Electronic – the relation between Op-Amp input current and input impedance

input-impedanceoperational-amplifier

How should I interpret these specs? They can't both be correct all the time, right? Looking at for example the CA3140 opamp.
the Zin (input impedance) is 1.5T Ohm, yet the input current is 10pA.
Solving this for voltage (Ohm's law, 1,5e12 * 10e-12) gives 15 volt, so at that input voltage, both these specs are correct.

  • But what happens when you give the opamp 100mV at its input?
  • Do I calculate the input current according to the Zin or do I assume
    that it's 10pA?

Assuming the opamp draws 10pA, the input impedance is now R = U / I = 0,1 / 10e-12 = 1e10 Ohm instead of 1.5e12.

Best Answer

Input resistance (1.5T\$\Omega\$ typical) is the change in input current for change in input voltage

\$R_{in}\$ = \$\frac {\Delta V_{in}}{\Delta I_{in}}\$

It is not clear whether this figure is intended to apply to differential input voltage or to common mode voltage or both.

Input current (10pA typical) is the current flowing into or out of the input pin.

If the input was an ideal current source, the input current would be constant so the input impedance would be infinite.

You can model the input (at DC) as a 1.5T ohm resistor to the other input (probably, given the disposition of the input protection diodes) and two +/-10pA current sources, one connected to each input.

Some op-amps (the ancient CA3140 is not one of them) have a rather high input resistance when the two inputs are close to each other in voltage but nonlinear networks across the inputs that turns into k-ohms if you apply more than a diode drop differentially. Not a problem in normal op-amp applications, but problematic if you're using it in applications where it might saturate (precision comparator, some precision rectifier circuits etc.).