**AIM**

To design and setup an RC integrator and differentiator circuits

**COMPONENTS AND EQUIPMENTS REQUIRED**

1. Capacitors

2. Resistors

3. Signal generator

4. CRO

**CIRCUIT DIAGRAM AND DESIGN**

**1.Â Â Â Â Â ****DIFFERENTIATOR WITH INPUT FREQUENCY 1 KHZ**

Input voltage is 20 v pp.

**DESIGN **

RC<0.0016T; Take R= 5.6 K (To avoid the loading R should be more than ten times the resistance of signal generator)Â Â SoÂ Â C= 220 p FÂ ( T=I msÂ because the input frequency is I kHz)

**2.Â Â Â Â Â ****INTEGRATOR**

**DESIGN**

RC>16 T

T= 1 ms, Take R=5.6 K to avoid loading effect of signal generator.

So C=2.2 uF (approximately)

**THEORY**

Â Â Â Â

Â Â Â Â Â Â Â Â Â Â A differentiator gives the derivative of input voltage as output.Â A differentiator using passive components resistors and capacitors is a high pass filter. The circuit is shown .It acts as a differentiator only when the time constant is too small. The voltage at output is proportional to the current through the capacitor. The current through the capacitor can be expressed asÂ Â C dv/dt. The output is taking across the resistor. So output will be RC dv/dt. Thus differentiation of input takes place.

Â Â Â Â Â Â Â Â Â When a square wave is applied at the input, during the positive half cycle capacitor charges. So initially the voltage across the resistor will be the applied voltage. As the capacitor charges, the voltage across resistor decreases.

Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Now consider the case of integrator. It is a low pass filter. Here the time constant of the circuit should be very large. Here output is taking across the capacitor. As the input square wave is applied, during the positive half cycle the voltage across capacitor increases from zero, to the maximum (peak value of applied voltage). During the negative half cycle, the capacitor starts to discharge and comes to zero. This process repeats for the remaining cycles and a triangular wave is obtained.

**PROCEDURE**

1. Set up the differentiator circuit.

2. Apply the square wave of 5V pp at 1 KHz.

3. Observe the output and plot it.

4. Do the above steps for differentiator also.

**WAVE FORMS**

**1.Â Â Â Â Â ****DIFFERENTIATOR**

**2.Â Â Â Â Â ****INTEGRATOR**

**RESULT **

Integrator and differentiator circuits were designed and set up .Output wave forms were plotted.

**SAMPLE** **VIVA QUESTIONS**

1.Â Â Â Â Â What is a high pass filter?

Â Â Â Â Â High pass filter is a circuit which passes frequencies above a cut off frequency.

2.Â Â Â Â Â What is the requirement of a high pass filter to act as differentiator?

Â Â Â Â Â Â Â The time constant should be very less.

3.Â Â Â Â Â What is low pass filter?

Â Â Â Â Â Â A circuit which passes all frequencies below a cut off is called a low pass filter.

4.Â Â Â Â Â What is the requirement of a low pass filter to act as integrator?

Â Â Â Â Â Â Â Time constant should be high

5.Â Â Â Â Â What is time constant of an RC circuit?

Â Â Â Â Â Â Â Â Â It is the time needed to charge the capacitor to 63.2% of input voltage.

6.Â Â Â Â Â Mention one application of the RC integrator and differentiator.

Â Â Â Â Â Â Â Â In amplitude modulation circuits

Â Â Â Â Â Â Â

## Comments