Basic Electronics, Chapter 5: Challenge 2 solution (Zener regulator and polarity protection)

Formulas: R_max = (Vin_min - Vz) / (IL_max + Iz_min)      I(R) = (Vin - Vz) / R      Iz = I(R) - IL
          Pz = Vz x Iz      PR = I(R)^2 x R      efficiency = Vout x IL / (Vin x I(R))

(a) 5.1 V Zener regulator, Vin = 11 to 13 V, load 0 to 30 mA, Iz_min = 5 mA
  1. Worst case for staying in regulation: lowest Vin (11 V) with the biggest load (30 mA):
       R_max = (11 - 5.1) / (0.030 + 0.005) = 5.9 / 0.035 = 168.6 ohm
  2. Round DOWN to E24 (150, 160, 180, 200 ...): 160 ohm.
     (180 ohm would be too large: Iz at 11 V and full load = 5.9/180 - 0.030 = 32.8 - 30 = 2.8 mA, below 5 mA.)
  3. Check with R = 160 ohm:
       Vin = 11 V, full load:  I(R) = 5.9 / 160 = 36.9 mA  -> Iz = 36.9 - 30 = 6.9 mA   (above 5 mA: good)
       Vin = 13 V, no load:    I(R) = 7.9 / 160 = 49.4 mA  -> Iz = 49.4 mA              (worst case for heat)
  4. Worst-case power:
       Zener:    Pz = 5.1 x 0.0494 = 252 mW    -> a 500 mW Zener is fine (50 % loaded)
       Resistor: PR = 0.0494^2 x 160 = 0.390 W -> use a 1 W resistor (a 1/2 W part would run at 78 % of its rating: too hot)
  5. Efficiency at 12 V and full load:
       I(R) = (12 - 5.1) / 160 = 43.1 mA   Pin = 12 x 0.0431 = 517 mW
       Pout = 5.1 x 0.030 = 153 mW         efficiency = 153 / 517 = 29.6 %
     About 70 % of the power is lost as heat. A linear regulator IC (Basic Electronics 9) wastes far less
     current at no load, and holds the voltage better.

(b) Series reverse-polarity diode, 12 V supply, 500 mA load
  1N4001 (Vf = 0.75 V at 0.5 A):
    Load voltage = 12 - 0.75 = 11.25 V
    Power lost in the diode = 0.75 x 0.5 = 0.375 W
    Share of total power (12 x 0.5 = 6 W) = 0.375 / 6 = 6.25 %
  1N5819 Schottky (Vf = 0.40 V at 0.5 A):
    Load voltage = 12 - 0.40 = 11.60 V
    Power lost = 0.40 x 0.5 = 0.20 W
    Share = 0.20 / 6 = 3.3 %
  The Schottky is better: it wastes almost half as much power, runs cooler, and the load gets 0.35 V more.
  Check: its 40 V reverse rating is plenty for a 12 V supply (the silicon part is needed only if the supply
  can have spikes above about 40 V). A shunt diode behind a fuse loses no voltage at all in normal use.

Checklist
  - Zener regulator: design for the LOW supply and HIGH load (keeps regulation); check the HIGH supply and NO load (heat).
  - Round R down so the Zener keeps its minimum current.
  - Check the power in both the resistor and the Zener.
  - Series protection diode: the load loses Vf and the diode dissipates Vf x I.