As shown in the given diagram of full wave bridge rectifier it consists of four diodes under the condition in which four diodes are connected the called bridge circuit. So due to this type of circuit is named bridge rectifier. A resistor is connected in the circuit where rectified output voltage appears called load resistor RL
Bridge Rectifier Current Path Negative Cycle
Bridge Rectifier Current Path Positive Cycle
Bridge Rectifier Output Waveform
PIV rating for diodes ≥ Vm
Because there are two paths for the load current, the average diode current is just half of the average load current: ID(avg) = I(avg)/2
Example 1: The full-wave bridge rectifier of 5.7 is supported by a 120 V source. If the load resistance is 10.8 Ω, find
(i) The peak load voltage
(ii) The DC voltage across the load
(iii) The DC load current
(iv) The average current in each diode
(v) The average output power
(vi) The rectifier efficiency
(vii) The ripple factor
(viii) The power factor
(i) Peak load voltage
- Vm = √2 VRMS = (1.414) (120) = 170 V
(ii) DC voltage across the load
- Vo(avg) = 0.636 x 170 = 108 V
(iii) DC load current
- Io(avg) = 108 / 10.8 = 10 A
(iv) Average current in each diode since the diodes carry the load current alternative half-cycle
- ID(avg) = Io(avg) / 2 = 10 / 2 = 5 A
(v) Average output power
- Po(avg) = Vo(avg) x Io(avg) = 108 x 10 = 1080 W
(vi) Rectifier efficiency
- η = 8 / π2 = 0.81 or 81%
(vii) Ripple factor
(viii) Power factor
- PF = P/S = (Vo(avg) x Io(avg)) / (VRMS x IRMS) = (108 x 10) / (120 x 10) = 0.9
Full-wave Bridge Rectifier with Inductive Load
Full-wave Bridge Rectifier with Inductive Load (a) Waveforms for (L = R) (b) Waveform for (L >> R)
Example 2: A full-wave bridge rectifier with an RL load is connected to a 120 V source the load resistance is 10 Ω and L >> R, find
(i) The average load voltage
(ii) The average load current
(iii) The maximum load current
(iv) The RMS value of load current
(v) The average current in each diode
(vi) The RMS current in each diode
(vii) The power supplied to the load
The peak load voltage is
- Vm = √ 2 VRMS = (1.414) (120) = 170 V
(i) Average load voltage- Vo(avg) = 0.636 x 170 = 108 V
(ii) Average load current
- Vo(avg) / R = 108 / 10 = 10.8 A
(iii) Maximum load current = average load current = 10.8 A
(iv) RMS value of load current = average load current = 10.8 A
(v) Average current in each diode
- ID(avg) = Io(avg) / 2 = 10.8 / 2 = 10.8 /2 = 5.4 A
(vi) RMS current in each diode
- ID(RMS) = Io(avg) / √2 = 10.8 / √2 = 7.6 A
(vii) Power supplied to the load
- I2RMS R = 10.82 x 10 = 1167 W
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