Given the following circuit:
what is the positive and negative peak values of Vo?
vi has a frequency of 1kHz, peak of 5V, and is a sine wave.
x(t) = Vsin(2(pi)*f*t)
KCL, KVL, etc.
The Attempt at a Solution
I made Vi to be 5sin(2(pi)1000t)
during its positive iteration it will have a peak voltage of 5V and the ideal diode D1 will be off ( Id1 = 0amps, Vd1 <0). The current across R2 (the resistor in series with the diode) will be 0 because it shares the current of D1. The current across R1 will also be 0 since it shares the current of D1:
Ir1 = Ir2 = 0amps.
Then the voltages of the resistors are as follow:
Vr1 = Ir1/r1 = 0 volts
Vr2 = Ir2/r2 = 0volts
I used KCL to find Vd1
Vi +Vd1 = Vr1 + Vr2
then Vd1 = -5 volts which agrees with Vd1 <0.
To find Vo I use KCL Vd1 + Vo = Vr2
then Vo = 5volts peak
but the answer is not 5 volts peak it is -2.5 volts peak. What have I done wrong? I also got a wrong answer for the negative iteration but maybe solving this issue will help me solve the other on my own. Notice, i treated Vo as an open and I connect both Vo and Vi to ground. Vo is supposed to be treated as an open right? Like a voltage meter? Thank you all.
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