Voltage Divider Earthing? What is happening to the *voltage of output #2?

[Eloc]

Hi,

I've only recently attempted to understand electronics beyond what I learnt as a kid at school.

I'm a little confused, probably because of some incorrect information, as to what exactly is happening to the 'second output' of a resistive voltage divider, i.e. a potentiometer, and what that implicates to earthing this output.

Isn't "voltage divider" a bit of a misnomer? Because it isn't exactly dividing the voltage (well at least not between the outputs), but simply 'reducing it' no? If the input voltage goes into lug #1 of a pot, the output from the wiper is a reduced voltage, but the output from lug #3 (to ground) isn't the rest of that input voltage ('voltage in' minus 'wiper voltage out') because it effectively has to go through the entire pot's resistance i.e. more resistance after the divide thus reducing current. So, lug #1 to lug #3 acts just like a fixed value resistor at all times.

The reason I ask this is because I have come across this exact use of a voltage divider as a 'dead battery simulator' here http://beavisaudio.com/Projects/DBS/ (scroll down to "As a Voltage Divider") where it says this:

"We also add a small resistor between the potentiometer and ground so that we can't accidentally turn the voltage sag all the way down and overload the power supply."

But surely the current going to ground (lug #3) has had to go through the resistance of the total value of the potentiometer (regardless of wiper position) anyway, so adding a resistor that is a fraction of the pot resistance is pointless; at least for the reason stated?

Am I missing something here?

Also, am I right in saying that a pot used in this configuration but with lug #3 not connected directly to ground, i.e. with more components before it is grounded, is never able to reduce the voltage output of lug #2 to zero volts; you've effectively got two 'circuits' in parallel with a resistor in front of both of them, so the minimum voltage to 'circuit #1' will be relative to the current draw (incorrect term?) of 'circuit #2'?

Sorry if this isn't particularly well articulated.

Any pointers are much appreciated.

[induction]

I think you're mixing the concepts of voltage and current, and you've got the voltage divider segments confused, but otherwise you've got the right idea.

If the input voltage goes into lug #1 of a pot, the output from the wiper is a reduced voltage, but the output from lug #3 (to ground) isn't the rest of that input voltage...

The rest of that voltage is between the wiper and lug #1.

Voltage drops are proportional to resistances. If V+ is connected to lug 3 (which is the usual convention), and ground is connected to lug 1, then the voltage between lugs 1 and 2 plus the voltage between lugs 2 and 3 equal the voltage between lugs 1 and 3. You can predict the value of the voltage at lug 2 based on the value of V+ and the position of the wiper (discounting tolerance issues). Assuming a perfect source, and a perfectly linear pot, V2 = V+ * pot rotation percentage = V+ * (resistance between 1 and 2)/pot value. This is true regardless of the value of the pot.

Also, am I right in saying that a pot used in this configuration but with lug #3 not connected directly to ground, i.e. with more components before it is grounded, is never able to reduce the voltage output of lug #2 to zero volts;

Exactly correct. That's the whole point. The extra resistor in the Beavis diagram effectively adds to the value of the pot while preventing you from turning the knob all the way down: Vout = V+ * 2.2k/12.2k = ~.18 * V+. Because a power supply of 0V is not useful for powering pedals. You are correct, though. That wouldn't overload the power supply. If it did, it would do so at any setting of the pot.

[ashcat_lt]

All of the current goes through both resistors, or rather through the "top resistor" and the parallel combination of the "bottom resistor" and the input to which the divider's output is connected.  It's actually split proportionately between those two parallel "bottom resistors", but as long as the input is comparatively Hi-Z, you can ignore it.

The entire input voltage is dropped across the two resistors, with each "holding up" an amount of that voltage proportionate to their resistance.  It's kind of like the "wiper" is a landing on a flight of stairs.  At the top of the stairs you have a given amount of potential energy.  At the landing, you have a bit less.  Some of it is used in getting down to the landing, the rest is there waiting to pull you down to ground.

It really is just Ohm's Law, though.  If your divider is 1K+1K, and you've got 9V, you've got 9V across 2K, equals 4.5 mA through both resistors.  4.5ma through 1K = 4.5V across each resistor.  Dig?

Edit - yes, BTW, Beavis is just plain wrong in that statement.  It's not the first time...     Turning that divider (without the static resistor) all the way down will give 0V to the circuit, but it won't hurt the supply any worse than the pot by itself.

[Eloc]

Thanks, I'm starting to fill in the gaps.

@induction

I think you're mixing the concepts of voltage and current

Yep sorry, I see where I've done that, i understand the difference between voltage and amperage, I've just mistakenly used the term 'current' interchangeably in a couple places. My bad.

...the voltage between lugs 1 and 2 plus the voltage between lugs 2 and 3 equal the voltage between lugs 1 and 3.

Ok, I see it now. My mistake initially is that when i see "voltage divider", and I see a diagram with "Voltage out" like this: http://upload.wikimedia.org/wikipedia/commons/d/db/Resistive_divider.png I imagine the potential differences being measured 'outside' of the potentiometer, I was imagining there being a "Voltage out 2" after 'R2', and I'm thinking "but the wiper position wouldn't affect that voltage" which is correct, but then I saw the instructions for the 'dead battery sim' and it seemed to imply that it would change that output; It's pretty simple electronics but I got myself into a confused loop. The voltages are actually measured across the resistors obviously.

@ashcat_lt

It's kind of like the "wiper" is a landing on a flight of stairs.  At the top of the stairs you have a given amount of potential energy.  At the landing, you have a bit less.  Some of it is used in getting down to the landing, the rest is there waiting to pull you down to ground.

That's a useful analogy thanks.

Edit - yes, BTW, Beavis is just plain wrong in that statement... Turning that divider... won't hurt the supply any worse than the pot by itself.

Thought so. It's tricky when your just getting your head around something and a piece of (mis)information contradicts you and makes you question the whole thing.

Thanks again.

[ashcat_lt]

Thought so. It's tricky when your just getting your head around something and a piece of (mis)information contradicts you and makes you question the whole thing.

Yep!  I'm afraid I have real problems with that Beavis site.  It seems to be geared toward newbies like yourself, but offers so much poor, misleading, or just plain wrong information...  Luckily, the circuits usually work fine, even if he doesn't explain the WHY correctly.

[PRR]

> I was imagining there being a "Voltage out 2" after 'R2'

Well, that point has a "ground" symbol on it, which we usually take to mean "Zero Volts".

And that the *other* end of "Vin" is also connected to that "ground".

Assume R1 and R2 are the same.

Then Vout to ground is half of Vin to ground.

(Assuming NO load on Vout.)

BEWARE any "circuit drawing" that does not show CIRCUITS. Complete loops of current. Here I would draw a battery from Vin to Ground. Also a meter from Vout to ground. NOW we have loops of current. No-loop "circuits" are meaningless. Electrons have to Go-Around. Yes, we often _draw_ parts with "conventions" like "grounds", and *assume* the reader understands the convention.

Or in Ashcat's thought; before you can fall down the stair you have to go up it. We don't have an infinite supply of bodies upstairs; we have to collect them at the bottom and hoist them back up. There's a complete LOOP of body-flow. The cited diagrams do not show the body-return hoist explicitly.