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.

Thanks for your explaination.

## 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
I just used the values of evaluation board.My application circuit is used for driving a laser diode,with its internal thermistor resistor and TEC. My temperature range is 15℃-35℃,Rth=10K @25℃,But I don't know the value at 15℃ and 35℃,I used the Rth=14.8K@15℃ and Rth=6.9K@35℃ which I looked up on the internet.
I disconnect the amp2 by remove the Z1 and use a 10K NTC thermistor.The value of Vout1 rises to nearby 5V and V(IN1N) drops when I heating the thermistor,Vout1 drops,and V(IN1N) drops to Vref/2 rapidly.
I also use a 16.9k instead of the thermistor,V(IN1N)=1.28V and Vout1=0,
when I use two 16.9K in parallel,V(IN1N)=1.253V and Vout1=1.84V.
besides,I use a 5.1K in series with a variable 10k，V(IN1N)=1.254V and Vout1 varies when I change the variale resistor.So V(IN1N) =Vref* (Rth+Rx)/(R+Rth+Rx) would satisfied?
As what I understood,since the proprety of the amp,V(IN1N)=V(IN1P) would satisfied?I want to use this application circuit to change the LD temperature . So my target temperature will change between 15℃ and 35℃ when I want to change the wavelength of the LD.Should it be set by change the voltage of IN2P.
Now I am more puzzled.What should I do ?It would be appreciated if you could give me a more detailed explanation.Thank you!

• I also find the application circuit of ADN8831,which seems looks like the ADN8834.I used ADN8830 ever,it seems different.But ADN8830 is not recommended,so I use ADN8834 instead.

• Hi @bai1390529449,

To make it easier for you to check if chopper 1 is functioning correctly, you may try configuring it to voltage follower / buffer. Isolate it first with chopper 2 (removing Z1). short IN1N to OUT1. inject voltage to IN1P. the OUT1 pin should follow the voltage you are injecting the IN1P.

Or, with the circuit values you have now, instead of using the thermistor, replace it with a variable resistor just for you to check if chopper 1 is good.

 RFB = 80600 RX = 7680 RTH = replace with diff values of resistor or use a variable resistor R = 17800 VREF = 2.5

you may try the following values in replace of RTH and the corresponding values of OUT1 should be observed:

 RTH OUT1 10000 1.2457 12000 1.3127 15000 1.4 16000 1.4272

This should work. If you still see problems, please drop me an email on noel.tenorio@analog.com so I can help you.

Regards!

• Hi,ntenorio!
I am bai1390529449 with the question of ADN8834 at the ADI Engineer Zone.
I am sorry for that reply to you so later because of the Spring Festival Holiday.
I configure it to voltage follower / buffer,disconnectting R,Rth,Rx and Rfb,short IN1N to OUT1,I change the value of IN1P devider,
 IN1P devider V(IN1P) V(OUT1) 20K,20K 1.255 1.255 20K,20K||80.6K 1.117 1.118 20K,20K||12K 0.689 0.690
I test with the latter option you give to me.
 Rth V(IN1N) V(OUT1) 10000 1.255 1.268 12000 1.255 0.697 15000 1.256 0.052 16000 1.256 0.002

It is diffrent with the values you give to me.

 RTH OUT1 10000 1.2457 12000 1.3127 15000 1.4 16000 1.4272
As  ,
when Rth+Rx>R,with Rth increase,the Vout1 should decrease.
What I still can not understand is that V(IN1N) should be calculated by Rth?Thank you!
• Hi bai1390529449,

Based on the first table you sent, the chopper amp1 is functioning properly as a voltage follower. That is good to see.

On the next concern, I am sorry, the second column should be the V(IN1N) voltage readout not OUT1.

 RTH OUT1 10000 1.2457 12000 1.3127 15000 1.4 16000 1.4272

However, on your results, I am wondering why your V(IN1N) voltage is almost not changing.

 Rth V(IN1N) V(OUT1) 10000 1.255 1.268 12000 1.255 0.697 15000 1.256 0.052 16000 1.256 0.002

It is diffrent with the values you give to me.

with the following values,

 RFB = 80600 RX = 7680 RTH = replace with diff values of resistor or use a variable resistor R = 17800 VREF = 2.5

You should get the following:

 RTH V(IN1N) VOUT1 10000 1.24577 1.288 12000 1.3127 0.7093 15000 1.40069 0.03213

If you get the same result, then you are good. we can proceed on the next stage.