# Resistive T-network

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.

• Oh... another Falstad user :) – Alexander von Wernherr Aug 3 '18 at 10:08
• Easy to use and comprehensive :) Sadly it fails at complex design – Simon Maghiar Aug 3 '18 at 10:09
• Simple calculation: use star-triangle transformation. – LvW Aug 3 '18 at 10:09
• Here you find the answer electronics.stackexchange.com/questions/333055/… – G36 Aug 3 '18 at 15:03

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.

• Thank you for the answer ! I didn't understood at the beginning but now is clear. – Simon Maghiar Aug 8 '18 at 17:53

simulate this circuit – Schematic created using CircuitLab

If we assume the op-amp is not limited by either the negative or positive supplies (not clipping), then V+ = V-

V+ = V- = 0V

Ir1 = (Vin - V-)/R1

Ir2 = (V- - V1)/R2

Ir2 = (-V1)/R2

Ir3 = (Vout - V1) / R3

Ir4 = Ir2 + Ir3

Ir4 = V1 / R4

Those are what can immediately gathered from our circuit.

Application of basic series parallel circuit with R2, R3, R4

V1 = Vout * (1/(1/R2 + 1/R4)) / (R3 + 1/(1/R2 + 1/R4))

V1 = Vout * (R2 + R4)/(R2*R4) / (R3 + (R2 + R4)/(R2*R4))

V1 = Vout * (101k/100M) / (100k + (101k/100M))

V1 = Vout * (101k/100M) / (100k*100M/100M + (101k/100M))

V1 = Vout * (101k/100M) / (100k*100M + 101k/100M)

V1 = Vout * (101k) / (100k*100M + 101k)

V1 = Vout * (101,000) / (100,000 * 100,000,000 + 101,000)

V1 = Vout * (101,000) / (10,000,000,101,000)

V1 = Vout * (101) / (10,000,000,101)

We calculated a value for V1 based in Vout.

Ir1 = Ir2

(Vin - V-)/R1 = (V- - V1)/R2

(Vin)/R1 = (-V1)/R2

(Vin)/R1 = (-Vout * (101) / (10,000,000,101))/R2

(Vin*R2)/R1 = (-Vout * (101) / (10,000,000,101))

Vin = (-Vout * (101) / (10,000,000,101))

-Vin * 10,000,000,101 / 101 = Vout

Standard form is Vout = -Vin * R2 / R1

Multiplying by 100k/101 (a factor of about 990.09), the equivalent R2 is 9.9 gigaohms

If this simulates a 10M feedback resistor then I have miscalculated somewhere