# Differentiator circuit - can output be larger than input?

I have a question regarding differentiator circuits, and whether or not their output voltage can be greater than their input voltage.

The voltage across the resistor in such a circuit corresponds to the derivative of the input signal:

$$V_{out} = RC \frac{dV_{in}}{dt}$$

I think this means that a high frequency input signal, which has a large rate of change of gradient produces a larger output signal than a low frequency signal, but I am unsure of whether or not it can exceed the input voltage.

What do I need to consider to understand this?

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• Would Electrical Engineering be a better home for this question? – Qmechanic Jan 28 '15 at 12:51
• It is a question I have from reading my physics lab script, but if you think it's more appropriate there... then ok. Can it be moved or do I need to post again? – Jacobadtr Jan 28 '15 at 12:59

The equation given is a low frequency approximation that holds only when

$$\omega RC \ll 1$$

The exact phasor equation for an RC 'differentiator' circuit is

$$V_{out} = \frac{j\omega RC}{1 + j\omega RC}V_{in}$$

Note that

$$V_{out} \le V_{in}$$

for all frequencies. However, when $\omega RC \ll 1$, we have that

$$V_{out} \approx j\omega RC\;V_{in}$$

which is the phasor equation for a differentiator. Since we've assumed $\omega RC \ll 1$ for this approximation, it follows that

$$V_{out} \ll V_{in}$$

for these frequencies in which the approximation holds.

Your formula is only true for an active differentiator, which includes (most simply) an op-amp (figure below). In this situation, there's a power supply, which sets the maximum output voltage, which can be higher than the input voltage. The properties of the op-amp will determine what happens if your formula predicts higher output voltages, but most likely it will saturate.

For the circuit that is only a capacitor and resistor, it's more properly thought of as a high-pass filter. There is a response something like a derivative, but it's not as simple. You can find more on wikipedia at https://en.wikipedia.org/wiki/Differentiator

• That equation comes from my lab-script where the circuit consists of only a capacitor and resistor (and signal generator + oscilloscope)... We are going to be thinking of the circuit as a high-pass filter, but seemingly as another property not a more accurate description. – Jacobadtr Jan 28 '15 at 12:49
• Nice(+1) Vout > Vin as long as 2 * pi * f * R * C > 1. But there is no reason for the opamp to saturate. (Just keep the input signal amplitude low.) – George Herold Jan 28 '15 at 14:00
• True, but (say) for a sharp enough step, it cam saturate for a little bit (depending on op-amp current etc) – Gremlin Jan 28 '15 at 14:09
• Sure, You can certainly cause it to saturate, but it's not required. Say a 1 kHz sine wave, C= 0.1uF and R = 10 k ohm. (And yeah, a square wave will be cause "problems".) – George Herold Jan 28 '15 at 14:17