Hello,
I was reading your files, in VOL 1 dc/ Divider circuits and Kirchhoff’s laws/Voltage divider circuits - 3/4’s down, I found a diagram like this, figure A.
Please look at figure A. The voltage can be reduced between terminals 3 and 4, and I have no problem with that. But I am having a problem with terminals 1 and 2 connected together? To me it’s creating unnecessary current throw potentiometer, and I do not understand its purpose.
Please look at Figure B. I opened the circuit between terminals 1 and 2. The voltage can be reduced between terminals 3 and 4 as before.
Please explain why 1 and 2 are shorted?
logmode

 

To understand the difference, look for an explanation of Thevenin equivalent circuits. The circuit to the left of terminals 3 and 4 can be replaced with a Thevenin resistance and a Thevenin voltage.

For circuit B, it is what it is. That is to say that the Thevenin voltage is the source voltage, and the Thevenin resistance is from the top of the potentiometer to the wiper at terminal 3.

For circuit A, it is slightly more complicated but not much. The Thevenin voltage is the voltage at the location of the wiper. The wiper breaks up the pot into two resistances, i.e. a voltage divier. For example, if the wiper is exactly in the middle (1/2 pot resistance above and 1/2 pot resistance below), the the Thevenin voltage is 1/2 of the source voltage.

The circuit A Thevenin resistance is found by first shorting the voltage source, and then "looking into" terminal 3 from the right to the left. Since the pot can be broken into resistances above and below the wiper, when you look into the wiper (terminal 3), you now see two resistances in parallel (because you shorted the voltage source. That parallel resistance is your Thevenin resistance.

You can now redraw either circuit A or circuit B and replace them with their respective Thevenin voltages and Thevenin resistances. Each circuit will likely have a different equivalent circuit now. I think you'll find that if the source voltage and potentiometer resistance is the same for both original circuits, then circuit A has a lower Thevenin voltage and a lower Thevenin resistance that does circuit B. Circuit A generally gives you more control of the current leaving the wiper.

It helps to understand the difference between the two circuits, because even though circuit A is usually preferred, a common failure of circuit A is to have an open circuit of the low side of the pot. The effect is to change the applied voltage and applied resistance, i.e. Thevenin equivalent voltage and resistance to the circuit downstream to the right.

 

Perhaps one requirement of a voltage divider might be to provide a variable output with zero volts as the minimum output. The right hand circuit can't achieve that requirement.

 

Hi,
I built the circuit and studied it. Thanks I have a much better understanding now. Please look at figure C.
Where ever the wiper touches is the voltage you get.
logmode

 

Hi,
I built the circuit and studied it. Thanks I have a much better understanding now. Please look at figure C.
Where ever the wiper touches is the voltage you get.
logmode

Right, as long as the current drawn by L is negligible.

 

Hi,
I found 11ma @30vdc for L.
OK Thanks
logmode

 

Hi,
Sorry to hijack the thread but I've got a voltage divider question too and I thought I might as well put it here. It's been three years since I've done any kind of electronics work so I'm a little rusty

I was reading through the worksheet :

http://www.allaboutcircuits.com/worksheets/e_divide.html

and I've run into problems with two of the questions I looked at.

Question 16
I understand that the wiper @25% sets the equivalent resistance of the pot to 1K25Ω . I then used the voltage divider formula and got Vout to equal 4.44V.
However the answer given is 1.29V.

Question 18
I'm having trouble calculating the equivalent resistance in this question. I've been scratching my head but I'm finding it hard to get my head around it. Are the two 5K resistors supposed to be added together in series and then, is the 100K in parallel with that?

I'm confused and I would greatly appreciate any help.

Thanks,
Damien

 

Hi. I found the answer ok.

 

See this also
http://forum.allaboutcircuits.com/showpost.php?p=463518&postcount=21

 

Hi, I cannot find the answer for Sausages Question 18.
I change potentiometers (R1, R2, & R3) into fixed voltage dividers ((R1A, & R1B), (R2A, & R2B), & (R3A, & R3B)). Found each potentiometer resistances. R1 X 50% = R1A, R1 X 50% = R1B, R2 X 50% = R2A, R2 X 50% = R2B, R3 X 50% = R3A, & R3 X 50% = R3B (Diagrams 1 and 2).
Look at the diagrams 2 and 3. I found total resistance for the parallel circuit and turned it into a series circuit. 1/ (1/ (R1B + R2A) + 1/R3) = 1/ (1/ (2500ohms + 2500ohms) + 1/100000) = 4762ohms. Then I added the resistance of the series circuit. R1A + parallel total resistance + R2B = total resistance. 2500 + 4762 + 2500 = 9762. Next found total current. 25/9762 = .002561a, which is the same for R1A, parallel circuit, and R2B. Then found the voltages, R1A is 2500ohms X .002561a = 6.4 v, parallel circuit is 4762 ohms X .002561a = 12.2v, R2B – 2500ohms X .002561a = 6.4v. Now I know the voltage of the parallel circuit is 12.2v. Now in the right leg of the parallel circuit, I added the resistance in the leg and divide the voltage by it to find current. 12.2v / 100000ohms = .000122a. Then took that current, multiplied it by resistance of R3A and found its voltage, .000122A X 50000ohms = 6.1v. Now subtract R1A and R3A voltages from total voltage and get R3 center tap voltage. 25v – (6.4v + 6.1v) = 12.5v (positive).
The forums answers are R total = 9762ohms, which is ok, but the other V out = -12.5v (negative) I can’t work out. What am I doing wrong?

logmode

 

Check your voltages source polarity and compare it with the polarity in the original question.

 

thanks, so the little part of the battery is the negative and the big part is positive. That means the ground is positive?

logmode

 

Ground has always 0V.

 

well done,
thank you
that was very helpful.

logmode