Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier Uncontrolled: R load, R-L load, R-C load Controlled Free wheeling diode Single-phase, full wave rectifier Uncontrolled: R load, R-L load, Continuous and discontinuous current mode Three-phase rectifier uncontrolled controlled Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Rectifiers DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches. Basic block diagram Input can be single or multi-phase (e.g. 3-phase). Output can be made fixed or variable Applications: DC welder, DC motor drive, Battery charger,DC power supply, HVDC AC input DC output Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Single-phase, half-wave, R-load + vs _ vo vo io vs w t p 2p Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Half-wave with R-L load + vs _ vo vR vL i Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 R-L load Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 R-L waveform w t vo vs, p 2p io b vR vL 3p 4p Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Extinction angle Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 RMS current, Power Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Half wave rectifier, R-C Load + vs _ vo iD p 2p 3p 4p Vm Vmax vs vo Vmin p /2 iD 3p /2 D Vo a q Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Operation Let C initially uncharged. Circuit is energised at wt=0 Diode becomes forward biased as the source become positive When diode is ON the output is the same as source voltage. C charges until Vm After wt=p/2, C discharges into load (R). The source becomes less than the output voltage Diode reverse biased; isolating the load from source. The output voltage decays exponentially. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Estimation of q Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Estimation of a Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Ripple Voltage Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Capacitor Current Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Peak Diode Current Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Example A half-wave rectifier has a 120V rms source at 60Hz. The load is =500 Ohm, C=100uF. Assume a and q are calculated as 48 and 93 degrees respectively. Determine (a) Expression for output voltage (b) peak-to peak ripple (c) capacitor current (d) peak diode current. p 2p 3p 4p Vm Vmax vs vo Vmin p /2 iD 3p /2 a q D Vo Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Example (cont’) Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Controlled half-wave + vs _ ig ia w t v vo a i g s + vo _ Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Controlled h/w, R-L load + vs _ i vo vR vL w t vs p 2p vo a io b Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Controlled R-L load Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Examples A half wave rectifier has a source of 120V RMS at 60Hz. R=20 ohm, L=0.04H, and the delay angle is 45 degrees. Determine: (a) the expression for i(wt), (b) average current, (c) the power absorbed by the load. 2. Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply. Determine the power factor absorbed by the resistance. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Freewheeling diode (FWD) Note that for single-phase, half wave rectifier with R-L load, the load (output) current is NOT continuos. A FWD (sometimes known as commutation diode) can be placed as shown below to make it continuos + vs _ io vo vR vL io vo= vs io + vo _ D2 is on, D1 is off vo= 0 + vs _ vo D1 is on, D2 is off Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Operation of FWD Note that both D1 and D2 cannot be turned on at the same time. For a positive cycle voltage source, D1 is on, D2 is off The equivalent circuit is shown in Figure (b) The voltage across the R-L load is the same as the source voltage. For a negative cycle voltage source, D1 is off, D2 is on The equivalent circuit is shown in Figure (c) The voltage across the R-L load is zero. However, the inductor contains energy from positive cycle. The load current still circulates through the R-L path. But in contrast with the normal half wave rectifier, the circuit in Figure (c) does not consist of supply voltage in its loop. Hence the “negative part” of vo as shown in the normal half-wave disappear. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

FWD- Continuous load current The inclusion of FWD results in continuos load current, as shown below. Note also the output voltage has no negative part. p 2p 3p 4p iD1 io output Diode current iD2 vo w t Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Full wave rectifier + vs _ is iD1 vo io Full Bridge D1 D2 D4 D3 Center-tapped (CT) rectifier requires center-tap transformer. Full Bridge (FB) does not. CT: 2 diodes FB: 4 diodes. Hence, CT experienced only one diode volt-drop per half-cycle Conduction losses for CT is half. Diodes ratings for CT is twice than FB Center-tapped D1 is + vs _ - vo + iD1 iD2 io vs1 vs2 D2 + vD1 - + vD2 - Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Bridge waveforms + vs _ is iD1 vo io Full Bridge D1 D2 D4 D3 p 2p 3p 4p Vm -Vm vs vo vD1 vD2 vD3 vD4 io iD1 iD2 iD3 iD4 is Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Center-tapped waveforms is + vs _ - vo + iD1 iD2 io vs1 vs2 D2 + vD1 - + vD2 - p 2p 3p 4p Vm -2Vm vs vo vD1 vD2 io iD1 iD2 is Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Full wave bridge, R-L load + vs _ is iD1 vo io vR vL vo vs io iD1 , iD2 iD3 ,iD4 is w t p 2p Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Approximation with large L Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

R-L load approximation vo vs io iD1 , iD2 iD3 ,iD4 is w t p 2p Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Examples Given a bridge rectifier has an AC source Vm=100V at 50Hz, and R-L load with R=100ohm, L=10mH determine the average current in the load determine the first two higher order harmonics of the load current determine the power absorbed by the load Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Controlled full wave, R load + vs _ is iD1 vo io T1 T4 T2 T3 Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Controlled, R-L load + vs _ is iD1 vo io vR vL p 2p vo Discontinuous mode b a +a io p 2p vo Continuous mode p+a a b io Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Discontinuous mode Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 Continuous mode Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Single-phase diode groups + vs _ vo vp vn io D1 D3 D4 D2 vo =vp -vn In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential. Therefore, the diode with its anode (+) at the highest potential will conduct (carry) id. For example, when vs is ( +), D1 conducts id and D3 reverses (by taking loop around vs, D1 and D3). When vs is (-), D3 conducts, D1 reverses. In the bottom group, the anodes of the two diodes are at common potential. Therefore the diode with its cathode at the lowest potential conducts id. For example, when vs (+), D2 carry id. D4 reverses. When vs is (-), D4 carry id. D2 reverses. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Three-phase rectifiers D1 io + van - D3 D5 vpn + vbn - n + vo _ D2 + vcn - vnn vo =vp -vn D6 D4 p 2p 4p Vm van vbn vcn vn vp vo =vp - vn 3p Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Three-phase waveforms Top group: diode with its anode at the highest potential will conduct. The other two will be reversed. Bottom group: diode with the its cathode at the lowest potential will conduct. The other two will be reversed. For example, if D1 (of the top group) conducts, vp is connected to van.. If D6 (of the bottom group) conducts, vn connects to vbn . All other diodes are off. The resulting output waveform is given as: vo=vp-vn For peak of the output voltage is equal to the peak of the line to line voltage vab . Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Three-phase, average voltage p/3 2p/3 Vm, L-L Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Controlled, three-phase + vo _ vpn vnn io T3 T2 T6 + vcn - n + vbn - + van - T5 T4 a van vbn vcn Vm vo Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

Output voltage of controlled three phase rectifier EXAMPLE: A three-phase controlled rectifier has an input voltage of 415V RMS at 50Hz. The load R=10 ohm. Determine the delay angle required to produce current of 50A. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003