# positive and negative terminal of voltage regulator

Discussion in 'General Electronics Chat' started by sharanbr, Jul 4, 2015.

1. ### sharanbr Thread Starter Active Member

Apr 13, 2009
76
1
I am sorry for posting such a basic question.

When I look at a power supply, I can only think of electrons flowing through a load.

Please see the attached figure. What really flows through the resistor in this figure?
I believe positive terminal of voltage source cannot source electrons.
Does it mean that ground here can source electrons to node B of resistor?

Also, in the figure, is the upper terminal of resistor A is at higher potential or terminal B?

Regards,

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2. ### ericgibbs Senior Member

Jan 29, 2010
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Have you considered that the positive terminal may have less electrons than the negative terminal.? so electrons will flow from negative to positive, via the resistor.

3. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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Inasmuch as apprehension of charge carrier behavior is complex, non-intuitive and unnecessary for these purposes, I advise you to focus upon charge flow without regard to the carriers! -- Moreover, please be warned that 'acquisition' of "extra-quantum level" carrier-centric paradigms will complicate your advanced studies with the necessity of unlearning faulty tuition...

With constructive intent
HP

Last edited: Jul 4, 2015
4. ### WBahn Moderator

Mar 31, 2012
17,441
4,696
Early in your learning it is quite natural to focus and mentally picture the flow of electrons in a circuit. The problem is that you have to constantly keep adjusting your work for the fact that electrons are negatively charged (by arbitrary definition thanks to Ben Franklin who had to basically make a 50/50 guess). Whether a positive amount of charge moves from right to left or a negative amount of charge moves from left to right, for nearly all purposes (there are exceptions) the effect is the same in that whatever is on the left is now more positively charged than it was and whatever is on the left is now less positively charged than it was. So, in almost all situations, we focus on the flow of charge and not on the form of the carrier that transports it. This is called conventional current and it won't take long before that is what will be natural for you to think in terms of.

But yes, physically, a battery (via chemical reactions inside) move electrons from the positive terminal and deposit them on the negative terminal. This results in the voltage difference we see between the terminals and, when connected to a circuit, that voltage difference results in electrons flowing from the negative terminal, through the circuit, and back to the positive terminal, where the chemical reactions once again move it from the positive terminal through the battery back to the negative terminal where it can go through the circuit again and again (but note that the same exact electron almost never makes even one complete circuit just due to the shear number of electrons available).

5. ### dl324 Distinguished Member

Mar 30, 2015
3,104
599
This is correct, electrons flow from the negative terminal on the battery through the resistor from B to A. You may find things less confusing if you think of conventional current flow, which flows from '+' to '-', instead of electron current.
Correct.
It might be less confusing if you connected the two grounds like this:

Using conventional current, a current of 1mA flows from the positive terminal of the battery through R1, generating a voltage drop of 1V across the resistor. Electrons flow in the opposite direction.
Yes

6. ### sharanbr Thread Starter Active Member

Apr 13, 2009
76
1
Dear All,

Thanks a lot. The answers are very useful.

One more question, from the figure above, since negative terminal of voltage source and resistor are connected to ground, I am assuming that the negative terminal does supply electrons. I get this doubt often as I see ground always as a point which sinks charges. Probably, I am confusing earth and ground a lot ... I don't know ...

7. ### ericgibbs Senior Member

Jan 29, 2010
2,413
369
Hi
Look at the circuit in post #5, no grounds are used.
Do not think of ground as a charge sink.

8. ### dl324 Distinguished Member

Mar 30, 2015
3,104
599
Having multiple ground symbols as in your original drawing are for convenience/clarity. All nets with the same name should be connected in the actual circuit. For a beginner, it's easier to visualize the loop with the grounds connected.

9. ### MikeML AAC Fanatic!

Oct 2, 2009
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For the purpose of analysis by Ohms Law, the current is 1mA from top to the resistor to the bottom of the resistor in all of these various circuits. I'm sure that there are several more ways this same circuit can be drawn.

Be sure to read the part of my signature line that deals with current...

10. ### MikeML AAC Fanatic!

Oct 2, 2009
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Here is an exercise for you:

The creators of this simulator program are engineers and scientists, not technicians (who sometimes cannot get beyond thinking about which direction electrons swim). In creating the simulator, they had to use consistent concepts of voltage and current (and which direction current flows). If you understand why the voltages and currents in this simulation are what they are, you will have come a long way in figuring out circuits...

I am using this simulation as an illustration. The simulator has found voltages at various nodes, and reports the magnitude of current that flows in the circuit elements. Note that there is a consistent method of naming the nodes (node1 to node5, mike) and the elements (V1 to V5, R1 to R5). By convention, all the voltages are reported by the simulator with respect to the ground node (the upside down triangle symbol).

Now, it is time for you to do some work:

1. Which direction does the current flow in each resistor R1 to R5? What is the magnitude of the current? The sign of the current? Check your work against the values reported by the simulator.

2. Why is the voltage at node5 negative?

3. How does current flow from the bottom of R3 to the negative pole of V3?

4. Why is I(R6) (the current in R6) ~= zeroA? Why is V(node4) ~= zeroV?

5. How does current get from the positive pole of V4 to the non-grounded end of R4?

Last edited: Jul 4, 2015
11. ### crutschow Expert

Mar 14, 2008
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A ground is neither a source nor sink for carriers.
It is simply a connection point that ties everything connected to ground together.

12. ### WBahn Moderator

Mar 31, 2012
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The "ground" symbol should really be called "common" or "reference". It's merely a node that we assign a voltage of 0V to. There truly is nothing beyond that. All of the nodes in the circuit that are connected to that symbol are simply the same node just as if we had left the symbol off and connected them all with wires on the schematic.

Even if it were truly shove-a-stick-in-the-earth ground, that wouldn't change anything. You can think of the earth as a huge reservoir of electrons from which we can add to or subtract from as need be. This is literally the case for high voltage DC monoline power transmission lines that run a thousand miles or more -- the electrons are take out of the ground at one end and put into the ground at the other. It does result in a net drift of electrons in the ground but the earth is such a massive store of electrons that the current is immeasurable at distances measured in dozens of feet from the grounding points.

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13. ### sharanbr Thread Starter Active Member

Apr 13, 2009
76
1
Ir1 = V1/R1 = 1/1000 = 1 ma
Direction - node1 to ground

Ir2 = 2/1000 = 2 ma
Direction - node2 to ground

Ir3 = V3/R3 = 3/1000 = 3 ma
Direction = node3 to node 4

I believe potential at node4 is 0 and hence no current flows through R6

I4 = 0 as no load is connected across voltage source

I5 = (0-5)/1000 = - 5 ma

Ir4 = 0 as no potential exists

All current directions are conventional current flow.

Here the reference is 5 voltage. Hence 5 volt is effectively 0 volt and voltage at node 5 is 0-5 = -5 volts

there is no potential difference across R6. Hence all current flows back to negative terminal of V3

Ir6 = -1/1000 = - 1 ma. So, I don't know why current is expected to be 0 here.

Again, not sure about V (node4) as voltage here is 1 volt

Sorry. I don't follow the question.

14. ### ian field Distinguished Member

Oct 27, 2012
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767
Conventional current was thought up before anyone had heard of the electron, so everyone perceived it as current coming out of the positive terminal (whoever decided which was to be the positive terminal had a 50:50 chance of getting it right).

Later as science developed, they realised that current flow was the movement of electrons - which carry a negative charge.

The idea of conventional current flow and current coming out of the positive terminal, was well established in my mind even before I got to school - way back then when things like that were tricky to keep clear in my mind, I kept reminding myself that its the negative cathode in a thermionic valve that emits the electrons - and those electrons are actually what the current flow is.

Its a bit of a history lesson - but that's part of how I remembered which one wasn't the other.

15. ### WBahn Moderator

Mar 31, 2012
17,441
4,696
The reasoning here is actually backwards. No current flows through R6 because there is no path through the ground node back to the other end of V3. Since no current can flow there can be no voltage across R6.

Just as all of the nodes connected to a common (ground) symbol are actually the same node, you can assign a name to a node (called a "net name") and all nodes with that same name are connected together. Rethink your answer with this in mind.

Your reasoning is fine, but you aren't expressing if correctly. The reference, by definition, is 0V, so it makes no sense to say that the reference is 5V (and if you DO say that, then the voltage at node5 would have to be 0V, do you see why). The V5 supply between COM and node5 is such that

(Vcom - Vnode5) = 5V
Vcom = 0V
Vnode5 = -5V

At this point you start going off the rails:

But you accurately described that the current in R6 would be zero previously, so where are you getting -1/1000 (which should be -1V/1000Ω -- units matter).

Again, this didn't give you problems earlier. Where is the 1 volt coming from.

One thing to note about nearly all simulators -- by longstanding convention simulators report current as the conventional current INTO a pin. When you ask about the current in a two-terminal device (such as a voltage source or resistor), it tells you the current into pin 1 of the device. For voltage sources pin 1 is the positive terminal. For a resistor it is merely one of the pins and, unfortunately, the standard symbol in most libraries doesn't distinguish between the two pins. So here the signs of the currents in the resistors is ambiguous. In my libraries I always add a small tick mark to the symbol on the pin 1 side so that I can better interpret the simulation results.

Sorry. I don't follow the question.[/QUOTE]

16. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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Be advised, however, that although the above may, for the intents and purposes at hand, be regarded as the 'net effect' - it is, nonetheless, a significant oversimplification - FWIW electron behavior in conductors, semiconductors and ionic fluids (e.g. electrolytic solutions and plasma) is crudely akin to 'domino action' (which is itself is a gross oversimplification) -- While the pedagogical merit of this level of 'detail' at the introductory level is in dispute, I know I would have greatly benefited (in the daze [sic] of my youth) had my tuition in this regard been reflexive of unadorned reality...

Respectfully, with best regards
HP

Last edited: Jul 4, 2015
17. ### ian field Distinguished Member

Oct 27, 2012
4,310
767
The person who decided current comes out the positive terminal (and/or which terminal to call positive) probably scuppered the invention of the time machine, so no one can go back in time and bump him off.

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18. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,710
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Indeed the 'dual perspective' corollary to non-intuitive carrier polarity assignment can be annoying! --- However it is the (well-nigh ubiquitous) confusion of charge with charge carriers that I see as most problematic...

À la 'holes' which, if such analogy (read crutch) must be employed, would more perspicuously be termed 'bubbles' --- Seldom does a month pass sans the necessity of my pointing out (i.e. insisting upon) the distinction between 'holes' and positrons! To folks who should know better!

Best Regards
HP

19. ### crutschow Expert

Mar 14, 2008
12,497
3,058
The first main practical effect of initially having reversed the positive and negative designations to the correct value from what they are now, would be when dealing with vacuum tubes.
In that case (with electrons being designated as having positive charge) then the first tube (valve) circuits would have used a negative supply (assuming cathode to common) and all the carriers (positive electrons) would have flowed from the common to the negative supply terminal.
Thus we would have needed to think in terms of current flowing from the bottom to the top of the schematic (assuming the common would still be placed at the bottom).

This would have likely continued with solid-state circuits, with negative power supplies and the carriers going from common to the supply (except notably for ECL circuits which generally used opposite polarity supplies).
Personally I think that would be harder to visualize then our present system of positive carriers from positive (supply) to negative, but perhaps that's just because I've thought top to bottom (sort of a water analogy) all my life.

Or would we have invented imaginary (now negative) carriers so we could still have the carriers going from top to bottom?

Last edited: Jul 4, 2015
20. ### MikeML AAC Fanatic!

Oct 2, 2009
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People who insist that all charge is carried by electrons have never worked on a ion implant machine (as used in the semiconductor industry).