Some noob questions

[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.