# How to reduce 24V to 0- 2.5V through a pot?

I have a 24V DC supply and I need to give a heater a 0-2.5V variable input. I have a very limited understanding on electronics but I've read here about voltage dividers. I used a 10k pot (and the 2 relevant resistors shown in a typical voltage divider diagram). I used the wiper output feeding the voltage divider and put on a voltmeter.

All was looking really good as I turned the pot: slowly increasing voltage. But right at the end, in the last few degrees of the pot,the voltage shot back up to 24V! (as though 24V had "leaked" through the divider). Could someone please tell me what I have done wrong? I even tried with other pots but all with the same effect.

The heater is mains powered, but the heater temperature is either controlled via a built in pot on the unit, or via a remote pot pluged into the side using either a 0-10V or 4-20mA input. We require this option but because the unit outputs 650 degrees Celsius, we want to limit this to around 150. So we need to restrict the input voltage to 0 - 2.5V.

• the heater is mains powered, but the heater temp is either controlled via a built in pot on the unit, or via a remote pot pluged into the side using either 0-10v or 4-20ma input. we require this option but because the unit outputs 650 degrees c, we want to limit this to around 150. so we need to restrict the input voltage to 0-2.5v Commented Feb 10, 2013 at 13:40
• You can build this divider with only 1 resistor and 1 potentiometer. Take a look at my answer. Commented May 7, 2014 at 7:11

Short answer: what you want can not be done this way.

As you assumed, a potentiometer itself forms a resistor divider. The two parts (let's call them upper and lower) divide the voltage proportionally to the two resistor values.

Now hen you connect something (in your case the heater) to the output of the potentiometer you in effect connect it in parallel to the lower resistor. Two resistors in parallel have a total resistance of 1/( 1/R1 + 1/R2). When one of the parallel resistors has a very low value this means that you can ignore the other one.

This is what happens in your setup: in effect you have a resistor divider consisting of

• the upper part of the potentiometer.

Your heater has a rather low resistance, so when you turn your potentiometer 'up' almost nothing happens, until the very end, when the resistance of the upper part of the potentiometer becomes comparable to that of the heater.

I don't know the resistance of the heater, but to create any heat it must be low. Combined with 24V this will result in a large current, for which your potentiometer is probably not designed. Don't be surprised when you see smoke coming out of your potentiometer.

Without switching, there is no way to do what you want effectively (that is, without wasting around 9 times the heat you get in your heater in the rest of the circuit). A simple solution using Pulse Width Modulation (PWM) could consist 555 chip circuit and a switching series transistor or FET.

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Based on new information:

You have a 24V DC source, a 10k potmeter, and you want it to produce 0 .. 2.5 V DC output? That's a very different ballgame, and an easy one.

The trick is not to reduce the output of the potmeter (because that suffers from the same load problem as I described above, although somewhat less because the resistors involved have a higher value), but the input. You can use the potentiometer itself as the lower resistor, you only need to add a top resistor (between the potentiometer and the 24V source).

The value is easy to calculate. You want 10k (your potentiometer) to get 2.5V. That leaves 24 - 2.5 = 21.5 for the top resistor. Each 1k drops 0.25V, hence 21.5V requires 21.5 / 0.25 = 86kOhm. 82k is a standard value.

• hi, thanks for the reply. maybe what you are saying is still correct but i think you have miss understood what im doing. the heater is mains powered, but the heater temp is either controlled via a built in pot on the unit, or via a remote pot pluged into the side using either 0-10v or 4-20ma input. we require this option but because the unit outputs 650 degrees c, we want to limit this to around 150. so we need to restrict the input voltage to 0-2.5v. i hope this explains things a bit better. thankyou Commented Feb 10, 2013 at 11:12
• also the remote pot plug also requires a 24v dc supply so i intended using this same supply to tap off to regulate the 0-2.5v input because it uses the same ov common Commented Feb 10, 2013 at 11:15
• thanks for the reply, i had a simlar answer from someone else, only they also added "However: This will only work if the load resistance is high. If the load resistance is comparable to the 10K resistance of the pot, then the combined resistance of the load and the pot will change the voltage divider ratio and increase the amount of current flow, increasing the voltage." is this true and what can be done in my situation? could i just put a very high resistor across the output? or is this not the answer? if yes, what size resistor? thankyou. Commented Feb 10, 2013 at 18:15
• as my 0-2.5v is only a input reference to a heater controller then i guess there is no load resistance as such.. is this correct? Commented Feb 10, 2013 at 18:21
• I expect so, but the final authority is the description of your device. Commented Feb 10, 2013 at 19:34

You need something like this:

U1A is operational amplifier (op-amp). It must be rail-to-rail kind of op-amp. LM358 is popular and not expensive rail-to-rail op-amp, you can use it.

Op-amp is required to make sure that your heater (I assume thats voltage controlled industrial device) gets low impedance voltage source at input, so whole voltage divider is not affected by regulator input resistance (R3 on my schematic).

You can change resistances, but don't go too low with total resistance (R1 + potentiometer), potentiometers don't like high currents.

You can add small capacitor (0.1uF for example) between divider and buffer to slow down regulation. This will work like low-pass filter and reduce noises from potentiometer.

In my circuit - resistance connected to source is 9.2k (8.2k resistor + 1k pot). If we connect that to 24V - our current will be:

I = U/R = 24V/9200ohm = 2.6mA

Power on 1k potentiometer, at 2.6mA:

P = I * I * R = 0,0068W = 68mW

If you just want to use the pot to input 4-20mA you need the pot and a 1.32k resistor in series. These both in parallel with a 1.32k resistor will give 4-20mA for most low impedance inputs Eg West 6100+ temperature controller etc.

simulate this circuit – Schematic created using CircuitLab