Ripple factor and Efficiency of rectifier

Ripple factor

The ratio of the RMS value (root mean square) of the AC component to the DC component of the output is defined as the Ripple factor and is denoted by γ.

Ripple factor, γ = VAC/VDC  | VDC is the average value of the DC output.

VRMS = √ VDC2 + VAC2    or    IRMS = √ IDC2 + IAC2

VAC = √ VRMS2 –  VDC2

Therefore the ripple factor equation is, γ = √ (VRMS2 – VDC2) / VDC  = √ ( VRMS / VDC) – 1

To calculate the ripple factor of a half wave and full wave rectifier, just substitute the RMS and Average value of the respective rectifier in the above equation.

Ripple factor of a Half wave rectifier

rms-value-half-wave-rectifieraverage-value-half-wave-rectifier

RMS Voltage of a half wave rectifier, VRMS = Vm / 2    | Vm is the peak voltage.

Average Voltage of a half wave rectifier, VAVG = Vm / π

Ripple factor, γ = √ (  [ (Vm/ 2) / (Vm / π) ]2 -1  ) = √  ( π / 2 )2 – 1  = 1.21

Ripple factor of a Full wave rectifier

rms-value-full-wave-rectifier average-value-full-wave-rectifier

RMS Voltage of a full wave rectifier, VRMS = Vm / √2

Average Voltage of a full wave rectifier, VAVG = 2 Vm / π

r = √ ( [ (Vm/√ 2 ) / (2 Vm / π) ]2 – 1 ) = √  ( π / (2 √ 2) )2 – 1  = 0.48

Efficiency

The ratio of DC output power to the AC input power of a rectifier is called as its efficiency. It is denoted by η.

Rectifier efficiency, η = DC output power/input AC power = PDC / PAC

Efficiency of a Half wave rectifier

IRMS = Im / 2   ,  PAC = IRMS2 (rf + RL) = ( Im /  2)2 (rf + RL)

Vm is the peak Current     |

PDC = IAVG2 RL = ( Im / π)2  RL

For a half wave rectifier, the Efficiency η = PDC / PAC = ( ( Im / π)2  RL )   /    ( ( Im /  2)2 (rf + RL) )

= 4 RL / π2 (rf + RL) = 0.405 RL / (rf + RL)

Therfore the maximum efficiecy = 40.5%

Efficiency of a Full wave rectifier

Similarly for a full wave rectifier,

IRMS = Im / √ 2   ,  PAC = ( Im / √ 2)2 (rf + RL)

PDC  = ( 2 Im / π)2  RL

For a half wave rectifier, the Efficiency η = PDC / PAC = ( ( 2 Im / π)2  RL )   /   ( ( Im / √ 2)2 (rf + RL) )

= 8 RL / π(rf + RL) = 0.810 RL / (rf + RL)

Maximum efficiecy = 81.0%

Hence we can see that, the efficiency of a full wave rectifier is double than that of a half wave rectifier.

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