To determine the maximum Collector to Base voltage ($V_{CB}$) that ensures the low-level injection condition is not violated in a PNP BJT biased in the inverse active region, we need to ensure that the excess minority carrier concentration is much smaller than the majority carrier concentration in the base.

Given Data:

• $N_E = 2 \times 10^{18} \text{ cm}^{-3}$ (Emitter doping concentration)
• $N_B = 5 \times 10^{17} \text{ cm}^{-3}$ (Base doping concentration)
• $N_C = 7 \times 10^{15} \text{ cm}^{-3}$ (Collector doping concentration)
• Room temperature $T = 300 \text{ K}$
• The intrinsic carrier concentration, $n_i$ at 300 K is approximately $1.5 \times 10^{10} \text{ cm}^{-3}$.

Steps to Solve:

1. Intrinsic Carrier Concentration: The intrinsic carrier concentration $n_i$ at 300 K is given as $1.5 \times 10^{10} \text{ cm}^{-3}$.

2. Equilibrium Minority Carrier Concentration: For the base region (which is the region of interest since the BJT is biased in inverse active mode), the minority carrier concentration (electrons in a PNP BJT base) under equilibrium conditions is given by: $n_{B0} = \frac{n_i^2}{N_B}$ Substituting the given values: $n_{B0} = \frac{(1.5 \times 10^{10})^2}{5 \times 10^{17}} = \frac{2.25 \times 10^{20}}{5 \times 10^{17}} = 4.5 \times 10^2 \text{ cm}^{-3}$

3. Low-Level Injection Condition: Low-level injection condition requires that the excess minority carrier concentration in the base, $\Delta n_B$, should be much less than the majority carrier concentration in the base ($p_B = N_B$). Mathematically: $\Delta n_B \ll N_B$

4. Excess Minority Carrier Concentration: The excess minority carrier concentration $\Delta n_B$ due to the applied $V_{CB}$ can be expressed using the Shockley equation: $\Delta n_B = n_{B0} \left(e^{\frac{V_{CB}}{V_T}} - 1\right)$ where $V_T = \frac{kT}{q}$ is the thermal voltage at room temperature (approximately 26 mV at 300 K).

5. Maximum $V_{CB}$: To ensure low-level injection, $\Delta n_B$ should be less than 10% of the majority carrier concentration $N_B$, i.e., $\Delta n_B < 0.1 \times N_B = 0.1 \times 5 \times 10^{17} = 5 \times 10^{16} \text{ cm}^{-3}$ Substituting for $\Delta n_B$: $4.5 \times 10^2 \left(e^{\frac{V_{CB}}{0.026}} - 1\right) < 5 \times 10^{16}$ Simplifying: $e^{\frac{V_{CB}}{0.026}} - 1 < \frac{5 \times 10^{16}}{4.5 \times 10^2} = 1.11 \times 10^{14}$ $e^{\frac{V_{CB}}{0.026}} < 1.11 \times 10^{14}$ Taking the natural logarithm on both sides: $\frac{V_{CB}}{0.026} < \ln(1.11 \times 10^{14}) \approx 32.23$ $V_{CB} < 32.23 \times 0.026 \approx 0.84 \text{ V}$

Answer:

The maximum Collector to Base voltage ($V_{CB}$) that ensures low-level injection is not violated is approximately 0.84 V.

Hence, the correct answer is (a) 0.84 V.

Would you like more details or have any questions?

Here are 5 relative questions:

1. What happens in a BJT when the low-level injection condition is violated?
2. How does the minority carrier concentration change with increasing $V_{CB}$?
3. What is the impact of temperature on the intrinsic carrier concentration $n_i$?
4. How would the calculation change if the BJT was in forward active mode instead of inverse active mode?
5. What are the typical applications where low-level injection conditions are critical in semiconductor devices?

Tip: Remember that ensuring low-level injection is critical for maintaining linear operation in many semiconductor devices.