basic question about current-to-voltage converters

Discussion in 'General Electronics Chat' started by Jiaqi Shen, Feb 5, 2005.

  1. Jiaqi Shen

    Thread Starter New Member

    Jan 30, 2005
    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.
  2. beenthere

    Retired Moderator

    Apr 20, 2004

    Your op amp has to put enough current through the feedback resistor to cancel the voltage at the input. That's why the inverting input is operating at virtual ground. As the op amp has to swing the output to push current through the resistor, then there will be some value of input current where the output voltage swing will hit the rail, and no longer be able to cancel the input current.

    The linitation is dual. The feedback resistor value and the op amp's output current limit. In any case, you run out of the op amp's range when it can't swing the output any farther, or supply/sink the necessary current.

    Resistors make nice current-to-voltage converters, too.