Electronic Design

What's All This Input Impedance Stuff, Anyhow?

A young teacher went to Temple. Several priests and pharisees asked him questions. They asked trick questions to try to fool him. The teacher very sweetly and correctly answered their many tricky questions. All those present were amazed at the teacher's wisdom in answering these difficult questions.

Then a pharisee asked him another trick question. He drew a circuit in the dirt. "Look at this op-amp circuit. It is often used as an example of a well-balanced circuit. Have you heard of this?"

The teacher agreed that he had seen this circuit and that it is a well known, well balanced circuit (see the figure). Then the crafty pharisee asked, "But what is the input impedance at the negative input that A1 has to drive? That is R1, right? Such as 1 kΩ." The teacher nodded. "But the impedance at the positive input, which A2 must drive, is R1 + R2. That is 101 kΩ , right?" The teacher nodded.

"Therefore, because these input impedances are grossly unequal, this is a sorely imbalanced circuit and not capable of doing good linear work." The pharisee rested his case.

The teacher stroked his beard. He questioned the pharisee, "So you are concerned that A2 can easily drive a light load, but A1 cannot easily drive a heavy load?" The pharisee jabbed that this was exactly the terrible problem.

More beard stroking. Then the teacher pointed, "But what if both inputs are driven together to +10 V dc? Both A1 and A2 have to drive the same 101-kΩ common-mode load. That does not sound like a poorly balanced circuit."

The pharisee was stunned. He had not thought of this. He said nothing. Much silence. He was beginning to suspect that he was in trouble. The teacher thought, and went on, "If the output impedance of A1 or A2 was perhaps 1Ω, you are still concerned that the gain would be different for the positive channel, and the gain for the negative channel would be lower. That is your concern?" The pharisee gulped and nodded.

The teacher said, "If you put a millivolt into the negative input, and you want the gain to be 100, and you want the output to be ­100 mV, the 1Ω will cause the output to be about 0.1% low, or actually 0.0999% low, because the 1Ω won't be able to drive the 1-kΩ input impedance with perfect gain, right? The output signal will only be ­99.9001 mV, right?" The pharisee gulped and nodded.

"And if A2 has to drive the positive input, the 1Ω will cause negligible error when driving 101 kΩ, right?" The pharisee nodded. But the teacher poked at the drawing in the dirt and made his point: "If the positive input moves 1.000 mV, and the positive input of A3 moves exactly 990.0892 µV, that's a very minor attenuation error caused by the 1Ω, right?" The pharisee thought and thought and nodded nervously.

The teacher observed, "But the 1Ω in series with the A1 output will cause the gain for the positive input to be hurt. The gain for the signal at the positive input of A3 won't be 101, but will be 100.9001, right?" Nod.

"And the 990.0892 µV X 100.9001 equals 99.9001 mV. Despite the alleged difference in input impedances, the gain for signals at the positive and negative inputs are exactly the same. If that were not true, the CMRR would not be so good, relying on the symmetry," the teacher said.

"Go, my son, and sin no more." The pharisee slunk away. And all who were attending were stunned by the wisdom of the teacher's answer.

Comments invited!

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P.O. Box 58090, Santa Clara, CA 95052-8090
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