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I'm using an HCPL-7840 (Isolation Amplifier) on a project.

Its datasheet says its equivalent input impedance is 500 kΩ, but impedance is frequency dependent and I'm going to use it in a DC application. So I guess for my application 500 kΩ is not really relatable.

So I did an experiment where I used a resistor divider, kept R1 fixed and tried out 4 different values for R2 (0, 12k, 24k, 51k), measuring the \$V_{\mathit {div}}\$ in each case and considering that there's this \$R_{\mathit {in}}\$ in parallel with R2:

circuit showing resistors added to chip pinout

In this experiment, my empirically calculated \$R_{\mathit {in}}\$ is around 25 kΩ during all 4 experiments.

  1. Any opinions if this is an OK method to find out the \$R_{\mathit {in}}\$ value @ 0Hz?
  2. Should I be able to read the input resistance with a multimeter between pins 2 and 3? (Currently, I can't)
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    \$\begingroup\$ Make sure you're not exceeding the differential voltage spec, stay within the common mode range, etc. \$\endgroup\$ Apr 24, 2022 at 21:21
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    \$\begingroup\$ it would be helpful if you tell us the Value of R1 in your test. Furthermore, I doubt that your test is relevant for if your R1 value is in the order of magnitude of 24k ( one of your R2 test value) this would impose on the + input more than 2.5V which would bring the amplifier into the non linear range. Please provide details about the value of your R1 , Why do you seek to know the input impedance ( briefly some details of your application) . Also, nowhere in the suggested document do I read about the input impedance being dependent on frequency, pls inform us where you read that ? \$\endgroup\$ Apr 24, 2022 at 22:45
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    \$\begingroup\$ The differential input of the isolation amp is an active circuit which outputs some current through the circuitry you are going to connect to inputs. If your circuit cannot stand that specified typically 0,5uA current your system will not work. Operational amplifiers want the same, but they generally try to push 10x less. You should design the circuit connected to the input so that the current push out of the isolation amp inputs do not cause substantial voltage change to the signal; for ex. the internal resistance of your signal source is less than 1000 ohms. \$\endgroup\$
    – user136077
    Apr 24, 2022 at 23:21

3 Answers 3

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Any opinions if this is an OK method to find out Rin value @ 0Hz?

Without the details it is hard to understand, specially for the case of \$R2=0\$. Besides, the method violated the maximum input voltage. Anyway, there is no resistor to be measured, so this is the information from the datasheet you should consider in the project:

enter image description here

But, considering the usual application for this device, it is hardly an important issue.

Should I be able to read the input resistance with a multimeter between pins 2 and 3?

Same as above. There is no resistor and your IC should be powered and with an input voltage within specs for you to measure the input current.

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Usually input resistance of this kind of amplifier is not very interesting. In any case, it is given as 500k\$\Omega\$ typical (no maximum specified). Impedance is the same as resistance in this case (for one thing, no frequency is specified). DC accuracy is more affected by bias current.

In the case of the HCPL-7840 there is -0.5 (typical) but up to -5uA flowing out of the input pins, with no maximum temperature coefficient specified. Input resistance is effectively the change in that input current with changes in input voltage (must be within the -200 to 200mV input range to be valid).

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\$R_{in}=\Delta V/ \Delta I\$ with input Vcm in the linear range and output in linear range with negative feedback.

Your test with Iin* ΔRin may not be satisfying this.

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