Please don\'t just tell me the answer, that won\'t help me. I need to understand
ID: 1921833 • Letter: P
Question
Please don't just tell me the answer, that won't help me. I need to understand WHY the answer is as such. I would just use ORCAD/PSPICE to analyze this, but I just started learning how to use that software and I keep getting messages about errors in my schematics, so the circuit won't simulate.
I understand that a diode acts as a "wave clipper". My guess is the answer is B. For the positive part of the sinuosoid, the diode is forward biased, acts as a short, and the Batter keeps Vout limited to 12V. For the negative part of the sinusoid, no such problem exists, so the Vout can reach it's full -24V.
Explanation / Answer
The answer (b).
Assuming ideal Diode, its forward bias voltage is zero.
Diode will start conduction when its anode potential is more than the cathode potential.
In the present case the battery voltage is 12V. and the input is a sine wave.
Diode will start conducting when 24sint > 12 ==> it will start conduction t = 300.
At this point output voltage(V0) is 10V
Diode will stop conducting when 24sint < 12 ==> Conduction will stop at wt = 1500.
So Output voltage V0 = 24sint V -- when t < 300
= 12 V -- when 300 < t < 1500
= 24sint V -- when t > 1500
If the diode is not ideal and has a forward bias voltage of Vb then the waveform will be clipped at 12+Vb