What is the duty cycle of the output waveform in Fig. 36-13b?

Question

Accepted Answer

Recall that the duty cycle is defined as the pulse width divided by the period. Duty cycle equals the conduction angle divided by 360°.

In Fig. 36-13b, the sine wave has a peak value of 10 V. Therefore, the input voltage is given by

[latex]v_{in} = 10 \sin heta[/latex]

The rectangular output switches states when the input voltage crosses +5 V. At this point, the foregoing equation becomes

[latex]5 = 10 \sin heta[/latex]

Now, we can solve for the angle [latex] heta[/latex] where switching occurs:

[latex]\sin heta = 0.5[/latex]

or

[latex] heta[/latex] = arcsin 0.5= 30° and 150°

The first solution, [latex] heta[/latex] = 30°, is where the output switches from low to high. The second solution, [latex] heta[/latex] = 150°, is where the output switches from high to low. The duty cycle is

[latex]D=\frac{conduction angle}{360°}=\frac{150°-30°}{360°}= 0.333[/latex]

The duty cycle in Fig. 36-13b can be expressed as 33.3%.

Question 36.4: What is the duty cycle of the output waveform in Fig. 36-13b...

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