Robotics Fundamentals 1, Chapter 10: Challenge 1 solution (power budget and fuse choice)

Formulas: rail load from the pack:  I(pack) = V(rail) x I(rail) / ( eta x V(pack) ),  eta = 0.90 for the buck
          runtime  t = 0.8 x V x C / P ;  fuse = 125 % to 150 % of the normal (largest sustained) current, below the wire rating, close to the battery
          wire drop  V = I x R(per metre) x ( 2 x one-way length ) ;  C-rate = I / C ;  sag = I x R(internal)
          linear regulator:  I(pack) = I(rail),  heat = ( V(in) - V(out) ) x I(rail)
Pack: 3S LiPo 11.1 V, 2200 mAh, R(internal) = 0.090 ohm.   Motors at the pack: 0.9 A average, 1.6 A worst working, 7.5 A stall, each.

(a) The 5 V rail:
        average   0.25 + 0.12 + 0.10 = 0.47 A         peak working   0.35 + 0.12 + 0.70 = 1.17 A
    From the pack through the 90 % buck:
        average   5 x 0.47 / 0.90 = 2.611 W,   2.611 / 11.1 = 0.2352 A
        peak      5 x 1.17 / 0.90 = 6.500 W,   6.500 / 11.1 = 0.5856 A

(b) Total pack current:
        average         2 x 0.9 + 0.2352 = 1.8 + 0.2352 = 2.035 A
        worst working   2 x 1.6 + 0.5856 = 3.2 + 0.5856 = 3.786 A
        stall           2 x 7.5 + 0.5856 = 15.0 + 0.5856 = 15.586 A
    Average power  P = 11.1 x 2.0352 = 22.59 W.

(c) Energy  E = 11.1 x 2.2 = 24.42 Wh,   usable  0.8 x 24.42 = 19.54 Wh.
    Runtime  t = 19.536 / 22.59 = 0.865 h = 51.9 minutes.

(d) Fuse:  1.25 x 3.786 = 4.73 A   to   1.50 x 3.786 = 5.68 A.   A 5 A fuse is inside the range (5 / 3.786 = 132 %).
    Wire must be rated ABOVE the fuse:  24 AWG (3.5 A) is not acceptable;  22 AWG (7 A) and 20 AWG (11 A) are.
    Voltage drop at 3.786 A for 0.4 m each way (0.8 m round trip):
        22 AWG:  R = 0.053 x 0.8 = 0.0424 ohm,   V = 3.786 x 0.0424 = 0.161 V   (1.4 % of 11.1 V)
        20 AWG:  R = 0.0333 x 0.8 = 0.02664 ohm, V = 3.786 x 0.02664 = 0.101 V  (0.9 % of 11.1 V)
    Choose 20 AWG: the lower drop and the larger margin below its 11 A rating (a stall of 15.6 A is three times the 5 A fuse: it opens in about a second or less, read its curve, which is the intended protection).
    Check the placement: the fuse goes in the positive lead next to the pack, and its voltage rating must be at least 12.6 V (the full pack).

(e) C-rate at stall:  15.586 / 2.2 = 7.08C   (a 20C pack copes).
    Voltage sag at stall:  15.586 x 0.090 = 1.40 V:  the pack's 11.1 V nominal gives about 9.70 V at the motors (a 12.6 % drop); the bulk capacitor at the driver and the stall rule limit how long that lasts.

(f) Linear regulator for the 5 V rail from 11.1 V:
        average   pack power = 11.1 x 0.47 = 5.217 W   (heat in the regulator = ( 11.1 - 5.0 ) x 0.47 = 2.867 W)     buck: 2.611 W
        peak      pack power = 11.1 x 1.17 = 12.99 W   (heat = 6.1 x 1.17 = 7.14 W: far too hot for a bare regulator)   buck: 6.500 W
    With the linear regulator the average power is 22.59 - 2.611 + 5.217 = 25.20 W and the runtime falls to 19.536 / 25.20 = 0.775 h = 46.5 minutes (against 51.9 min with the buck).
    The buck saves 2.6 W on the average and avoids a regulator that dissipates 7 W at the servo's peak: use the buck.