Author Topic: Some noob questions  (Read 14166 times)

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Offline Mr DTopic starter

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Some noob questions
« on: July 16, 2018, 07:25:36 pm »
***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?

-------------------------------------------------

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.
« Last Edit: July 16, 2018, 07:27:31 pm by Mr D »
 

Offline Ice-Tea

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Re: Some noob questions
« Reply #1 on: July 16, 2018, 07:36:27 pm »
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.

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #2 on: July 16, 2018, 07:45:22 pm »
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?
 

Offline Ice-Tea

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Re: Some noob questions
« Reply #3 on: July 16, 2018, 07:54:14 pm »
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...

Offline Old Printer

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Re: Some noob questions
« Reply #4 on: July 16, 2018, 07:59:45 pm »
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).
 

Offline Terry01

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Re: Some noob questions
« Reply #5 on: July 16, 2018, 08:00:36 pm »
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.   ;D
Sparks and Smoke means i'm nearly there!
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #6 on: July 16, 2018, 08:06:40 pm »
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?

 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #7 on: July 16, 2018, 08:11:20 pm »
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?
 

Offline bsudbrink

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Re: Some noob questions
« Reply #8 on: July 16, 2018, 08:38:20 pm »
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.
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #9 on: July 16, 2018, 08:52:12 pm »
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?
 

Offline Terry01

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Re: Some noob questions
« Reply #10 on: July 16, 2018, 09:03:32 pm »
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  :)
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Offline bsudbrink

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Re: Some noob questions
« Reply #11 on: July 16, 2018, 09:05:10 pm »
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.
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #12 on: July 16, 2018, 09:09:04 pm »
@Terry: i've read quite a few times that voltage absolutely doesn't flow. But you're talking a flow of volts.

???
 

Offline PA4TIM

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Re: Some noob questions
« Reply #13 on: July 16, 2018, 09:11:04 pm »
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.  ;)

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Online rstofer

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Re: Some noob questions
« Reply #14 on: July 16, 2018, 09:18:09 pm »

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.

Quote

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.
 

Offline PA4TIM

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Re: Some noob questions
« Reply #15 on: July 16, 2018, 09:22:47 pm »
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.
« Last Edit: July 16, 2018, 09:37:28 pm by PA4TIM »
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Offline Mr DTopic starter

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Re: Some noob questions
« Reply #16 on: July 16, 2018, 09:28:20 pm »

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?
 

Online rstofer

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Re: Some noob questions
« Reply #17 on: July 16, 2018, 09:28:54 pm »
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.
 

Offline PA4TIM

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Re: Some noob questions
« Reply #18 on: July 16, 2018, 09:36:47 pm »
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.

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Online rstofer

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Re: Some noob questions
« Reply #19 on: July 16, 2018, 09:39:18 pm »

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.
« Last Edit: July 16, 2018, 09:44:32 pm by rstofer »
 

Offline PA4TIM

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Re: Some noob questions
« Reply #20 on: July 16, 2018, 09:41:21 pm »


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  >:D
« Last Edit: July 16, 2018, 09:45:52 pm by PA4TIM »
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Offline Mr DTopic starter

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Re: Some noob questions
« Reply #21 on: July 16, 2018, 09:47:04 pm »
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??
 

Online rstofer

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Re: Some noob questions
« Reply #22 on: July 16, 2018, 09:50:14 pm »
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.
« Last Edit: July 16, 2018, 09:58:43 pm by rstofer »
 

Offline bsudbrink

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Re: Some noob questions
« Reply #23 on: July 16, 2018, 09:51:23 pm »
Nothing "pulls".  Everything "pushes".  Except maybe inductors but don't try to understand those yet.
 

Offline PA4TIM

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Re: Some noob questions
« Reply #24 on: July 16, 2018, 09:52:32 pm »
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.

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« Last Edit: July 16, 2018, 09:59:22 pm by PA4TIM »
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