Basic Electronics, Chapter 9: Challenge 3 solution (power, brown-outs and wiring)

Formulas: linear: P = (Vin - Vout) I, Iin = Iout + Iq;   buck: Iin = Pout / (eta Vin);   runtime = Q x 0.8 / I;   rise = P x theta
          terminal voltage V = Voc - I R(total);   hold-up C = I t / dV;   wire drop = I x R/m x 2 x length;  wire loss = I^2 x R/m x 2 x length;  C-rate = I / Q

(a) 2S LiPo 7.4 V, 1.8 Ah; motors 0.8 A at the battery; 5 V rail at 0.5 A
    Pout = 5 x 0.5 = 2.5 W
    Linear (Iq = 5 mA):    Iin = 0.5 + 0.005 = 0.505 A      loss = (7.4 - 5) x 0.5 + 7.4 x 0.005 = 1.2 + 0.037 = 1.24 W       efficiency 5 / 7.4 = 68 %
    Buck at 88 %:           Iin = 2.5 / (0.88 x 7.4) = 0.384 A    loss = 2.5 / 0.88 - 2.5 = 0.34 W
    Total battery current:  linear 0.8 + 0.505 = 1.305 A       buck 0.8 + 0.384 = 1.184 A
    Runtime at 80 % usable:  linear 1.8 x 0.8 / 1.305 = 1.10 hours       buck 1.8 x 0.8 / 1.184 = 1.22 hours  (about 10 % longer)
    Temperature rise of the linear part: 1.2 W x 65 C/W = 78 C above ambient (a 7805 in free air gets very hot: in a closed robot at 30 C it would be 108 C;
        it needs a heatsink or it will shut down).   An AMS1117 at 100 C/W: 1.2 x 100 = 120 C rise: not acceptable.
    Conclusion: a buck converter wins at 0.5 A from 7.4 V; a linear regulator is for the small, quiet loads.

(b) Brown-out
    The logic and the motors share the pack: Voc = 5.4 V, R = 1.15 ohm, start-up current 3 A.
    Voltage at the logic = 5.4 - 3 x 1.15 = 5.4 - 3.45 = 1.95 V.   That is below the 2.7 V brown-out threshold: the processor resets (and the motors, unloaded by the
    dip, release and the voltage recovers: the board reboots and starts the motors again: a reset loop).
    With a soft start limiting the motors to 1 A:  V = 5.4 - 1 x 1.15 = 4.25 V:  well above 2.7 V, so there is no reset.   (The current in total is 1 A plus the logic's
    own current; including the 70 mA logic: V = 5.4 - 1.07 x 1.15 = 4.17 V; still fine.)
    Hold-up capacitor for 10 ms at 70 mA, dV = 1.1 V:   C = I t / dV = 0.07 x 0.010 / 1.1 = 636 uF   ->   choose 1000 uF (the next larger standard value)
    Why it does not cure a stalled motor: the capacitor supplies only C dV / I = 1000e-6 x 1.1 / 0.07 = 15.7 ms; a stalled motor holds the voltage down for as long as
    it is stalled (seconds), so the logic loses its supply again. The cure is to limit the start-up and stall current, to separate the supplies or to use a buck-boost for the logic;
    the capacitor (with a Schottky diode so that the dip does not drain it back into the battery) only covers brief spikes.

(c) Wire for a 3 A load 1.5 m from the battery (3 m of wire out and back)
    22 AWG (53 mohm/m):   R = 0.053 x 3 = 0.159 ohm     drop = 3 x 0.159 = 0.477 V (6.4 % of 7.4 V)     loss = 9 x 0.159 = 1.43 W
    18 AWG (20.9 mohm/m): R = 0.0209 x 3 = 0.0627 ohm   drop = 3 x 0.0627 = 0.188 V (2.5 %)             loss = 9 x 0.0627 = 0.56 W
    22 AWG carries the current (rated 7 A) but the drop costs the motors 0.48 V: use 18 AWG for motor and battery wiring.
    Fuse: it must exceed the peak of 3.8 A (and the start-up blip) and stay below the rating of the thinnest wire (7 A for 22 AWG; much more for 18 AWG):
        1.25 x 3.8 = 4.75 A  ->  choose the standard 5 A fast-blow fuse, fitted close to the battery's positive terminal.
    C-rate for 3 A from the 1.8 Ah pack:  3 / 1.8 = 1.67C: any LiPo pack copes easily (a 20C pack could give 36 A: which is why the fuse matters).

Checklist
  - Compare regulators by the input current and the heat, not the output voltage alone.
  - Brown-out voltage = open-circuit voltage - (start-up current x total resistance); then subtract any regulator dropout.
  - Reduce the start-up current first (soft start); a capacitor only bridges milliseconds.
  - Fuse above the peak, below the wire rating; use thicker wire for motor current to limit the drop.