Clippers and Clampers, Assignments of Electronics

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Electrical Science-2 (15B11EC211)

Unit-

Clipper and Clamper

Lecture-

Outline

• Content

▫ Introduction to clippers

▫ Classifications

▫ Series clipper

▫ Parallel clipper

▫ Biased clipper

▫ Examples

▫ Combination clipper

▫ Summary of clippers

▫ Reference

Why clipper used?

• The purpose of clippers is to protect a circuit against too high or too low voltages.
• A simplest clipper circuit is consist of one diode, one resistance and a AC input source.

V (^) in

Classification of clippers

• According to non- linear devices used ▫ Diode clippers and Transistor clippers
• According to circuit configuration ▫ Series clippers and Parallel clippers
• According to biasing ▫ Biased clippers and Unbiased clippers.
• According to level of clipping ▫ Positive clippers, Negative clippers and combination clippers

Series…

If, V (^) i > V (^) d, diode ON, V 0 =V (^) i - V (^) d

V (^) i R V^0

F.

B

V (^) d

Where, V (^) d=0 V (for ideal diode) V (^) d=0.7 V (practical diode or silicon diode)

Red line represent Practical diode and Blue line for Ideal diode

Input and output waveform

Parallel

Where, V (^) d=0 V (for ideal diode) V (^) d=0.7 V (practical diode or silicon diode)

Parallel configuration

R

D V

V

i

The parallel variety has the diode in a branch parallel to the load.

If ,V (^) i > V (^) d, positive half cycle diode ON, V 0 =V (^) d

R

V

V

i

F.

B

Vd

Biased positive clipper

• We know that the biasing is a process in which we connect a battery (reference voltage V (^) R ) with the diode. Same as here we just attach battery with diode in clipper circuit.
• When the input signal voltage is positive.
• If V (^) i < V (^) R + V (^) d , diode is reversed biased and does not conduct. Therefore, V (^) o = V (^) i
• If V (^) i > V (^) R + V (^) d , diode is forward biased and thus, V (^) o = V (^) R +V (^) d

Biased negative clipper

• When the input signal voltage is negative.
• If V (^) d > -V (^) R -V (^) i , diode is reversed biased and does not conduct.

Therefore, V o = V i

• If V (^) d < -V (^) R -V (^) i , diode is forward biased and thus, V (^) o= - V (^) R -V (^) d

V (^) i^ V^0 V (^) R

• If supply voltage (V (^) i ) greater than V (^) R volts will turn the diode on and conduction can be established through the load resistor.
• Forward diode is replaced by a V (^) d
• You can see, input output waveform as given below
• Where, V (^) m is peak value of applied input voltage

R

F.

B

V V^0

i >V^ R +V^ d

V

R

V

d

The output voltage V 0 =V (^) i -V (^) R -V (^) d

V (^) m

V (^) R+ V (^) d

Diode on

Diode off

V (^) i

T/2 (^) T

V (^) m –V (^) R -V (^) d

V (^) o

T/2 T

KVL

V (^) i =V (^) R diode change the state

Example

• Determine the output waveform for the sinusoidal input of given fig.1 consider ideal diode.

Fig. 1 Solution: For ideal diode V (^) d = Step 1: The output is again directly across the resistor R. Step 2: The positive region of Vi and the dc reference voltage (V (^) R ) are both applying “pressure” to turn the diode on. The result is that we can safely assume the diode is in the “on” state for the entire range of positive voltages for Vi. Once the supply goes negative, it would have to exceed the reference voltage of 5 V before it could turn the diode off.

Fig. 1

Practice problem: Find the output voltage for the network examined in above example if the applied signal is the square wave as,

Fig. 3

Ans.

Example 2

• Determine the output waveform of given network

Solution:

- Step 1: In this example the output is defined across the series combination of the 4-V supply and the diode, not across the resistor R.

Practice problem: Repeat above using a silicon diode with V (^) d = 0.7 V.

Ans

Combination clipper

• When a portion of both positive and negative of each half cycle of the input voltage is to be clipped (or removed), combination clipper is employed. The circuit for such a clipper is shown in below fig.