4
\$\begingroup\$

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.

\$\endgroup\$
6
\$\begingroup\$

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.).

\$\endgroup\$
  • \$\begingroup\$ It may be worth noting that many devices specify that input currents will always be within a certain range, but specify nothing about how they will vary with voltage within that range. A device might behave as a rather low resistance within certain narrow voltage ranges without ever sourcing or sinking very much current. \$\endgroup\$ – supercat May 15 '15 at 20:38
  • \$\begingroup\$ @supercat Do you have a specific example you're thinking of here? Certainly happens if you go outside the common mode range and change the bias of a bipolar op amp. Some of the precision bipolar amplifiers have bias cancellation circuits that can do interesting things. \$\endgroup\$ – Spehro Pefhany May 15 '15 at 20:40
  • 1
    \$\begingroup\$ I was thinking of things like rail-to-rail op amps which use different circuitry to handle the upper and lower parts of the input range. Both parts of the circuit may be trimmed to get the input current as close to zero as positive, but if one part happens to have a small positive current and the other a small negative current, and if the switch-over is somewhat abrupt, the change in voltage required for that change in current may be quite small. \$\endgroup\$ – supercat May 15 '15 at 21:42
2
\$\begingroup\$

If you had a 1V RMS signal driving a 1kohm load you'd expect to see a current drawn from your signal of 1mA RMS. If, incidentally, a constant 1A DC were also injected, this current would flow into the voltage source causing an AC current and a DC current to flow into or out of the voltage source.

None of this has any bearing on the 1kohm input - the voltage it sees is 1V RMS from the voltage source.

However, if the voltage source were not zero ohms in impedance (but say 10 ohms), the incidental current of 1A DC would see a total impedance of 10||1000 = 9.901 ohms and this would produce a DC offset voltage across both voltage source and 1kohm input of 9.901 volts.

However, there would still be an AC current of 1mA RMS taken by the 1kohm.

Hope this makes sense. There is no relationship between input impedance and leakage/bias currents. What your question describes is somewhat equivalent to the input leakage current having nowhere to go except form a voltage across the input impedance but this is of no consequence because if the pin is open circuit (no voltage source and its associated impedance) you can't expect to make any sense of what the input does let alone predict the output voltage of the op-amp.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.