The diode current-voltage relation is I = [exp(1000 V/T) - 1] mA, with V in volts and T in kelvin. At T=300 K the measured current is 5 mA. If V has an uncertainty ±0.01 V, find the resulting current uncertainty.
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Question

The diode current-voltage relation is I = [exp(1000 V/T) - 1] mA, with V in volts and T in kelvin. At T=300 K the measured current is 5 mA. If V has an uncertainty ±0.01 V, find the resulting current uncertainty.

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G1 Physics · Accepted Answer

At I=5 mA, exp(1000 V/T)=I+1=6. Hence dI/dV=(1000/T)exp(1000 V/T)= (1000/300)×6 =20 mA V⁻¹. Therefore ΔI≈20×0.01=0.2 mA. Correct answer: C.

The diode current-voltage relation is I = [exp(1000 V/T) − 1] mA, with V in volts and T in kelvin. At T=300 K the measured current is 5 mA. If V has an uncertainty ±0.01 V, find the resulting current uncertainty.