Question

A full wave rectifier circuit is fed from a transformer having a center tapped secondary winding. The rms

voltage from either end of secondary to center tap is 30 V. if the diode forward resistance is 2 Ω and that

of the half secondary is 8 Ω, for a load of 1 kΩ, calculate,

(a) Power delivered to load

(b) % regulation at full load

(c) efficiency of rectification

(d) T.U.F. of secondary.​

Answer 1

Answer:

Peak inverse voltage (PIV) is 2Vmax in center tapped full wave rectifier (but it is Vmax in full wave bridge rectifier). By using a center tapped transformer we are creating two AC sources which are out of phase by 180° ( Vao and Vbo ) but with same amplitude of Vmax.

Answer 2

Answer:

a) Power delivered to the load is 0.714 W.

b) Percentage regulation at full load is 0.93%.

c) Efficiency of rectification is 0.80.

d) Transformer utilisation factor of the secondary is 0.802.

Explanation:

Given,

The RMS voltage from either end of secondary to the centre tap, $V_{rms}$ = 30V

• RMS is defined as the square root of the mean square. 

The forward resistance, $R_{f}$ = 2Ω

The half secondary resistance, $R_{s}$ = 8 Ω

The load resistance, $R_{l}$ = 1 KΩ = 1000 Ω

Maximum voltage, $V_{m}$ =?

As we know,

• $V_{m} = V_{rms} \sqrt{2}$
• $V_{m} = 30\sqrt{2}$ = 42.42 V

Maximum current,

• $I_{m} = \frac{V_{m} }{R_{f} + R_{s} + R_{l} }$
• $I_{m} = \frac{42.42}{2 + 8 + 1000}$
• $I_{m} = \frac{42.42}{1010}$ = 0.042 A = 42 × 10⁻³A

The current supplied by the transformer,

• $I_{DC} = \frac{2 I_{m} }{\pi }$  
• $I_{DC} = \frac{2 \times 42 \times 10^{-3} }{3.14 }$ = 26.75 × 10⁻³A

a)  Power delivered to load:

• $P_{l} = I_{DC} ^{2} R_{l}$
• $P_{l} = (26.75 \times 10^{-3} )^{2} \times 1000$ = 0.714 W

b) % regulation at full load:

No load voltage,

• $V_{NL} = \frac{2V_{m} }{\pi }$
• $V_{NL} = \frac{2\times 42.42 }{3.14 }$ = 27 V

Full load voltage,

• $V_{FL} = I_{DC} \times R_{l}$ = 26.75 × 10⁻³ × 1000 = 26.75 V

Therefore,

• % regulation at full load = $\frac{V_{NL} - Vx_{FL} }{V_{FL} } \times 100$ = $\frac{27 - 26.75}{26.75}\times 100$ = 0.93%

c) Efficiency of rectification: 

• E = DC output/AC input

• $E =\frac{8}{\pi ^{2} } \times\frac{1}{1 +\frac{R_{f}+ R_{s} }{R_{l} } }$ =  $E =\frac{8}{3.14 ^{2} } \times\frac{1}{1 +\frac{2+ 8 }{1000} } }$ = 0.80

d)  T.U.F. of secondary.​

Transformer utilisation factor of secondary

First, we have to calculate the AC rating,

AC rating = V (rms) × I (rms) =  $V_{rms}\times \frac{I_{m} }{\sqrt{2} }$ = $30 \times\frac{42 \times 10^{-3} }{\sqrt{2} }$ = 0.89 W

Now,

T.U.F = DC power output/ AC rating

T.U.F = $\frac{0.714}{0.89}$ = 0.802