written 7.8 years ago by | • modified 7.8 years ago |
Mumbai University > Electronics ana telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 5M
Years: May 2015
written 7.8 years ago by | • modified 7.8 years ago |
Mumbai University > Electronics ana telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 5M
Years: May 2015
written 7.8 years ago by |
The circuit which removes a portion of input signal is called clipping circuit or clipper.
The figure1.3 depicts clipper circuits. The term clipper refers to diode based network which has ability to "clip" off portion of input waveform without affecting the remaining part of alternating waveform.
There are two types of clippers namely series and parallel. In series clipper, diode is connected in series with the load. In parallel clipper, diode is in parallel to the load.
As shown in figure1, diode conducts in the positive half of input waveform and hence this part will be outputted. In the negative half of waveform, diode does not conduct and hence this part is clipped off and hence output will only have positive half of the waveform.
In the parallel clipper, diode conducts inthe positive half and hence the input will be grounded. During negative half, diode does not conduct and hence it will be open at this time and hence negative half is available at the output port as shown.
Diode clamper:
The circuits which are used to add certain amount of d.c.level to an A.C signal.
The figure1.4 depicts clamper circuit. As the name suggest "clamp" means lift or clinch. The clamping network will shift the DC input level of signal to the other position. This clamper circuit consists of diode, capacitor and resistor.
During the interval from time 0 to T/2, diode is in ON state. In this interval, output voltage is directly across short circuit and hence output voltage $V_o$ it 0 volt.
During the interval from time T/2 to T, diode is in open circuit mode and does not conduct. Hence applying Kichhoff voltage law in the inner loop, -V-V-V_(o) = 0. Hence $V_o$ = -2V.We get the waveform as shown in the figure1.4.