AD8429 offset voltage

Hello,

in my circuit I have a problem, with the offset voltage. I dont how is the best way to solve it.

Below you can see my circuit.

I have measurement the volteges Up, Un, Uref and Uout with different values of Rg. The relay are reed relay. My aim was it, to make a offset compensation  with the Uref voltage. But if the gain (small Rg) is to high, than I get an offset voltage, which is bigger I have calculated. Thats the reason I measurment the voltages to find out, where the big offset come from. For this measurement the Uref voltage is constant.

v Rg Up Un Uref Uout
1 10M 123,5mV 126mV -4,1mV -6,2mV
10 666,6 123,4mV 125,8mV -4,09mV -29mV
100 60,6 123mV 175mV -4,09mV -250mV
circa 1000 6 123mV 139mV -4,09mV -2,3V
circa 2000 3 124,7mV 140mV -4,1mV -3V

1.The output voltage is: (Up - Un) * v + Uref. For the first and the second line thats ok. But the other lines it is funny.

2. I understand the voltage Up, the bias current is constant and flow throw the 1Meg resistance and create the offset voltage of circa 123mV.

3. But I dont understand, why the voltage Un is change from v=100. Is this normal for the AD8429 ?

4. Have you any idea, where problem should be? Is it a measuremnt error?

Thank you in advance.

Best regards

Dirk

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  • Hello Anna,

    I have replaced the AD8429 with the AD8421. It works fine. I dont get problems with the saturation.

    But their is one point, I need more informations.

    The maximum rating of the input voltage is ca. +-35V with UCC=+-5V.

    I have BAV199 diodes at the input, which  limit the signal (shown in the picture)

    ,

     

    I want to know, what is happening with power dissipation about the Rg resistance.

    Assume, the input voltage is to high and the diodes limit it, to u_diff.

    For example u_diff = u_p - u_n = 5Vp - -5Vp = 10Vp = 7.07V and Rg = 10 ohm.

    I have found in the datasheet, that the voltage about Rg is the 2*u_diff.

    I get P = U^2 / Rg = (2*7.07V) ^ 2 / 10 ohm = 20 watt.

    What happen by this condition?

    1. If the INA destroyed ?

    2. Limit the INA the current through Rg?

    3. What can be the  maximum current through Rg?

    4. Did I need an additional protection at the input, like in the picture?

    Thank you very much for yout help.

    Best regards

    Dirk

Reply
  • Hello Anna,

    I have replaced the AD8429 with the AD8421. It works fine. I dont get problems with the saturation.

    But their is one point, I need more informations.

    The maximum rating of the input voltage is ca. +-35V with UCC=+-5V.

    I have BAV199 diodes at the input, which  limit the signal (shown in the picture)

    ,

     

    I want to know, what is happening with power dissipation about the Rg resistance.

    Assume, the input voltage is to high and the diodes limit it, to u_diff.

    For example u_diff = u_p - u_n = 5Vp - -5Vp = 10Vp = 7.07V and Rg = 10 ohm.

    I have found in the datasheet, that the voltage about Rg is the 2*u_diff.

    I get P = U^2 / Rg = (2*7.07V) ^ 2 / 10 ohm = 20 watt.

    What happen by this condition?

    1. If the INA destroyed ?

    2. Limit the INA the current through Rg?

    3. What can be the  maximum current through Rg?

    4. Did I need an additional protection at the input, like in the picture?

    Thank you very much for yout help.

    Best regards

    Dirk

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