Basic Electronics, Chapter 5: Challenge 3 solution (rectifier and flyback)

Formulas: Vpk = 1.414 x Vrms      bridge peak = Vpk - 1.4 V      half-wave peak = Vpk - 0.7 V
          ripple Vr = I / (f_r x C)      mean Vdc = Vpeak - Vr/2      minimum = Vpeak - Vr
          f_r = 100 Hz for a bridge and 50 Hz for a half-wave on 50 Hz mains
          flyback: i(t) = (I0 + Vc/R) x e^(-t R/L) - Vc/R      zero current at t = (L/R) ln(1 + I0 R / Vc)

(a) 12 Vrms transformer, bridge, 2200 uF, 200 mA load
  Vpk = 12 x 1.4142 = 16.97 V
  Bridge peak output = 16.97 - 1.4 = 15.57 V
  Ripple Vr = 0.2 / (100 x 2200e-6) = 0.2 / 0.22 = 0.909 V
  Mean output = 15.57 - 0.909 / 2 = 15.12 V
  Minimum output = 15.57 - 0.909 = 14.66 V
  Capacitor: the voltage across it never exceeds 15.6 V; with the 1.5 to 2 x rule choose 25 V (23 V minimum).
  Diodes: average current = 200 mA / 2 = 100 mA each (each conducts half the time);
          peak reverse voltage = Vpk = 17 V. A 1N4001 (1 A, 50 V) is suitable. Allow for the switch-on surge.

(b) Half-wave, same capacitor and load
  Peak output = 16.97 - 0.7 = 16.27 V
  Ripple = 0.2 / (50 x 2200e-6) = 0.2 / 0.11 = 1.818 V   (exactly twice the bridge ripple, since f_r is halved)
  Mean = 16.27 - 1.818 / 2 = 15.36 V      Minimum = 16.27 - 1.818 = 14.45 V
  The ripple is twice as large for the same capacitor, so the half-wave needs twice the capacitance to match the bridge.
  The diode carries the whole 200 mA average, and its peak reverse voltage with a capacitor is about
  2 x Vpk = 34 V (the capacitor holds +Vpeak while the AC swings to -Vpk): still inside a 1N4001's 50 V.

(c) Capacitor for 0.5 V ripple, bridge, 200 mA
  C = I / (f_r x Vr) = 0.2 / (100 x 0.5) = 0.004 F = 4000 uF
  Next E6 value up: 4700 uF
  Actual ripple with 4700 uF: 0.2 / (100 x 4700e-6) = 0.2 / 0.47 = 0.426 V  (under the 0.5 V target)

(d) Relay coil: 12 V, R = 240 ohm, L = 240 mH, I0 = 50 mA; tau = L / R = 0.24 / 240 = 1 ms
  With a 0.7 V flyback diode (Vc = 0.7 V):
    Vc / R = 0.7 / 240 = 2.92 mA
    i(t) = (50 + 2.92) mA x e^(-t / 1 ms) - 2.92 mA
    For 5 mA: e^(-t/tau) = (5 + 2.92) / 52.92 = 0.1497    t = -1 ms x ln(0.1497) = 1.90 ms
  With a 12 V Zener in series with the diode (Vc = 12.7 V):
    Vc / R = 12.7 / 240 = 52.9 mA
    Zero current at t = 1 ms x ln(1 + I0 R / Vc) = 1 ms x ln(1 + 12 / 12.7) = 1 ms x ln(1.945) = 0.665 ms
    (and to 5 mA: e^(-t/tau) = (5 + 52.9) / 102.9 = 0.5628, so t = 0.575 ms)
  Voltage the transistor sees when it switches off = supply + clamp = 12 + 12.7 = 24.7 V,
  so use a transistor rated for at least 40 V (comfortable margin).
  Result: the Zener clamp cuts the release time from 1.9 ms to about 0.58 ms (to 5 mA), at the price of a
  higher voltage stress on the transistor.

Checklist
  - Bridge: two diodes conduct, so subtract 1.4 V; ripple frequency is twice the mains frequency.
  - Use C = I / (f_r x Vr) and round up to a standard value; the voltage rating needs a margin over the peak.
  - Flyback: the clamp voltage sets both the release time (higher is faster) and the transistor's voltage stress.