# Differentiator Circuit

So I have a circuit like this:

With the output equation as: $V_{out} = -(\frac{R_2}{R_1})\frac{s}{s+\frac{1}{R_1C}} V_{in}$

So the question is to derive the time-domain equation in terms of the input and also to show that the circuit performs the function of a differentiator.

So here's my work:
$V_{out} = -(\frac{R_2}{R_1})\frac{s}{s+\frac{1}{R_1C}} V_{in}$ $\mathcal{L^{-1}}(V_{out}) = -(R_2C)\mathcal{L^{-1}}(\frac{s}{sR_1C+1} V_{in})$ If we assume that $R_1=0$ then we have $\mathcal{L^{-1}}(V_{out}) = -(R_2C)\mathcal{L^{-1}}(s V_{in})$ The $sV_{in}$ implies a differentiator since $\mathcal{L^{-1}}(s) = \delta'(t)$

But I'm pretty sure that is either totally wrong or mostly wrong. Where am I going wrong on this derivation for the time-domain equation?

The most straightforward way to work this with R1 = 0 is to see that the time-domain current through the capacitor is just:

$i_C = C \dfrac{dv_c}{dt} = C\dfrac{dv_{in}}{dt}$

due to the virtual ground at pin 2 of the (ideal) op amp.

It follows that:

$v_{out} = -R_2C\dfrac{dv_{in}}{dt}$

Now, the circuit pictured is not a differentiator; it is a high-pass filter. However, when the frequency is much lower than $\frac{1}{R_1C}$, the frequency-domain output is approximately:

$V_{out} = -R_2CsV_{in}$

But differentiation in the time domain is multiplication by s in the complex frequency domain. So, the circuit is an approximation of a differentiator when the input frequencies of interest are much lower than $\frac{1}{R_1C}$.

• A real op-amp will of course have a pole in it's open-loop response and this interacts with R2 & C to form a bandpass filter with very low damping & consequent ringing. R1 helps reduce this effect in practical circuits. – MikeJ-UK Feb 22 '13 at 9:02