Subtracting voltages

[chroot]

Can anyone think of a circuit that takes two input voltages and produce an output voltage equal to the difference in input voltages? In other words, a "voltage subtractor?"

And don't answer "op-amp!" I don't want to use an op-amp. The circuit also must function with only one positive supply.

Any ideas? Can it be done? What did people do in the pre op-amp days?

- Warren

 

[russ_watters]

Unless I'm reading the question wrong, this is precisely what happens when you put a battery into a device backwards. If you go -+,-+ with AA batteries, you get 3V. -+,+- gives you 0V.

 

[chroot]

russ,

I'm not dealing with batteries (which are chemical devices actively capable of maintaining a potential difference across their terminals). You can flip batteries upside down and they don't care, and continue maintaining 1.5V across their terminals from - to +.

I'm dealing with two aribtrary voltages, from two temperature sensors, referred to the same ground. I want to get the difference of the two, without having to use an op-amp. And no, you cannot "flip one upside down," since they are referred to the same ground.

- Warren

 

[Tyger]

Are these thermistors?

You may just have to isolate one from the ground and change the connections.

 

[chroot]

Tyger,

I *can* use separate ground voltages for both temp sensors. One could be grounded at 0V, the other at, say 3V.

If one sensor is at 30 degrees C, it will produce 0.3V above its ground.

If the other is 50 degrees C, it will produce 0.5V above its ground.

If I used 0V and 3V as grounds, the sensor voltages would thus be appproximately, say, 0.3V and 3.5V. How could I then subtract them and get 0.2V?

- Warren

 

[wimms]

What is intended precision, frequency range (or latency), voltage range, linearity? What are impedances?

Charging capacitor between inputs and then switching (with FETs) it over to output with one leg grounded would do it. But it'd be discrete sampler, and cap discharges with time. so depends on application.

Substraction can lead to negative voltages. What do you mean by only working with positive volts?

Some sort of differential amplifier. But I'd think you'd find suitable opamp faster and cheaper than trying to reinvent it.

 

[chroot]

Yup... op-amps are really the only way to go. The switch-mode cap circuit would be a huge pain in the ass.

I was just hoping there was some tricky way, using only passive components, to subtract two voltages. This is obviously not true! Oh well.

- Warren

 

[Integral]

If these were simply 2 thermocouples you could connect them in series with positive ends (or negative) connected. This should do exactly what you want. (Thermocouples act just like batteries, with a varible voltage which depends on the temperature of the junction.)

 

[wimms]

Originally posted by chroot
Yup... op-amps are really the only way to go. The switch-mode cap circuit would be a huge pain in the ass.

I was just hoping there was some tricky way, using only passive components, to subtract two voltages. This is obviously not true!

Well, I can only say that scrapping switch-mode cap you deny yourself a lot of fun.

On other note, if your subtractor circuit can have floating ground relative to your sensors, then simplest way to have difference is to connect subtractor ground to one of sensors and use other sensor as input voltage. You have free subtraction. But I suppose you need common ground, or you wouldn't be asking..

I *can* use separate ground voltages for both temp sensors. One could be grounded at 0V, the other at, say 3V.

If you can feed one sensor with negative voltage (common ground) and thus extract signal as negative voltage, then simple average sum of the two will always be 1/2 of their difference. job for simplest resistor network.

 

[PHIMP]

Chroot...why did you not want to use op-amps when they are the most obvious (and therefore easiest) way to go?

 

[petmar]

wait a min here...

what about using a wheatstone bridge?
http://www.electronics2000.com/basics/wheat01.jpg
Rs is originally set to be equal to Rx, and the sum of these two should be equal to R1+R2. when this is tipped one way by the sensor arm resistance changing (Rs), then the difference in voltage is expessed in the G (galvanometer) arm.
just a thought...

 

[Richard Torregano]

Try a wheatstone bridge circuit.

Richard