2
\$\begingroup\$

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.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Oh... another Falstad user :) \$\endgroup\$ – Alexander von Wernherr Aug 3 '18 at 10:08
  • 1
    \$\begingroup\$ Easy to use and comprehensive :) Sadly it fails at complex design \$\endgroup\$ – Simon Maghiar Aug 3 '18 at 10:09
  • 1
    \$\begingroup\$ Simple calculation: use star-triangle transformation. \$\endgroup\$ – LvW Aug 3 '18 at 10:09
  • \$\begingroup\$ Here you find the answer electronics.stackexchange.com/questions/333055/… \$\endgroup\$ – G36 Aug 3 '18 at 15:03
4
\$\begingroup\$

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: -

enter image description here

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: -

enter image description here

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.

|improve this answer|||||
\$\endgroup\$
  • 1
    \$\begingroup\$ Thank you for the answer ! I didn't understood at the beginning but now is clear. \$\endgroup\$ – Simon Maghiar Aug 8 '18 at 17:53
2
\$\begingroup\$

schematic

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

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.