The current voltage relation of diode is given by I = (e^(1000 V/T) - 1) mA, where the applied volta

Q: Why do we use differentiation for finding errors?

Differentiation gives us the linear approximation of a function at a specific point. For very small changes (errors), the tangent line (differential) is a highly accurate representation of the actual change.

Q: Can I differentiate the equation without taking the log first?

Absolutely! If you differentiate $I = e^{1000V/T} - 1$ directly, you get $dI = e^{1000V/T} \cdot (1000/T) dV$. Since $e^{1000V/T} = I + 1$, you arrive at the exact same expression. Taking the log just makes the algebra feel a bit cleaner.

Q: Why is the error in current dependent on the current itself?

This is the hallmark of an exponential relationship. The slope of an exponential curve increases as the value increases. Therefore, at higher currents, the same small voltage error will cause a much larger current error.

Q: Do we need to convert the current from mA to Amperes?

In this specific problem, the equation is given such that $I$ is already in mA. If you look at the options, they are also in mA. So, keeping $I = 5$ directly gives $dI$ in mA.

Q: What if the voltage error was negative, like $-0.01$ V?

The differential $dI$ would also be negative, meaning the current would be less than expected. Usually, when we talk about 'error' or 'uncertainty', we consider the maximum absolute error, which is why we often use the positive magnitude.

Question

The current voltage relation of diode is given by $I = (e^{1000 V/T} - 1)$ mA, where the applied voltage $V$ is in volt and the temperature $T$ is in kelvin. If a student makes an error measuring $\pm 0.01$V while measuring the current of $5$ mA at $300$K, what will be the error in the value of current in mA?

Options:

(A) 0.2 mA

(B) 0.02 mA

(C) 0.5 mA

(D) 0.05 mA

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are designing a sensitive electronic circuit. You have a diode, and its current-voltage relationship is governed by the exponential function:

$$I = (e^{\frac{1000 V}{T}} - 1) 	ext{ mA}$$

We are operating this diode at a specific state where the current $I$ is $5 	ext{ mA}$ and the temperature $T$ is $300 	ext{ K}$. However, our voltmeter isn't perfect; it has a measurement error of $\pm 0.01 	ext{ V}$. Our goal is to determine how this tiny uncertainty in voltage propagates through the exponential relationship to create an error in the measured current.

### The Power of Differentials

To find the error in current ($dI$) resulting from a small error in voltage ($dV$), we need to use the concept of differentials. Geometrically, this is equivalent to drawing a tangent line at our operating point on the $I-V$ curve. For very small changes, the curve behaves almost like this straight tangent line.

Before we rush into differentiating the exponential function directly, let's make our algebraic lives a bit easier. We can rearrange the equation by moving the $-1$ to the left side:

$$I + 1 = e^{\frac{1000 V}{T}}$$

### Logarithmic Simplification

Now, to bring that exponent down to the ground level, we take the natural logarithm on both sides. This is a classic mathematical maneuver that simplifies exponential equations beautifully:

$$\ln(I + 1) = \frac{1000 V}{T}$$

With the equation in this linear-looking form, we can differentiate both sides with respect to their respective variables. The derivative of $\ln(I + 1)$ is $\frac{1}{I + 1} dI$, and on the right side, the derivative of $V$ is simply $dV$:

$$\frac{1}{I + 1} dI = \frac{1000}{T} dV$$

### The Final Calculation

Now we have a direct, linear relationship between the current error $dI$ and the voltage error $dV$. Let's isolate $dI$:

$$dI = (I + 1) \left(\frac{1000}{T}ight) dV$$

It is time to substitute our known operating conditions into this differential equation. We plug in $I = 5 	ext{ mA}$, $T = 300 	ext{ K}$, and $dV = 0.01 	ext{ V}$:

$$dI = (5 + 1) \left(\frac{1000}{300}ight) (0.01)$$

$$dI = 6 	imes \frac{10}{3} 	imes 0.01$$

$$dI = 20 	imes 0.01 = 0.2 	ext{ mA}$$

**The error in the value of current is $0.2 	ext{ mA}$.**

Notice a fascinating physical insight here: the error in current ($dI$) is directly proportional to $(I + 1)$. This means that in exponential devices like diodes, the higher the operating current, the more violently sensitive the device becomes to tiny voltage fluctuations!

Select Answer:

Visualized Solution

Visualizing the Diode Characteristics

The Concept of Differentials

Rearranging the Equation

Applying Logarithm

Differentiating the Expression

Substituting Known Values

Calculating the Error

Conclusion and Physical Insight

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Power of Differentials

Logarithmic Simplification

The Final Calculation

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