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ADA4922-1 Offset voltage and Output Load


I am using ADA4922-1 to convert a single ended square wave of 1 KHz frequency from single ended to differential. The input square wave have an arbitrary DC offset. In order to remove the DC offset i am using a high pass filter between signal source and input of ADA4922-1. The high pass filter is a capacitor of 1uf in series and a resistor of 300K in parallel. The problem is that ADA4922-1 itself is introducing an offset of about 0.3V at the input. Even when there is no input signal, 0.3V apear at the input port while the output of ADA4922-1 is correspondingly at +/-0.3V.

Supply voltage is +/- 12V with 100 nF coupling capacitor. Both outputs have an RC low pass filter R=41 ohms, C=3.9 nF. The output drives an ADC with input impedance of 1 Gohms

Please suggest a solution to remove this DC offset. Also suggest should i add a load resistor as the ADC impedance is very high?

Thankyou & Regards


  • What you are seeing is the effect of the bias current. It is spec'd at typically 2uA which when flowing through your input shunt resistance of 300k would generate 600mV of offset. So, your part is actually twice as good as the spec

    High pass filtering like you are trying to do would require very low bias currents due to your high resistor value.

    It's usually easier to avoid the offset by running the AC coupled signal as a current into an inverting amplifier where the bias current would be removed by the feedback. It still assumes that your signal current would be much higher than the bias current though.



    If you don't want to change the amplifier, then a first order fix would be to add a 300k resistor to the reference input (assuming it is currently just tied to ground). You would still get an offset on the output, but your differential output would be much closer to zero. This assumes that your ADC has a fairly large common mode range as the bias current can be as high as 3.5uA resulting in more than 1 V of common mode on the output. Also make sure to bypass the resistor with a cap unless you don't care about the additional noise.

    p.s. why did you pick such a low knee frequency for your high pass filter? Increasing the cut-off  would allow you to reduce the resistor significantly.

  • Thankyou KRZ for quick and detailed reply. Well i did try to reduce the input shunt resistance and increase cutoff frequency, but due to extra loading to the source, the signal drops significantly. I shall try to use the 300k resistor to the reference and let you know about the outcome.

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