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Wheatstone Bridge (thermistor) to Op Amp

Struan Wallis

Mar 21, 2017
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Hi,

I'm trying to make a thermometer using a thermistor in a Wheatstone bridge going to an LM741.

However, whilst I'm getting an output of roughly 1.5v from the op amp, it doesn't change with temperature or the variable resistor changing.

The fixed resistors are 330k and the variable goes up to 100k and is in series with a 220k. The thermistor is 330k at 25°c and has a beta value of 4840k.

Any help would be appreciated.
 

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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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That circuit will cause the output to be either as positive or as negative as it can be.

As negative as it can be, for the 741, is not the negative supply rail.

It's hard to read the resistor values, but I suggest they are such that the voltage on the non inverting (+) input is always lot than that on the inverting (-) input.

You probably want to reduce the open loop gain (by using negative feedback) to get an output voltage proportional to the input differential voltage.
 

Struan Wallis

Mar 21, 2017
3
Joined
Mar 21, 2017
Messages
3
That circuit will cause the output to be either as positive or as negative as it can be.

As negative as it can be, for the 741, is not the negative supply rail.

It's hard to read the resistor values, but I suggest they are such that the voltage on the non inverting (+) input is always lot than that on the inverting (-) input.

You probably want to reduce the open loop gain (by using negative feedback) to get an output voltage proportional to the input differential voltage.

Thanks for your help, it makes sense.

Do you have any suggestions for the size of the resistors?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
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25,510
The values must be such that the ratios of the two resistors must be very close.

Without feedback, the variable resistance must cause the potential of one input to be able to be adjusted above and below the other.

So for the arms to be balanced, if one side is 10k and 15k, the other side must be in the same ratio (e.g. 100k and 150k)

If the former side is variable, you might choose a 10k resistor for the upper resistor, and for the lower resistor a combination of an 8k2 resistor and a 5k potentiometer.

To minimise wasted power, you wouldn't make the values too small (1 ohm and 1.5 ohms).

To minimise noise and issues with leakage you wouldn't make them too high (10M and 15M could be a problem).

To minimise problems caused by input offset currents in your op amp, it is desirable to keep the impedance at each input similar. This gets a little more complex, but maintaining similar values in each arm is a good start (e.g. 100k and 150k in both or 100k/150k and 120k/180k). When you add feedback you can correct for different input impedances but it's easier not to.

Wheatstone Bridges are normally used for differential measurement, often with the goal of balancing both arms or determining which unknown value is large or smaller than the other.

If you want to make an absolute measurement you typically compare a value with a reference and look at the analog difference. Whilst the circuit is similar the output is treated quite differently.
 

Struan Wallis

Mar 21, 2017
3
Joined
Mar 21, 2017
Messages
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Thank you so much for the help, I'll try what you've mentioned and let you know if it works.
 
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