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BY applying virtual ground i will get Vo=2v now applying nodal at non inverting terminal i will get (V-2v)/R=I now from this i will get V/I=-R. but the answer given is Option B mean only R

whats mistake i am doing?


2 Answers 2


Input resistance -R is right as long as the circuit is not saturated and does not oscillate. If your signal source is an ideal voltage source and the opamp is ideal, the voltage gain really is 2. Input resistance -R means that the opamp supplies current to the signal source.

Negative resistance means that the circuit has a tendency to be unstable. If the input signal source has reactive parts, the circuit can oscillate. If there's a series resistance greater than R, the circuit drifts to max. output limit + or - because it supplies the needed input through the feedback to the + input. The circuit gets saturated. When saturated, the input resistance is R.

Addition due the comment: enter image description here

Because the opamp actively pushes current trough the feedback resistance R to the singnal source, the behaviour depends heavily on the internal resistance R1 of the signal source. If R1 > R the circuit is unstable. Uout pulls U1 and small Uo cannot resist. For example if R=10kohm, R1=11kohm, Uo=sinusoidal 0,2Vpp, the opamp has +15V and -15V supplies and behaves otherwise ideally, then the output voltage is constantly either +15V or -15V.

The signal source sees in this case a resistor R who has the other end connected to +15V or -15V. This is what I mean by saying "input resistance is =R"

Bigger peak to peak amplitude Uo can flip the state of the output. The circuit is like a Schmitt Trigger which is designed to convert continuously changing voltages to 2 state square waves.

Let R1 be less than R. For example R=10kohm and R1=9kohm. Let Uo still be small, say 0,2Vpp. If you simulate the circuit, you will see something not very usual: U1=2Vpp and Uout=4Vpp. The circuit seems to amplify the signal already in the input of the active amplifier. 0,2Vpp has grown to 2Vpp. That's how negative resistance works. Between A and B theres -10kohm as the load for the signal source. That amplifies.

Is it real or some kind of illusion created by the simulation? If there are long wires that can work as resonant circuit at the usable frequency band of the opamp, then the circuit probably oscillates in practice. You can simulate this by replacing R1 by an LC resonant circuit. Let R=10kohm, L=1mH and C=1uF, Uo can be 0V (=wire) the opamp can be nearly anyone and the supplies of it are +/- 15 volts. In a couple of seconds you see that Uot starts to swing between the + and - maximum output voltages. The frequency is about 5kHz.

  • \$\begingroup\$ i am really not getting your second paragraph.can you little explain more how can we get input resistance R? \$\endgroup\$
    – Rohit
    Commented Aug 25, 2017 at 3:24
  • \$\begingroup\$ @Rohit the answer is augmented. \$\endgroup\$
    – user136077
    Commented Aug 25, 2017 at 18:41

\$\small V_O=V-IR\$ ... (1)

\$\large \frac{ 0-V}{R}=\frac{V-V_O}{R}\$

hence \$\small V_O=2V\$ ... (2)

From (1) and (2):

\$ \large \frac{V}{I}=\small - R\$

  • \$\begingroup\$ if you apply this kcl will violet check once thats where i get confused. \$\endgroup\$
    – Rohit
    Commented Aug 24, 2017 at 10:16
  • \$\begingroup\$ The voltage at \$\small V^+\$ is \$\small V\$; due to the negative feedback the voltage at \$\small V^-\$ is also \$\small V\$. Current through the upper two resistors is \$\small I=\frac{0-V}{R}=\frac{V-V_O}{R}\$ Current through the lower resistor is \$\small \frac{V-V_O}{R}\$. \$\endgroup\$
    – Chu
    Commented Aug 24, 2017 at 10:22
  • 1
    \$\begingroup\$ i don't know how to explain you.... ): \$\endgroup\$
    – Rohit
    Commented Aug 24, 2017 at 10:24
  • 1
    \$\begingroup\$ OK, if \$\small V_O=2V\$ then the current through the lower \$\small R\$ from right to left is of magnitude \$\frac{V}{R}\$. But \$\small I's\$ positive direction is from left to right, hence \$\small I=-\frac{V}{R}\$. If you're applying KCL to the output node, then don't forget that there is current flowing out of the op amp. \$\endgroup\$
    – Chu
    Commented Aug 24, 2017 at 10:42
  • \$\begingroup\$ ... the op amp is actually supplying \$\small 2I\$. \$\endgroup\$
    – Chu
    Commented Aug 24, 2017 at 10:49

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