Newbie Q re physical current flow

dc

Member
2001-12-26 9:04 pm
NYC
I've been trying to teach myself electronics for several months now, but continue to run into the same problem with current flow. I've bought and read a number of different books and even a couple of cd-rom programs, hoping that one of them would state things in just the right way so I would understand. But each of these sources seems to treat current flow in a slightly different manner and I have not been able to reconcile them. So, I'm posting, admitting to the world my lack of knowledge and inability to understand electronics fundamentals *GASP*.

On to the question:

In a basic DC circuit with one battery and one resistor, does the current flow out of the positive terminal, or out of the negative terminal? I've read that electrons flow out of the negative, and push their neighbors further down the circuit, while those electrons nearest the positive terminal are attracted by the high positive potential (or relative lack of electrons) effectively being sucked back into the battery. So, the negative terminal has a greater potential than the positive terminal (is this correct usage of the term "greater potential"?) and it pushes electrons through the circuit while the positive terminal effectively pulls the electrons back into the battery. Seems to make sense to me, so far.

But, if this is what is physically happening in a circuit, what is the explanation behind physical current (sometimes called conventional current), which, from what I've read, seems to flow from the positive terminal back to the negative? Is physical current purely abstract; not representing anything that's actually physically occurring within the circuit? (if so, seems to me to be a bit of a misnomer). Or is it somehow tracking the flow of positive charges? I've read that the preferred method for circuit analysis is to try to follow physical current flow, but that direction you assign for current flow doesn't matter, as long as your analysis is consistent. Is this true?

Related question: Does current flow from a positive DC voltage source, such as a +12V to ground (for instance, in a voltage divider network or a common emitter circuit)? In which direction does current flow between ground and a negative DC voltage source (-12V)? In other words, I can see how current could flow from a voltage source to ground, but I don't understand how current could emanate from ground and flow into a voltage source...

I'm guessing that there's a simple, one sentence answer to all of the above, but I'm just not seeing it. Any help would be greatly appreciated.


Thanks,

brad
 
Hi dc,

Back when scientists in the 18th century were investigating electricity they had to be able to consistantly reproduce each others experiments. To do that meant they had to be able to say "hook the + lead to this and the - lead to that and something happens" and have everyone else know what they meant

So somebody (I believe it was Ben Franklin) took a battery and said this terminal is plus and this terminal is minus. Since nobody really understand electricity yet, he had a 50/50 chance of getting it right, but missed. After a century or so the convention was cast in concrete and it was too late to change it when they figured out electron flow.

As far as electronics are concerned none of this really matters as long as everyone uses the same convention. Electrons flow from negative to postive and "holes" flow from positive to negative. Unless you are dealing with semiconductor design this really doesn't have any impact on what you are doing. Besides, audio work is mostly about AC anyways.

Phil
 
As haldor said: The convension is that current flows from plus to minus allthough the physical electrons flows the other way.

Since ground (0 V) is "more positive" than -12 V then the above rule applies here too. Therefore the current flows from the ground terminal to the -12 V terminal.

In both cases I mean through a load like a resistor or so connected between the terminals.

Hope this makes sense :confused:

/Marcus
 

dc

Member
2001-12-26 9:04 pm
NYC
Thanks for the replies.

What you've written makes sense and is in keeping with what I've read, but, I still don't quite understand one thing. Below I've attempted to describe my frustrations using a voltage divider example:

1. Imagine there is a voltage divider circuit with +12V at the top, then three resistors of equal ohmic value connected in series (R1, R2 and R3). R3 is connected to ground.

2. R1 drops 4V. (Is it wrong to imagine that there are 8V left after the current leaves R1 (12-4=8)?)

3. R2 drops 4V. (4V left after R2)

4. R3 drops 4V and the voltage is at 0 as it should be as it enters the ground.

Using physical (conventional) current to analyze this circuit, current is flowing from the +12V voltage source to the ground.

But, if you look at electron flow and what is actually happening in the circuit, does this mean that the electrons are being "sucked" out of the ground (usually a bolt through the bottom of the case of the amplifier)? It just doesn't seem quite right to me that electrons would be sucked out of the case.... Wouldn't that make the case charged (doesn't losing electrons give something a positive charge)? Because ground serves as a reference point, both the + and - leads need to be connected to ground, right? So maybe the + and - connections to ground serve to cancel each other out, leaving the case (or star ground) with neither a + nor a - charge?

I can see it if both ends of the circuit are attached to terminals of a battery (one terminal pushing electrons out and the other terminal sucking them back in), but when one end is attached to ground, it doesn't seem like it would work.

Also, when one part of the circuit is said to have "greater potential" is this generally the part of the circuit which is positive in relation to the rest of the circuit?

Thanks again.

brad
 
hi

(Is it wrong to imagine that there are 8V left after the current leaves R1 (12-4=8)?)
No!

Wouldn't that make the case charged (doesn't losing electrons give something a positive charge)
now you got it mixed up electrones goes from -to+

I've read that the preferred method for circuit analysis is to try to follow physical current flow, but that direction you assign for current flow doesn't matter, as long as your analysis is consistent. Is this true?
yes(but I always imagine that current goes from + to - then the diodsigns will make sense:)

Also, when one part of the circuit is said to have "greater potential" is this generally the part of the circuit which is positive in relation to the rest of the circuit?
Yes (but if "the rest of the circuit" should be true then it should state "greatest potential".
So maybe the + and - connections to ground serve to cancel each other out, leaving the case (or star ground) with neither a + nor a - charge?
Remember voltage is relative so your ground is positive relative to your -.
This relative thinking is important .

Keld
 

dc

Member
2001-12-26 9:04 pm
NYC
Keld,

Thanks for the response!


quote:
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Wouldn't that make the case charged (doesn't losing electrons give something a positive charge)


No!


now you got it mixed up electrones goes from -to+

Either I'm not understanding you here, or I wasn't very clear and made you misunderstand me. I meant to say that if a + voltage source, such as a +12V source, is connected to ground, then electrons are flowing from ground to the +12V voltage source(from a position which is negative, relative to the +12V voltage source to the part of the circuit with the greatest potential), right? So, that must mean that the electrons are being taken from the ground? Once those electrons are taken from the ground, wouldn't that leave the ground with a positive charge?

It's this idea of taking electrons from the ground that makes me uncomfortable.

However, it seems that if there were also a - voltage source (a total of two rails; the +12V circuit mentioned above and a mirror image -12V voltage divider circuit) things might balance out. For, as you pointed out, ground is positive relative to a -12V voltage source. If that's true and that -12V voltage source is also connected to ground through 3 resistors of equal ohmic value (as in my previous post), this negative voltage source would contribute electrons to ground (electrons flowing here from the -12V voltage source through the voltage divider into the relatively more positive ground; the part of this voltage divider circuit with the greatest potential). It seems that electrons in the -12V voltage divider circuit would be contributing electrons to ground at the same rate that the +12V voltage divider circuit is taking them out. Thus, electrons would in effect, be flowing from the -12V voltage source, through 3 resistors to the ground and then through another 3 resistors to the +12V voltage source.

Am I even close???
 

dc

Member
2001-12-26 9:04 pm
NYC
So, in sum:

1. Ground will always be neutral (0 - neither negative nor positive) regardless of what is connected to it;

2. It's best to assign current flows from positive voltages to negative voltages when analyzing circuits, despite the fact that

3. Electrons actually flow from a negative voltage to a positive voltage.

If my statements in other posts were all correct, then my questions have been answered. Thank you all very much for taking the time to respond!

brad
 
The thing about ground is that it is part of the closed chain of components and interconnects that form the current path. It ALWAYS has to be a closed loop. Even in the supply. It is easier to see with a battery (you mensioned that you could see how it worked with a battery but not ground).

The ground is of course "refilled" with electrons by the transfomer or some other power source.

Imagine two series connected batteries of 12V each. If the connetion between the batteries is choosen to be ground then it is. Then you have a +/-12V supply.

But if the negative connection of the "lower" battery is choosen to be ground, then you have a 0 V/+12 V/+24 V supply.

Did I mess things up now?

/Marcus
 
Maybe this helps ...

Let's forget about those silly electrons and state that current flows from higher potential to lower potential, OK? I.e. from the + terminal of a battery to the - terminal.

If you attach a resistor over the battery, and float the lot in free space, then current leaves +, goes through the resistor (voltage drops), enters the - terminal, and is by the battery's action pushed up to + again. Right: within the source current goes from - to +, that's why the source has to work. (incidentally, that's why in most amplifiers the power reservoir caps are in the signal path: they carry the actual AC signal current!)

If you take the same setup and add an additional wire between battery - terminal and a convenient point you call ground or earth, then the current loop remains exactly the same (no electron thingies going in or out of earth), only now the related voltages can be referenced: e.g. 0 V at the - terminal, 12V at the +. If you add a wire from + terminal to earth you have a -12V supply.

Think in loops. Every current always is in a loop. Otherwise things get charged, which does not belong in audio south to my electrostatic loudspeakers


:)
 
dc said:
Keld,

I meant to say that if a + voltage source, such as a +12V source, is connected to ground, then electrons are flowing from ground to the +12V voltage source(from a position which is negative, relative to the +12V voltage source to the part of the circuit with the greatest potential), right? So, that must mean that the electrons are being taken from the ground? Once those electrons are taken from the ground, wouldn't that leave the ground with a positive charge?


Am I even close???

I think I understand what is tripping you up. The plus terminal is only positive compared to ground when the negative terminal of the voltage source is connected to ground.

If you connect the plus terminal to ground (directly or through your resistors) with the negative terminal disconnected (floating), the postive terminal will be at ground potential and the negative terminal will be at -12V. No current will flow through the resistors (except for a tiny burst of current to charge up the internal capacitance of the battery when you first make the connection).

When you connect the negative terminal to ground, current will flow through the resistors and the postive terminal will measure +12V. The ground connection is irrelevant to the current flow, the point is that current flows from one terminal of the generator to the other. Either terminal (or somewhere in-between can be connected to ground) without changing this.

If you have a voltmeter (or a 12 V lightbulb) you can try this for yourself and see how it works.

Phil
 

dc

Member
2001-12-26 9:04 pm
NYC
Haldor,

Your last reply was new to me, so it took me a while to figure out what you meant. I'd like to restate your message to see if I've got it right.

1. A circuit is attached to a 12V battery. The potential difference between the + terminal and the - terminal is 12V, regardless of the direction assigned to current flow.

2. The battery is removed and replaced with an AC source. The AC comes through a wall outlet and is connected to a step-down transformer with a center tap. The center tap is connected to chassis ground and the secondaries are fed through a bridge rectifier, and then connected to filter caps to reduce ripple remaining on the line. After the caps, there is a total of 24V potential difference between the secondaries. Because the center tap is equidistant from either secondary and the center tap is connected to ground, one secondary becomes a +12V rail and the other one is a -12V rail.

3. If only the +12V is attached to, say, a common-emitter circuit, and the -12V is left unattached, then there is no closed circuit, and current will not flow? Based on schematics of partial circuits I've seen in textbooks, I was under the impression that ground could serve as a return, thus closing the circuit and allowing current to flow. After all, ground, at 0V has less potential than the +12V rail, no?

Is it accurate to read Haldor's last post to mean that current will only begin to flow once both the + and - leads are somehow connected, whether or not via the ground? When both leads are connected, there is a 24V potential difference, regardless of whether the circuit is connected to ground, correct? If the center tap is not placed in the center of the transformer, but placed so that spacing between one secondary and the center tap is 3 times larger than between the center tap and the other secondary (75/25), does this mean that the resulting rail voltages will be +18V and -6V? How does "floating" the circuit change things?

I was under the impression that connecting the +12V to chassis ground would create a closed circuit, allowing current to flow.... And that the same would be true with the -12V. I guess that would mean that there would be two loops, one for the +12V/ground and one for the -12V/ground. Now that I type it out, it doesn't seem right.

I think a lot of my confusion stems from the voltage divider circuit shown in the opening pages of _The Art of Electronics_ and another schematic for a common-emitter transistor from the _Tab Guide to Understanding Electricity and Electronics_. Both of these show a + voltage source connected to ground through a series of resistors. I realize that these are partial circuits, but was confused trying to imagine from where the current was coming and to where it was going. I didn't realize that a - voltage was also required to be connected to create a potential difference and allow current to flow.

Many thanks,

brad
 
dc said:
2. The battery is removed and replaced with an AC source. The AC comes through a wall outlet and is connected to a step-down transformer with a center tap. The center tap is connected to chassis ground and the secondaries are fed through a bridge rectifier, and then connected to filter caps to reduce ripple remaining on the line. After the caps, there is a total of 24V potential difference between the secondaries. Because the center tap is equidistant from either secondary and the center tap is connected to ground, one secondary becomes a +12V rail and the other one is a -12V rail.

3. If only the +12V is attached to, say, a common-emitter circuit, and the -12V is left unattached, then there is no closed circuit, and current will not flow? Based on schematics of partial circuits I've seen in textbooks, I was under the impression that ground could serve as a return, thus closing the circuit and allowing current to flow. After all, ground, at 0V has less potential than the +12V rail, no?

Is it accurate to read Haldor's last post to mean that current will only begin to flow once both the + and - leads are somehow connected, whether or not via the ground? When both leads are connected, there is a 24V potential difference, regardless of whether the circuit is connected to ground, correct? If the center tap is not placed in the center of the transformer, but placed so that spacing between one secondary and the center tap is 3 times larger than between the center tap and the other secondary (75/25), does this mean that the resulting rail voltages will be +18V and -6V? How does "floating" the circuit change things?

I was under the impression that connecting the +12V to chassis ground would create a closed circuit, allowing current to flow.... And that the same would be true with the -12V. I guess that would mean that there would be two loops, one for the +12V/ground and one for the -12V/ground. Now that I type it out, it doesn't seem right.

I think a lot of my confusion stems from the voltage divider circuit shown in the opening pages of _The Art of Electronics_ and another schematic for a common-emitter transistor from the _Tab Guide to Understanding Electricity and Electronics_. Both of these show a + voltage source connected to ground through a series of resistors. I realize that these are partial circuits, but was confused trying to imagine from where the current was coming and to where it was going. I didn't realize that a - voltage was also required to be connected to create a potential difference and allow current to flow.

2. Correct.

3. +12 to Ground thru a load (a resistor will do) is a loop and current will flow.
-12V to Ground thru a load is a loop and current will flow

As long as there is a difference in the voltage (potential difference) between the terminals and a load connecting them, current will flow.

Finally, let me restate that you dont want to use chassis ground as the return path of the power. Chassis ground is connected to the Earth terminal of the power point. All power should return thru the Neutral terminal which is connected via the transformer.