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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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  • \$\begingroup\$ Would Electrical Engineering be a better home for this question? \$\endgroup\$ – Qmechanic Jan 28 '15 at 12:51
  • \$\begingroup\$ 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? \$\endgroup\$ – Jacobadtr Jan 28 '15 at 12:59
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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.

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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.

enter image description here

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

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  • \$\begingroup\$ 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. \$\endgroup\$ – Jacobadtr Jan 28 '15 at 12:49
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    \$\begingroup\$ 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.) \$\endgroup\$ – George Herold Jan 28 '15 at 14:00
  • \$\begingroup\$ True, but (say) for a sharp enough step, it cam saturate for a little bit (depending on op-amp current etc) \$\endgroup\$ – Gremlin Jan 28 '15 at 14:09
  • \$\begingroup\$ 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".) \$\endgroup\$ – George Herold Jan 28 '15 at 14:17
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@Jacobadtr, it is so simple...

In an RC circuit (a resistor and a capacitor in series) the two voltages - VR (across the resistor) and VC (across the capacitor) are complementary. If you apply a constant input voltage VIN across the whole network, VC increases while VR decreases so that their sum stays constant through time. If you take the increasing VC (from zero to VIN) as an output, you get an integrating circuit (as in the funny picture below); if you take the decreasing VR (from VIN to zero) as an output, you get a differentiating circuit. Now it is obvious neither of the two partial voltages can exceed the total input voltage...

enter image description here

(the voltages are represented by red bars whose length is proportional to the value of the voltage)

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