I am given this figure below:


And I am trying to find Vout and the gain.

From looking at this source: https://www.electronics-tutorials.ws/opamp/opamp_5.html I have determined the gain to be 10k/1k * (v+ - v-). I was just hoping for some confirmation on this answer that I got...

I also tried to find Vout, however, in doing so I neglected the 1k resistor to get: Vout = -10k * C * d/dt(Vin)

I only neglectected the 1k resistor as I am not sure how to go about finding Vout with that resistance there. Is neglecting the 1k resistor wrong?

I appreciate any help, thank you.

  • \$\begingroup\$ Since there are no terminals labeled "v+" and "v-" in your schematic, your formula could be right or could be wrong, but there's no way for us to read your mind and figure out which terminals you think those labels refer to. \$\endgroup\$
    – The Photon
    Apr 3, 2019 at 0:10
  • \$\begingroup\$ On some op-amps, "v+" and "v-" are the names of the power supply terminals. If that's the case for your op-amp, then the formula is almost certainly wrong. \$\endgroup\$
    – The Photon
    Apr 3, 2019 at 0:12
  • 1
    \$\begingroup\$ This is not Differential Amplifier. This is Differentiator Amplifier, the correct link is here. Sunnyskguy already answered but I believe this paper from TI may help you learning to design real life practical differentiator circuit with op-amp. \$\endgroup\$
    – Unknown123
    Apr 3, 2019 at 2:35
  • \$\begingroup\$ You have a DC-blocking high-pass-filter circuit, with gain of 10x for higher frequencies. \$\endgroup\$ Apr 3, 2019 at 3:50

3 Answers 3


The circuit you have shown is not, technically, a differential amplifier. Sure, it has two inputs, \$V_{IN}\$ and 0V, but the fact that one of them is zero means that the input is not differential. The input here is "single-ended", and this configuration would be better described as just an "inverting amplifier". The presence of a capacitor means that behaviour is frequency dependent, so an even better description would be "single ended input, inverting, high pass filter".

There are a number of ways to describe this circuit's behaviour. The first way is by inspection. Start with the basic premise that the inverting amplifier configuration consists of two impedances \$Z_1\$ and \$Z_2\$:


simulate this circuit – Schematic created using CircuitLab

Gain, the ratio of \$v_{OUT}\$ (output amplitude) and \$v_{IN}\$ (input amplitude), in general is given by:

$$ \frac{v_{OUT}}{v_{IN}} = -\frac{Z_2}{Z_1} $$

On the right, \$Z_1\$ is formed by the series pair \$R_1\$ and \$C_1\$, and \$Z_2\$ is a simple resistance \$R_2\$ so that:

$$ \frac{v_{OUT}}{v_{IN}} = -\frac{Z_2}{Z_1} = -\frac{R_2}{R_1 + Z_{C1}} $$

From our knowledge of capacitors, we know that C1's impedance varies with signal frequency. At very high frequency a capacitor's impedance approaches 0Ω, and at very low frequency, its impedance goes off to ∞Ω. The combined impedance \$R_1 + Z_{C1}\$ will therefore be, at low frequency:

$$ Z_{1(LF)} = R_1 + \infty \Omega = \infty \Omega $$

So at low frequency, gain is:

$$ \frac{v_{OUT}}{v_{IN}} = -\frac{R_2}{\infty \Omega} = 0 $$

At high frequency, when \$Z_{C1} = 0\$, gain becomes:

$$ \frac{v_{OUT}}{v_{IN}} = -\frac{R_2}{R_1} $$

Ideally, we want an expression to find gain at any frequency, not just "very high" or "very low". To do that, we usually work in the frequency domain, in terms of \$j\omega\$, where \$j = \sqrt{-1}\$ and \$\omega\$ is angular frequency (in radians per second).

In the frequency domain, we can apply Ohm's and Kirchhoff's laws as usual, but all signals and impedances are expressed as complex values. The complex impedance of a resistor R is simply its resistance (purely real), and the impedance of a capacitance \$C\$ is \$\frac{1}{j\omega C}\$ (purely imaginary).

$$ \begin{aligned} Z_2 &= R_2 \\ \\ Z_1 &= R_1 + \frac{1}{j\omega C_1} \\ \\ \frac{v_{OUT}(\omega )}{v_{IN}(\omega )} &= \frac{Z_2}{Z_1} \\ \\ &= \frac{R_2}{R_1 + \frac{1}{j\omega C_1}} \\ \\ &= \frac{j\omega C_1R_2}{1 + j\omega C_1R_1} \end{aligned} $$

That expression contains all the information we need to find gain and phase relationship across the entire spectrum of input frequencies. To find gain (the ratio of output and input amplitudes), we take the magnitude of this complex value:

$$ \begin{aligned} gain &= \left| \frac{j\omega C_1R_2}{1 + j\omega C_1R_1} \right| \\ \\ &= \frac{\left| j\omega C_1R_2 \right|}{\left| 1 + j\omega C_1R_1 \right|} \\ \\ &= \frac{\omega C_1R_2}{\sqrt{ 1 + {(\omega C_1R_1)}^2 }} \\ \\ \end{aligned} $$

Using that formula, you can plug in any value for frequency ω to find gain at that particular frequency.

If you want to work in units of Hertz (Hz), use the relationship \$\omega = 2\pi f\$, or \$f=\frac{\omega}{2\pi}\$.


Your estimate at high f is correct Av = -Rfb/Zin(f) = -10 above f >> 1/2πRinC for Rin=1k

For this configuration the transfer function is the ratio of the Feedback Impedance to the inverting Input impedance as Vin- is a virtual ground.


The output impedance of the signal source must be added to the 1k input resistor to calculate the gain at frequencies the capacitor and amplifier can pass.


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