Question

In

the reverse-bias region the saturation current of a silicon diode
is about 0.1 μA (T = 20°C). Determine its approximate value if the
temperature is increased 40°C.

Answer 1

The approximate value of the reverse saturation current at a temperature 40°C higher is 0.8 μA.

Step-by-step explanation:

The reverse saturation current of a silicon diode is given as 0.1 μA at 20°C. To determine its approximate value at a temperature 40°C higher, we can use the formula:

I2 = I1 * (T2 / T1)^3

where I2 is the new current, I1 is the initial current, T2 is the new temperature, and T1 is the initial temperature.

Plugging in the values:

I2 = 0.1 μA * (40°C / 20°C)^3 = 0.8 μA

Therefore, the approximate value of the reverse saturation current at a temperature 40°C higher is 0.8 μA.

Answer 2

The approximate value of the reverse saturation current of a silicon diode at an increased temperature can be calculated using the equation I = Io * exp((qV)/(kT)), where Io is the reverse saturation current, q is the electronic charge, V is the voltage across the diode, k is Boltzmann's constant, and T is the temperature in Kelvin.

Step-by-step explanation:

In the reverse-bias region of a silicon diode, the saturation current is about 0.1 μA at a temperature of 20°C. To determine its approximate value when the temperature is increased by 40°C, we can use the equation:

I = Io * exp((qV)/(kT)), where I is the current, Io is the reverse saturation current, q is the electronic charge, V is the voltage across the diode, k is Boltzmann's constant, and T is the temperature in Kelvin.

Since the saturation current is independent of the voltage and only depends on the temperature, we can use the same value of 0.1 μA as Io. Plugging in the appropriate values, we have:

I = (0.1 μA) * exp((qV)/(k * (20°C + 40°C + 273.15)))

I ≈ 0.1 μA * exp((qV)/(k * 333.15 K))

where K is Kelvin, q is the charge of an electron, and k is Boltzmann's constant.