Electronics and Communication Engineering - Analog Electronics - Discussion

How at 2v phase is pi/6?

I still don't get how pi/6 comes, as it says 8v p-p so for +ve half cycle peak voltage is 4v so phase is pi/2, and for ref volt 2v the phase should be pi/4 not pi/6.

And for -ve half cycle, does the comparator remain off? Why?

How can you tell that the total time period is 2*pi ? In the question it is not mentioned.

In this the total time period is 2pi, so (Ton+Toff) = 2pi.

We know that duty cycle = Ton/(Ton+Toff).

Now we should find Ton period i.e., here it is given that p-p voltage is 8v. So for +ve half cycle the amplitude is (0 to 4) we and phase is (0 to pi) right! It is given that ref voltage is 2v. So when ever the voltage crosses ref voltage comparator comes to ON state.

At 2v the phase is pi/6 (mark 2 volts in the above wave plot and draw a horizontal line towards wave. That point shows the phase on theta axis). So phase is pi/6. Therefore the Ton is (pi- (pi/6+pi/6) ) which gives us the duty cycle as 1/3.

Please explain with more details.

Since Vref=2V, so the comparator can be assumed to set it as a threshold voltage. Now let us the compare the phase with the amplitude. When voltage exceeds 2V, phase is just pi/6. Again after reaching the peak value it returns to pi/6 at 5*pi/6 step.

So the time t which it exceeds the threshold can be considered the ON time and the rest as OFF. Thus duty cycle = (pi-pi/6-pi/6) /pi which gives 1/3.

Since frequency is constant. So phase is linearly proportional to time. i.e duty cycle = Ton/Toff satisfies the above expression.

Same doubt. Why should we take pi/6?

How that pi/6 will come in formula?

How that pi-pi/6-pi/6 will come in the problem?

What is the impact of Vref in the prob?