Lecture (04) Uncontrolled Rectifier Circuits

Transcription

1 Lecture (04) Uncontrolled Rectifier Circuits By: Dr. Ahmed ElShafee ١ Dr. Ahmed ElShafee, ACU : Spring 2018, EPC403 Power Electronics introduction Power rectifiers converts AC to DC which uses power diodes Called uncontrolled rectifiers as it gives continuous fixed output voltage as long as input AC voltage has fixed amplitude. Can be classifies according to Single/three phases Half wave, full wave, or bridge Basic assumption, that we use ideal diodes we ignores any internal resistance/voltage drop through diode is forward modes, which is accepted as power rectifiers works on high AC voltage/current. ٢

2 Single phase rectifier circuits As mentioned before can be classified Half wave/ full wave Full wave rectifier can be classified Center tapped Bridge ٣ The single phase half wave rectifier/half ware Very basic Not used in industrial applications as it causes DC component in AC source which may damage transformers and generators in electric grids ٤

3 ٥ Secondary coil voltage During positive cycle, diode is in forward mode, and lad current (ignore voltage drop across diode) During the ve cycle, diode will operate in reverse mode, then the output voltage in 0, the voltage drop across the diode in reverse mode ٦

4 Make sure that you choose diode that have Peak Inverse Voltage (PIV) > V AK The load voltage have two components DC, and AC. The DC component is the average voltage; ٧ The DC load current is The consumed DC power is ٨

5 The effective voltage (Root Mean Square Value) The effective current ٩ Ignoring consumed power by diode, the load effective consumed power is As mentioned before the rectified voltage have DC component and AC component (ripples), the effective AC voltage is Rectifier efficiency (the ratio between the output DC power and the input AC power.) ١٠

6 Rectifier Form Factor (FF) RMS value and average value. Ripple Factor (ratio between AC and DC components) ١١ Transformer utilization factor (ratio between consumed DC power by load and transformer secondary coil power) ١٢

7 Example 01 For the single phase half wave rectifier shown, if primary transformer coil is 220 volts, and transformation ration 5, load resistance is 10 ohm. Calculate rectifier efficiency, form factor, ripple factor, peak inverse voltage. ١٣ ١٤

8 ١٥ Note that transformer utilization factor is /TUF=1/0.286=3.496 Which mean the AC source power should be equals or greater than 3.5 times of DC load power, to ensure the required power is being transformed to load Half wave rectifier rectify only half of wave, which means transformer holds a DC current which in turn saturate the transformer and damage its coils. ١٦

9 Single phase full wave center tap rectifier A center tapped transformer, 2 diodes are used as shown. The 1 st secondry coil gives voltage shifted by 180 from 2 nd coil. That means a certain cycle, is D1 is forward, D2 is reveresd and vice versa ١٧ ١٨

10 The average output voltage The average output current ١٩ Effective output voltage The effective current The PIV on each diode is twice then half wave rectifier ٢٠

11 Single phase full wave bridge rectifier It requires 4 diodes In period (0 to pi), D1, and D2 are in forward mode. Current flow from source through D1 to load the back to source through D2 ٢١ In period (pi to 1pi), D3, and D4 are in forward mode. Current flow from source through D3 to load the back to source through D4 ٢٢

12 Equations that used for center tapped are used in bridge rectefire calculations Exception: PIV in bridge is half of center tapped The voltage drop through diodes in doubled but we agreed to ignore it. ٢٣ ٢٤

13 Example voltage supply with 10 ohms load. Find average voltage/current, effective voltage, average power, output power. Form factor, ripple factor, max PIV ٢٥ ٢٦

14 ٢٧ Filtration and smoothing circuits Rectified wave has ripples, which expressed as ripple factor, and form factor. To get steady DC voltage a filtration circuits are needed (called filters), to prevent ripple from reaching loads. Filters may based on coil, capacitor, or both which is used after rectifier, we called it DC filters. ٢٨

15 AC rectifier of LC may be used before rectifier, but it caused AC signal disturbance. DC filter of capacitor always used after rectifier, ripple factor is decreased by increasing capacitor value ٢٩ ٣٠

16 Load voltage is calculated as follows F presents the rectified wave frequency, which is twice the AC supply voltage Ripple factor is ٣١ Three phase rectifier The maximum DC voltage can be produced from single phase rectifier is Vm (2/pi). This is suitable for low ower applications (tens of kilo watts) if more power is needed 3 phase rectifier is used. 3 phase rectifier classified into Half wave Full wave And as expected full wave rectifier produce twice of voltage produced by half wave rectifier. ٣٢

17 Figure show three phase half wave recteifier, a diode is used with each phase, source should be 4 wires(star). Load is connected between common node of all diodes and nutral. ٣٣ ٣٤

18 Da; supply current in period between (pi/6, and 5pi/6) when phase a output voltage is greater than other phases Db; supply current when its voltage is greater than other phases. Db; supply current when its voltage is greater than other phases. Each diode supply current for 120 degree (2/3 pi) ٣٥ Average out voltage is calculated as follows ٣٦

19 Three phase bridge rectifier (full wave) is used in applications with high power ٣٧ ٣٨

20 Current flow on the following pairs Ab (D1,D6) Bc (D3,D2) Ca (D5,D4) Ba (D3,D4) Cb (D5,D6) Ac (D1,D2) ٣٩ so three phase bridge rectifier, considered to be to serial half wave rectifiers, so output voltage is twice then half wave rectifier. ٤٠

21 Applications Most important applications are batteries charger DC machines driver Induction motors drivers UPS ٤١ Figure shows batteries charger. Diode operates in forward mode in +ve signal half Assuming that Vs > E Connection time is period between (α and β), ٤٢

22 ٤٣ consider Then When Vs < E the diode works in reverse mode, ٤٤

23 Charging current can be calculated Average charging current ٤٥ Example 03 E=12V, capacity = 100 wh, average charging current = 5 Amp Vs= 120v, transformer ration 2:1 Calculate: diode connection angle, current limiting resistor value, max power of limiting resistor, charging time, rectifier efficiency, PIV ٤٦

24 Charging angle Charging current ٤٧ Limiting resistor ٤٨

25 Effective current ٤٩ Limiting resistor power Average charging power Charging time ٥٠

26 Efficiency Peak inverse voltage As noticed circuit efficiency doesn t exceed 18% Using full wave charger will enhance circuit efficiency ٥١ Thanks,.. ٥٢ Dr. Ahmed ElShafee, ACU : Spring 2018, EPC403 Power Electronics