Thermistor and Op-Amp Circuit Design

[songoku]

Homework Statement

A thermistor T is included in the circuit shown. Calculate the resistance of resistor R such that the output of the op-amp will change polarity at 6^oC if at that temperature the thermistor has resistance 3 kΩ

Homework Equations

The Attempt at a Solution

The inverting input = 6/14 V, so if the output changes polarity, the value of non-inverting input should be less or greater than 6/14 V, depends on whether it changes from positive to negative or vice versa.

My question is how to determine whether it changes polarity from positive to negative or vice versa? Because in my opinion, the answer will be different. To change from positive to negative, the non-inverting input should be less than 6/14 V while changing from negative to positive, the non-inverting input should be greater than 6/14 V.

The answer is 2.25 kΩ. Based on that, I think the polarity changes from negative to positive. Here's my working:
\frac{3}{3+R}>\frac{6}{14}

Solving that, I got R < 4. How to determine the exact value of R?

Hence, basically I have two questions:
1. How to determine whether it changes polarity from positive to negative or vice versa?
2. How to determine the exact value of R?

Thanks

 

[NascentOxygen]

The inverting input = 6/14 V

Check your working. This is not right.

My question is how to determine whether it changes polarity from positive to negative

When the voltage on the + input exceeds the voltage on the - input, the op amp o/p will be ++.
Where ++ is something close to the V+ rail that powers the op amp. When that condition is no longer met, the o/p drops to the V- rail.

 

[songoku]

Check your working. This is not right.

The 6 kΩ and 8 kΩ form potential divider and the voltage at 6 kΩ is the same as the voltage at (-) so the voltage at (-) will be 6/14 times voltage of power supply, which is V. Hence, the voltage at (-) = 6/14 V ?

When the voltage on the + input exceeds the voltage on the - input, the op amp o/p will be ++.
Where ++ is something close to the V+ rail that powers the op amp. When that condition is no longer met, the o/p drops to the V- rail.

Maybe I don't understand the question and I also don't get your hint. The question doesn't specify the change polarity, whether it is from positive to negative or vice versa. I suppose I can pick value of R to make the (+) input higher than (-) input so the output is the positive of power supply voltage; or I can pick another value to make (+) input lower than (-) input so the output will be the negative of power supply voltage. In my opinion, there are two possibilities, which, I guess, is incorrect.

Are there really two possibilities or I am totally wrong?

Thanks

 

[NascentOxygen]

The 6 kΩ and 8 kΩ form potential divider and the voltage at 6 kΩ is the same as the voltage at (-) so the voltage at (-) will be 6/14 times voltage of power supply, which is V. Hence, the voltage at (-) = 6/14 V ?

Imagine for a moment that the 8kΩ resistor was changed to 800kΩ. Would you expect that to push the voltage at its junction with the 6kΩ resistor almost to V or almost to 0? Your maths must agree with what happens in practice.

Yes, the question doesn't indicate whether the thermistor increases or decreases its R as temp increases. But you really don't need to know to be able to answer the question, i.e., determine the value for R. The op amp is used as a comparator, so that if the + input is higher, op amp o/p is high, and if the voltage on the + input is lower, then the op amp o/p is low. There is no ambiguity; o/p is directly controlled by the input. All you have to determine is the switch-over voltage.

 

[songoku]

Imagine for a moment that the 8kΩ resistor was changed to 800kΩ. Would you expect that to push the voltage at its junction with the 6kΩ resistor almost to V or almost to 0? Your maths must agree with what happens in practice.

Not sure if I understand you correctly, but the answer will be 0? Because 800 kΩ >> 6 kΩ so the voltage at 6 kΩ will be very small. Honestly, I do not know the connection of your question and the mistake about 6/14 V

Yes, the question doesn't indicate whether the thermistor increases or decreases its R as temp increases. But you really don't need to know to be able to answer the question, i.e., determine the value for R. The op amp is used as a comparator, so that if the + input is higher, op amp o/p is high, and if the voltage on the + input is lower, then the op amp o/p is low. There is no ambiguity; o/p is directly controlled by the input. All you have to determine is the switch-over voltage.

Switch - over voltage here means that the value of voltage where (+) input is the same as (-) input?

 

[NascentOxygen]

Not sure if I understand you correctly, but the answer will be 0? Because 800 kΩ >> 6 kΩ so the voltage at 6 kΩ will be very small. Honestly, I do not know the connection of your question and the mistake about 6/14 V

The voltage at the junction of 6k and 8k will (let's take some wild guesses), have to be either 6/14 V or 8/14 V. For the arrangement here, it's not 6/14 V.

Switch - over voltage here means that the value of voltage where (+) input is the same as (-) input?

Yes. The question uses the phrase "change polarity", so it's not requiring you to know whether from V+ to V-, or from V- to V+. So instead of setting up an inequality 3k / (3k +R) > ...
set it to be 3k / (3k+R) = ...

BTW, you managed that potential divider voltage correctly, using the 3kΩ and R resistances.

 

[songoku]

The voltage at the junction of 6k and 8k will (let's take some wild guesses), have to be either 6/14 V or 8/14 V. For the arrangement here, it's not 6/14 V.

How to know which one is correct? Is the output taken always with respect to the ground? By using 8/14 V, I got the answer but I still want to understand the concept here.

Another question, if the (+) input is the same as (-) input, what will be output of comparator? Zero?

Thanks

 

[gneill]

How to know which one is correct? Is the output taken always with respect to the ground? By using 8/14 V, I got the answer but I still want to understand the concept here.

Ground is the common reference point for potentials, and in this circuit it is also a common reference for the op-amp since its positive and negative supply voltages are referenced to ground. So to find the potentials for the voltage dividers in this circuit the reference is also to ground.

Another question, if the (+) input is the same as (-) input, what will be output of comparator? Zero?

Theoretically, for an ideal op-amp, if the inputs are at the same potential then the output will be zero. Vout = (V⁺ - V^-)A, where A is the open loop gain of the op-amp.

Given the very large open-loop gain of an op-amp, such a balance point would be extremely delicate. The output might briefly switch erratically between the supply rails if the temperature passes slowly through the 'transition zone'. It depends upon the thermal mass of the thermistor, thermal noise in the components, offset currents in the op-amp, and any other random environmental factors. It's not easy to keep an open-loop op-amp in its linear region!

 

[NascentOxygen]

How to know which one is correct? Is the output taken always with respect to the ground? By using 8/14 V, I got the answer but I still want to understand the concept here.

You could determine the current that flows through that series combination of resistors. Knowing the current, you can calculate the voltage across each, and hence determine the voltage at their junction.

Another question, if the (+) input is the same as (-) input, what will

It practically never will be. Well, if you connected both to the same voltage, that would make them equal. In that case, there would still be offset voltage differences, so the output may still be at one extreme or the other. But remember that the op amp is just a linear amplifier with very high gain, so if you could carefully tweak the input voltages, with a bit of care you could show that o/p = 100000.Vin, where Vin is the difference between the two input voltages.

 

[songoku]

Ground is the common reference point for potentials, and in this circuit it is also a common reference for the op-amp since its positive and negative supply voltages are referenced to ground. So to find the potentials for the voltage dividers in this circuit the reference is also to ground.

Theoretically, for an ideal op-amp, if the inputs are at the same potential then the output will be zero. Vout = (V⁺ - V^-)A, where A is the open loop gain of the op-amp.

Given the very large open-loop gain of an op-amp, such a balance point would be extremely delicate. The output might briefly switch erratically between the supply rails if the temperature passes slowly through the 'transition zone'. It depends upon the thermal mass of the thermistor, thermal noise in the components, offset currents in the op-amp, and any other random environmental factors. It's not easy to keep an open-loop op-amp in its linear region!

You could determine the current that flows through that series combination of resistors. Knowing the current, you can calculate the voltage across each, and hence determine the voltage at their junction.


It practically never will be. Well, if you connected both to the same voltage, that would make them equal. In that case, there would still be offset voltage differences, so the output may still be at one extreme or the other. But remember that the op amp is just a linear amplifier with very high gain, so if you could carefully tweak the input voltages, with a bit of care you could show that o/p = 100000.Vin, where Vin is the difference between the two input voltages.

Ok I think I get it. Thank you for the help