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A full-wave rectifier circuit with a 1-kOhm load 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,( [ Select ] in Volts) of the rectified output? For what [ Select ] of a cycle does each diode not conduct? What is the average output voltage

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krlosdark12

Answer:

a) 13.5

b) positive cycle downer diode off negative cycle upper diode off

c) 8.60V

Explanation:

For make the process easier we need to convert the input Vrms to peak voltage, for a sine wave the peak voltage is given by:

[tex]V_p=Vrms*\sqrt{2}\\V_p=170V[/tex]

The voltage of the extremes of the transformer output is:

[tex]V_o=\frac{V}{6}\\\\V_o=28.3V[/tex]

because this is a center-tapped transformer the output voltage is divided by 2. So the peak voltage of the resistor is:

[tex]V_r=\frac{V_o}{2}-0.7\\V_r=13.5V[/tex]

The voltage of the downer diode will have a displacement of 180 degrees because the reference point is on the center tap. so for the positive cycle of the signal, the voltage is negative on the downer diode so it will not conduct, and for the negative cycle, the upper diode won't conduct.

the average voltage is given by:

[tex]V_{avg}=2*\frac{V_r}{\pi}\\\\V_{avg}=8.60V[/tex]

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