Bandwidth of a differentiator

Hi,

How to calculate the bandwidth of the following circuit? What is the general formula?

That is an AC coupled detector circuit, not a differentiator. The low frequency response is determined by the value of C and the 1k and 2.2k resistors. There is no theoretical upper frequency limit.

crutschow said:

That is an AC coupled detector circuit, not a differentiator. The low frequency response is determined by the value of C and the 1k and 2.2k resistors. There is no theoretical upper frequency limit.

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Hi,

When I apply square wave at the input I get spikes at the output. That is, +ve spike for for the rising edge of square wave and -ve for the falling edge of the square wave. So, it should be differentiator at least for the square wave. What is your opinion on that?
And secondly, could you please mention the formula for calculating the bandwidth of the given circuit?

Muashr
You are correct, the capacitor and resistor on the input form a basic high pass filter, which is effectively a differentiator.

I assume by "bandwidth", you mean the 3dB bandwidth, the point where the output is 0.707 times the input volts.
Just considering the CR circuit for the moment it should be possible to calculate the 3dB bandwidth by finding the frequency at which the reactance of the capacitor and the resistance of the resistor form a potential divider such that the output voltage is equal to 0.707 times the input voltage.

The diode and resistor add a bit of complication.
If the diode is a perfect diode, the circuit can possibly be analysed by just considering the 2.2K resistor to be in parallel with the 1K resistor, IF we are just considering square waves.
If the diode is not perfect, ie it has some forward volt drop, for low input voltages it may not conduct at all and so the circuit becomes an "all stop" filter! ie an open circuit.

JimB