Question E7G03:
What is the typical input impedance of an op-amp?  

Choices:  
A: 100 ohms  
B: 10,000 ohms  
C: Very low  
D: Very high    

Correct Answer is  D: Very high  

To calculate the input impedance conceptually:
$$Z_{in} = \frac{V_{in}}{I_{in}}$$
For an ideal op-amp, $I_{in} \to 0$, so:
$$Z_{in} \to \infty$$

Why D (Very high) is correct:
- Op-amps are designed so their inputs draw negligible current, minimizing loading on the previous stage.
- Typical values:
  - Bipolar-input op-amps: $Z_{in} \sim 10^5$ to $10^7\ \Omega$
  - JFET/CMOS-input op-amps: $Z_{in} \sim 10^9$ to $10^{12+}\ \Omega$
- Rule of thumb for circuit analysis: treat op-amp input current as approximately zero ($I_{in} \approx 0$), i.e., very high input impedance.

Why the other choices are incorrect:
- A: 100 ohms — Far too low; would heavily load the source and is not characteristic of op-amp inputs.
- B: 10,000 ohms (10 kΩ) — Still too low for typical op-amp inputs; even older bipolar op-amps are usually in the megaohm range.
- C: Very low — Opposite of the defining property; op-amp inputs are high impedance, not low.

Useful rules of thumb:
- Ideal op-amp model: $Z_{in} \to \infty$, $I_{in} \to 0$.
- Configuration effects:
  - Non-inverting input sees the op-amp’s very high $Z_{in}$.
  - Inverting configuration’s input impedance is set by the input resistor, but that’s due to the feedback network, not a low op-amp input.