When VD < 0, the diode will be off, and at this moment, VD = 4 V - = 4 - 10/(10+10) = 4 - υo(t) υS(t) υS(t)/2 Since VD < 0 now, so 4 - which means > 8 V υS(t)/2<0, υS(t) Conclusion: when > 8V, υS(t) diode will be off, and the output = υo(t) υS(t) 10/(10+10) = 5 sin(2.000πt) All conditions, the diode will be on, and the output will be = 4 V. υo(t)
When VD < 0, the diode will be off, and at this moment, VD = 4 V - υo(t) = 4 - υS(t) 10/(10+10) = 4 - υS(t)/2 Since VD < 0 now, so 4 - υS(t)/2<0, which means υS(t) > 8 V Conclusion: when υS(t) > 8V, diode will be off, and the output υo(t) = υS(t) 10/(10+10) = 5 sin(2.000πt) All conditions, the diode will be on, and the output will be υo(t) = 4 V.
Part%20II_Tutorials.docx