--- title: "SCR analogy; half-wave rectifier for 1 A average" subject: "Advanced Electronics" chapter: 7 key: "semester_4/advanced_electronics:scr-analogy-hwr-120v" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-analogy-hwr-120v site: BCT Engineering --- # SCR analogy; half-wave rectifier for 1 A average Advanced Electronics · Chapter 7: Introduction to Power Electronics - Asked in: 2071 Shrawan, 4+2+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-analogy-hwr-120v - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7 ## Question Explain the two transistor analogy of SCR with necessary mathematical expression. For the half-wave controlled rectifier with a load R = 50Ω, input voltage is a 120V-rms ac voltage. Assume that the drop across the SCR is 1.5V when it is conducting. (a) What should be the firing angle if it is desired to deliver an average current of 1A to the load? (b) What is the average power delivered to the load under the condition of (a)? ## Short answer An SCR is modeled as a PNP and NPN transistor pair with regenerative feedback, latching when α1+α2→1. For a 120V rms half-wave rectifier with R=50Ω and 1.5V SCR drop, to deliver average current 1A, firing angle α≈29.8°; average load power is about 137.5W. ## Answer #### Two-transistor analogy of SCR An SCR (PNPN) can be viewed as a PNP transistor Q1 (P1-N1-P2) and an NPN transistor Q2 (N1-P2-N2) connected so that the collector of each feeds the base of the other. ```text A | Q1 (PNP) | \ | +-- IB1 = IC2 IC1 | | | G --+-- Q2 (NPN) | K ``` ```text IC1 = α1·IA + ICBO1 IC2 = α2·IK + ICBO2 IK = IA + IG, IA = IC1 + IC2 IA = α1·IA + α2(IA + IG) + ICBO1 + ICBO2 IA = (α2·IG + ICBO1 + ICBO2) / (1 − (α1 + α2)) ``` With IG = 0, α1 + α2 is small and only leakage flows (forward blocking). A gate current increases α2, current increases, α1 rises, and when **α1 + α2 → 1** the anode current becomes very large (limited by the load). The regenerative action latches the SCR ON, and the gate loses control. #### Numerical Given: R = 50 Ω, Vs = 120 V rms, so Vm = 120√2 = 169.71 V; SCR drop VT = 1.5 V. Taking the peak load voltage as Vm − VT = 168.21 V: **(a) Firing angle for Iav = 1 A** ```text Iav = (Vm − VT)(1 + cos α)/(2πR) 1 = 168.21 (1 + cos α)/(2π × 50) 1 + cos α = 314.16/168.21 = 1.8677 cos α = 0.8677 α = 29.8° ``` (If the 1.5 V drop is subtracted over the conduction interval exactly, α ≈ 29.0°; the difference is small.) **(b) Average power to the load** Power in a resistor depends on the RMS value: ```text Vrms = ((Vm − VT)/2)·√[ (π − α)/π + sin2α/(2π) ] = 84.10 × √[ 0.8344 + 0.1373 ] = 82.90 V P = Vrms²/R = 82.90²/50 P ≈ 137.5 W ``` **Answer: α ≈ 29.8°, average load power ≈ 137.5 W** (the DC component alone is Vdc·Idc = 50 V × 1 A = 50 W). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation represents the two-transistor analogy? A. IA = (α2·IG + ICBO1 + ICBO2)/(1 − (α1 + α2)) B. IA = (α1·IG + ICBO)/(1 + α1 + α2) C. IA = IG/(α1 + α2) D. IA = α1·α2·IG Answer: A. Obtained from IC1 + IC2 with IK = IA + IG. (Key point: Analogy equation) 2. Taking SCR drop into account, what peak load voltage is used? A. 120 V B. Vm − VT = 169.71 − 1.5 = 168.21 V C. 169.71 V D. 120 − 1.5 V Answer: B. Vm = 120√2 = 169.71 V. (Key point: Peak voltage) 3. What condition gives Iav = 1 A? A. cos α = 1, α = 0° B. α = 90° C. 1 + cos α = 2πR·Iav/(Vm − VT) = 1.8677, so α ≈ 29.8° D. 1 − cos α = 1.8677 Answer: C. Half-wave average current formula solved for cos α = 0.8677. (Key point: Firing angle) 4. Why is power computed with Vrms and not Vdc·Idc? A. Vdc·Idc is correct and gives 137.5 W B. Power equals Vm^2/R C. Average power is zero for half-wave D. Power in a resistor depends on the RMS value, giving about 137.5 W Answer: D. P = Vrms²/R; the 50 W from Vdc·Idc is only the DC component. 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