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Half-Wave &Full Wave Rectifiers, Filtering, Regulated Power Supply
Half-Wave Rectifier Equivalent DC Output Voltage
Example Given: v in (RMS) = 110 V (60 HZ) Turns Ratio 10:
Find: v (^) out(DC Effective) v in (Peak) = 1.414 v in (RMS) = 1.414 x 110 = 155.5 V v out (Peak) = 1/10 v in (Peak) = 1/10 x 155.5 = 15.6 V vDiode = 15.6 - 0.7 = 14.9 V V out (DC Effective) = 0.318 vDiode = 0.318 x 14.9 ≈ 4.7 VDC
Exercise # Given: v in(RMS) = 110 V (60 HZ) Turns Ratio 5: Find: V out(DC Effective)
Answer: 9.7 VDC
Exercise # Given: v in(RMS) = 120 V (60 HZ) Turns Ratio 5: Find: V out(DC Effective)
Answer: 10.6 VDC
Full-Wave Center-Tapped Rectifier Equivalent DC Output Voltage
Example
Given: v in (RMS) = 110 V (60 HZ) Turns Ratio 10:
Find: V out(DC Effective) v in (Peak Center) = 1.414 v in(RMS) = 1.414 x 110 = 155.5 V v out (Peak) = (1/2) (1/10) v in (RMS) = 1/20 x 155.5 = 7.8 V vDiode = 7.8 - 0.7 = 7.1 V V out (DC Effective) = 0.636 vDiode = 0.636 x 7.1 ≈ 4.5 VDC
Exercise #
Given: v in(RMS) = 110 V (60 HZ) Turns Ratio 5: Find: V out(DC Effective)
Answer: 9.5 VDC
Exercise #
Given: v in(RMS) = 120 V (60 HZ) Turns Ratio 5: Find: V out(DC Effective)
Answer: 10.4 VDC
Filtering
v ripple(peak- peak) = I out(DC) / 2 f C
I out(DC) = V out(DC) / RLoad
Two Steps
- Assume V out (DC) (Without filtering, i.e., use peak of the rectified wave , NOT the DC average value.)
- Solve for I out(DC) and v ripple(peak- peak)
- Recalculate V out(DC) Load = V out (DC) (without filtering) - [ v ripple(peak- peak)] / 2 Example
Given: v in (RMS) = 110 V (60 HZ) Turns Ratio 10: RLoad = 100 Ω C = 1000 μF f = 60 Hz
Find: V out(DC) Load
From Full-Wave Bridge Rectifier (from Example page 3, above) V out (DC) (without filtering) = 14.2 VDC I out(DC) = V out (DC) / RLoad = 14.2/100 = 0.142A = 142 mA v ripple(peak- peak) = I out(DC) / 2 f C = 0.142 / (2 x 60 x 1000 x 10-6^ ) = 1.18 V V out (DC) Load = V out (DC) (without filtering) - [ v ripple(peak- peak)] / 2 = 14.2 - (1.18) / 2 = 13.6 VDC Exercise
Given: v in (RMS) = 120 V (60 HZ) Turns Ratio 5: RLoad = 240 Ω C = 470 μF f = 60 Hz V out (DC) (without filtering) = 32.5 VDC (from Problem, page 3, above. Note 32.5 not 32.5 x .636 = 20.7) Find: V out(DC) Load
Answer: I out(DC) = 136 mA v ripple(peak- peak) = 2.4 V V out (DC) Load = 31.3 VDC
Regulated Power Supply
Scanned Images: Electronic Devices, Ali Aminian & Marian Kazimierczuk, Pearson-Prentice Hall, 2004
