Problem 1 : full wave rectifier
Calculate the average output voltage from a half-wave and full wave center tapped rectifier ifVipeak= 169.7 V.
Solution problem 1
Full wave center tapped rectifier:if Vs peak = 169.7 V, then Vs2 peak = 84.9 V
Vavg = 0.636 x (1/2) Vs peak
= 0.636 x 84.9 = 54 V
Half-wave rectifier:
Vavg = 0.318 x Vs peak = 0.318 X 169.7 V = 54 V
As you can see from this example, even though the peak output of the full wave center tapped rectifier was half that of the half-wave rectifier, the average output was the same because the full wave center-tapped rectifier doubles the number of half-cycles at the output, compared to a half wave rectifier.
Problems 2 : Full & Half wave rectifier
Calculate the average output voltage from a half-wave and full wave center tapped rectifier if K, peak = 169.7 V.Solution problem 2
Half wave rectifier :Vavg = 0.318 x Vs peak
= 0.318 X 169.7 V = 54 V
Full wave center tapped rectifier:
If Vs peak = 169.7 V then Vs1 peak and Vs2 peak = 84.9 V
Vavg = 0.636 x (1/2) Vs peak
= 0.636 x 84.9 V = 54 V
As you can see from this example, even though the peak output of the full wave center tapped rectifier was half that of the half wave rectifier, the average output was the same because the full wave center tapped rectifier doubles the number of half-cycles at the output, compared to a half-wave rectifier.
Problem 3 : Center-tapped rectifier
Calculate the average output voltage from a center tapped rectifier and a bridge rectifier if Vs peak = 169.7 V.Solution problem 3
Center-tapped:Vavg = 0.636 x (1/2) Vs
=0.636 X 84.9 V
= 54 V
Bridge:
Vavg = 0.636 X Vs
= 0.636 X 169.7 V
= 107.9 V
As you can see from this example, unlike the center-tapped rectifier that only connects half of the peak secondary voltage across the load, the bridge rectifier connects the total peak secondary voltage across the load.