Basic Electronics, Chapter 3: Challenge 2 solution (design and load a divider)

Goal: 9 V in, about 2.5 V out, about 1 mA in the divider, E24 resistors.

Step 1: first estimate
  R_total = Vin / I = 9 V / 1 mA = 9 k ohm
  R2 = Vout / I = 2.5 V / 1 mA = 2.5 k ohm
  R1 = 9 k - 2.5 k = 6.5 k ohm

Step 2: nearest E24 values
  E24 includes 2.4 k and 6.2 k (and 2.7 k, 6.8 k). Try R1 = 6.2 k, R2 = 2.4 k:
  Vout = 9 x 2.4 / (6.2 + 2.4) = 21.6 / 8.6 = 2.5116 V   (0.5 % high: good)
  I    = 9 / 8.6 k = 1.047 mA
  (6.8 k and 2.7 k would give 9 x 2.7 / 9.5 = 2.558 V: further out.)
  Choice: R1 = 6.2 k, R2 = 2.4 k.

Step 3: loaded with RL = 10 k, Thevenin method
  Vth = 2.5116 V   (the unloaded output)
  Rth = R1 || R2 = (6.2 k x 2.4 k) / 8.6 k = 14.88 / 8.6 = 1.7302 k ohm
  Vout = Vth x RL / (Rth + RL) = 2.5116 x 10 / (1.7302 + 10) = 25.116 / 11.7302 = 2.1411 V

Step 4: direct check
  R2 || RL = (2.4 k x 10 k) / 12.4 k = 1.9355 k ohm
  Vout = 9 x 1.9355 / (6.2 + 1.9355) = 17.419 / 8.1355 = 2.1411 V     Matches.

Step 5: effect of the load
  Change = (2.1411 - 2.5116) / 2.5116 = -0.1476  ->  -14.8 %
  The 10 k load is too heavy for a divider whose output resistance is 1.73 k.

Step 6: load for a 2 % error
  Vout/Vth = RL / (Rth + RL) >= 0.98
  RL >= 0.98 x Rth / 0.02 = 49 x Rth = 49 x 1.7302 k = 84.8 k ohm
  So the load must be at least about 85 k ohm (roughly 35 x R2).

Step 7: rescale both resistors to a tenth (R1 = 620 ohm, R2 = 240 ohm, both E24)
  Unloaded Vout is unchanged: 9 x 240 / 860 = 2.5116 V (the ratio is the same)
  Rth = 620 || 240 = (620 x 240) / 860 = 173.0 ohm
  With RL = 10 k:  Vout = 2.5116 x 10000 / (173.0 + 10000) = 2.469 V
  Change = (2.469 - 2.5116) / 2.5116 = -1.7 %    (was -14.8 %)
  Gain: a stiff output that the load barely disturbs.
  Price: ten times the current. I = 9 / 860 = 10.47 mA, and the power wasted
  is 9 V x 10.47 mA = 94 mW (was 9.4 mW). On a battery that is a tenfold drain.
  This trade-off (accuracy against wasted current) is the heart of divider design.

Checklist
 [ ] first estimate from the chosen current, then snap to E24 and recalculate
 [ ] Rth = R1 || R2, Vth = unloaded output
 [ ] check the loaded answer by the direct parallel method
 [ ] error under x %  needs  RL >= (1 - x)/x times Rth