Basic Electronics 11, Chapter 11: Challenge 2 solution (read a datasheet table)

Formulas: R(min) = (V(pin) - V(f, min)) / I(max)          I = (V(pin) - V(f)) / R          P = V(f) x I
          P(allowed at T) = P(rated) - slope x (T - 25)          average current = peak x duty

Rule: design for the worst case. A series resistor must keep the current under its limit when the LED's forward voltage is at its MINIMUM (the largest current).

(a) Minimum resistor for a 3.3 V Pico pin, current never above 10 mA
    V(f, min) = 1.7 V (the datasheet's lowest forward voltage)
    R(min) = (3.3 - 1.7) / 0.010 = 1.6 / 0.010 = 160 ohm
    The E12 values: 150 (too small: gives 10.7 mA at V(f) = 1.7 V), 180 (the next value up)
    Choose 180 ohm.

(b) Currents with 180 ohm
    V(f) = 1.7 V (minimum):   I = (3.3 - 1.7) / 180 = 8.89 mA     (the worst case, under 10 mA)
    V(f) = 1.9 V (typical):   I = (3.3 - 1.9) / 180 = 7.78 mA
    V(f) = 2.2 V (maximum):   I = (3.3 - 2.2) / 180 = 6.11 mA
    Power in the LED:   at V(f) = 1.7 V:  1.7 x 8.89 mA = 15.1 mW          typical: 1.9 x 7.78 = 14.8 mW          maximum: 2.2 x 6.11 = 13.4 mW
    Worst case about 15 mW (the power is largest when V(f) is about half the supply: 1.65 x 1.65 / 180 = 15.1 mW), against a 75 mW rating: lots of margin.
    (The LED is also a bit dimmer than at 20 mA; the Pico's default pin drive is 4 mA, selectable up to 12 mA, so this 8.9 mA needs the higher drive setting.)

(c) 70 degrees C ambient
    Allowed power = 75 - 1.0 x (70 - 25) = 75 - 45 = 30 mW
    At V(f) = 1.9 V:   I(max) = 30 / 1.9 = 15.8 mA            (the full-rated 20 mA at 1.9 V is 38 mW: too much at 70 C)
    The power rating reaches zero at:   25 + 75 / 1.0 = 100 degrees C
    So in a hot enclosure the LED must be driven well below its 25 C current; the Pico circuit above (15 mW) is still under the 30 mW limit.

(d) Pulsed 100 mA at 10 % duty, 1 kHz
    Average current = 100 mA x 0.10 = 10 mA          (under the 30 mA continuous limit)
    Period = 1 / 1000 = 1 ms          pulse width = 0.10 x 1 ms = 100 microseconds
    The peak (100 mA) equals the peak rating, with no margin; the conditions (1 kHz, 10 %) must also be kept, so it is within the ratings only if neither is exceeded.
    In practice, keep a margin (for example 80 mA) and remember that the power at 70 C is also limited by the derating of part (c).

(e) Three LEDs from three Pico pins
    Total = 3 x 8.89 mA = 26.7 mA (worst case)     under the roughly 50 mA for all pins     OK
    Each pin carries 8.89 mA, within a pin's limit when set to the 12 mA drive strength.

Lessons
    - Use the minimum column for "too much current" and the maximum column for "not enough voltage".
    - The typical column is for expectations, not guarantees.
    - The absolute maximum (30 mA) is not a target: design for about a third of it.
    - Always check the derating when the part will be warm.