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Three-diode string voltage regulation

Short answer

Bias current I=7.9 mA; small-signal string resistance 3rd=19 Ω. ±10% supply change causes ΔVo≈±19 mV. Connecting 1 kΩ load draws 2.1 mA, dropping output by −40 mV.

A string of three diodes is used to provide a constant voltage of about 2.1 V. Calculate the change in this regulated voltage caused by (i) a ±10% change in the power supply voltage; (ii) connection of a 1 kΩ load resistance. Assume η = 2. [Figure: supply 10 ± 1 V, series resistor R = 1 kΩ, three diodes in series to ground giving output Vo, with load RL = 1 kΩ connectable across the output]

A forward-biased diode drops almost constant voltage, so three diodes in series with a resistor form a simple voltage regulator. Changes are found with the small-signal (incremental) resistance of the string.

Given

V+ = 10 V ± 10 % (±1 V), R = 1 kΩ, three diodes (0.7 V each, Vo ≈ 2.1 V), η = 2, VT = 25 mV.

Bias current and diode resistance

I = (10 - 2.1)/1 kΩ = 7.9 mA
rd (each) = η VT / I = 2 × 25 mV / 7.9 mA = 6.33 Ω
3 rd = 19.0 Ω

(i) Supply change of ±10 %

The signal ΔV+ = ±1 V divides between R and the string resistance:

ΔVo = ΔV+ × 3rd / (R + 3rd) = ±1 × 19.0 / (1000 + 19.0)
    = ±18.6 mV

Output changes by about ±19 mV (±0.9 % of 2.1 V), against a ±10 % change in supply: line regulation = 18.6 mV/V.

(ii) Connecting RL = 1 kΩ

The load draws about 2.1 V / 1 kΩ = 2.1 mA from the string, so the string current drops by 2.1 mA:

ΔVo = -ΔI × 3rd = -2.1 mA × 19.0 Ω = -40 mV

Output falls by about 40 mV (≈ 2 %). (More exactly, the diodes' rd rises a little as I falls, so the true drop is slightly larger.)

Load regulation = ΔVo/ΔIL = -19 mV/mA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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