Basic Electronics 10, Chapter 10: Challenge 2 solution (PWM and interrupts)

Formulas: D = compare / top          V(avg) = D x V          steps = f(clk) / f(PWM)          bits = floor(log2 steps)
          RC filter ripple dV = V D (1 - D) / (R C f)          time constant tau = R C          settling to 1/256: ln(256) tau = 5.55 tau
          pulse rate = rpm / 60 x pulses per rev          CPU load = rate x ISR time          one count = 60 / (pulses per rev x window) rpm

(a) analogWrite(pin, 90) on an Uno
    D = 90 / 255 = 0.3529 = 35.3 %
    On a 5 V pin:             V(avg) = 0.3529 x 5 = 1.765 V
    On a 6 V motor supply:    V(avg) = 0.3529 x 6 = 2.118 V      (through a driver, which also drops a little)

(b) PWM at 25 kHz
    Pico (125 MHz):    125e6 / 25e3 = 5000 steps    log2(5000) = 12.29    -> 12 bits
    ESP32 (80 MHz):    80e6 / 25e3 = 3200 steps     log2(3200) = 11.64    -> 11 bits
    Uno Timer 1, 8-bit phase-correct, 16 MHz:   16e6 / 510 = 31 373 Hz = 31.4 kHz     (255 levels)
    Doubling the frequency would halve the steps and cost one bit: 25 kHz is a good silent choice on all three.

(c) RC filter, R = 4.7 k, C = 2.2 uF, 5 V, D = 0.5 (D(1 - D) = 0.25)
    Time constant:  tau = 4700 x 2.2e-6 = 10.34 ms          cut-off  f(c) = 1 / (2 pi tau) = 15.4 Hz
    Ripple at 490 Hz:     5 x 0.25 / (4700 x 2.2e-6 x 490)     = 1.25 / 5.067  = 0.247 V peak to peak
    Ripple at 31 372 Hz:  5 x 0.25 / (4700 x 2.2e-6 x 31372)   = 1.25 / 324.4  = 3.85 mV peak to peak
    Settling to 1 part in 256:  t = tau x ln(256) = 10.34 ms x 5.545 = 57.3 ms
    The fast PWM makes the ripple 64 times smaller (31 372 / 490), at no cost in settling time.

(d) Encoder: 7 pulses per rev on the motor shaft, motor 9000 rpm, gearbox 30:1
    Motor speed in rev/s:  9000 / 60 = 150 rev/s
    Pulse rate:            150 x 7 = 1050 interrupts per second
    CPU load of a 12 us ISR:   1050 x 12e-6 = 0.0126 = 1.26 %
    Counts in a 50 ms window:  1050 x 0.050 = 52.5 (so 52 or 53 counts per window)
    One count = 60 / (7 x 0.050) = 171.4 rpm of the motor, which is 171.4 / 30 = 5.71 rpm at the output shaft
        (a single count of error is 1 / 52.5 = 1.9 % of the reading: about right; a 100 ms window would halve it)
    Output shaft speed:    9000 / 30 = 300 rpm
    Motor rpm from counts: counts x 60 / (7 x 0.050) = 52.5 x 171.4 = 9000 rpm

Expected readings
    (a) a meter on a PWM pin (DC range) averages to about 1.76 V; a scope shows 35 % high
    (c) a scope on the capacitor shows 2.5 V with 0.25 V of ripple at 490 Hz, and a flat line (4 mV) at 31.4 kHz
    (d) in the program the count in a 50 ms window is 52 or 53, alternating, as the window and the pulses do not line up

Wiring and program checklist
  - PWM output: pin -> driver input, with a 100 k pull-down to GND at the driver so the motor is off while the board boots.
  - Encoder: sensor output to an interrupt-capable pin (Uno D2 or D3; any GPIO on a Pico or ESP32), with the sensor's own pull-up if it is open-collector.
  - ISR: count only, shared variable volatile, copy it with interrupts off on an 8-bit chip.
  - Filter test: measure the capacitor voltage with a meter; it should be D x 5 V once settled.