SCR analogy; half-wave rectifier for 1 A average

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.

      A
      |
     Q1 (PNP)
      |   \
      |    +-- IB1 = IC2
     IC1   |
      |    |
  G --+-- Q2 (NPN)
           |
           K

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

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:

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).