BamaBino
Well-known member
Isn't the circuit output a Half-wave rectified Sin wave?
Therefore the average voltage would be Vpeak/Pi?
Therefore the average voltage would be Vpeak/Pi?
Complex Imaginary Test 3 Problem 38
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BamaBino
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I used the integral method and a table of standard integral for this problem and came up with the same answer.
To answer your question, a rectified sinusoid has a Vave = 2*Vpeak/pi (Power Reference Manual PG 27-4 Eq27.17).
EDIT: I also used the integral method on #71 but you have to make sure to convert the Vrms to Vpeak and then integrate.
To answer your question, a rectified sinusoid has a Vave = 2*Vpeak/pi (Power Reference Manual PG 27-4 Eq27.17).
EDIT: I also used the integral method on #71 but you have to make sure to convert the Vrms to Vpeak and then integrate.
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BamaBino
Well-known member
vdubeee,
You came up with the same answer as me or the CI3 book?
Vave = 2*Vpeak/pi is for full-wave rectified sinusoid. this circuit is half-wave rectifier, right?
You came up with the same answer as me or the CI3 book?
Vave = 2*Vpeak/pi is for full-wave rectified sinusoid. this circuit is half-wave rectifier, right?
I thought it was a full-wave because the top diode conducts when the voltage is positive, then the bottom diode conducts when the voltage is negative (reverse-biased). I checked my old Electronics books and for a half-wave rectifier they are only showing one diode for the rectifier circuit.
I got an answer very close to CI book using the integration method.
I think the way CI drew the rectifier is somewhat confusing. With the way they drew the circuit, you just have to think through the operation as the voltage signal changes. Hopefully, I am not leading us both astray but I think that is correct.
EDIT #1: If it is indeed a half-wave rectifier, then your equation is correct.
EDIT #2: Look at Problem 71. I think that is the half-wave rectifier.
I got an answer very close to CI book using the integration method.
I think the way CI drew the rectifier is somewhat confusing. With the way they drew the circuit, you just have to think through the operation as the voltage signal changes. Hopefully, I am not leading us both astray but I think that is correct.
EDIT #1: If it is indeed a half-wave rectifier, then your equation is correct.
EDIT #2: Look at Problem 71. I think that is the half-wave rectifier.
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Bama,
I ran the circuit as shown in PSpice and found that the circuit as drawn with a sinusoidal voltage input just passes the input signal and does no rectification. So I think this goes back to what several of us have found out and that is CI has done a bad job on laying out the problem for it to be solved in the manner they intended.
There are a few simple modifications to the diagram that would accurately illustrate a full-wave diode rectifier as they mention in this circuit is in the solution.
I ran the circuit as shown in PSpice and found that the circuit as drawn with a sinusoidal voltage input just passes the input signal and does no rectification. So I think this goes back to what several of us have found out and that is CI has done a bad job on laying out the problem for it to be solved in the manner they intended.
There are a few simple modifications to the diagram that would accurately illustrate a full-wave diode rectifier as they mention in this circuit is in the solution.
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