A full wave PN junction diode rectifier uses a load resistance of 1200 Ω. The internal resistance of each diode is 9 Ω. Find efficiency.

Steps to Solve

1. Calculate Total Resistance in Circuit

First, we need to find the total resistance. For a full-wave rectifier, two diodes conduct in each half cycle. Therefore, the total resistance $R_t$ is given by the sum of the load resistance $R_L$ and twice the internal resistance of one diode $R_D$.

$$ R_t = R_L + 2R_D $$ Given:

• $R_L = 1200 , \Omega$
• $R_D = 9 , \Omega$

So,

$$ R_t = 1200 , \Omega + 2 \times 9 , \Omega = 1200 + 18 = 1218 , \Omega $$

2. Calculate Load Current

Now, we can find the load current $I_L$ using Ohm's Law. The total voltage $V$ across the load needs to be known; however, for efficiency calculation, if the output voltage is expressed relative to input voltage $V_{in}$, we can use $V$ directly.

Assuming a constant input voltage $V$:

$$ I_L = \frac{V}{R_t} $$

3. Calculate Power Across Load and Diodes

The power delivered to the load $P_{load}$ and power lost in the diodes $P_{diodes}$ can be calculated as follows:

• Power across load: $$ P_{load} = I_L^2 \times R_L $$

• Power loss in diodes can be calculated as: $$ P_{diodes} = I_L^2 \times 2R_D $$

4. Calculate Efficiency

Efficiency ($\eta$) is the ratio of the power delivered to the load to the total power supplied, typically expressed as a percentage.

$$ \eta = \frac{P_{load}}{P_{total}} \times 100 = \frac{P_{load}}{P_{load} + P_{diodes}} \times 100 $$

Substituting the power equations:

$$ \eta = \frac{I_L^2 R_L}{I_L^2 R_L + I_L^2 (2R_D)} \times 100 $$

This simplifies to:

$$ \eta = \frac{R_L}{R_L + 2R_D} \times 100 $$

5. Final Calculation of Efficiency

Now substituting the values for $R_L$ and $R_D$:

$$ \eta = \frac{1200}{1200 + 2 \times 9} \times 100 $$ $$ \eta = \frac{1200}{1218} \times 100 $$

Calculating this,

$$ \eta \approx 98.54% $$