Question

A silicon diode has a reverse saturation current of \(10^{-12}\text{ A}\) at 300 K. The approximate increase in temperature such that saturation current doubles is

• A. \(5^\circ\text{C}\)
• B. \(10^\circ\text{C}\)
• C. \(15^\circ\text{C}\)
• D. \(20^\circ\text{C}\)

Hint:

Remember this fundamental rule of electronics: The reverse saturation current of any semiconductor diode doubles for every \(10^\circ\text{C}\) rise in temperature. This rule is independent of the initial current value (\(10^{-12}\text{ A}\)).

Answer 1

The Correct Option is B

Concept: The reverse saturation current (\(I_0\)) of a semiconductor diode is highly dependent on temperature because thermal energy continuously generates minority charge carriers. For both silicon and germanium diodes, the reverse saturation current increases exponentially with temperature. As an empirical rule of thumb, it approximately doubles for every \(10^\circ\text{C}\) rise in temperature: \[ I_0(T_2) = I_0(T_1) \cdot 2^{\frac{T_2 - T_1}{10}} \] where \(\Delta T = T_2 - T_1\) represents the total change in temperature in degrees Celsius or Kelvin.

Step 1: Apply the operational rule for current doubling. The problem requires the reverse saturation current to double its value: \[ I_0(T_2) = 2 \cdot I_0(T_1) \] Substitute this condition into the exponential rule formula: \[ 2 \cdot I_0(T_1) = I_0(T_1) \cdot 2^{\frac{\Delta T}{10}} \]

Step 2: Solve for the temperature change (\(\Delta T\)). Divide both sides of the equation by \(I_0(T_1)\): \[ 2^1 = 2^{\frac{\Delta T}{10}} \] Equating the exponents since the bases are identical: \[ 1 = \frac{\Delta T}{10} \quad \Rightarrow \quad \Delta T = 10^\circ\text{C} \] Thus, the temperature must increase by approximately \(10^\circ\text{C}\) for the reverse saturation current to double, matching option (B).