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Diode current at 100°C

Short answer

The voltage at 1 mA drops by 176 mV from 20°C to 100°C. At constant applied voltage, the current rises exponentially: I2 = 1 mA × exp(176 mV / (1.6 × 32.15 mV)) ≈ 30.6 mA, using VT = kT/q at 100°C.

A diode conducts 1mA at 20°C. If it is operated at 100°C, What will be its current? Given data are η = 1.6 and negative temperature coefficient value = -2.2mV/°C.

Idea: at a fixed current the diode voltage falls by about 2.2 mV for each °C rise. If the voltage across the diode is held constant, this fall appears as an exponential rise in current.

Given

I1 = 1 mA at T1 = 20 °C, T2 = 100 °C, η = 1.6, temperature coefficient = -2.2 mV/°C.

Step 1: change in voltage for the same current

ΔV = (-2.2 mV/°C)(100 - 20) = -176 mV

So at 100 °C the voltage needed for 1 mA is 176 mV lower. At the original voltage the diode is 176 mV "more forward biased" than needed for 1 mA.

Step 2: thermal voltage at 100 °C

VT = kT/q = 8.617×10⁻⁵ × 373.15 = 32.15 mV
η VT = 1.6 × 32.15 = 51.4 mV

Step 3: current at constant voltage

I2 = I1 · exp(|ΔV| / (η VT))
   = 1 mA × exp(176 / 51.4)
   = 1 mA × exp(3.42) ≈ 30.6 mA

Answer: I ≈ 31 mA (about 30 times larger) at 100 °C for the same applied voltage.

Assumption: VT is taken at 100 °C. Using VT = 25 mV throughout would give ≈ 81 mA, so state the VT you use.

✓ Answer checked by Claude Opus 5.5 (high effort)

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