Question

A germanium diode carries a current of 1 mA at room temperature when a forward bias of 0.15v is applied. Estimate the reverse saturation current at room temperature.

Answer 1

Explanation:

follow me and make my answer brainlist plz plzzzzzzz plzzzzzzz

Answer 2

Answer:

The reverse saturation current at room temperature for the given Germanium diode is approximately $3\mu A$

Explanation:

p-n Junction Diode:

• It is a two-terminal semiconductor device that primarily allows current in one direction.
• When an n-type semiconductor is joined with the p-type semiconductor a p-n junction is obtained.
• This diode can be biased in two ways, Forward bias and reverse bias.
• In forward bias, the positive terminal of the voltage source is connected to the p-side and the negative terminal to the n-side. There is a minimum threshold voltage required to overcome the depleted region.
• In the reverse bias, the positive terminal of the voltage source is connected to the n-side and the negative terminal to the p-side. In this case, the depletion region will increase as the number of charges on either sides increases until it cancels the external voltage, this means no current flows in the diode.
• When a large reverse voltage is applied to the diode, the barrier created by the depletion region breaks down hence allowing the flow of current. This reverse voltage is called the breakdown voltage.
• An equation that relates the flow of current through the diode as a function of voltage is expressed as,

                         $I=I_{0}(e^{\frac{qV}{kT} - 1})$

where $I$ is the current flowing through the diode, $I_{0}$ is the reverse saturation current, q the electronic charge, V is the applied voltage, k the Boltzmann constant and T temperature in Kelvins.

Step 1:

Given current carried by the diode to be 1mA, the applied voltage to be 0.15v, at room temperature (293K). We have to find the reverse saturation current $I_{0}$.

                     $I=1mA = 10^{-3} A\\V=0.15v\\q=1.6 \times 10^{-19} C\\kT = 1.38 \times 10^{-23} \times 298 \approx 4.11 \times 10^{-21} J$

Step 2:

First, let's calculate the power of e in the given equation for current,

                     $\frac{qV}{kT} = \frac{1.6 \times 10^{-19} \times 0.15}{4.11 \times 10^{-21}} \approx 5.84$

Step 3:

Substituting all the values in the equation for current,

                     $10^{-3} = I_{0}(e^5 - 1)\\I_{0} = \frac{10^{-3}}{(e^5 - 1)} \\I_{0} \approx 0.003 \times 10^}-3} = 3 \times 10^{-6} A\\I_{0} = 3\mu A$

Therefore, the reverse saturation current for the given Germanium diode is 3$\mu A$.