Question

Q) Develop a practical differentiator that will differentiate signals with frequencies up to 500 Hz. The gain at 100 Hz should be 1 . if the operational amplifier used in the design has unity gain frequency of 1 MHz, what is the upper cutoff frequency of differentiator? (solve neatly)

Answer 1

At high frequencies the gain of the ideal differentiator is very high. This high gain makes the circuit unstable. Thus to avoid this resistance R1 is added in series with capacitor C1 and a capacitor C2 is added in parallel with resistance R2. Here R1 and C2are called as compensating components. Input source loading is avoided through R1 though C1 gets shorted. At high frequency, the reactance of C2 is small so high frequency noise gets bypassed and not appeared at output.

The 0db frequency of the diffrentiator is 100 Hz.  

The 0db frequency will be given by,  

Let C1 be 1 uF.  

Then,

  

Therefore,  

  

The higher frequency is 500 Hz.

Which is given by,

  

  

Practically,  

  

The gain of the practical differentiator is given by,  

  

Where, Z0 is the impedence of R2 parellel to C2 and Zin is the impedance of R1 and C1 in series.  

  

So,

Dividing by , we get,  

Substituting,

we get,

Dividing by , we get,

Substituting,

we get,  

  

From the ideal differentiator, the 0 dB frequency

The higher frequency  

Substituting these ,

If the unity gain frequency of the op amp is 1 MHZ.

The higher cutoff frequency is given by,

  

C1 = 1uf and C2 = 198.7 nF.  

So we need to find R2.  

Since ,

  

  

Since,

  

Therefore,