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Question:

The current voltage relation of a diode is given by I = (e^(1000V/T)) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ±0.01V while measuring the current of 5mA at 300 K, what will be the error in the value of current in mA?

0.5 mA

0.05 mA

0.02 mA

0.2 mA

Solution:

5 = e^(1000V/T) => e^(1000V/T) = 5 (1)
Again, I = e^(1000V/T)
dI/dV = e^(1000V/T) * (1000/T)
dI = (1000/T) * e^(1000V/T) dV
Using (1) ΔI = (1000/T) * 5 * 0.01 = 5000/T * 0.01 = 50/T
For T = 300K, ΔI = 50/300 = 1/6 mA ≈ 0.2 mA