.  So, in practical terms, Ohm's Law is applied almost universally.


The one overruling fact, though, is that if you cannot directly measure the value (as in the 50uA meter movement mentioned above) then you must have the other two values for Ohm's Law to be used.  This is fundamental mathematics - whichever way you write V=IR (you can also see it written E=IR), there can ONLY ever be ONE unknown.

[drussell]

I believe I do understand some of Mr. D's confusion from many years ago when I was a beginner...

You do, indeed, need to understand that Ohm's Law is a constant LAW and that you can convert and figure out the voltage drop across a known resistance as the current just as well as you can use the current through a circuit with a known resistance to infer the voltage.  (edit: or the current and voltage to calculate the resistance...)

It simply depends on the instrument you are using to actually measure the quantity you're observing.  A small current measured by a galvanometer of some sort or other current meter like you would find in a VOM, (typically the 5k, 20k, 30k or 50k ohm-per-volt meter movement) in series with (for voltage) or parallel with (for current) measurement of that quantity actually measures by physical deflection due to the current inducing a magnetic field.

On the other hand, as mentioned previously, the analog to digital converter (ADC) in a typical DMM actually measures the voltage potential (or Electro-Motive Force, EMF) being pushed/pulled between those two points, despite the fact that the actual input resistance may be very high (10,000,000 ohms, 10,000,000,000 ohms or even more on some bench meters) or even essentially completely open.

They're not using the same method to measure the specific quantity but it can always be calculated if you know one of the two other variables.  (Well, for DC, anyway...  )

[Mr D]

Hi folks,

During my summer holiday to the Alps i re-read through this thread a few times, and also thought about Ohm's law quite a bit and i think i got my head around it.

I've also discovered the MathsTutorDVD lectures and they're really helpful.

Now i think the next thing for me to understand is ground.

I have some questions about this, but first i need to clear something else up: While searching Youtube for info about ground, i came across this guy's vids:



Can anyone tell me: why is he dipicting flow from neg. to pos. in this video?

[Mr. Scram]

Hi folks,

During my summer holiday to the Alps i re-read through this thread a few times, and also thought about Ohm's law quite a bit and i think i got my head around it.

I've also discovered the MathsTutorDVD lectures and they're really helpful.

Now i think the next thing for me to understand is ground.

I have some questions about this, but first i need to clear something else up: While searching Youtube for info about ground, i came across this guy's vids:



Can anyone tell me: why is he dipicting flow from neg. to pos. in this video?

When electricity was discovered they found out that it flows from one end to the other. They couldn't actually see or measure which way it flowed however, so they picked a direction and went with it. When science finally advanced enough to actually measure the direction of the flow, it turned out that the convention is actually wrong. Electrons flow from the negative pole to the positive and not the other way around. It doesn't really matter though, because the traditional model works perfectly fine as it is unless you go into some detail most people will never deal with. It's too much trouble to swap everything around, so people just work with the "wrong" model.

"Conventional current" is the name of the traditional model we all know and tend to work with where electricity flows from positive to negative. This turns out to be wrong, but works perfectly fine. "Electron flow" is the name for using a model with the actual direction of the electrons. People don't use this as often and when you see a generic schematic, you can assume it to be conventional current.

[Mr D]

Thanks!

But how then do i make sense of a component like a diode, in which current can only flow in one direction?

If electrons in reality only flow from neg. to pos., how do the electrons get through the diode and all the way back to the positive terminal?

[james_s]

Thanks!

But how then do i make sense of a component like a diode, in which current can only flow in one direction?

If electrons in reality only flow from neg. to pos., how do the electrons get through the diode and all the way back to the positive terminal?

Provided the diode is wired so that the cathode is facing the negative terminal of the battery, the electrons flow through the diode similar to how they would flow through a wire. If the diode is flipped around they don't flow at all, aside from a very small amount called leakage current.

Unless you are working with vacuum tubes you can pretty much forget about the direction electrons are actually flowing and just go with conventional flow, which assumes they go from positive to negative.

[Mr. Scram]

Thanks!

But how then do i make sense of a component like a diode, in which current can only flow in one direction?

If electrons in reality only flow from neg. to pos., how do the electrons get through the diode and all the way back to the positive terminal?

You have to remember that the entire convention of conventional current is in reverse. This means that diodes in reality actually work opposite of what we think they do. So if you reverse the flow of electricity but also reverse the direction the diode stops electricity flow, everything is working out again. Electricity is stopped in the direction you'd expect and lets it through in the right direction too.

If you reverse everything it's basically all the same again.

[james_s]

Hence my suggestion to not even worry about the details and just use conventional flow. There's a reason it's called conventional flow, all of the schematic symbols are drawn with the assumption that flow is from positive to negative. It's easier at this stage to just operate under this assumption rather than complicating it with details that are irrelevant to most hobby type circuits.

[rstofer]

You will notice that in the first 10 seconds of that battery video (all I watched, actually), the author specifically states that he will be looking at electron flow.  Unless there is some chemical reason later in the video (and there might be), that is a regrettable choice.

We use conventional current flow (+ to -).  The current flows in the direction of the arrow in diodes and transistors.  The US Navy used to teach electron flow.  I don't know if they have changed.  Probably not...

As long as you are consistent in marking polarity through components, it really doesn't matter which system you use but you will confuse the vast majority of readers if you use electron flow.

[Mr D]

OK, i'll take it on faith that it's not important to focus on and won't dwell on it much longer.

It just seems unintuitive to me that a circuit would work correctly if that actual flow would be the reverse of what we imagine when designing it. For example, a complicated garden sprinkler system with valves wouldn't (presumably) work if the water flow was reversed.

[james_s]

A sprinkler is an open loop, water flows through the pipes, comes out the sprinklers and lands on the lawn. Your goal with the sprinklers is to take water from a source and relocate it to the lawn. With electricity you need a circuit, the electrons are flowing in a loop around and around, you can't see them anyway so what difference does it make which direction they're flowing? The reason conventional flow is backwards in the first place is that for the first hundred years or so of working with electricity it was impossible to tell the actual direction of flow. As long as everything is consistent it makes no difference at all. The reason designing a circuit works when the flow is backwards is that all the symbols are drawn backwards too. The direction the actual electrons are moving is completely irrelevant because you can't see them anyway.

[Mr D]

OK, for the moment i'll ignore it.

So that brings me to ground.

Until now i've only been thinking about closed circuits involving batteries and resistors.

But how does ground tally with this?

Attached are images of two circuits. According to EveryCircuit they both produce the same current. But how can this be?

In one, the positive and neg. terminals are not connected (because the flow leaves the resistor and enters another enormous resistor called the Earth?!)

Until now i have been presuming that to have a circuit with flow, the pos. and neg. terminals must be connected.

[Mr. Scram]

OK, for the moment i'll ignore it.

So that brings me to ground.

Until now i've only been thinking about closed circuits involving batteries and resistors.

But how does ground tally with this?

Attached are images of two circuits. According to EveryCircuit they both produce the same current. But how can this be?

In one, the positive and neg. terminals are not connected (because the flow leaves the resistor and enters another enormous resistor called the Earth?!)

Until now i have been presuming that to have a circuit with flow, the pos. and neg. terminals must be connected.

You're close to the answer, but take the wrong turn just before the end. What makes you think Earth is a giant resistor? You might want to look at what's needed to make a properly earthed connection before you answer that. An extra hint is that telegraph wires often are singular wires, not pairs.

[Mr D]

So if you take a 9 volt battery, attach a wire to each terminal, then hold one terminal in one hand, and the other terminal in the other hand, then your body, is it not, acting as a resistor? And the amount of current flowing through your body will determined by Ohm's law.

Now, replace your body with the earth under your feet, stick one wire in the ground, and the other wire in the ground a meter away, then the planet
Earth is also acting like a resistor?

Or is this not what is meant by "ground" in a circuit?

[Mr. Scram]

So if you take a 9 volt battery, attach a wire to each terminal, then hold one terminal in one hand, and the other terminal in the other hand, then your body, is it not, acting as a resistor? And the amount of current flowing through your body will determined by Ohm's law.

Now, replace your body with the earth under your feet, stick one wire in the ground, and the other wire in the ground a meter away, then the planet
Earth is also acting like a resistor?

Or is this not what is meant by "ground" in a circuit?

Have you looked at what's needed to make a properly earthed connection and why telegraph lines tend to be single wires?

There's a relevant difference between your body and Earth.

[Mr D]

Have you looked at what's needed to make a properly earthed connection and why telegraph lines tend to be single wires?

Have i looked at the answer to my question? No, and i wouldn't know where to find it. That's why i'm asking questions here.

[Mr. Scram]

Have i looked at the answer to my question? No, and i wouldn't know where to find it. That's why i'm asking questions here.

It's generally expected you try to answer questions yourself by doing some research. The questions that you can't answer yourself or require some insight can then be asked on something like a forum. Entering "what's needed to make a properly earthed connection" in search engine immediately yields useful results, such as the link below. I'm not trying to be a hard-ass here, but you'll really learn more if you investigate yourself too and I certainly wouldn't suggest searching for something if it's impossible to find.

If you follow the link you'll see that a properly earthed connection is a sizeable copper rod stuck into not too dry ground deep enough. This suggests there's a conductive path to the Earth itself.



Following the telegraph hint we find a page that says the following: "There were some very good reasons that the closed circuit system was almost universally used on Morse wire circuits in the USA and Canada. Single wire, ground or earth return circuits were normally used, which had half the resistance of a metallic loop, since the earth return contributed no resistance to the circuit if it was more than a few miles in length. The only circuit resistance was due to the line wire resistance and the resistance in the instrument windings or other equipment in circuit."

Apparently, the Earth effectively doesn't have much resistance at all even when compared to a wire circuit. Now, explaining this is where the forum might help. We already established there's a relevant difference between your body and Earth. This difference is the size. You probably know that larger diameter wires have less resistance than thin wires. Even though the Earth is made up of not wonderfully conductive materials, it is massively large. This is why it ends up being a pretty good conductor. It's essentially a wire with a really, really huge diameter.

http://www.housingforhealth.com/housing-guide/electrical-earth-connection-2/
http://www.morsetelegraphclub.org/wirechief/#closed

[The Soulman]

The "earthing" symbols used in the two schematics are very common (pun intended) and depict
a connection to a common conducting surface, this can be a ground plane on a pcb, or the metal frame of a car, or...

[Mr D]

OK, but i'm still not much the wiser!

The earth has effectively no resistance?

So, when the current reaches the bottom left corner of the rectangle in the attached schematic, where does the current want to go? Upwards to the neg. terminal or down into the ground? Which has less resistance?

See here at 4:30:


To quote: "A circuit is a closed loop that carries electricity.....in order to have a circuit, it has to come back to where it starts from"

How is a circuit that runs a wire from the pos. terminal of a source straight down in to the ground a closed loop?

[Mr D]

According to the attached schematic, having both the pos. and neg. terminals attached to ground will let the current flow.

But what if the two connections to ground are 10cm or 1km from each other? Does it matter?

Is it the case that the Earth is electrically neutral (or at least much more neutral then a power source), so current will flow from pos. into ground, and from ground towards neg.?

So in fact a closed circuit isn't needed?

Would current flow if the two wires (from pos. & neg.) where attached to two different but identical planets, 100 million kilometers apart?

[james_s]

You're confusing "ground" with "earth", they are not necessarily the same thing. When you see ground symbols on a schematic it is normally indicating "these nodes all tie together at that same point" and that point happens to be called ground. It's the common reference point for any voltage measurements shown for the circuit. Schematics are drawn this way for clarity as it reduces the rat's nest of wires in the drawing. Often you'll have various points for Vcc or various voltages which may be drawn in numerous points around the circuit. Again those symbols mean those nodes connect together. The "+5v" symbol drawn on a dozen ICs means each of those ICs power pin connects to the same +5V rail, it's neater than drawing a wire that goes all over the whole circuit.

Yes the earth has some resistance, so does a wire, everything (except for superconductors) has resistance, current can still flow.

[rstofer]

According to the attached schematic, having both the pos. and neg. terminals attached to ground will let the current flow.

But what if the two connections to ground are 10cm or 1km from each other? Does it matter?

Is it the case that the Earth is electrically neutral (or at least much more neutral then a power source), so current will flow from pos. into ground, and from ground towards neg.?

So in fact a closed circuit isn't needed?

The earth being 'neutral' or not has nothing to do with anything.  You can model the resitance between any two points on the earth as millions or billions of parallel resistors forming a grid.  There are test instruments for measuring earth resistance.

http://support.fluke.com/find-sales/Download/Asset/2633834_6115_ENG_A_W.PDF

On high voltage substations, we build a 2'x2' mesh of 4/0 wire under the entire area and we tie all of the metal structures to that grid.  A few Ohms of ground resistance doesn't sound like much until you drop 100,000 amps into it and the voltage goes sky high.  This is a really big deal!  Just ask the cows that die when they stand on the ground near a HV tower that experiences a fault.  The voltage difference across their legs is enough to kill them.  The grid extends a few feet out from the fence line to ensure that a person touching the fence is standing over the grid.  There's a lot of engineering in this stuff.

Would current flow if the two wires (from pos. & neg.) where attached to two different but identical planets, 100 million kilometers apart?

In the perfect vacuum of space?  Probably not.  But the vacuum isn't perfect so probably yes.

The force is so tiny it couldn't be measured and is swamped by the gravitational force.  Not important to EEs
https://www.quora.com/Is-there-any-electromagnetic-attraction-between-planets-and-the-sun

[Mr D]

Thanks for trying but, urrrghhhh, it's not making any sense to me.

Just when i thought i had my head around Ohm's law (which i do), this ground thing looms it's ugly head.

Let me try to precisely express what i'm not understanding:

In the attached schematic, both the pos. & neg. terminals lead to this undefined "ground". Apparently according to James S this is not real earth, but, for example the metal case of my device.

So the current will travel from pos., down the circuit, into the metal case, then back into the neg. terminal, right? So, we have a complete circuit.

But in such a device, the case is often earthed, right? So why would the current choose to travel back into the neg. terminal when there's an easier path (with virtually no resistance) to real earth?

Or to turn the question round the other way: I take a long nail, wet my hand and jam into the positive hole of a wall socket. I'm likely to get killed, right?

But why? The current can flow into my body, but not back into the negative terminal, so there's no complete circuit so no current flow, right?

Until now i've been dealing with simple circuits: a battery, one or two resistors, and that's it. Can you guys not at least partly understand how this idea of a ground leading somewhere undefined (at least in my mind) might be somewhat confusing? Is it the case of the device? Or real earth. Or an earthed case ........ or.....?

Cheers! ......... D

[Mr D]

But if i was to stop trying to intuitively understand it, is this a correct summary of what i need to know about ground:

Every time i see a ground symbol in a schematic, i just have to imagine a wire travelling from that symbol to a single bus, with that bus (that has zero resistance) leading back to the negative terminal?

[The Soulman]

Thanks for trying but, urrrghhhh, it's not making any sense to me.

Just when i thought i had my head around Ohm's law (which i do), this ground thing looms it's ugly head.

Let me try to precisely express what i'm not understanding:

In the attached schematic, both the pos. & neg. terminals lead to this undefined "ground". Apparently according to James S this is not real earth, but, for example the metal case of my device.

So the current will travel from pos., down the circuit, into the metal case, then back into the neg. terminal, right? So, we have a complete circuit.

Yes! 

But in such a device, the case is often earthed, right? So why would the current choose to travel back into the neg. terminal when there's an easier path (with virtually no resistance) to real earth?

No, The current path needs to lead back to the power source/battery, additional connection to earth makes no difference in this circuit.

Or to turn the question round the other way: I take a long nail, wet my hand and jam into the positive hole of a wall socket. I'm likely to get killed, right?

Yes! 

But why? The current can flow into my body, but not back into the negative terminal, so there's no complete circuit so no current flow, right?

To be accurate, there is no positive or negative terminal in a wall outlet because its AC not DC, the propper term
would be phase and neutral.
The neutral is always connected to a ground connection "somewhere" (but never in your house or in any appliance).
Therefore current can flow from phase (thru you...) to ground.

Until now i've been dealing with simple circuits: a battery, one or two resistors, and that's it. Can you guys not at least partly understand how this idea of a ground leading somewhere undefined (at least in my mind) might be somewhat confusing? Is it the case of the device? Or real earth. Or an earthed case ........ or.....?

Cheers! ......... D

meh, don't worry, if you read all of this back in a couple of months you will also have a good laugh. 

[The Soulman]

But if i was to stop trying to intuitively understand it, is this a correct summary of what i need to know about ground:

Every time i see a ground symbol in a schematic, i just have to imagine a wire travelling from that symbol to a single bus, with that bus (that has zero resistance) leading back to the negative terminal?

Good enough, also in a schematic almost all voltages (unless noted differently) are referenced to ground (connect black probe of voltmeter/multimeter to ground).

So don't worry if you see negative voltages somewhere, it is a matter of perspective.

[rstofer]

Of course it's not as simple as that!  There are at least 3 ground symbols in common use.

https://www.rapidtables.com/electric/Ground_Symbols.html

The one we care most about is the COMMON ground - the triangle with the end pointing down.  This is electrical COMMON whether it connects to EARTH ground by way of CHASSIS ground or not.

This is an arbitrary point and it is not always the (-) end of a battery.  I could put a voltage divider around the outside of a 12V battery, ground the center point, call it common ground and the result is +6 and -6 volts relative to common ground.  Think about a center tapped transformer.  6.3V either side of the center tap and 12.6V across the ends.

When you study circuit analysis, ground will usually be obvious and, for homework problems, is generally the lower left corner of the circuit.  But it doesn't have to be!  You can pick ANY point as 'ground'.  Some choices will work better (easier) than others but they will all work.

When you get to op amps and look at single supply variants, you will see that arbitrary midpoint of the power supply created in every design (except for bizarre boundary designs).  It is important to bias the op amp to mid range such that signals can vary above and below ground (AC signals).  Again, this is where w2aew's op amp videos come in handy because he sometimes uses single supply op amps.  For dual supply op amps, ground is at the midpoint between the supplies (usually).

Further in the op amp stuff, you will find out about a 'virtual ground'.  An interesting situation where the op amp tries to create enough feedback on the (-) input to keep the difference between the inputs at 0V.  If the (+) input is grounded (0V) then the (-) input is a virtual ground and also 0V.  This is an IMPORTANT bit to understand.  Kirchhoff's Current Law comes into play here.

The law is simple:  The sum of the currents entering a node is exactly equal to the sum of the currents leaving the node.  In short, current can't pile up nor can it spill on the floor.  Or, what goes in comes out...


13 minutes into this video, you will see a voltage divider on the (+) input of an op amp.  It is probably described much earlier.  I just clicked an arbitrary time and the schematic showed what I wanted to talk about



That voltage divider puts the (+) input at half the supply voltage so the negative feedback will hold the (-) input at that same level.

Why don't you look at the Electrical Engineering path over at Khan Academy?  I did the entire course in a day or so and it's quite good.  True, it was all review but the explanations are pretty good.  If you catch the math for capacitors and inductors, that's great.  If not, there are other tutorials you can work through later.  Don't worry too much about the math, most hobbyists don't need it to enjoy the hobby.  But it is elegant!

What you WILL see is the range of applications of Ohm's Law and how it applies everywhere.

[rstofer]

Or to turn the question round the other way: I take a long nail, wet my hand and jam into the positive hole of a wall socket. I'm likely to get killed, right?

But why? The current can flow into my body, but not back into the negative terminal, so there's no complete circuit so no current flow, right?

You sort of understand the idea of a circuit.  From 'circus' meaning, I imagine, 'around'.  A circuit is complete when there is a path around.  The path may be low resistance or high resistance.  This determines how much current will flow when there is an applied voltage.  But always a circuit - a complete path.

So, when you jam that nail into the wall outlet, assuming you hit the narrow blade, the phase, there is a voltage applied to your body but unless some part of your body has a low resistance path to neutral (ground) there won't be any current flow.  That's the key point!  There is no 'circuit'.  Now, grab a water pipe with your other hand and you're dead meat.

This idea of a short circuit to ground is a lot more complex and I don't want to get into it.

Until now i've been dealing with simple circuits: a battery, one or two resistors, and that's it. Can you guys not at least partly understand how this idea of a ground leading somewhere undefined (at least in my mind) might be somewhat confusing? Is it the case of the device? Or real earth. Or an earthed case ........ or.....?

Cheers! ......... D

We have common or circuit grounds all the time without ever referencing them to chassis ground or earth ground.  Battery circuits will almost never have an earth ground reference unless it comes in from the scope probe.  Why would it?  Circuits will be complete without needing an external conductor to earth.

OTOH, it is common to earth ground all exposed non-current-carrying metallic surfaces of electrical appliances (exemption for double insulated).  That's why the BNC connectors on test equipment are tied to earth ground.  Hook up you scope to your battery powered robot and now things are earth referenced.

Those animations you posted earlier showing a path through real earth are not helpful.  Yes, they are correct but they are a distraction.  We really can't talk about that path with out a lot more understanding of fields and such.  Not important at this point.

Think 'circuit', 'circus', 'around'.  There is a complete path from (+) to (-).  If not, no current will flow.  Voltage may be present at some points but without a circuit, no current flows around.

[Mr D]

Many thanks so far guys, i'm starting to develop the smallest inkling of understanding.

But still some things not making sense.

Attached is a pic with two, independent circuits.

To my mind, the connection of the ground at different points in the circuits should make no difference whatsoever, as they're simply connections to a bus that itself isn't connected to anything, not even to the negative battery terminal.

However, what i observe is that the voltage potential is different in the circuits. How can the voltage potential be effected when the ground is going to a dead-end street (bus leading nowhere)?

[james_s]

Any time you have two or more identical symbols with the same name on a schematic it means those points are all connected together. If you look at the actual physical circuit, "ground" will typically just be a trace like any other on a board that ties all the indicated points together. It may also be connected to the metal housing, but it doesn't have to be. It might connect to earth ground but again it doesn't have to. All the schematic tells you is that those points connect to each other, and calling it "ground" is common for a return path used by a large number of nodes in a circuit. It is usually tied to the negative terminal of the battery but not always, until the 1950s many cars were positive ground, doesn't really make any difference which way the current is flowing as long as every part of the circuit is in agreement. Terms like "ground" are really just imaginary concepts, but you have to pick some point in a circuit to call ground because any voltage measurement has to be the potential between two nodes. If you say a circuit should have 3.5 Volts on a certain pin of a component, that's assumed to be 3.5V relative to the node called "ground", otherwise you'd have to ask "Relative to what?" for every measurement.

[james_s]

Many thanks so far guys, i'm starting to develop the smallest inkling of understanding.

But still some things not making sense.

Attached is a pic with two, independent circuits.

To my mind, the connection of the ground at different points in the circuits should make no difference whatsoever, as they're simply connections to a bus that itself isn't connected to anything, not even to the negative battery terminal.

However, what i observe is that the voltage potential is different in the circuits. How can the voltage potential be effected when the ground is going to a dead-end street (bus leading nowhere)?

The voltage potential is different because when you measure a voltage, you are measuring it relative to ground.  If you define a different point as ground, you will see a different voltage relative to it.

The two circuits are indeed identical from a functional stantpoint, the only difference is in describing quantities. If you are standing in the street level of a building that has a basement below, you could call the place where you are standing "ground" and say that the floor above you is positive 1, and the basement floor is negative 1. Now you could instead decide that it makes sense to call the basement floor ground, then the lobby is positive 1 and the floor above that is positive 2. The building configuration has not changed, only the perspective from which you describe it. In order to specify the floor level you have to agree on what your measurement is relative to and in most cases it's relative to the ground, ie what you're standing on when you're outside the building. It doesn't have to be though, you could call the top floor 0 and assign negative numbers to any lower floors but it would likely confuse somebody else if you asked them to meet you on the negative 7th floor because the general assumption is that floor 1 is the ground level floor.

[Mr D]

So are you saying that in reality, in that last pic i posted, the two circuits are actually identical, and the reason the EveryCircuit app shows a different voltage potential in the two circuits is that it's imagining putting the black lead of it's DMM in a different place?

So wherever you choose as "ground", if you put your black DMM lead there, the DMM will call that point 0v and work out the rest accordingly?

[james_s]

Yes, that's exactly it. Whatever we call "ground" is the 0V or reference point, to which all other measurements in the circuit are relative.

You can also take a voltage measurement between any two points you want, so you might say "the voltage across resistor R3 should be 0.8V" and that would be perfectly valid, but if you were to just say "the voltage at the top of R3 is 0.8V" that implies relative to the node called ground, not necessarily across R3. Note that my "R3" example is completely arbitrary and not in any way related to the specific circuit(s) you have posted.

When dealing with devices powered by the national electrical grid, ground also happens to be connected to earth for safety and practical reasons but with something powered by an isolated source like a battery or a transformer "ground" can be anything you want as long as you are consistent.

[Mr D]

Great, many thanks to all who replied, i feel i've come a long way in one day as this morning it made no sense to me whatsoever and now i feel like i have a bit of a handle on it!

And now to bed!

And to be continued!

[Brumby]

It is unfortunate that terms like "Earth", "Ground" and "Chassis" are all too often used to simply make reference to a common connection.  When the familiar definition of these terms is not applicable, it really can be confusing.

This common connection is also frequently given the role as the reference point for other measurements.  While this is not a guaranteed relationship, it would be rare to come across a schematic which did not do this.

The only time you can really be certain of a common connection being your reference point is when you have something described as 0V - such as a 0V power rail.

In practical terms, this reference point is (as you have worked out) where you would put the black lead from your DMM.

As you become more familiar with schematics and circuit topologies you will be able to make better judgements about reference points and measurements - but the first steps are understanding the simple things.  Each of these can be mysterious and confusing, until it 'clicks'.  Those are wonderful moments.

[Mr D]

Thanks!

I feel like i've got just enough understanding to start looking at some real circuits.

Could anyone tell me: in the circuit below, how come i see no power source?

Or is that triangular element in the middle the power source? But it's not a battery, so what is it?!

Yeah yeah, this is a super-nooby question!!

[rstofer]

The triangle is an op amp and the power supplies (probably +-15V) are not shown.  This is typical of all op amp circuits used in tutorials.  There might be an exception when the circuit uses a single rail op amp.  Then the + power input will have some voltage and the - power input will have a ground symbol.

That' an awfully advanced circuit considering...

Here's a circuit that shows the power supplies:

http://www.learningaboutelectronics.com/Articles/How-to-connect-a-LM741-op-amp-chip-to-a-circuit

[Mr D]

OK!

I'm not trying to understand the circuit, i'm trying to understand and make sense of the elements and general structure!

So is the circuit only missing the power supply to make it a complete, viable circuit?

[rstofer]

It looks complete.  The tricky part is that incandescent lamp in the upper left corner.  That device is used as a non-linear resistor.  When cold, it has very low resistance which means the op amp has very high closed loop gain.  When the circuit is oscillating normally, the resistance will increase (because the current increased) until everything balances out with a value of Rf/2 so the op amp closed loop gain should be around 2.

[Mr. Scram]

OK!

I'm not trying to understand the circuit, i'm trying to understand and make sense of the elements and general structure!

So is the circuit only missing the power supply to make it a complete, viable circuit?

Power supplies, plural. Most op-amps require a postive and negative supply. Have you checked out Dave's op-amp video? You might not understand all or even most, but it helped me when I was completely unfamiliar with what they are and how they work. Knowing the common configurations helps a lot when looking at schematics.

[Mr D]

Haha, you're getting way ahead of me there! (Mr Rstofer)

I'll check of Dave's vid.

OK, i need some more fundamentals filled out:

I understand how a simple circuit with a bulb works. It's a closed loop with a resistor (the bulb) converting energy into light, so the circuit is useful.

But if i imagine an oscillator, how does the oscillating voltage get on to the next part of the circuit? How does it move on to an amplifier, while still keeping the closed loop back to the negative battery terminal?

Or is induction used to have several closed loops of circuit communicate information (voltage difference) between each other while being part of a bigger schematic consisting of several such closed loops?

[rstofer]

Vout is shown on your schematic and while I don't want to even think about the filters (the Z equations), at the end of the day, Vout is produced from the op amp  power supplies through a bunch of transistors inside the op amp.

So, Vout and Gnd (part of power supplies not shown) goes on to whatever comes next.

When I said +15 and -15V, those voltages are relative to something.  That something is the ground created where the two supplies get connected together to essentially create a 30V supply with a center tap to ground.

Vout is relative to ground (not shown).   But the op amp itself doesn't care about ground, doesn't need to know about ground and isn't connected to ground.  That's why op amp videos spend so much time on this fact.

[Mr D]

Ok, so I guess a circuit is always a single interconnected network, with multiple ground points? So the goal is to start with a power source with a steady voltage difference and eventually output either light through a bulb, or audio via an oscillating voltage difference to a speaker, or information through a 8 seg LED, or whatever. But always there's the current running to ground after the bulb or speaker or 8seg display?

But then i don't see how such a complicated scheme could work. You change one resistor value somewhere and won't that have a knock-on effect thoughout the rest of the circuit?? How can this be manageable in a complicated circuit? Or are there ways to sandbox parts of the circuit against this sort of chaos?

[Brumby]

OK!

I'm not trying to understand the circuit, i'm trying to understand and make sense of the elements and general structure!

So is the circuit only missing the power supply to make it a complete, viable circuit?

To answer this question directly: Yes.

The actual power situation is also a bit obscure - with this describing what's around, but not shown:

So, Vout and Gnd (part of power supplies not shown) goes on to whatever comes next.

When I said +15 and -15V, those voltages are relative to something.  That something is the ground created where the two supplies get connected together to essentially create a 30V supply with a center tap to ground.

Vout is relative to ground (not shown).

The only hint is the symbol at bottom left which indicates a connection to the "ground" point - which is the 0V point of the +15V / 0V / -15V power supply.  The +15V and -15V power connections to the op-amp are implied - because it is known that an op-amp will need some sort of power supply.  Also, the output (voltage) will be referenced to something - because it has to be referenced to something - and this is commonly the 0V point.


These connections are omitted in many cases to simplify the schematic, so that the important aspects of it - the signal path, biasing and so on - aren't cluttered by connections we know must exist.

Sometimes you will see the full set of connections for everything; sometimes you will see nothing of the power connections ... and sometimes you might have short lines going nowhere with labels, such as this:



While not explicitly shown those +12V points are connected together and go to the +12V point of the power supply.  Same for the -12V points.

Don't worry about the V_SRC "battery".  That is just a way of showing an input voltage, often used in teaching situations.

[rstofer]

Yes, making a component change will change how the circuit operates, to some extent.  The thing is, real circuits are made up of independent modules interconnected in some way

Take LEDs connected to a uC.  The dropping (ballast) resistor is calculated at some point in the process and it is based on two factoids:  The V_f of the diode at the desired I_f (essentially intensity).  Then we calculate the series resistor to drop V_cc the proper amount (V_cc - V_f) when I_f is flowing.  If we want to be pedantic, there is a bit of voltage drop inside the uC to account for.

Having done all that, nothing changes even if I change clock frequency or the input voltage to the analog to digital converter (ADC).  They are in different modules.  Now, if I multiplex the displays (common with 7 segment displays), the calculation is a lot more involved because each segment is only illuminated for some percentage of time.  So the current needs to be higher and the resistor needs to be lower.  And so it goes.  Ohm's Law at work.

We don't tend to make things interdependent by having series resistors if we can avoid it.  Sometimes they are necessary and we know when they are.  Watch w2aew's transistor videos.  Pay attention to how he biases the base in the Common Emitter Amplifier.  And how gain is controlled by the ratio of the collector and emitter resistor.  Of course, the collector resistor also has to represent the load being driven - a subject for later on.

[Brumby]

But always there's the current running to ground after the bulb or speaker or 8seg display?

I'm not sure what you mean by "after".  Electricity (except for static discharge) flows through a circuit.  Any output to a bulb, speaker or display will put a current through that circuit element (typically) to ground.  It's not really "after" - it's an essential part of the single circuit function of making the speaker cone move or the bulb to light up.

But then i don't see how such a complicated scheme could work. You change one resistor value somewhere and won't that have a knock-on effect thoughout the rest of the circuit?? How can this be manageable in a complicated circuit? Or are there ways to sandbox parts of the circuit against this sort of chaos?

This is a good question - and leads us into the topic of impedance.  Actually working through this in a "complicated circuit" can become a bit of a rabbit hole - so be careful about jumping in too deep!  However, you don't need a bagful of maths to understand the principles.

Interaction between stages (where a "stage" is a group of components that do a specific function within a larger circuit) is, indeed, a consideration in circuit design and the primary factor involved is impedance.

Let's look at a circuit that processes a signal.  It will have an input and it will have an output.  That output will, in turn, be used as an input to a following circuit.  Now the question is - how much will the following circuit affect the circuit we are looking at when that input is connected to our output?

The answer is impedance.

If the output of our circuit has a *"high" impedance, then the signal it produces will look like it is coming through a *high value resistor and it will not be capable of providing much current, before the voltage drops down to an *"unusable" level.  If it has a *"low" impedance, then the signal will look like it is coming through a *low value resistor and it will be able to provide a *fair bit of current with the voltage dropping slightly.

If the input of the following circuit has a *"low" impedance then it is going to look like a *low value resistor that is going to want to pull a *significant amount of current, but if it has a *high impedance, then it will not need much current for it to be able to process the signal.


The simplest way to look at this is with the following diagram.  It is very simplistic and will not reflect what you will actually see in schematics - but it demonstrates the principle.



Here, Z₁ will be the output impedance of the previous circuit and Z₂ the input impedance of the circuit that follows.  How these impedances actually come about are the subject of some maths - maths that can be bewildering - but, for the sake of simplicity, just think of it like this and you will get the "feel".


Plugging in High/Low Input/Output combinations, you should notice the preferred approach is for the output impedance of one circuit be low compared to the input impedance of the following circuit as this will minimise the effect of connecting the following circuit to the one we are looking at.  The greater this ratio, the less the interaction will be.

This principle applies not only to electronics, but to an extremely wide range of physics as well.  However, sticking to electronics, this matter of impedance is pervasive.  It is everywhere.  If we take a simple example of a PA system, we have to consider the output impedance of the microphone and the input impedance of the pre-amp; the output impedance of the pre-amp and the input impedance of the power amp ... and then the output impedance of the power amp and the speaker impedance.

But this issue of impedances also exists within any circuit.  Each section of circuit that performs a particular job will have an input impedance and an output impedance.  For example, inside the pre-amplifier, you might have a transistor or two to boost the signal from the microphone to a level that makes it easier to work with.  This signal might then go through some tone control circuitry.  Then it might need a circuit that looks like an amplifier, but doesn't increase the signal level.  Such circuits are often used as a "buffer" where its input impedance is such that it does not affect the previous circuit (which may have a *high impedance) but provides a *low impedance output for whatever is subsequently connected.  (It is not uncommon to see unity gain buffer amplifiers - and that is their specific purpose.)

As these chain along the circuit, these impedances must be taken into consideration to ensure the signal is passed along in a useful manner.

* These terms are relative.  (This is important!)  What's painfully high in one situation might be insanely low in another.


This issue of impedances also extends into digital circuits and is the effective issue behind them as well.  We see them expressed more as the current capability of an output or the specification for an input which makes life easier as it hides away the internals (unless you really want to go deep) and just provides the numbers you need to know for a practical circuit.