Hey, it's a similar question again.
This time, I think my teacher got a wrong answer.
Here's the problem:
_________________________________________________________
Draw the output voltage waveform, and find \(V_{AVG}\).
The diode is considered ideal.
\(V_i=30sin\omega t\)
_________________________________________________________
Diode is ON when \(V_i>-11.3V\), in this case \(V_o=-11.3V\)
Diode is OFF when \(V_i<-11.3V\), in this case \(V_o=V_i\)
\(V_o\) waveform:
\(\theta_1=\pi + arcsin \frac{-11.3}{-30}\)
\(\theta_2=2\pi - arcsin \frac{-11.3}{-30}\)
OK, here's what I've done. I pushed the waveform up +11.3.
\(V_{AVG}=\frac {1}{2\pi}[\int_{\theta_1}^{\theta_2}18.7sin\theta \cdot d\theta]-11.3\)
\(V_{AVG}=-16.8V\)
The teachers answer is -15.8. I don't know how he got there but can anyone confirm that I'm right.
This time, I think my teacher got a wrong answer.
Here's the problem:
_________________________________________________________
Draw the output voltage waveform, and find \(V_{AVG}\).
The diode is considered ideal.
\(V_i=30sin\omega t\)
_________________________________________________________
Diode is ON when \(V_i>-11.3V\), in this case \(V_o=-11.3V\)
Diode is OFF when \(V_i<-11.3V\), in this case \(V_o=V_i\)
\(V_o\) waveform:
\(\theta_1=\pi + arcsin \frac{-11.3}{-30}\)
\(\theta_2=2\pi - arcsin \frac{-11.3}{-30}\)
OK, here's what I've done. I pushed the waveform up +11.3.
\(V_{AVG}=\frac {1}{2\pi}[\int_{\theta_1}^{\theta_2}18.7sin\theta \cdot d\theta]-11.3\)
\(V_{AVG}=-16.8V\)
The teachers answer is -15.8. I don't know how he got there but can anyone confirm that I'm right.
