Chapter 29, Problem 68A

To determine

To find the current of a silicon diode at ${104}^{\text{o}}\text{C}$ if the reverse bias potential difference remains constant.

Answer to Problem 68A

The diode current at ${104}^{\text{o}}\text{C}$ is $\text{8}\text{.192 μA}$ .

Explanation of Solution

Given data:

For silicon, diode current at $0^{\text{o}}\text{C}$ , $I=1.0\text{ nA}$

The reverse bias potential difference remains constant.

For each $8^{\text{o}}\text{C}$ increase in temperature, current doubles.

Calculation:

According to the problem,

Diode current at $0^{\text{o}}\text{C}$ , $I=1.0\text{ nA}$

The reverse bias potential difference remains constant, so the current varies only due to temperature.

For each $8^{\text{o}}\text{C}$ rise in temperature, current doubles.

To calculate diode current at ${104}^{\text{o}}\text{C}$ ,find the number of $8^{\text{o}}\text{C}$ increases in temperature.

Number of $8^{\text{o}}\text{C}$ increases, $n=\frac{{104}^{\text{o}}\text{C}}{8^{\text{o}}\text{C}}=13$

Therefore,

The diode current at ${104}^{\text{o}}\text{C}$$=I\times 2^n$

  $\begin{array}{l}=1.0\text{ nA}\times 2^{13}\\ =8192\text{ nA}\\ =8.192\text{ μA}\end{array}$

Conclusion:

Thus, the diode current at ${104}^{\text{o}}\text{C}$ is $8192\text{ nA}$ or $\text{8}\text{.192 μA}$$\left(\text{Since},1\text{ nA}=\frac{1}{1000}\text{μA}\right)$