Applications of op amps

Applications of Op-Amps

Dr. C.SARITHA

Lecturer in Electronics

S.S.B.N. DEGREE & PG.COLLEGE

ANANTAPUR

OVERVIEW Introduction Definitions Circuit Diagrams Derivations Applications Conclusion

Integrator The circuit in which the output wave form is

the integral of input wave form is known as an integrator

Such type of circuit is obtained by using basic inverting amplifier configuration where we use a capacitor in feed back

Circuit diagram

Vs

IcIR

Explanation Input is applied to inverting terminal of the

op-amp. Non inverting terminal is grounded. If sin wave is applied terminal then the

output will be cosine wave.

For an ideal op-amp Ri=infinite

R0=0

For an ideal op-am

input current=output current

IR=IC

IR=Vi-Vs/Ri

The capacitor current Ic=c(d/dt(VS-V0)

Ic=-c(d/dtV0-VS)

Input current =Output current

(Vi-VS)/Ri=-C(d/dtV0-VS)

Vi/Ri-VS/Ri =-C(d/dtV0-VS)

V0=-AVS

VS=-V0/A

Substitute this Vs in the above equation

Vi/Ri+V0/ARi=-C(d/dt(V0+VS/A)

=-C(d/dtV0)-C(d/dtVS/A)

V0/A and V0/ARi

Are very less compare to V0 and hence this terms are neglected.

Vi/Ri=-C(d/dtV0)

Integrating on both sides b/w 0 to t

t t

oVi/Ridt=- o Cd/dtV0

1 t

Ri 0 Vidt= -CV0

V0= -1 t

RiC 0 Vidt

The output of the integrator is the integral of input voltage with time constant that is V0 is directly proportional to integral of Vidt and inversely proportional to the time constant.

The input is sine wave the output become cosine wave

Input=

Output=

Similarly the input is square wave the output become Triangle waveInput=

Output=

Applications Analog computer A to D converters Many linear circuits Wave Shapping circuit

Differentiator The differentiator is the circuit whose output

wave form is the differential input wave form.

The differentiator may be constructed form the basic inverting amplifier

Here we replace the input resistor by a capacitor.

Circuit Diagram

The capacitor current Ic=CI(d/dtVi-VS)

Current through the feed back resistor

IR=(VS-V0)/RF

Input current=Output Current

IC=IR

C(d/dt(Vi-VS)=(VS-V0)/RF

CiRF(d/dt(Vi-VS)=VS-V0

V0=-CiRF(d/dt(Vi-VS)+VS

Gain A=-V0/VS

VS=-V0/A

Substitute VS in the above equation

-CiRFd/dt(Vi-VS)+VS=V0

-CiRFd/dt(Vi-VS)-VO/A=V0

V0/A is very small and hence neglected

VO=-CiRFd/dt(Vi)

The input is cosine wave the output become sine waveInput=

Output=

Similarly the input is Triangle wave the output become Square waveInput=

Output=

Active Filters Filter is a circuit which gives the DC from

the given input AC. The filter which constructed by using an op-

amp is known as the active filter. (Because the op-amp is an active component)

Types of filters High pass filter:

It allows the high frequency signals and filter them (convert them into DC).

Low pass filter:

It allows the low frequency signals and filter them (convert them into DC).

Low pass filters

RC

Working A first order filter consists of a single RC

network connected non inverting terminal of the op-amp

At low frequency the capacitor appears open and the circuit acts like an inverting amplifier with a voltage a gain of (1+R2/R1)

…Continues As the frequency increases the capacitive

reactance the capacitive reactance decreases causing a decrease in the voltage at the non inverting input and hence at the output.

Frequency Response Here fcutoff the value of input signal

frequency at which the output decrease to 0.707 times its low frequency value

High pass filters

Working A first order active high pass RC filter also

consist of a RC network connected to the non inverting terminal of the op-amp in case low pass.

Here R and C are inter changed.

…Continues In this case at low frequencies the

reactance of the capacitor is infinite and it blocks the input signal. Hence the output is zero.

As we increase the frequency, capacitive reactance decreases, and out put increase

Frequency Response Here fcutoff the value of input signal

frequency at which the output increase to 0.707 times its low frequency value