Input impedance of an op Amp

Question

I noticed that the input impedance of an op Amp is extremely high. Why is that so?

Answer 1

It's one of the rules. For an ideal opamp goes that

• input impedance is infinite (input current is zero)
• gain is infinite
• output impedance is zero
• input offset voltage is zero

If these requirements aren't met several basic opamp circuits wouldn't work. Take for instance the inverting amplifier.

The transfer function is

\$ \mathrm{V_{OUT} = - \dfrac{R_f}{R_{IN}} \cdot V_{IN}} \$

as derived in this answer. The proof relies on the infinite input impedance, but you can't explain the transfer function based on both inputs equal, because that's not a property of the opamp! So-called proofs that start from the fact that the inverting input is at ground are invalid.

Note that FET input opamps do better than their BJT input counterparts. The former will have pA input current, whereas for the latter this may be several \$\mu\$A.

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Further reading:
Opamps for everyone

Answer 2

A very high input impedance gets us closer to an ideal op-amp. The characteristics of an ideal op-amp are:

• Infinite bandwidth
• Infinite gain
• Infinite input resistance

The ideal op-amp exists because using it as a basis for analysis provides several worthwhile shortcuts that simplify the math involved. The infinite input resistance is important because it ensures that no current goes into the op-amp. This simplifies the analysis of feedback op-amp circuits.

Also, high input impedance is desirable in most circumstances. It allows a signal with very weak drive to be correctly read by the op-amp and amplified. If it had low input impedance the op-amp would draw down the voltage of the weak signal and not properly amplify it.

Answer 3

The higher the input impedance, the less likely that the opamp itself will effect the input signal.

Think of input and output impedance as a voltage divider made from two resistors. The input impedance of the opamp is the lower resistor, while the output impedance of whatever device is feeding the opamp is the high resistor.

The best case is when the output impedance is super low while the input impedance is super high. In this case the voltage divider is barely bringing the voltage down.

Worst case is when the output impedance is super high and the input impedance is super low. Then the signal might be divided down to 1/100th of the original voltage, or worse!

So, it's better if the input to the opamp has the highest impedance possible.

There are cases where you might want a lower input impedance. But in these cases you usually just put a load resistor on the input signal and don't rely on the opamp to do that for you. Your load resistor (or terminating resistor, or whatever) will have higher tolerances than what the opamp will give you normally.

Answer 4

The answers are all good, but there is a more basic and more important concept behind that, in part explained by David.

The Op-amp, at least the most common type of op-amp, is a voltage-input, voltage-output component: it means that it takes a voltage as input, does some operation and spits out another voltage.

Reading a voltage requires attaching an instrument to the voltage source, as you would do with a voltmeter: this instrument has to be put in parallel (has to read the same voltage) with the source itself, and will see a current on it that depends on the input resistance of the instrument itself, in this case the op-amp. The current will also flow on the output resistance of the source, as shown in the picture:

The current on the resistor will cause a drop on the input voltage to the op-amp, that won't be anymore equal to the source: we don't want that.

So if the op-amp has an infinite resistance, the current on the input will be zero and the read voltage will be Vg, as desired. Of course that's an ideal condition, but the higher the better.