Basic Electronics, Chapter 5: Challenge 1 solution (LED resistors)

Formulas: R = (Vs - Vf) / I     round R UP to the next E12 value     I(actual) = (Vs - Vf) / R(chosen)
          P(R) = I^2 R = (Vs - Vf)^2 / R      P(LED) = Vf x I
          multiplexing: average current = peak current x duty cycle

(a) Green LED (Vf = 2.1 V), 5 V supply, about 12 mA
  R = (5 - 2.1) / 0.012 = 2.9 / 0.012 = 241.7 ohm
  E12 values around it: 220, 270. Round UP: 270 ohm.
    (220 ohm would give 13.2 mA: allowed, but we decided to round up so the current never exceeds the target.)
  Actual current: I = 2.9 / 270 = 10.74 mA
  Resistor power: P = I^2 R = 0.01074^2 x 270 = 31.1 mW   (a 1/4 W resistor is plenty)
  LED power:      P = Vf x I = 2.1 x 0.01074 = 22.6 mW

(b) Red LED (Vf = 1.9 V) on a 3.3 V pin, about 8 mA
  R = (3.3 - 1.9) / 0.008 = 1.4 / 0.008 = 175 ohm
  Next E12 value up: 180 ohm
  Actual current: I = 1.4 / 180 = 7.78 mA
  Spread between LED samples (same 180 ohm resistor):
    Vf = 1.7 V:  I = 1.6 / 180 = 8.89 mA
    Vf = 2.1 V:  I = 1.2 / 180 = 6.67 mA
  The current moves from 6.7 to 8.9 mA (about 2.2 mA, roughly +/-15 %) for a +/-0.2 V change in Vf,
  because only 1.4 V sits across the resistor. At 5 V (3 V across the resistor) the same +/-0.2 V
  would change the current by only about +/-7 %. Less headroom means less stable current.

(c) Four red LEDs (2.0 V each) in series on 12 V at 15 mA
  Total LED drop: 4 x 2.0 = 8.0 V
  R = (12 - 8.0) / 0.015 = 4.0 / 0.015 = 266.7 ohm  -> E12 up: 270 ohm
  Actual current: 4.0 / 270 = 14.81 mA
  Resistor power: 4.0^2 / 270 = 59.3 mW  (1/4 W is fine)
  Four blue LEDs: 4 x 3.2 = 12.8 V, which is MORE than the 12 V supply, so no current can flow
  properly: it cannot work. (Three blue LEDs need 9.6 V: R = (12 - 9.6)/0.015 = 160 ohm, E12 up: 180 ohm.
  For four you need at least 12.8 V plus some headroom for the resistor, for example 15 V.)

(d) Multiplexed 4-digit display
  Duty cycle per digit = 1/4 = 25 % (one digit on at a time).
  Average = peak x duty, so peak = 5 mA / 0.25 = 20 mA (the LED's maximum: do not go higher).
  Resistor on 5 V with red LEDs (Vf = 2.0 V): R = (5 - 2.0) / 0.020 = 150 ohm (an E12 value; no rounding).
  Frame rate 100 Hz, so one frame = 10 ms and each digit is on for 10 ms / 4 = 2.5 ms.
  Perceived brightness matches a steady 5 mA LED. (Peak power per segment: 2.0 x 0.020 = 40 mW for 2.5 ms.)

Checklist
  - Use Vs - Vf, not Vs, for the voltage across the resistor.
  - Round R up to E12, then recompute the real current with the value you will fit.
  - Series LEDs: add the Vf values and check that the supply is larger.
  - Parallel LEDs: one resistor per LED.
  - Multiplexing multiplies the peak current by the number of groups (here 4), but never beyond the LED maximum.