Basic Electronics, Chapter 9: Challenge 1 solution (the motor, PWM and the bridge)

Formulas: V = I R + Ke w     T = Kt I (Ke = Kt in SI)     w0 = (V - I0 R) / Ke     I(stall) = V / R     T(stall) = Kt (I(stall) - I0)
          at load current I:  w = (V - I R) / Ke,  T = Kt (I - I0),  P(out) = T w,  P(in) = V I
          max power at I = (V + I0 R) / (2 R);  max efficiency at I = sqrt(I0 I(stall)),  eta(max) = (1 - sqrt(I0 / I(stall)))^2
          gearbox: speed / N, torque x N x eg
          ripple dI(p-p) = V D (1 - D) / (L f);  shoot-through current = V / (R1 + R2);  t(off) = Rg Ciss ln(Vdrive / Vth)

(a) The 12 V motor: R = 3.0 ohm, I0 = 0.15 A, 6000 rpm at 12 V
    w0 = 6000 x 2 pi / 60 = 628.3 rad/s
    Ke = (12 - 0.15 x 3.0) / 628.3 = 11.55 / 628.3 = 0.018382 V.s/rad
         in V per krpm: 0.018382 x 1000 x 2 pi / 60 = 1.925 V/krpm            Kt = 18.38 mN.m per A
    Stall current   = 12 / 3.0 = 4.0 A
    Stall torque    = 0.018382 x (4.0 - 0.15) = 0.07077 N.m = 70.8 mN.m
    At I = 0.8 A:
        speed   w = (12 - 0.8 x 3.0) / 0.018382 = 9.6 / 0.018382 = 522.2 rad/s = 4987 rpm
        torque  T = 0.018382 x (0.8 - 0.15) = 11.95 mN.m
        P(out)  = 0.011949 x 522.2 = 6.24 W        P(in) = 12 x 0.8 = 9.60 W        efficiency = 6.24 / 9.60 = 65.0 %
        (loss in the winding I^2 R = 0.64 x 3 = 1.92 W; friction I0 x E = 0.15 x 9.6 = 1.44 W; 6.24 + 1.92 + 1.44 = 9.60 W)
    Maximum power:        I = (12 + 0.15 x 3.0) / (2 x 3.0) = 12.45 / 6 = 2.075 A
        speed = (12 - 2.075 x 3) / 0.018382 = 5.775 / 0.018382 = 314.1 rad/s = 3000 rpm (half of the no-load speed)
        torque = 0.018382 x (2.075 - 0.15) = 35.4 mN.m (half of the stall torque)
        P(out) = (2.075 - 0.15) x (12 - 6.225) = 1.925 x 5.775 = 11.1 W         efficiency = 11.1 / (12 x 2.075) = 44.6 %
    Maximum efficiency:   I = sqrt(0.15 x 4.0) = 0.775 A       eta = (1 - sqrt(0.15 / 4.0))^2 = (1 - 0.1936)^2 = 65.0 %
        speed = (12 - 0.775 x 3) / 0.018382 = 5027 rpm      torque = 0.018382 x 0.625 = 11.5 mN.m
        (the motor of this challenge is happiest around 0.8 A, which is 20 % of its stall current: run it there)
    With the 30:1 gearbox, eg = 0.65, at 0.8 A:
        output speed = 4987 / 30 = 166 rpm
        output torque = 11.95 mN.m x 30 x 0.65 = 233 mN.m = 2.4 kg.cm
        output power = 6.24 x 0.65 = 4.06 W
        output stall torque = 70.8 x 30 x 0.65 = 1380 mN.m (the gears must be strong enough, or the program must stop the motor on a stall)

(b) PWM with L = 2.5 mH
    Electrical time constant:  tau = L / R = 2.5 mH / 3.0 ohm = 0.83 ms
    Ripple at D = 0.5, V = 12:   dI = 12 x 0.5 x 0.5 / (L f) = 3 / (0.0025 f)
        f = 1 kHz:    3 / 2.5 = 1.2 A p-p    = 8 times the 0.15 A no-load current: the current reverses every cycle (bad: heat, whine)
        f = 20 kHz:   3 / 50 = 0.06 A p-p    = 0.4 of the 0.15 A: smooth, and above hearing
    D = 0.7:   average voltage = 0.7 x 12 = 8.4 V
        no-load speed = (8.4 - 0.15 x 3.0) / 0.018382 = 7.95 / 0.018382 = 432.5 rad/s = 4130 rpm   (about 69 % of 6000 rpm: nearly proportional to D)

(c) The H-bridge
    Shoot-through, one leg (two 20 mohm MOSFETs in series across 12 V):  I = 12 / (0.02 + 0.02) = 300 A (limited only by the supply, wires and the transistors' own pulse rating:
        a few microseconds of this destroys the parts, and the supply droops, resetting the logic)
    Conduction loss at 6 A: two MOSFETs in the path, P = I^2 (R1 + R2) = 36 x 0.04 = 1.44 W
    Dead time:  t(off) = 47 ohm x 3.3 nF x ln(5 / 2) = 155.1 ns x 0.9163 = 142 ns
        dead time = 2 x 142 ns = 284 ns   as a share of a 50 us period = 0.284 / 50 = 0.57 %   (negligible at 20 kHz)

Checklist
  - Convert datasheet units (V per krpm, mN.m per A) to SI before using the formulas.
  - Stall current = V / R is the design current for the driver, the wires, the fuse and the battery.
  - Ripple depends on L and f: raise f until the ripple is a small fraction of the average current, and above 20 kHz so that it is silent.
  - Shoot-through is the one catastrophic bridge fault: dead time and a driver chip that provides it.