powerelectronicsuncontrlled rectifier4th sem ca1.pptx

uncontrlled rectifier

1. RAMKRISHNA MAHATO GOVERNMENT ENGINEERING COLLAGE, PURULIA NAME- KUNAL BISWAS DEPARTMENT- ELECTRICAL ENGINNERING ROLL NO- 35001622015 REG. NO- 223500110119 PAPER NAME - ELECTRIC MACHINE-I PAPER CODE- PC-EE401 SEMESTER- 4 TH YEAR- 2 ND SESSION- 2022-2026  

2. C ontent : ▶ Single Phase Half Wave Uncontrolled Rectifier (with R load, RL load and RL with FD) ▶ Single Phase Full Wave Uncontrolled Rectifier. Centre Tapped (Mid Point) Rectifier Bridge Rectifier ▶ Three Phase Full Wave Uncontrolled Rectifier. 3 – φ Bridge Rectifier 1

3. Single Phase Half Wave Rectifier: ▶ During each “positive” half cycle of the AC sine wave, the diode is forward biased as the anode is positive with respect to the cathode resulting in current flowing through the diode. ▶ Since the DC load is resistive (resistor, R), the current flowing in the load resistor is therefore proportional to the voltage (Ohm´s Law), and the voltage across the load resistor will therefore be the same as the supply voltage, V s (minus V f ), that is the “DC” voltage across the load is sinusoidal for the first half cycle only so V out = V s. ▶ During each “negative” half cycle of the AC sinusoidal input waveform, the diode is reverse biased as the anode is negative with respect to the cathode. 2

4. Single Phase Half Wave Rectifier: ▶ Therefore, NO current flows through the diode or circuit. Then in the negative half cycle of the supply, no current flows in the load resistor as no voltage appears across it so therefore, V out = 3

5. Single Phase Half Wave Rectifier (R Load): 4 ▶ 𝐴𝑣. 𝑜𝑢𝑡𝑝𝑢𝑡 𝑉 𝑜 𝑙𝑡𝑎𝑔𝑒 𝑉 = 1 2𝜋 𝜋 ∫ 𝑉 𝑚 𝑠𝑖𝑛𝜔𝑡𝑑 (𝜔 𝑡) = 𝑉 𝑚 2𝜋 −𝑐𝑜𝑠𝜔𝑡 𝜋 0 = 𝑉 𝑚 𝜋 ▶ 𝑅𝑀𝑆 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑉 𝑟𝑚𝑠 = 1 𝜋 2 2 2𝜋 ∫ 𝑉 𝑚 𝑠𝑖𝑛 𝜔𝑡𝑑(𝜔𝑡) 1 / 2 = 𝑉 𝜋 1−𝑐𝑜𝑠2𝜔𝑡 𝑚 2𝜋 2 ∫ 𝑑 (𝜔 𝑡) 1 / 2 𝑉 = 𝑚 2 ▶ 𝐴𝑣. 𝐿𝑜𝑎𝑑 𝐶𝑢𝑟 𝑟 𝑒𝑛 𝑡 𝐼 = 𝑉 = 𝑉 𝑚 𝑅 𝜋𝑅 ▶ 𝑅𝑀𝑆 𝐿𝑜𝑎𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 𝑟𝑚𝑠 ▶ 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 = 𝑉 𝑠 ∗𝐼 𝑟𝑚𝑠 = 𝑉 𝑟𝑚𝑠 = 𝑉 𝑚 𝑅 2𝑅 𝑉 𝑟𝑚𝑠 ∗𝐼 𝑟𝑚𝑠 = 2𝑉 𝑠 2𝑉 𝑠 = 0.707 𝑙𝑎𝑔

6. Single Phase Half Wave Rectifier (RL Load): ▶ Current I continues to flow even after source voltage V S is negative because of the presence of inductance L in load. ▶ After + ve half cycle, diode remains ON, so – ve half cycle appears across load current until I decays to zero at ωt = β. ▶ When I = 0 at ωt = β; V L = 0, V R = and V S appears as reverse bias across diode D. ▶ At β, diode voltage V D jumps from to V M sin β where β > π. ▶ Here β = γ is the conduction angle of the diode. 5

7. Single Phase Half Wave Rectifier (RL Load): ▶ 𝐴𝑣. 𝑜𝑢𝑡𝑝𝑢𝑡 𝑉 𝑜 𝑙𝑡𝑎𝑔𝑒 𝑉 = 2𝜋 1 𝛽 ∫ 𝑉 𝑚 sin 𝜔 𝑡 𝑑 𝜔𝑡 = 𝑉 𝑚 2𝜋 1 − cos 𝛽 𝑅 ▶ 𝐴𝑣. 𝑂𝑢𝑡𝑝𝑢𝑡 𝐶𝑢𝑟𝑟𝑟𝑒𝑛𝑡 𝐼 = 𝑉 = 𝑉 𝑚 2 𝜋𝑅 1 − cos 𝛽 6

8. 1 – φ Full Wave Rectifier – Centre Tapped ▶ Also called Mid – point rectifier. ▶ The turns ration from each secondary to primary is taken as unity for simplicity. ▶ When “A” is +ve w.r.t mid – point O, D1 conducts for π radians. ▶ When “B” is +ve w.r.t mid – point O in the next half cycle, D2 conducts for the other π radians. ▶ Peak Inverse Voltage (PIV) for both D1 and D2 is 2 V S and hence it is called 1 – φ 2 – pulse diode rectifier. 7

9. 1 – φ Full Wave Rectifier – Centre Tapped ▶ 𝐴𝑣. 𝑂𝑢𝑡𝑝𝑢 𝑡 𝑃 𝑜 𝑤 𝑒 𝑟 𝑉 𝜋 𝑚 = 1 ∫ 𝜋 𝑉 sin 𝜔 𝑡 𝑑 𝜔𝑡 = 2𝑉 𝑚 𝑅 ▶ 𝐴𝑣. 𝑂𝑢𝑡𝑝𝑢 𝑡 𝐶𝑢 𝑟 𝑟 𝑒𝑛 𝑡 𝐼 = 𝑉 = 2𝑉 𝑚 𝑅 𝜋𝑅 ▶ 𝑅𝑀𝑆 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑉 𝑟𝑚𝑠 = ∫ 1 𝜋 𝜋 2 𝑉 𝑚 𝑠𝑖𝑛 2 𝜔𝑡𝑑(𝜔𝑡) 1 / 2 𝑉 2 = 𝑚 = 𝑉 𝑆 ▶ 𝑅𝑀𝑆 𝑂𝑢𝑡𝑝𝑢𝑡 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 𝑟 𝑚 𝑠 = 𝑉 𝑟𝑚𝑠 = 𝑉 𝑠 𝑅 𝑅 𝑟𝑚𝑠 ▶ 𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑙𝑜𝑎𝑑 = 𝑉 𝑟𝑚𝑠 ∗ 𝐼 𝑟𝑚𝑠 = 𝐼 2 𝑅 ▶ 𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 𝑝𝑓 = 𝑉 𝑆 .𝐼 𝑟𝑚𝑠 𝑉 𝑟𝑚𝑠 .𝐼 𝑟𝑚𝑠 = 1 8

10. 1 – φ Full Wave Bridge Rectifier ▶ On the positive half cycle of transformer secondary supply voltage, diodes D1 and D2 conduct, supplying this voltage to the load. ▶ On the negative half cycle of supply voltage, diodes D3 and D4 conduct supplying this voltage to the load. ▶ It can be seen from the waveforms that the peak inverse voltage of the diodes is only V m ▶ The average output voltage is the same as that for the centre - tapped transformer full- wave rectifier. ▶ 𝑃𝑒𝑎𝑘 𝑅𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑣𝑒 𝐷𝑖𝑜𝑑𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 𝑚 = 𝑉 𝑚 𝑅 ▶ 𝐴𝑣. 𝑂𝑢𝑡𝑝𝑢 𝑡 𝑉 𝑜 𝑙𝑡𝑎𝑔𝑒 𝑉 𝜋 = 2𝑉 𝑚 ; 𝑅𝑀𝑆 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑉 𝑟𝑚𝑠 = 2 𝑉 𝑠 9

11. 1 – φ Full Wave Bridge Rectifier 𝐷𝐴 ▶ 𝐴𝑣. 𝐷𝑖𝑜 𝑑 𝑒 𝐶 𝑢𝑟 𝑟 𝑒 𝑛𝑡 𝐼 = 2𝜋 1 𝜋 ∫ 𝐼 𝑚 𝜋 sin 𝜔𝑡 𝑑 𝜔𝑡 = 𝐼 𝑚 ; ▶ 𝑅𝑀𝑆 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑉 𝑟𝑚𝑠 = 2𝑉 𝑆 2𝜋 ▶ 𝑅𝑀𝑆 𝐷𝑖𝑜𝑑𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 𝐷𝑟𝑚𝑠 = ∫ 1 𝜋 2 𝐼 𝑚 si n 2 𝜔 𝑡 𝑑 𝜔𝑡 1 / 2 𝐼 = 𝑚 2 1

12. 3 – φ Bridge Rectifier: ▶ Two series diodes are always conducting while four diodes are blocking. ▶ One of the conducting diodes is odd numbered while the other is even numbered. ▶ Each diode conducts for 120 º. ▶ Current flows out from the most +ve source terminal through an odd numbered diode through the load followed by the even numbered diode and then back to the most –ve source terminal. ▶ Output has less ripples and the diodes are numbered in accordance to their conductance. ▶ The bridge uses both the +ve and –ve halves of the i/p voltage. ▶ Ripple frequency is 6*f. ▶ Upper set of diodes constitutes the +ve group while the lower set constitutes the –ve. ▶ Transformer Primary – Secondary is in Delta – Star configuration. 11

13. 3 – φ Bridge Rectifier: ▶ The diode with the most +ve voltage will be conducting. ▶ B is chosen as reference. ▶ During 0º - 30º, the voltage at C is highest (arbitrarily). Hence D5 is conducting as it is the most +ve. ▶ Between 30º and 150º, A becomes the most +ve and hence conducting. ▶ During 150º - 270º, B being most +ve conducts. ▶ The cycle repeats itself. ▶ Each diode conducts for 120º. 𝑂 ▶ 𝑂𝑢𝑡𝑝𝑢 𝑡 𝑉 𝑜 𝑙𝑡𝑎𝑔𝑒 𝑉 = 𝜋 / 3 ∫ 𝜋 / 3 1 2𝜋 / 3 𝑉 𝑚 sin 𝜔𝑡𝑑 𝜔 𝑡 = 3𝑉 𝑚 𝜋 = 0.955 ∗ 𝑉 𝑚 ▶ 𝑅𝑀𝑆 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑉 𝑟𝑚𝑠 = 𝜋 / 3 ∫ 𝜋 / 3 1 2 𝜋 / 3 𝑚 2 2 𝑉 𝑠𝑖𝑛 𝜔𝑡𝑑(𝜔𝑡) 1 2 = 3𝑉 𝑚 𝜋 = 0.9558 ∗ 𝑉 𝑚 12

14. 3 – φ Bridge Rectifier: 13

15. 3 – φ Bridge Rectifier: 𝑅 ▶ 𝐼 = 𝑉 = 0.9558 ∗ 𝐼 ; 𝐼 𝑚 𝐷(𝑎𝑣𝑔.) 0(𝑎𝑣𝑔.) = 𝐼 / 3 𝑠 ▶ 𝑅𝑖𝑝𝑝𝑙𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 6 ∗ 𝑓 ; 3 𝐷𝑖𝑜𝑑𝑒 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = 2𝜋 = 120° ▶ 𝑅𝑖𝑝𝑝𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 𝑅𝐹 = 2 𝑛 2 −1 = 0.404 ▶ 𝑅𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 3 𝜋 2 ∗ 1 . 9558 = 99.82% ▶ 𝑅𝑀𝑆 𝑙𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 𝑠 = 𝜋 ∫ 𝜋 / 3 2 2 𝜋 / 3 𝐼 2 𝑠𝑖𝑛 2 𝜔𝑡𝑑(𝜔𝑡) 𝑚 1 / 2 = 0.7804 ∗ 𝐼 𝑚 ▶ 𝑇𝑈𝐹 = 𝑃 𝑑𝑐 𝑉𝐴 𝑟𝑎𝑡𝑖𝑛𝑔 = 2 ∗ 6 3 𝜋 3 ∗ . 78 4 = 0.9541 6 ▶ 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑉𝐴 𝑟𝑎𝑡𝑖𝑛𝑔 = 3 ∗ 𝑉 𝐼 = 3 ∗ 𝑉 𝑚 ∗ 0.7804 ∗ 𝐼 𝑠 𝑠 𝑚 14

16. References ▶ http://www.eng.uwi.tt/depts/elec/staff/rdefour/ee33d ▶ Power Electronics, P S Bimbhra. ▶ ht t p s : / / ww w . sl i de s h a r e . net / nit i shku m ar54943600 / 3 -phas e - d i o d e - rect i fi e r spow e r - electronics ▶ http://www.ee.co.za/article/rectifier-transformers-technology-update.html 15