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I am currently writing a lab report for an introductory lab in electrical engineering on the subject of op-amps. One question that I'm stuck at is

With the op-amp operating in its linear regime, what voltage would you expect at the negative (inverting) input terminal, with respect to ground? How does this compare with your measurement of \$V_n\$?

I think the goal is to test the idealness of an inverting op-amp with the data that I have here: Op-amp idealness

The op-amp circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Supply voltages:
\$V_{cc+}=15.664\$ V
\$V_{cc-}=-15.143\$ V

I was taught that the golden rule of an ideal inverting op-amp is \$V_n=V_p\$. But what is \$V_p\$? It can't be the input voltage because of the golden rule. How do I answer those questions with the given data?

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    \$\begingroup\$ Was there a schematic with the question? It would be very useful. \$\endgroup\$
    – Transistor
    Commented Feb 8, 2020 at 22:27
  • \$\begingroup\$ @Transistor Yes, I just added it. \$\endgroup\$
    – Sckizel
    Commented Feb 8, 2020 at 23:20
  • \$\begingroup\$ "But what is Vp?" By inspection, voltage Vp--the voltage at the op amp's non-inverting input--is zero volts. (NB: Ground potential is zero volts, and the op amp's non-inverting input is connected directly to ground potential.) \$\endgroup\$ Commented Feb 9, 2020 at 1:24

1 Answer 1

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If you read the datasheet, Vp-Vn = Vio (input offset voltage), which should be close to ~3mV as the spec says. Ideally, you may think it should be zero.

Ip and In is the input bias current, Iib, which should be ~30pA. Your measurement seems to high, but your equipment may not go that sensitive.

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