A 230 V, 60 Hz voltage is applied to the primary of a 5:1 step-down, center-tap transformer use in a full wave rectifier having a load of 900 Ω. If the diode resistance and secondary coil resistance together has a resistance of 100 Ω, determine (a) d.c. voltage across the load,

Question

A 230 V, 60 Hz voltage is applied to the primary of a 5:1 step-down, center-tap transformer use in a full wave rectifier having a load of 900 Ω. If the diode resistance and secondary coil resistance together has a resistance of 100 Ω, determine (a) d.c. voltage across the load, (b) d.c. current flowing through the load, (c) d.c. power delivered to the load, (d) PIV across each diode, (e) ripple voltage and its frequency and (f) rectification efficiency.

Accepted Answer

The voltage across the two ends of secondary = [latex]\frac{230}{5}[/latex]= 46 V

Voltage from center tapping to one end, [latex]V_{rms} = \frac{46}{2}[/latex] = 23 V

(a) The d.c. voltage across the load, [latex]V_{d.c.} = \frac{2V_{m}}{\pi} = \frac{2 × 23 × \sqrt{2}}{\pi}[/latex] = 20.7 V

(b) The d.c. current flowing through the load, [latex]I_{d.c.} =\frac{V_{d.c.}}{(r_{s} + r_{f} +R_{L})} = \frac{20.7}{1000}[/latex] = 20.7 mA

(c) The d.c. power delivered to the load,

[latex]P_{d.c.} = (I_{d.c.})^{2} × R_{L} = (20.7 × 10^{– 3})^{2}[/latex] × 900 = 0.386 W

(d) PIV across each diode =2[latex]V_{m} = 2 × 23 × \sqrt{2}[/latex] = 65 V

(e) Ripple voltage, [latex]V_{r, rms} = \sqrt{(V_{rms})^{2} – (V_{d.c.})^{2}}[/latex]

= [latex]\sqrt{(23)^{2} – (20.7)^{2}}[/latex] = 10.05 V

Frequency of ripple voltage = 2 × 60 = 120 Hz

(f) Rectification efficiency, η = [latex]\frac{Pd.c.}{Pa.c.}=\frac{(V_{d.c.})^{2}/R_{L}}{(V_{rms})^{2}/R_{L}}=\frac{(V_{d.c.})^{2}}{(V_{rms})^{2}}[/latex]

= [latex]\frac{(20.7)^{2}}{(23)^{2}}=\frac{428.49}{529}[/latex] = 0.81

Therefore, percentage of efficiency = 81%

Question 18.10: A 230 V, 60 Hz voltage is applied to the primary of a 5:1 st...

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