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My Application circuit like this

I tested the voltage of IN1N,OUT1,and use the pin TMPGD drives a LED,

V(IN1N)= V(1N1P),V(OUT1)= V(IN2P),As

I changed the valtage of TempSet  V(IN2P),Vout1 changed,according to the fomula,Rth changed,

but V(1N1N) still equals to Vref/2,As the voltage divider,

V(IN1N) =Vref* (Rth+Rx)/(R+Rth+Rx)

V(IN1N) not changed,Rth Neither.

So it is  contradict.I can not understand.

Besides,when I tested the voltage of IN1N,the LED always twinkle.

## Top Replies

• Hope you are doing great.

Based on what I can see, you are trying to check the chopper amplifiers functionality first. What I can suggest here is try to test the stage one by one and…

• Hi,ntenorio!

Thanks for your reply.What confusing me is just that the values between  V(IN1N) and Rth.

As I tested,V(IN1N) = Vref/2 is always established.

But the value V(IN1N) you give me varies with Rth.So I want to ask is that the value is you tested or calculated by the formula V(IN1N) =Vref* (Rth+Rx)/(R+Rth+Rx).

If is you calculated,I hope you could test a real value.If is you tested,I hope you can help me analyze my question.Thanks a lot.

Besides,I also find another question as same to me.https://ez.analog.com/cn/mems/f/q-a/95406/adn8834

• Yes. The OUT1 voltage will be varying depending on the IN1N. It will be the thermal loop function. OUT1 will only be equal to the TEMPSET (IN2P) if the target temperature is reached. Therefore, the thermistor should be placed directly attached to the material you want to maintain temperature. when the material has higher temp than the target, the RTH will have lower resistance, (higher OUT1, then lower OUT2) moving forward, give cooling current.

That is how the thermal loop works. Hope you understand it better now.

• V(OUT1) varies depending on the ININ,as I tested, V(ININ) doesn't change,but V(OUT1) change.

As what I tested firstly,I used a LD with internal TEC and thermistor, when I changed V(IN2P),V(OUT1) is always equal to V(IN2P).It means that the temperature  has reached the temperature I set?

But V(IN1N) is always equal to V(IN1P) = Vref/2.Why the value V(IN1N) doesn't change?

When I used V(IN1N) = Vref/2,and derive the formula of V(OUT1).So V(IN1N) should be a constant and that I tested before is correct?

• But V(IN1N) is always equal to V(IN1P) = Vref/2.Why the value V(IN1N) doesn't change?

This is correct. That is the ideal characteristic of op-amp. the virtual short. meaning, equal voltages at the two terminals. in the case of the chopper amplifier 1, we are varying the RTH to vary the voltage input in IN1P - to achieve the OUT1 voltage we desire. However, it is normal that we can measure same voltages on the two input pins. Your derivation is correct.

• OK！I know.Does it mean that the design is OK and I can use it directly?

• I think it's fine. since your target temp is 25degC, and with the thermistor you use, it is already good that the V_TEMPSET is set at 1.25V.

Regards.

• But I want to change the temperature,and the V(OUT1) will change varies with V(IN2P)(V_TEMPSET).Would it works well?

• You mean, you want to change the target temperature? To do that, you just need to change the TEMPSET  (IN2P) voltage level That should work.