Question

4.71 A full-wave rectifier circuit with a 1-kΩ operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 transformer having a center-tapped secondary winding. It uses two silicon diodes that can be modeled to have a 0.7-V drop for all currents. What is the peak voltage of the rectified output? For what fraction of a cycle does each diode conduct? What is the average output voltage? What is the average current in the load?

Answer 1

Answer:

$V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

Explanation:

The peak voltage after the 6 to 1 step down is $V_{p} = \frac{120}{6} \sqrt{2} = 28,3V$. Then, the peak voltage of the rectified output is $V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi} = \frac{55,2 V}{\pi } = 17,6 V$V_{d}[/tex] and according to the statement, the diodes can be modeled to be $V_{d} = 0,7 V$. Then, the peak voltage in the load is $V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$.

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The average current in the load is calculated as:

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$