# A basic question about current-to-voltage converters

• posted

I have a basic question seeming quite simple: An op amp with an open loop gain A, and a feedback resistor Rf, are connected as a current-to-voltage converter which can be found in many common applications. If the input impedance of the op amp is very large and can be regarded ideally "infinite" (as the case of FET-input op amp), the input resistance of the current-to-voltage converter can be estimated as Rf/A (Why?). With this presumption, can I estimate the voltage at the virtual ground pin as I*Rf/A (I denotes the input signal current)? How can I estimate the maximum feedback resistance Rf which the op amp can load? Does that value depend on the open loop gain A? If yes, please tell me the way or where I can find a definite answer. Thanks. Jiaqi

• posted

With the non-inverting input at zero volts and your inverting input at some value, V, then your output voltage is V*A, right? This produces a current through Rf of (V*A - V) / Rf, the voltage difference between the output and the inverting input divided by the feedback resistor. So:

I = (V*A-V)/Rf = V*(A-1)/Rf, so V = I*Rf/(A-1)

So I think the voltage at the virtual ground node (the inverting input) is I*Rf/(A-1), not I*Rf/A. This makes the input impedance Rf/(A-1), I think.

For DC considerations, your Rf depends on your maximum allowed output voltage and your maximum input current. If you have a maximum output of 10V and your maximum current will be 1mA, then your Rf is limited to 10V/1mA*[(A-1)/A] or about 10k Ohm.

Jon

• posted

current

I think the output voltage should be -V*A rather than V*A, and the input current I should be expressed as (V-(-V*A))/Rf=V*(A+1)/Rf. So the input voltage V is I*Rf/(A+1) rather than I*Rf/(A-1). The input impedance is Ri=Rf/(A+1) rather than Rf/(A-1). Am I right, Jon?

Jiaqi

• posted

The input impedance (Rf/A) is almost zero, because the opamp tries to keep the inverting input at gnd potential by means of the output voltage through the feedback resistor. With a smaller resistor the compensation current will need less voltage swing on the O/P.

With this presumption, can

Your are right with the voltage at the inverting input, additionally is superimposed the offset voltage. Rf is (almost) independent of the open loop gain. It is only dependent on the max output voltage swing. If your input is +1uA, with 1Meg Rf the O/P voltage will be -1V, so with a voltage swing of +/-13V the max. Rf to prevent saturation would be 13Meg.

welcome

```--
ciao Ban
Bordighera, Italy```
• posted

I'm a hobbyist and get my signs wrong! Yes, I like that better.

The output would indeed be -V*A.

Jon

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