   ## Efficiency of single-phase half-wave rectifier

The efficiency of single phase half-wave rectifier is given by the ratio of the output dc power to the total amount of input power supplied to the circuit. It is also called conventional efficiency.

Now,

Therefore:

#### Ƞ = Pdc/Pin = (4/ π2)RL/(RL + R0) = 0.406/(1+ R0/RL) = 40.6%/(1 + R0 /RL)

Here,

IL = rms value of load current

ILM = maximum value of load current

IL(dc) =average value of load current

Rs = transformer secondary resistance

rd = diode forward resistance

R0 = Rs + rd

## 2 thoughts on “Efficiency of single-phase half-wave rectifier”

1. Burke brush says:

But this web tutorial states that a single diode used in a simple hi-low dimmer switch for a light bulb will be almost 100% efficient. That is, a 100W bulb on 120VAC would be reduced to a 50W output using a half-wave rectifier. You can’t be saying that 60% of the energy coming in to the rectifier is lost. Where does the energy go?

http://www.ibiblio.org/kuphaldt/electricCircuits/Semi/SEMI_3.html#xtocid141882

2. arish says:

great