ADN8834

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

    •  Analog Employees 
    Feb 1, 2021 +1 suggested

    Hi ,

    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…

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  • 0
    •  Analog Employees 
    on Feb 1, 2021 5:11 AM

    Hi ,

    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 see if it is working. Example, try to test the chopper amplifier 1 first before connecting it to chopper amplifier 2. That way, it would be easier for you to identify if the stages are working. 

    However, on your circuit, what value of RFB did you use? can you share to me your actual schematic so I can check?

    Regards. 

  • Thanks for your reply!

    The vlue of RFB is 80.6K. When I try to show my schematic,I can not find the upload option.I used  values of the Evaluation Board of ADN8834.

  • 0
    •  Analog Employees 
    on Feb 12, 2021 5:09 PM in reply to bai1390529449

    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!
  • 0
    •  Analog Employees 
    on Mar 1, 2021 8:34 AM in reply to bai1390529449

    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. 

  • 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

  • 0
    •  Analog Employees 
    on Mar 1, 2021 12:10 PM in reply to bai1390529449

    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.

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  • 0
    •  Analog Employees 
    on Mar 1, 2021 12:10 PM in reply to bai1390529449

    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.

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