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?

Best regards

Dirk

• Hi, Dirk.

I'll replicate your circuit in my simulators and will get back to you soon.

Regards,

Anna

• Hello,

thank you very much for your supports.

I use +/-5,2V for the supply.

• Hi, Dirk.

Sorry for the late response. I replicated your circuit in multisim and had used the values of Un, Up and Uref from your table. My Uout were not the same as yours. Maybe there's something wrong with your Un and not the part. This is because if you assume that you have a -2.5mV offset due to the AD8429 offset current, IOS*1Meg (if this is the A grade, you could expect up to ±100mV according to the IOS spec) and Vref is -4.1mV, then your output in gain=1 would be -6.6mV, gain=10 would be -29.1mV, gain=100 would be -254.1mV, and gain=1000 would be ~-2.5V. These are all very close to your Uout numbers.

If the difference between the input voltages was actually -52mV at gain=100, then the output would be at ~-5.2V (or saturated if the ±5V supply was used). That is not the case, so it is more likely that the Un measurement is wrong. Also, I would not expect the input current to change unless the input stage is overloaded by common-mode or differential voltage outside of the operating range of AD8429.

Make sure the input resistance of the meter is set to high impedance, or it would disturb the input voltages by loading the 1Meg. Even so, the meter can load the input network/cause slow settling/introduce charge injection. Typically, a better measurement is just to alternately short the 1Meg resistor at each input and measure the changes in output voltage. If you define positive bias current as flowing into the inputs, and Ios = (Ib+) – (Ib-), then the combinations are:

Both inputs shorted to GND: Vout = Vos,rto

+IN shorted only: Vout = Vos,rto + (Ib-)*1Meg*gain

-IN shorted only: Vout = Vos,rto – (Ib+)*1Meg*gain

Neither input shorted: Vout = Vos,rto – Ios*1Meg*gain

The gain must be low enough that the output doesn’t saturate. The measured numbers for Ios and (Ib+) – (Ib-) should match.

A source or sensor with low differential output impedance would overdrive the 1Meg resistors by setting a stiffer differential voltage, greatly reducing these errors. If Up and Un aren’t normally driven this way and the AD8429 offset current is not acceptable for the application, consider the AD8421, a low noise, low power in-amp with A-grade offset current of 2nA max. AD8421 also has lower current noise, so it would provide much lower total noise when there is high source resistance.

Please let us know if you have further concern.

Regards,

Anna

• Hello Anna,

I need some time to check the things and to start a new measurement.

I write back, if I had done.

Best regards,

Dirk

• Hello Anna,

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

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