Can someone explain to me why this negative feedback resistive T-network simulates a 10M ohm resistance ?
The overall closed loop gain is -100 because 10M/100k = 100.
Electronic – Resistive T-network
feedbackgain
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Best Answer
The simple way to look at is that the 100 kohm resistor from the op-amp output and the 1 kohm resistor form a 100:1 potential divider (approximately). This means that the op-amp gain is 100 times higher than it would be if there was only a single 100 kohm resistor as the feedback element. This makes the 100 kohm resistor, in effect, 100 times in value or 10 Mohm.
If you need more understanding, take a look at this picture: -
You know that the gain at the op-amp output is -100 and it shouldn't be a surprise to find that the gain at the junction of the 9.9 Mohm and the 100 kohm (in the feedback path) is -1.
The 9.9 Mohm resistor is (approximately) equivalent to the 100 k / 1 k potential divider in the original circuit and gives approximately the same result as this: -
Of course you could convert the op-amp output to a current source in parallel with a 100 kohm resistor and note that the 100 kohm then becomes in parallel with the 1 kohm resistor. Then you could convert back to a voltage source (about 100 times smaller than the original op-amp output) in series with a 0.99 kohm resistor but you'd come to the same conclusion.