The current voltage relation of diode is given by I = (e1000v/T – 1) mA, where the applied voltage V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA?
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The diode current is given by I = (e1000V/T – 1) mA. We need to find the error in current, ΔI, when there is an error in voltage, ΔV = ±0.01 V, at I = 5 mA and T = 300 K.
For small changes, the error propagation is ΔI ≈ (dI/dV) * ΔV. First, find the derivative dI/dV.
dI/dV = d/dV[e1000V/T – 1] = (1000/T) * e1000V/T.
At T=300K, 1000/T = 1000/300 = 10/3. Since I = 5 mA, and I = e1000V/T – 1 ≈ e1000V/T (for I>>1mA), so e1000V/T ≈ 5.
Therefore, dI/dV ≈ (10/3) * 5 = 50/3 mA/V.
Now, ΔI ≈ (50/3) * (0.01) = 50/3 * 1/100 = 50/300 = 1/6 ≈ 0.1667 mA.
The magnitude of error is approximately 0.2 mA.
Final Answer: 0.2 mA