Basic Electronics, Chapter 8: Challenge 1 solution (levels, margins and pull-ups)

Formulas: NML = VIL - VOL     NMH = VOH - VIH     a connection works if VOH(driver) >= VIH(receiver) and VOL(driver) <= VIL(receiver)
          divider: Vout = Vin x R2 / (R1 + R2)     rise time (10 % to 90 %) = 2.2 x R x C     I2C rise time (30 % to 70 %) = 0.8473 x R x C
          pull-up limits: Rp(min) = (Vcc - VOL) / IOL     Rp(max) = tr / (0.8473 x C)

(a) Noise margins (values from the table in section 1)
    74HC at 3.3 V driving a 74HC input at 3.3 V:   VIL = 0.99, VIH = 2.31, VOL = 0.1, VOH = 3.2
        NML = 0.99 - 0.1  = 0.89 V        NMH = 3.2 - 2.31 = 0.89 V
    74HC at 5 V driving a 74LS input:              driver VOL = 0.1, VOH = 4.4;  receiver (74LS) VIL = 0.8, VIH = 2.0
        NML = 0.8 - 0.1 = 0.7 V           NMH = 4.4 - 2.0 = 2.4 V           works, with big margins
    ESP32 pin (3.3 V) driving a 74HC at 5 V:       driver VOH = 2.64, VOL = 0.33;  receiver VIH = 3.5, VIL = 1.5
        NMH = 2.64 - 3.5 = -0.86 V        NEGATIVE: fails (a HIGH may not be read as HIGH). Not allowed.
    ESP32 pin driving a 74HCT at 5 V:              receiver VIH = 2.0, VIL = 0.8
        NMH = 2.64 - 2.0 = 0.64 V         NML = 0.8 - 0.33 = 0.47 V         works (but keep the wire short: the margins are modest)
    74LS output driving a 74HC input at 5 V:       driver VOH = 2.7, VOL = 0.5;  receiver VIH = 3.5, VIL = 1.5
        NMH = 2.7 - 3.5 = -0.8 V          fails: this is why the 74HCT family was invented.  (NML = 1.5 - 0.5 = 1.0 V is fine)

(b) 5 V to 3.3 V divider
    Target Vout = 3.3 V:  R2 / (R1 + R2) = 3.3 / 5 = 0.66  ->  R2 = 0.66 R1 / 0.34 = 1.94 k for R1 = 1 k.
    E12 values: 1.8 k -> Vout = 5 x 1.8 / 2.8 = 3.214 V  (OK, below 3.3 V)
                2.2 k -> Vout = 5 x 2.2 / 3.2 = 3.438 V  (above 3.3 V: avoid; still under the 3.6 V absolute maximum, but with no margin)
    Choose R2 = 1.8 k:  Vout = 3.21 V.   (An E24 value, 2.0 k, would give 3.33 V.)
    Rise time: the source resistance seen by the pin is R1 || R2 = 1000 x 1800 / 2800 = 643 ohm.
        15 pF load:    2.2 x 643 x 15e-12  = 21 ns          fine for signals up to several MHz
        100 pF wire:   2.2 x 643 x 100e-12 = 141 ns         limits the signal to about 1 to 2 MHz; for a longer wire use a buffer chip (74LVC) instead of a divider
    A LOW stays 0 V; a divider only works in this direction (5 V to 3.3 V), and loads the 5 V output with 5 V / 2.8 k = 1.8 mA.

(c) Open-drain bus
    Rp(min) = (5 - 0.4) / 3 mA = 4.6 / 0.003 = 1.53 k            (smaller than this and the sensor cannot hold the wire down to 0.4 V)
    Rp(max) = 1 us / (0.8473 x 200 pF) = 1e-6 / 1.6946e-10 = 5.90 k   (larger and the wire rises too slowly)
    Choose the E12 value 3.3 k (between the limits):
        rise time = 0.8473 x 3.3 k x 200 pF = 0.56 us
        current when pulled low = (5 - 0.4) / 3.3 k = 1.39 mA   (well under the sensors' 3 mA)
    Mistaken push-pull device driving HIGH while a sensor holds the wire LOW:
        total resistance = 65 + 130 = 195 ohm       current = 5 V / 195 = 25.6 mA   (far above the sensor's 3 mA rating: it can be damaged)
        wire voltage = 5 V x 130 / 195 = 3.33 V     neither a valid LOW (needs below VIL) nor a valid HIGH: the bus is corrupted.
        (the real on-resistance of the sensor output rises with current, so the real answer varies; the lesson is that two outputs fighting waste current and give an invalid level)

Checklist
  - Always check BOTH inequalities (HIGH and LOW), and the over-voltage limit of the receiver.
  - 3.3 V output into a 5 V CMOS input: use a 74HCT/74AHCT at 5 V. 5 V output into a 3.3 V input: divider, 74LVC or MOSFET shifter.
  - Open-drain pull-up: between (Vcc - VOL)/IOL and tr/(0.8473 C).
  - Never connect two push-pull outputs to one wire.