Question

Consider the diode circuit shown in the figure below:



The cut in voltage of diode $D$ is equal to $V_{on}=0.7V$. If the input to the circuit $V_{in}\left(t\right)=3\sin \left(2\pi t\right)V$, then the average value of the output voltage $V_{out}\left(t\right)$ will be equal to _____ $V$.

Answer 1

The output waveform for the above circuit can be drawn as


where $\theta =si{n}^{-1}\left(\frac{1.7}{3}\right)=0.60rad$

To find the average value of the output we have to find the area $A_1$ and $A_2$ for the above figure

$Area{A}_1={\int }_{{\theta }_1}^{\pi -{\theta }_1}(3sin\theta -0.7)d\theta =2{\int }_{{\theta }_1}^{\pi /2}(3sin\theta -0.7)d\theta$

$=2[-3cos\theta -0.7\theta {]}_{0.60}^{1.57}=2[-1.099+3\times 0.824+0.42]=3.586V$
Now, area $A_2$ can be calculated as
$A_2=1[2\pi -(\pi -{\theta }_1-{\theta }_1\left)\right]$
$=1(\pi +2{\theta }_1)$
now ${\theta }_1=4.34V$
$\therefore$ Average $=\frac{3.586+4.34}{2\pi }=1.26V$