JEE Electronic Devices PYQs with Video Solutions, Download PDF

Solution & Explanation

1. Identify the Majority Carrier Concentration

In an n-type extrinsic semiconductor, each donor impurity atom donates one free conduction electron. At room temperature ($$300 \,\, \text{K}$$), assuming complete ionization of the dopant atoms, the concentration of electrons ($$n_n$$) is approximately equal to the donor impurity concentration ($$N_D$$):

$$n_n \approx N_D = 4.5 \times 10^{22} \,\, \text{m}^{-3}$$

2. Apply the Mass Action Law

According to the Mass Action Law , under thermal equilibrium conditions, the product of the free electron concentration and the hole concentration is a constant for a given semiconductor material at a fixed temperature, and it is equal to the square of its intrinsic carrier concentration:

$$n_n \cdot p_n = n_i^2$$

3. Calculate the Hole Concentration ($$p_n$$)

Rearranging the Mass Action Law formula to solve for the minority carrier concentration (holes):

$$p_n = \frac{n_i^2}{n_n}$$

Substitute the given numerical values into the equation:

$$p_n = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}}$$

$$p_n = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}}$$

$$p_n = 0.5 \times 10^{32 - 22} = 0.5 \times 10^{10}$$

$$p_n = 5.0 \times 10^{9} \,\, \text{m}^{-3}$$

Correct Option Key: Option A ($5.0 \times 10^{9} \,\, \text{m}^{-3}$)

Solution & Explanation

1. Calculate the Total Circuit Current ($$I_S$$)

Since the Zener diode is operating in its breakdown region, the voltage across it remains fixed at its Zener breakdown voltage:

$$V_Z = 6 \,\, \text{V}$$

The voltage drop across the series protective resistor ($$R_S$$) is the difference between the unregulated input supply voltage ($$V_{\text{in}}$$) and the Zener voltage ($$V_Z$$):

$$V_{R_S} = V_{\text{in}} - V_Z = 10 \,\, \text{V} - 6 \,\, \text{V} = 4 \,\, \text{V}$$

Using Ohm's Law, the total source current ($$I_S$$) passing through the series resistor is:

$$I_S = \frac{V_{R_S}}{R_S} = \frac{4 \,\, \text{V}}{100 \,\, \Omega} = 0.04 \,\, \text{A} = 40 \,\, \text{mA}$$

2. Calculate the Load Current ($$I_L$$)

Because the load resistor ($$R_L$$) is connected in parallel with the Zener diode, the voltage drop across it is exactly equal to the Zener voltage ($$6 \,\, \text{V}$$).

Using Ohm's Law, the current drawing into the load branch ($$I_L$$) is:

$$I_L = \frac{V_Z}{R_L} = \frac{6 \,\, \text{V}}{300 \,\, \Omega} = 0.02 \,\, \text{A} = 20 \,\, \text{mA}$$

3. Apply Kirchhoff's Current Law (KCL) to Find $$I_Z$$

The total supply current entering the parallel junction splits between the Zener diode branch and the load branch:

$$I_S = I_Z + I_L$$

Rearranging the equation to isolate the target Zener branch current parameter:

$$I_Z = I_S - I_L$$

$$I_Z = 40 \,\, \text{mA} - 20 \,\, \text{mA} = 20 \,\, \text{mA}$$

Correct Option Key: Option B ($$20 \,\, \text{mA}$$)

Solution & Explanation

1. Evaluate Statement I (Diffusion Current)

In a newly formed $$p\text{-}n$$ junction:

• The $$p\text{-region}$$ has a high concentration of holes (majority carriers), while the $$n\text{-region}$$ has a high concentration of electrons.
• This distinct concentration imbalance causes holes to diffuse across the junction from the $$p\text{-side}$$ to the $$n\text{-side}$$, and electrons to diffuse from the $$n\text{-side}$$ to the $$p\text{-side}$$.
• Since the conventional current direction aligns with the motion of positive charge carriers (holes), the resulting diffusion current flows from the $$p\text{-region}$$ to the $$n\text{-region}$$ .

Therefore, Statement I is TRUE .

2. Evaluate Statement II (Drift Current)

As diffusion continues, uncovered immobile ions are left behind at the boundary layer:

• Positive donor ions gather on the $$n\text{-side}$$ and negative acceptor ions gather on the $$p\text{-side}$$, establishing a built-in electric field directed from the $$n\text{-side}$$ toward the $$p\text{-side}$$.
• This internal electric field sweeps minority carriers across the junction (holes from the $$n\text{-side}$$ to the $$p\text{-side}$$ and electrons from the $$p\text{-side}$$ to the $$n\text{-side}$$).
• The net flow of this drift current runs from the $$n\text{-region}$$ to the $$p\text{-region}$$ , directly opposing the diffusion path. In an unbiased equilibrium state, these two opposing current components balance each other out exactly ($$I_{\text{diffusion}} = I_{\text{drift}}$$).

Therefore, Statement II is TRUE .

Correct Option Key: Option B (Both Statement I and Statement II are true)

Solution:

Step 1: Identify the given values and parameters

• Intrinsic carrier concentration ($$n_i$$) = $$1.5 \times 10^{16} \text{ m}^{-3}$$
• Electron concentration ($$n_e$$) after doping = $$4.5 \times 10^{22} \text{ m}^{-3}$$

Step 2: Apply the Mass-Action Law

According to the law of mass action for semiconductors under thermal equilibrium, the product of the free electron concentration ($$n_e$$) and the hole concentration ($$n_h$$) is a constant and is equal to the square of the intrinsic carrier concentration ($$n_i$$):

$$n_e \cdot n_h = n_i^2$$

Step 3: Rearrange the formula to solve for hole concentration ($$n_h$$)

$$n_h = \frac{n_i^2}{n_e}$$

Step 4: Substitute the values and calculate

$$n_h = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}}$$

$$n_h = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}}$$

$$n_h = 0.5 \times 10^{10} = 5.0 \times 10^9 \text{ m}^{-3}$$

Since the majority carriers are electrons ($$n_e \gg n_h$$), this confirms the semiconductor becomes an n-type extrinsic semiconductor with a hole concentration of $$5.0 \times 10^9 \text{ m}^{-3}$$.

Correct Answer: Option A