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Problems 1. Find the time required to deliver a charge of 200Ah through a single-phase diode HWR with an output current of 100A rms and with sinusoidal input voltage. Assume diode conduction over a half- cycle. 2. A single phase 230 V, 1kW heater is connected across single phase 230 V, 50 Hz supply through a diode. Calculate the power delivered to the heater element. Find also the peak diode current and input power factor. Problem * Asingle phase full bridge diode rectifier is supplied from 230 V, 50 Hz source. The load consists of R= 10 Q and a large inductance so as to render the load current constant. Determine: a) Average values of output voltage and output current, b) Average and rms values of diode currents, c) Rms values of output and input currents and supply power factor. N-4: For an SCR, the gate-cathode characteristics has a straight-line slope of 130. For trigger source voltage of 15V and allowable gate power dissipation of 0.5 watts, compute the gate-source resistance. Solution. Here V I, = 0.5 W and £ —130 This gives I, = 10.5/180] = 0.062 = 62 mA Gate voltage, V,, = 180 x 62 x 103 = 8.06 V For the gate circuit, £,=1,R, + V, = 0.062R, + 8.06 = 15 a 2 _ 158.06 += ppgg = H1L94 2. Solution. (a) The slope of load line gives the required gate source resistance. From the load line, series resistance required in the gate circuit is 120 Q. (b) Here Vi, = 0.4 W For the gate circuit, E,=RL et Vy 0.4 15 = 1201, ae & or 12017 -151, +0.4=0 Its solution gives I, = 38.56 mA or 86.44 mA 0.4x10° 9856.7 10.37 V 3 or _ 04x10" _ 4 go7v. "86.44 EH a) At a=n/2 Vin Vac = on (1+ cosa) = 0.1592¥,, Vac 0.1592V,, RR Vin a sin2a Vins =e t= = 0.3536, _ Vems _ 9.3536Vj, fms = RR P. Vae 1 nat = 2 © — 20.27% Poms Vis Lims 16 Given V, = 230 V, f = From Eq. (6.1), 50 Hz, R=10 Q, a = 60° . [5 + 9¢ =e eco) = a (1 + cos 60°) = 77.64 1/2 ~ E) +5 sin 120° = 145.873 V 3) 2 = 14.587 A = 77.64 x 7.764 = 602.8 W V, Jy, = 145.873 x 14.587 = 2127.85 W . ficiency = Ste __6028_ _ 9 5559 0, 9g age P,. 2127.85 ae 602.8 I, 280 14587 7 01797 =V,, = v2 .V, = V2 x 280 = 325.22v, icici 18 Q.# A single phase 230 V, 1 kW heater is connected across 1-phase, 230 V, 50Hz supply through an SCR. For a firing-angle delays of 45° and 90°, calculate the power absorbed in the heater element. 6-4 6-5 6-6 6-7 In the single-phase converter of Fig. 6-9, the input voltage has a square waveform with amplitude of 200 V at a frequency of 60 Hz. Assume L, = 3 mH and I, = 10 A. (a) Obtain analytical expressions for u and V, in terms of V,, L,, @, f,, and a. Why is a independent of a, unlike in Eq. 6-24. (b) Obtain the i, and v,, waveforms and calculate the commutation interval u and V, for the following values of the delay angle a: 45° and 135°. SINGLE-PHASE CONVERTERS Consider the single-phase, half-controlled converter shown in Fig. P6-5, where v, is sinusoidal. (a) Draw v,, i,, and vy, waveforms and identify the devices conducting for various intervals for the following values of a: 45°, 90°, and 135°. (b) Calculate DPF, PF, and %THD for Vy = 3V4o. where V,, is the de output at a = 0. (c) Repeat part (b) for a full-bridge converter. (d) Compare results in parts (b) and (c). Figure P6-5 In terms of V, and J, in the single-phase converter of Fig. 6-5a, compute the peak inverse voltage and the average and the rms values of the current through each thyristor. The single-phase converter of Fig. 6-9 is supplying a de load of 1 kW. A 1.5-kVA-isolation transformer with a source-side voltage rating of 120 V at 60 Hz is used. It has a total leakage ©) i kee , , a Given, ee a iLesmN i fie devalion fey obich ge fo Te that Noe wnaerin bel oi is loc Oat - RV 4 8p0— Landilt) - 0: 062 dtl (a) Sev t>o, ly ayy | ma fi belay (ad ee > £— looo(s) — 0'062.541ts). = O ee S 4 Be esp = 12.90'32. ==—_——— Se i ania 9 fy |= a are faa ¥ Ge eet) fee ae OE a GB =-6:03 peso ace me + sity Behar i res TOs Tale lis ike bY j baking TF, S Stfor | Se Eee is = AG au b od ithe f te Sore ay 08 b-e — Ln aon te IDET. Step 4: THD formula THD = — ._—_. But since total RMS = fundamental RMS: 1 THD = —____ @ Final result: * Harmonic analysis: only 1st harmonic (fundamental sine) e THD = 0 (no distortion) Would you like me to now do the same process for a non-sinusoidal wave (like a square or half-wave rectified sine), so you can actually see harmonics appear and get a nonzero THD? QOgVSE Step 1: Harmonic analysis (Fourier series) The Fourier series expansion is: sin(wt) + t) + : sin(5wt) +... So it has: ® Fundamental (1st harmonic) ® Odd harmonics: 3rd, 5th, 7th, etc. # No DC, no even harmonics Step 2: RMS values 4 Fundamental amplitude: — T 4 Xiyms = == = 0.9003 TV 2 Total RMS of square wave = 1 (since it's +1 all the time). Step 3: THD formula Substitute values: _y _ v1 — 0.8105 ~ 0.9003 —<0.9008 __ 0.4368 on OF, tod 0.9003 a 48.5% @ Result for a square wave: ¢ Fundamental: strong (90% of total RMS) e Harmonics: odd harmonics present * THD # 48.5% 
