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 4, 2021 3:38 AM in reply to bai1390529449

    Hi @bai390529449,

    I see, you are using the evaluation board values. Are you using just the circuit values, or you are using the actual evaluation board?  

    I have here a clearer definition of the variables in the equation:

    VOUT1 = output of the chopper amplifier 1
    RTH = actual resistance of the thermistor AT A SPECIFIC TEMPERATURE
    RX = linearizing resistor
    R = RX + RTH (RTH @ the target (center) temperature)

    The linearizing resistor Rx and the RFB depends on the temperature range you want to maintain, and the corresponding thermistor resistance at those boundaries. the OUT1 output should swing from 0V to 2.5V (0.3V to 2.2V for practical values). from the evaluation board values, it will depend on the thermistor. Can you share what is your target temp range temperature  and the target temperature to be maintained? You should also know the corresponding thermistor resitance at the boundaries.  You may focus first on the chopper amplifier 1. On the evaluation board circuit, try disconnecting it first on the chopper 2 amp. To vary the IN1N voltage, try to heat and cool the thermistor, and observe the change in the OUT1 voltage. Once you're good, you can now proceed with the chopper amplifier 2. 

    Regards.

  • Hi,ntenorio
    Thanks for your reply again!
    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!

Reply
  • Hi,ntenorio
    Thanks for your reply again!
    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!

Children
  • 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.

  • 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