I'm a beginner studying *The Art of Electronics*, and on page 25 they introduce the differentiator. Basic circuit like so:

They give the complete equation:

$$

I = C \frac{d}{dt}(V_{in} – V) = {V\over R}

$$

I understand this so far. But then they say: *if we choose R and C small
enough so that* \$\frac{dV}{dt} \ll \frac{dV_{in}}{dt}\$,

*then…*

$$

C\frac{dV_{in}}{dt} \approx \frac{V}{R}

$$

This I don't follow. Can someone elaborate or explain a bit more? I see why the above equation makes it a differentiator — V is proportional to \$\frac{dV_{in}}{dt}\$. But why does a small R and C cause the one derivative to me much less than the other?

## Best Answer

If you rearrange the first equation, you get

\$C\dfrac{dV_{in}}{dt} = \dfrac{V}{R} + C\dfrac{dV}{dt}\$

So if you reduce

Cenough, you'll make the derivative term on the right insignificant compared toV/R, and you'll get your second equation.Alternately, if you reduce

R, you'll make theV/Rterm larger, and again the right-hand-side derivative term will become insignificant, and you'll get the desired result.So I'd say it's not that you must decrease

RandCtogether, but you do have do some combination of reducingRand reducingCto make the circuit work like a differentiator.