Opamp integrator, differentiator

Opamp differentiator circuit

Definition

Opamp Differentiator is a circuit which provides output proportional to the differential of input signal. If Vi is the input signal applied to a differentiator then the output is Vo = K*dVo/dt where K is proportionality constant.

Opamp differentiator operation

The opamp differentiator is as shown below

Opamp differentiator

Opamp differentiator

It is obvious from the circuit shown above that negative feedback is provided from output to inverting terminal.Using the concept of virtual ground the inverting terminal will be at zero potential(Since the non inverting terminal of opamp is at ground potential). the differentiator circuit can be redrawn as follows

Applying KCL at inverting node of opamp, we get

                                                             (0-Vout)/R + Ic = 0               

                                                                    Ic = Vout/R

where Ic = C*d(0-Vin)/dt. Hence we get Vout = -R*C*dVin/dt.

If we apply a periodic triangular signal to opamp differentiator the output will be a periodic square wave. 

Opamp integrator circuit

Integrator is a circuit which provides output proportional to the integral of input signal.If Vi is the input signal applied to a integrator then the output is \fn_phv V_{o} = K*\int V_{i}*dt where K is proportionality constant.

Opamp integrator operation

The opamp integrator is as shown below

Opamp integrator

Opamp integrator

It is obvious from the circuit shown above that negative feedback is provided from output to inverting terminal.Using the concept of virtual ground the inverting terminal will be at zero potential(Since the non inverting terminal of opamp is at ground potential). the integrator circuit can be redrawn as follows 

Applying KCL at inverting node of opamp, we get

                                             (0-Vout)/R + Ic = 0               

                                                    Ic = Vout/R

where Ic = (-1/C)*\dpi{120} \bg_white \int V_{in}*dt . Hence we get

                                              Vout = – (1/R*C)* \dpi{120} \bg_white \int V_{in}*dt

If we apply a period square signal to opamp differentiator the output will be a periodic triangular wave. 

Recent Posts