The output voltage is given by
Vout = - 1/ (RfCf) [dVin / dt]
Time constant = - RfCf
The negative sign indicates that there is a phase shift of 180 degree between input and output. The main advantage of such an active differentiator is the small time constant which gives perfect differentiation.
Sometimes a compensation resistance is needed to connect to the non-inverting terminal to provide the bias compensation. The compensation resistance values is given by Rcomp = (Rf parallel with R1 ).
CIRCUIT DIAGRAM
INTEGRATOR
Integrator design
The output voltage is given by
Vout = - 1/ (RfCf) Vin (t) + Vo (0)
Time constant = Rf Cf
1) To find Cf
The gain value is given by A = (Rf / R1) / (1 + jω RfCf) ------------------- (1)
The corner frequency is fc = 1 / 2RfCf -------------------------------- - (2)
Choose, fc = 100Hz and
Rf = 10KΩ
By substituting all in equation (2), calculate the value of Cf .
2) To find R1
Let Gain (A) = 1 and substitute all remaining values in equation (1), then find the value of R1.
MODEL GRAPH
INTEGRATOR
CIRCUIT DIAGRAM
DIFFERENTIATOR
DESIGN
Differentiator design
The gain value is given by A = - jω RfC1 / (1 + jω R1C1)2 ------------------- (1)
The lower corner frequency is fa = 1 / 2R1C1 ----------------------------------- (2)
The upper corner frequency is fb = 1 / 2RfC1 ----------------------------------- (3)
Always assume fa < fb < fc and Rf C1 < T. Where T is time constant.
Design procedure
1. Choose fa as the highest frequency of the input signal. i.e. fa = 100Hz
2. Choose C1 to be less than 1 micro Farad and calculate the value of R1.
Choose C = 1micro Farad and from equation (2) and Calculate R1.
3. Choose fb as 10 times fa which ensures that fa < fb. That is fb =10 fa. Now find Rf.
4. To find Cf , use RfC1 = R1C1 and Rcomp = R1 parallel with Rf .
MODEL GRAPH
DIFFERENTIATOR
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