# Voltage divider with adjustable output

#### shortbus

Joined Sep 30, 2009
9,863
I know how to calculate for a voltage divider. And found many online calculators for doing it. But what do you do when you want one with a specific high and low value with it being adjustable between the two limit's? How do you decide what value pot to use? Another of my dumb questions.

#### Brownout

Joined Jan 10, 2012
2,390
Solve the divider equation for voltage.

#### #12

Joined Nov 30, 2010
18,224
Start with how much current you want in the voltage divider. Find a standard value of pot that will create your range of voltage (with a little adjustment of the proposed current). Add a resistor on each end to use up the excess voltage.

Wait...the voltages load down when I connect it to anything. The old timers just used 10X the load current and called that, "good enough".
Nowadays, you can add a voltage follower like an op-amp and get some very precise results.

#### Bernard

Joined Aug 7, 2008
5,784
For pot, I reach into my pot box & pull out one with leads-- a 10K. 5 v supply & I want output from 1V to 3 V. V across pot 3 -1 = 2V; then I = 2/ 10000 = .2 mA. Top R = E/I =2 V/ .0002 A = 10 k; bottom R = 1 V/ .0002 = 5 k. Output feeding into hi Z comparator.

#### Brownout

Joined Jan 10, 2012
2,390
Output feeding into hi Z comparator.
Prolly want a unity gain buffer.

#### cmartinez

Joined Jan 17, 2007
7,970
If your power load is minimal (say, you want to feed the voltage divider's output into a comparator's input) what you do is place 2 resistors in series with trimpot, with the trimpot in the middle. That way you can easily calculate an specify upper and lower limits for your arrangement.

#### cmartinez

Joined Jan 17, 2007
7,970
BTW, why don't you tell us the voltage source, and the upper and lower limit of the voltage divider's output that you want?

#### Brownout

Joined Jan 10, 2012
2,390
Wait...the voltages load down when I connect it to anything. In that case, treat the load as a parallel resistor to your pot and low side resistor (if load terminates to ground) or to pot and high side resistor (if load terminates to VCC)

#### shortbus

Joined Sep 30, 2009
9,863
Cmartinez got it right, this is for a comparator circuit. I guess what I'm asking is there some "rule" to do this? I was going to figure the high value of the reference voltage needed and then the low value and subtract but that gives weird pot values. I don't want someone to do it for me, have to learn this stuff for later use. Just a 'rule of thumb' type answer on how people in the know go about it. I've looked everywhere to find the answer but either not phrasing the question right or something.

As an example let's say, and these are off the cuff numbers, just an example. 12VDC in, with an output of between 6VDC and 2VDC. What would the process be to design that volt divider?

Most of the circuits using a LM339 or LM393 don't show any resistors from the voltage divider to the input pins, some do. Is there a rule for choosing when to use input resistors? Or do you normally just make the divider resistor values high enough to eliminate the in put resistors?

I know this is probably a basic thing taught in school for electronics design, but it is hard to find the basics. Every resource I have and the net doesn't address this type of thing, or at least I can't seem to find it. Any help is much appreciated.

• cmartinez

#### Alec_t

Joined Sep 17, 2013
13,448
12VDC in, with an output of between 6VDC and 2VDC. What would the process be to design that volt divider?
I'd do it like this:-
1) Pick a pot value < one tenth the comparator input impedance. Let's say 10k.
2) We want 6V-2V = 4V across the pot.
3) There needs to be 12V - 6V = 6V across the top series resistor, hence this resistor value = pot value times 6/4 = 15k.
4) There will be 2V across bottom series resistor, so this resistor value = pot value times 2/4 = 5k.

• shortbus and ErnieM

#### ErnieM

Joined Apr 24, 2011
8,356
Well, it depends on a lot of things:

1) What voltage supplies it?
3) How accurate do you want the two end points? Can there be some slop (little higher or lower on the ends) or must it be exact?

I am guessing the max voltage is below the supply voltage, and the min voltage is above ground. For that you need three resistors, call them Rt Rp and Rb for top, pot, and bottom.

The max voltage is just
Vmax = Vin * (Rp + Rb)/ (Rt + Rp + Rb)

And the bottom voltage is:
Vmin = Vin * (Rb)/ (Rt + Rp + Rb)

You can pick a standard pot value to get 2 equations in two unknowns and use some algebra to solve it. Or, given all that, I would use excel to solve the formulas and just guess at values till it seems good enough.

Edit: missed your update but this still stands.

• shortbus

#### Brownout

Joined Jan 10, 2012
2,390
And the bottom voltage is:
Vmin = Vin * (Rb)/ (Rt + Rp + Rb)
This reduces to Vmin = Vin * (Rb)/(Rt + Rb)

#### gerty

Joined Aug 30, 2007
1,303
I'd use an adjustable voltage regulator and just be done with it

#### ErnieM

Joined Apr 24, 2011
8,356
This reduces to Vmin = Vin * (Rb)/(Rt + Rb)
That depends on the way you connect the pot. I had the each end going to either Rt or Rb, and the wiper as the voltage output.

A sketch would have helped but I can't do those where I am now.

• shortbus

#### #12

Joined Nov 30, 2010
18,224
As an example let's say, and these are off the cuff numbers, just an example. 12VDC in, with an output of between 6VDC and 2VDC. What would the process be to design that volt divider?
I like working with 1 ma. Good noise immunity, not much heat.
4 volt range. 5k would need 800ua except a 5k pot is never exactly 5k
Measure it...4.8k
well within the guarantee spec of 10%
4V/4.8k = 833.333ua
Need to waste the bottom 2 volts and the top 6 volts.
2V/833 E-6 = 2.4k to waste 2 volts
and 3 of them to waste 6 volts is 7.2k
Got any?
Only 7.5K?
You know how to do parallel resistors. That's my first method.
Ernie has a point, too. A wiper wiping all by itself can get nasty spots as it ages. A design where the wiper is connected to one end of the pot is an important way to keep from blowing the speakers out every time you get a dirty volume control.

• shortbus

#### ErnieM

Joined Apr 24, 2011
8,356
Ernie has a point, too. A wiper wiping all by itself can get nasty spots as it ages. A design where the wiper is connected to one end of the pot is an important way to keep from blowing the speakers out every time you get a dirty volume control.
Sir I believe that is your point, not mine, and a much better way than I was proposing. Though in the short term either should work.

• #12

#### shortbus

Joined Sep 30, 2009
9,863
Well, it depends on a lot of things:

1) What voltage supplies it?
3) How accurate do you want the two end points? Can there be some slop (little higher or lower on the ends) or must it be exact?

I am guessing the max voltage is below the supply voltage, and the min voltage is above ground. For that you need three resistors, call them Rt Rp and Rb for top, pot, and bottom.

The max voltage is just
Vmax = Vin * (Rp + Rb)/ (Rt + Rp + Rb)

And the bottom voltage is:
Vmin = Vin * (Rb)/ (Rt + Rp + Rb)

You can pick a standard pot value to get 2 equations in two unknowns and use some algebra to solve it. Or, given all that, I would use excel to solve the formulas and just guess at values till it seems good enough.

Edit: missed your update but this still stands.
Accuracy not all that important. Just needs to be close like in horse shoes instead of close as in hand grenades. Thanks for the way of doing it.

#### shortbus

Joined Sep 30, 2009
9,863
I'd do it like this:-
1) Pick a pot value < one tenth the comparator input impedance. Let's say 10k.
2) We want 6V-2V = 4V across the pot.
3) There needs to be 12V - 6V = 6V across the top series resistor, hence this resistor value = pot value times 6/4 = 15k.
4) There will be 2V across bottom series resistor, so this resistor value = pot value times 2/4 = 5k.
Thanks Alec. This and Ernie's will be going in my save file. The 10k is enough to work with the 339 inputs then?

#### #12

Joined Nov 30, 2010
18,224
Whatcha do is look up the Ibias current for the chip
25 na
25 na x 10k = 250 uv of error
(Not the exact formula, but a quick estimate.)

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#### cmartinez

Joined Jan 17, 2007
7,970
Cmartinez got it right...
See @#12 ? I got one thing right!

Can I have my special title now??? I was thinking along the lines of VDG = "Voltage Divider Guru" ... or even better, "The Ayatollah of Electro-la" • shortbus