## 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.

#### Ƞ = P_{dc}/P_{in} = power in the load/input power

Now,

#### P_{dc }= I_{L(dc)}^{2}R_{L } = (I_{LM}/ π)^{2}.R_{L }= I_{LM}^{2}.R_{L}/ π^{2}

#### P_{in }= I_{L}^{2 }(R_{L }+ R_{0}) = (I_{LM}/2)^{2} (R_{L }+ R_{0}) = I_{LM}^{2} (R_{L }+ R_{0})/4

Therefore:

#### Ƞ = P_{dc}/P_{in }= (4/ π^{2})R_{L}/(R_{L }+ R_{0}) = 0.406/(1+ R_{0}/R_{L}) = 40.6%/(1 + R_{0 }/R_{L})

Here,

I_{L} = rms value of load current

I_{LM }= maximum value of load current

I_{L(dc)} =average value of load current

R_{L }= load resistance

R_{s} = transformer secondary resistance

r_{d }= diode forward resistance

R_{0 }= R_{s }+ r_{d}

**Related topic**

**Single-phase half-wave rectifier****AVERAGE AND RMS VALUE OF SINGLE-PHASE HALF-WAVE RECTIFIER****Frequency Component of Half-Wave Rectifier Voltage and Current****Ripple Factor of single phase Half-Wave rectifier****Peak Inverse Voltage (PIV) of single phase half wave rectifier****Peak current of single phase half wave rectifier****Transformer Utilization Factor (TUF) of single phase half wave rectifier****Advantage and Disadvantage of single-phase half-wave rectifier**

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

great