Lecture 4 - Three-phase circuits, transformer and transient analysis of RLC circuits. Figure 4.1
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1 Lecture 4 - Three-phase circuits, transformer and transient analysis of RLC circuits Power supply to sizeable power converters are often from three-phase AC source. A balanced three-phase source consists of three voltages at the mains frequency given by a c 4π/3 2π/3 a b v = sinωt = 0 an max Figure 4.1 2π vbn = max sin ωt = 2 π /3 3 4π vcn = max cos ωt = 4 π / These voltages are potentials with respect to a neutral point, commonly assumed at ground (zero) potential, but this assumption is not necessary. The neutral point may or may not be available. The line to line voltages are Lecture 4 Three-phase 4-1 F. Rahman
2 vab = van vbn = π 3max sin ωt+ = 6 3 π /6 vbc = vbn vcn = π 3max sin ωt = 2 3 π /2 4.2 vca = vcn van = 7π 3max sin ωt = π /6 a c ca bc an a ab b Figure 4.2 Note that the line to line voltage phasors are displaced by 30 from the phase voltages. If a balanced three-phase load is connected to a three-phase supply, the currents in the load circuit are given by Lecture 4 Three-phase 4-2 F. Rahman
3 For a Y-connected load and a I al I a v a I L I a an j Ze φ n n v c v b Figure 4.3 I a an = = φ = I φ Z bn Ib = = 2 /3 = I 2 /3 Z ( π φ) ( π φ) cn Ic = = 4 /3 = I 4 /3 Z ( π φ) ( π φ) 4.3 Note that for a Y-connected circuit, the line currents are the same as the phase currents and the line-line voltages are 3 times the phase voltages. i.e., I La = I a, etc, and =, and so on. ab a Lecture 4 Three-phase 4-3 F. Rahman
4 For a Δ-connected load, a a I La I ca I ab c b c b I bc Figure 4.4 I I I ab bc ca ab 3 π /6 = = = 3 /6 = 3I /6 Z ( π φ) ( π φ) bc 3 π / 2 = = = 3 / 2 = 3I / 2 Z ( π φ) ( π φ) ca 3 7 π /6 = = = 3 7 /6 = 3I 7 /6 Z ( π φ) ( π φ) 4.4 For the Δ-connected circuit, the line-line voltage are the same as the load phase voltages while the line currents are 3 times the phase currents; Lecture 4 Three-phase 4-4 F. Rahman
5 i.e., ( ) ILa = Iab Ica = 3I 4.5 The average power in a three-phase circuit is given by P = 3I cosφ = 3 I cosφ 4.6 L L The angle φ is the angle of the phase impedance, so that the power factor of a three-phase circuit supplied from a sinusoidal source is given by the power factor of each phase. The above statement on power factor is based on the assumption that the three-phase circuit caries only sinusoidal currents in each phase. When non sinusoidal currents flow, the power factor PF is given by PF where Psup plied 3I1 cosφ = = 3I 3I s s s I1 = cosφ 4.7 I I 1 = RMS value of the fundamental current of each phase, Amp I s = RMS value of the total supply current of each phase, Amp φ = phase angle of the fundamental current of each phase with respect to the line-neutral voltage of the corresponding phase, rad = RMS supply voltage/phase, olts Lecture 4 Three-phase 4-5 F. Rahman
6 Transformer relationships: i 1 i 2 v 1 N 1 N 2 v 2 Figure 4.5 N = N and NI 1 1= NI 2 2 (4.8) These ideal relationships do not include the leakage flux, magnetizing current and the iron loss (due to hysteresis and eddy currents). v 1 v 2 v 1 v 2 Figure 4.6 The dot convention is used to indicate which terminals of the transformer rise and fall in potential together, with the same polarity of voltage. In the above example, the terminals with dots become more positive (or negative) with respect to the other terminal, together. Lecture 4 Three-phase 4-6 F. Rahman
7 The core flux is determined from the applied voltage to the input terminals according to dϕ v1 = N1 dt (4.9) For a sinusoidal input voltage v 1, ϕ is also sinusoidal. Thus ϕ = ϕ sinωt max v = N 2π fϕ cosωt = N 2π fb A cosωt 1 1 max 1 max c (4.10) 2π fn1bmaxac 1 = = 4.44fN1BmaxAc (4.11) 2 where A c is the cross-section area of the transformer core and B max is the peak core field (flux density). Normally, B max is determined by the core material. It is in the range of T. If the hysteresis and eddy current losses can be ignored, an increase of f means a proportionate decrease of N 1 for the same flux levels (i.e., the same utilisation of the core), resulting in a nearly proportionate decrease in the size of the core. Transformers for high frequency operation use ferrite or iron power cores which are formed by injection molding. These cores have high resistivity and very narrow hysteresis width which result in very low hysteresis and eddy current losses. Consequently, high switching frequencies up to a few hundred khz (or even MHz) can be used, resulting in very compact (i.e., light Lecture 4 Three-phase 4-7 F. Rahman
8 weight) cores for power converter circuits. Ferrites and metallic compounds have low hysteresis and eddy current losses, so that these are normally used for core material in power converter circuits operated at high frequency. Role of transformers in power electronics 1. It provides for isolation of high and low voltage circuits, such as between control circuits, power circuits and sensor circuits. 2. It matches the output voltage of a converter to the requirements of a load, by taking advantage of the transformer turns-ratio. 3. In some applications, the limited voltage rating of devices in a converter, the output of which must be connected to a high voltage AC system, forces a transformer to be used. 4. In some DC-DC converters, a transformer allows operation of a converter with a suitable duty cycle (i.e., not too small) when the converter supplies a low DC output voltage from a much higher DC input voltage. 5. Use of transformers at high frequency rather than at 50Hz, reduces the overall size of the transformer and converter. Lecture 4 Three-phase 4-8 F. Rahman
9 Transient response of R-L-C circuit with a diode LC circuit with blocking diode S D L v o i C Figure 4.7 I max cmax = 2 Figure 4.8 Waveforms in the circuit of figure 4.7 Lecture 4 Three-phase 4-9 F. Rahman
10 Assuming that the supply is a DC source, di 1 v = = L + idt + dt C Assuming co = 0 at t = 0, co 4.12 i( t ) = C sinωt = Imax sinωt L where Imax = C 1 and ω = L LC and di = cosωt = at t = dt L L and vc(t) = (1 - cosωt) 4.16 Because of the reverse blocking property of the diode, current i will flow for one half cycle only. Thus, at T t = t 1 = [ LC] 2 = π, vc( ) = 2, where T = 2π, the period of ω oscillation of the circuit. The peak current through the diode circuit obviously occurs when t = t Lecture 4 Three-phase 4-10 F. Rahman
11 The DC source and the diode can often adequately represent the diode rectifier which delivers a rectified but pulsating DC output. The L may represent the inductance of the filter (or of the AC source) and the C may represent the capacitor of the filter. If the rectifier DC source is from a 50 Hz AC source, which means that the voltage varies slowly with time, the transient response of the LC circuit may still be adequately represented by the analysis given above. In other words, the peak voltage across the capacitor will be determined by the level of the input voltage at the point in time when it is turned on. The worst condition is for the case when the pulsating DC supply is turned on at its peak. LC circuit with blocking diode and resistance S D L v o i C Figure 4.9(a) Lecture 4 Three-phase 4-11 F. Rahman
12 i c i cmax 0 2 cmax 0 R increases Figure 4.9(b) As before, di 1 = L + Ri + idt + co 4.17 dt C This second order equation will have a characteristic equation the roots of which are given by s,s R R 1 = ± 2L L LC 4.18 These roots will have negative real parts and depending on values of R, L and C, these may have imaginary parts (ie under-damped response). Assuming under-damped response, the solution for i(t) is Lecture 4 Three-phase 4-12 F. Rahman
13 where αt [ ω ω ] i( t ) = e A cos t + A sin t r 2 r R α = L 2 2 r 0 ω = ω α LC and ω 0 = 4.22 When the capacitor is already charged up to some voltage, the capacitor charging current will flow when the supply voltage exceeds the capacitor voltage. When the supply source is ideal, the current will be found from solving the differential equation above. Note that if the inductance is small, the capacitor current can become very peaky. The capacitor voltage v c is normally zero just before the switch S is first turned on. After that, the capacitor charges up and holds a steady DC voltage across it. With a sufficiently large value of C, v c will vary within a small range as pulsates. Note that when the capacitor voltage is non zero, the diode D will conduct only when the the pulsating voltage exceeds the voltage of the capacitor. Otherwise, the diode is reverse biased. Lecture 4 Three-phase 4-13 F. Rahman
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