Voltage divider question

Thread Starter

KatherineTurner

Joined Nov 24, 2016
13
I have a simple voltage divider circuit below.




Pin 1 and Pin 2 are both reading 4V.

I'm trying to figure a way to adjust Pin 1's reading with an additional potentiometer without affecting the 4V reading on Pin 2 and vice versa, but without success.

Ultimately, I would like to keep R1 at 100 ohm but R2 can be varied as long as I am able to adjust one of the Pin readings while keeping the other one at 4V. (eg. Pin 1 adjusted to 3V, while Pin 2 reads 4V or Pin 2 adjust to 3V, while Pin 1 reads 4V)

My question is that is the above criteria possible on this voltage divider? If so, how can it be done with potentiometers and switches?
 

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Reloadron

Joined Jan 15, 2015
7,501
Place a potentiometer across your current divider circuit. One end to your supply (8 Volts) and the other end to your Ground. So the pot is in parallel with your R1 and R2. The pot wiper is Pin 1. The pot wiper will have a variable voltage between 0 and 8 volts. I would also increase the values or R1 and R2 unless you have 100 ohms for a reason. As is your divider (R1 & R2) current is 8 / 200 = 40 mA. I would consider using 1 K resistors. I would also use a 1K or 10K pot.

Ron
 

Rufo4004

Joined Nov 24, 2017
2
I have a simple voltage divider circuit below.

I'm trying to figure a way to adjust Pin 1's reading with an additional potentiometer without affecting the 4V reading on Pin 2 and vice versa, but without success.
A: This is not possible. Pin 1 and Pin2 are IN THE SAME NODE and they ALWAYS have the same voltage. Pin 1 and 2 are the same thing.
Ultimately, I would like to keep R1 at 100 ohm but R2 can be varied as long as I am able to adjust one of the Pin readings while keeping the other one at 4V. (eg. Pin 1 adjusted to 3V, while Pin 2 reads 4V or Pin 2 adjust to 3V, while Pin 1 reads 4V)

My question is that is the above criteria possible on this voltage divider? If so, how can it be done with potentiometers and switches?
A: Whatever you connect to PIN 1 or 2 will change the voltage divider. As somebody suggested you above, use two separated dividers. Instead of two resistors and a pot you may use a single pot (consider the power drained through the pot at 8V), this way you can have a full range (0 to 8V) at the pot wiper. Adding resistors in differents ways allows you to limit the range. This kind of circuits are ok as voltage reference when the output is connected to a high impedance and stable input circuit (like an operational amplifier input). If you are planning to use this divider with or without a pot to supply power to another circuit (like a small DC motor) this WILL NOT WORK.

Beginning with electronics? Welcome!!!. Hope this help.

 
Last edited:

Thread Starter

KatherineTurner

Joined Nov 24, 2016
13
What is the load for these outputs, what are they to be connected to?
These two pins will not connect to anything. I'm basically just playing around with the components and observing the numbers at both pins.


A: Whatever you connect to PIN 1 or 2 will change the voltage divider. As somebody suggested you above, use two separated dividers. Instead of two resistors and a pot you may use a single pot (consider the power drained through the pot at 8V), this way you can have a full range (0 to 8V) at the pot wiper.

Beginning with electronics? Welcome!!!. Hope this help.
But could this cause a short at the pot between 8V and ground if I turn the wiper all the way down to 0 ohm? I'm thinking adding a pot and connect the first pot terminal to 8V and the third terminal to ground and the second terminal (wiper) is connecting to the first terminal, so it's acting like a variable resistor ranging from 0V to 8V. Am I getting this correct?

Yea, I am pretty new to electronics :p
 

Rufo4004

Joined Nov 24, 2017
2
These two pins will not connect to anything. I'm basically just playing around with the components and observing the numbers at both pins.

But could this cause a short at the pot between 8V and ground if I turn the wiper all the way down to 0 ohm? I'm thinking adding a pot and connect the first pot terminal to 8V and the third terminal to ground and the second terminal (wiper) is connecting to the first terminal, so it's acting like a variable resistor ranging from 0V to 8V. Am I getting this correct?

Yea, I am pretty new to electronics :p
A: Correct, it will be a short. You will see the pot burning. If you just want to do a voltage divider do not connect the wipe to any other pot's terminals. In this case just connect your multimeter between 0V and the wipe to see the readings while moving the pot's shaft. Note that you can burn the pot even leaving the wipe in the air. You need to be aware the Ohm value and the power supported by the pot. Calculate the power in the pot (using 8V and the pot resistance). It must not be higher than the specification in the pot's datasheet. If not, you may cook and destroy the pot... I would recommend you to read more about basic circuits with resistors.

(
).
 

Danko

Joined Nov 22, 2017
1,829
Ultimately, I would like to keep R1 at 100 ohm but R2 can be varied as long as I am able to adjust one of the Pin readings while keeping the other one at 4V. (eg. Pin 1 adjusted to 3V, while Pin 2 reads 4V or Pin 2 adjust to 3V, while Pin 1 reads 4V)
My question is that is the above criteria possible on this voltage divider? If so, how can it be done with potentiometers and switches?
Yes, you can. Exactly what you ask.
Divider.png
 

hobbyist

Joined Aug 10, 2008
892
Hi, yes,
Looking at your diagram above,

I hook up my ohmeter to terminals #1 and #2
Turn the knob counterclockwise, and the resistance measures minimum.
Turn clockwise and resistance is maximum.

NOTE:

Now that I'm looking at the drawing to the right of it, in my case it would be the wiper doing the movement.
I was thinking the construction was the resistive track being in the front of the device, where the terminals are, but now that I'm lookinjg at the drawings and the results I'm getting, the resistive track must be on the back side of the device, away from the terminals. I guess.

I had the wrong perspective of the construction of the track with respect to the location of the terminals.

The drawing shows clearly it is the wiper doing the movement.

Thanks for bringing that to my attention.
 

MrChips

Joined Oct 2, 2009
30,707
It's all relative, really.
However, physically, the track is kept stationary while the wiper moves.
I suppose one can conceive an application where the wiper is fixed and the track moves. Right now I cannot think of any such situations.
 

hobbyist

Joined Aug 10, 2008
892
However, physically, the track is kept stationary while the wiper moves.
Yeh, I know, I recognize that now, since I seen that diagram on the right, For some reason I always pictured the horseshoe shape being in the front, where the terminals are located. I kind of looked at it the way it's drawn up in schematics. No wonder I got the orientation backwards.
 
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