00:01
Let's discuss this question.
00:02
So here we have to draw a circuit diagram of a simple positive chamber with input and output waveforms for an input peak to peak of vpp, that is 40 volts and we need to assume the germanium diode is huge.
00:17
And secondly we need to draw the circuit diagram of a clipper that will produce an output waveform.
00:23
So here for the positive half cycle, the battery voltage will require is equal to.
00:28
And for the negative half cycle the battery voltage required is equal to we need to find that so let's start for a here the diagram will be this is v initial and here we can say this will be 20 and this will be the wave this will be to 5 and this will be 5 the waveform will be like this so here we can say the this is 40 volt that is vpp is equal to 40 volt and here we have minus 20 this is time t so here we can say the circuit diagram will be here we have a capacitance here this is d here we have a resistance this will be rl and here we can say this is v0 and this is v i that is v i that is v i that is v initiation.
01:56
So here we can say therefore here germanium diode is used.
02:16
So here vf b will be equal to 0 .2 volt.
02:24
Now here we can say so output voltage waveform will be if this is the diagram here we can say this is p so here the waveforms.
02:43
So here the waveform will be this is 40 volt here we have vm that is equal to 20 and here we can say this is 0 .2 volt so here the diagram will be like this this will be 2 pi and this will be t and this will be v0 so this is the diagram so here we can say so in case of positive passes the input signal to the output low when diode is rb and the input signal diode is equal to fb so here we can say for b this would be the circuit diagram this is the resistance and this is d1 that is diode 1 and this is v1 similarly here we have diode 2 and this is v2 so here this is v0 and here we can say this is r.
04:30
So as given the output voltage is 3 .7 at positive when input input is max that is 3 .7b3...