Before the differential ADC, I have a fully differential amplifier as the front end. I want to simulate and measure differential input impedance.

My working range is max. 20 kHz.

Why am I using the 500/2k=1/4 ratio? Because I want to include the incoming signal in the range.

ADS131E08S is the ADC I use and in the datasheet I read that the input impedance is 200 MΩ. For this reason, I installed a 200 MΩ resistor as a load.

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(V(Vin+)-V(Vin-) ) / (I(E2:3)- I(E1:3))

Generally, there is a concept of high input impedance at the input of cards that read an ADC. Let's assume the differential op-amp input goes to the connectors. In this case, is the op-amp input really a value like 2 kΩ?

Here are the possibilities:

  1. The conclusion is correct, you should do something to increase the input impedance.The fully differential amplifier structure has low impedance, should I use an input similar to the instrument amplifier?
  2. I misunderstand the concept of differential impedance.
  3. I made an input calculation error.
  4. Everything is wrong.

Can someone help sort out the confusion in my mind?


2 Answers 2


I'm not an expert in this fully diff. Amps. However, I see an inverting configuration with a 2kohm series resistance at the input, and there's a virtual ground at the input of this amplifier, which means the impedance is very low at low frequencies at that node. So the 2k is very expected.

It seems to me you though the 2k is in series with a really high input impedance, but this is not the case in this situation due to the virtual ground.

You might expect a high input if you use a non-inverting configuration, but I'm not sure how those would go these fully-diff amps. I guess you might need 2 of them to make such a topology.

  • \$\begingroup\$ I changed it to real rates, resistor values. Can you edit your answer? \$\endgroup\$
    – Electronx
    Mar 27 at 22:52
  • \$\begingroup\$ @electronx The only thing I'll have to edit is that the input impedance is now 2kohm. What is it that you don't get that the input impedance of an inverting amplifier is the input resistor you select? \$\endgroup\$
    – Designalog
    Mar 28 at 6:11

Fully differential amplifiers have a "virtual short" between their inputs due to how they operate (equivalent to the concept of "virtual ground" on inverting single ended operational amplifiers).

So the differential input impedance of your circuit is simply 4K (R2 in series with R3).


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