Q8Analog Electronics

Question

Draw the neat diagram of full wave bridge rectifier circuit and derive the average (or DC) value, efficiency (\eta) and Peak inverse voltage (PIV) for full wave bridge rectifier.

Answer 1

During the positive half cycle, diodes D1 and D2 are forward biased and conduct, while D3 and D4 are reverse biased. During the negative half cycle, D3 and D4 conduct, and D1, D2 are reverse biased. In both cases, the current through the load $R_{L}$ flows in the same direction, giving full-wave rectification.

For input $v = V_{m} sin θ$ , the average over a full cycle (but current flows for all $2 π$ ):

V_{d c} = \frac{2 V _{m}}{π} \approx 0.637 V_{m}

Rectification efficiency is the ratio of DC power delivered to load to the total input AC power:

η = \frac{P _{d c}}{P _{a c}} = \frac{( 2 V _{m} / π ) ^{2} / R _{L}}{( V _{m} / 2 ) ^{2} / ( R _{L} + r _{d} )} \approx \frac{8}{π ^{2}} \approx 81.2%

In the bridge rectifier, when D1 and D2 conduct, diodes D3 and D4 are reverse biased. The peak inverse voltage across each reverse-biased diode is:

PIV = V_{m}

This is a major advantage of the bridge rectifier over the center-tap full-wave rectifier, where $PIV = 2 V_{m}$ . The bridge rectifier requires lower-rated diodes, making it more economical and practical for high-voltage applications.

γ = \frac{V _{r m s (a c)}}{V _{d c}} = (\frac{V _{r m s}}{V _{d c}})^{2} - 1 = (\frac{V _{m} / 2}{2 V _{m} / π})^{2} - 1 \approx 0.482

Q8Analog Electronics

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