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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, Controlled Continuous and discontinuous current mode

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## Chapter 2 AC to DC CONVERSION (RECTIFIER)

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**Chapter 2AC 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, • Controlled • Continuous and discontinuous current mode • Three-phase rectifier • uncontrolled • controlled Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**AC input**DC output 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 Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**+**vs _ + vo _ w t vs p 2p vo io Single-phase, half-wave, R-load Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**i**+ vR _ + vo _ + vs _ + vL _ Half-wave with R-L load 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**vs,**io b w t vo vR vL 2p 0 4p p 3p R-L waveform Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**Extinction angle**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**Vm**+ vs _ iD + vo _ vs Vo 2p 3p 3p /2 4p D p /2 p vo Vmax Vmin iD Half wave rectifier, R-C Load a q Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**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. Operation 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**Estimation of a**Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**Ripple Voltage**Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**Capacitor Current**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**Vm**vs Vo 2p 3p 3p /2 4p D p /2 p vo Vmax Vmin iD a q 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. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**Example (cont’)**Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**ig**v ia s + vs _ w w w t t t vo v i g a Controlled half-wave + vo _ Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**i**+ vR _ + vo _ + vs _ w + vL _ t vs p 2p vo io b a Controlled h/w, R-L load 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**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**+**vR _ + vL _ io + vo _ + vs _ io io vo= 0 + vo _ vo= vs + vs _ + vo _ io D2 is on, D1 is off D1 is on, D2 is off 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 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**output**vo io iD1 Diode current iD2 t w 0 2p 3p p 4p 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. Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**D1**D3 is is iD2 iD1 iD1 io + vs _ + vo _ D4 Full Bridge D2 D1 + vs1 _ + vD1 - + vs _ - vo + + vs2 _ io + vD2 - D2 Center-tapped Full wave rectifier • 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 Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**D1**D3 is iD1 io + vs _ + vo _ D4 Full Bridge D2 Vm vs 4p 3p 2p p Vm vo vD1 vD2 -Vm vD3 vD4 -Vm io iD1 iD2 iD3 iD4 is Bridge waveforms Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**is**iD1 iD2 Vm vs D1 4p + vs1 _ 3p + vD1 - 2p p Vm + vs _ - vo + vo + vs2 _ io vD1 + vD2 - -2Vm D2 Center-tapped vD2 -2Vm io iD1 iD2 is Center-tapped waveforms Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**io**is iD1 + vR _ + vo _ + vs _ w t + vL _ vs p 2p iD1 , iD2 iD3 ,iD4 io vo is Full wave bridge, R-L load 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**vs**w t p 2p io iD1 , iD2 vo iD3 ,iD4 is R-L load approximation 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**is**T1 iD1 io T3 + vs _ + vo _ T2 T4 Controlled full wave, R load Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**io**is iD1 + vR _ + vo _ + vs _ + vL _ io b p a 2p +a p vo Discontinuous mode p+a io b p 2p a vo Continuous mode Controlled, R-L load Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**Discontinuous mode**Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**Continuous mode**Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**D1**io D3 vp + vs _ + vo _ D4 vn D2 vo =vp -vn Single-phase diode groups • 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**Vm**Vm van vbn vcn vp vn vo =vp - vn 3p 4p p 2p 0 Three-phase rectifiers D1 io + van - D3 D5 vpn + vbn - n + vo _ D2 + vcn - vo =vp -vn vnn D6 D4 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**vo**vo p/3 Vm, L-L 0 2p/3 p/3 Three-phase, average voltage Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003**io**Vm + van - T3 T5 vpn + vbn - n + vo _ T2 + vcn - vnn T6 T4 Controlled, three-phase T1 a vbn van vcn 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

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