Basic Electronics, Chapter 7: Challenge 1 solution (quantisation and amplifier gain)

Formulas: LSB = Vref / 2^N     code = floor(Vin / LSB)     rms noise = LSB / sqrt(12)     SNR = 6.02 N + 1.76 dB
          non-inverting gain = 1 + Rf/Rg     closed-loop gain = A / (1 + A/G)     bandwidth = GBW / G
          summing amplifier: Vout = - Rf x ( V1/R1 + V2/R2 + V3/R3 )

(a) LSB values
      10-bit, 5 V:    LSB = 5 / 1024   = 4.883 mV
      12-bit, 3.3 V:  LSB = 3.3 / 4096 = 0.8057 mV
      16-bit, 3.3 V:  LSB = 3.3 / 65536 = 50.35 microvolts
    12-bit converter, Vin = 1.000 V:
      1.000 / 0.0008057 = 1241.2  ->  code 1241            (the converter truncates)
      voltage that code stands for = 1241 x 0.8057 mV = 0.99985 V      (the step covers 0.99985 V to 1.00066 V,
                                                                         so the reading is 0.15 mV low)
      rms quantisation noise = 0.8057 mV / sqrt(12) = 0.8057 / 3.464 = 0.233 mV
      ideal SNR = 6.02 x 12 + 1.76 = 74.0 dB

(b) Bridge sensor 0 to 80 mV, Rg = 1 kohm
    Ideal gain limit: 3.3 V / 0.080 V = 41.25, so Rf / Rg must be 40.25 or less (1 + Rf/Rg <= 41.25).
    E12 values near Rf = 40.25 k:  39 k (gain 40, output 3.20 V) and 47 k (gain 48, output 3.84 V: too big, the ADC saturates).
    Choose Rf = 39 kohm:    gain = 1 + 39 / 1 = 40        full-scale output = 40 x 0.080 = 3.20 V
    Sensor-side resolution = 0.8057 mV / 40 = 20.1 microvolts per count   (3.2 V of output covers 3971 counts)
    Exact gain with A = 100,000:   b = Rg/(Rf + Rg) = 1/40
        gain = A / (1 + A x b) = 100000 / (1 + 100000/40) = 100000 / 2501 = 39.984    (0.04 % below 40: negligible)
    Bandwidth = GBW / G = 1 MHz / 40 = 25 kHz  (the -3 dB frequency)
    Gain at 10 kHz (single-pole model): 40 / sqrt(1 + (10/25)^2) = 40 / 1.077 = 37.1     (down 7 %, -0.6 dB)
    Note: any signal faster than a few kilohertz needs less gain per stage or a faster amplifier.

(c) Summing amplifier: Rf = 10 k, V1 = 1.0 V through 10 k, V2 = 0.5 V through 20 k, V3 = 2.0 V through 10 k
      currents into the virtual ground:  1.0 / 10 k = 0.100 mA,  0.5 / 20 k = 0.025 mA,  2.0 / 10 k = 0.200 mA     total 0.325 mA
      Vout = - 10 k x 0.325 mA = - 3.25 V        (needs a negative supply: on one supply the stage would saturate at 0 V)
    Minus the average of three inputs:
      Vout = - (Rf / R) x (V1 + V2 + V3), so Rf / R = 1/3 for the average.
      With Rf = 10 k:  R = 30 k for all three inputs (30 k is 3 x 10 k in series if you have no 30 k).
      Check: V1 = 1.0, V2 = 0.5, V3 = 2.0 -> sum 3.5 V -> Vout = - 3.5 / 3 = - 1.17 V.

Checklist
  - Convert the sensor's span to the ADC's span: gain = ADC full scale / sensor full scale, then round DOWN to a standard value.
  - Gain of the non-inverting stage is 1 + Rf/Rg (not Rf/Rg).
  - Bandwidth = GBW / gain: check it against the fastest signal.
  - Sum currents, not voltages, at the virtual ground of the inverting stage.