Question 7.4: A symmetrical p^+-n-p^+ bipolar transistor has the following...
A symmetrical p^+-n-p^+ bipolar transistor has the following properties:
Emitter Base
\\ \space \\ A = 10^{-4} cm² \quad \quad \quad \quad \quad N_a = 10^{17} \quad \quad \quad \quad \quad \quad Nd = 10^{15} cm^{-3} \\ \space \\ W_b = 1 μm \quad \quad \quad \quad \quad τ_n = 0.1 μs\quad \quad \quad \quad \quad \quad τ_p = 10 μs \\ \space \\ \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad μ_p = 200\quad \quad \quad \quad \quad \quad μ_n = 1300 cm² >V-s \\ \space \\ \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad μn = 700 \quad \quad \quad \quad \quad \quad μ_p = 450 cm² /V-s
(a) Calculate the saturation current I_{ES} = I_{CS}.
(b) With V_{EB} = 0.3 V and V_{CB} = -40 V, calculate the base current I_B, assuming perfect emitter injection efficiency.
(c) Calculate the base transport factor B, emitter injection efficiency γ, and amplification factor β, assuming that the emitter region is long compared with L_n.
In the base,
p_n=n_i^2/n_n=(1.5 \times 10^{10})^2/10^{15}=2.25 \times 10^5 \\ \space \\ D_p=450(0.0259)=11.66, \space L_p=(11.66 \times 10^{-5})^{1/2}=1.08\times 10^{-2} \\ \space \\ W_b/L_p=10^{-4}/1.08\times 10^{-2}=9.26\times 10^{-3} \\ \space \\I_{ES}=I_{CS}=qA(D_p/L_p)p_nctnh(W_b/L_p) \\ \space \\ =(1.6\times 10^{-19})(10^{-4})(11.66/1.08\times 10^{-2})(2.25\times 10^5)ctnh \space 9.26\times 10^{-3} \\ \space \\ =4.2\times 10^{-13}A \\ \space \\ \Delta p_E=p_ne^{qV_{EB}/kT}, \Delta p_C\approx 0 \\ \space \\ \Delta p_E=2.25\times 10^5\times e^{(0.3/0.0259)}=2.4\times 10^{10} \\ \space \\ I_B=qA(D_p/L_p)\Delta p_E \tanh(W_b/2L_p)
or
I_B=\frac{Q_b}{\tau_p}=qAW_b\Delta p_E/2\tau_p=1.9\times 10^{-12}A
In the emitter,
D_n=700(0.0259)=18.13\\ \space \\ L_n=(18.13\times 10^{-7})^{1/2}=1.35\times 10^{-3}\\ \space \\ I_{En}=\frac{qAD_n^E}{L_n^E}n_p^Ee^{qV_{EB}/kT} \\ \space \\ I_{Ep}=\frac{qAD_p^B}{L_p^B}p_n^Bctnh\frac{W_b}{L_p^B}e^{qV_{EB}/kT} \\ \space \\ \gamma =\frac{I_{Ep}}{I_{En}+I_{Ep}}=\Big[1+\frac{I_{En}}{I_{Ep}}\Big]^{-1}\\ \space \\ \gamma =\Big[1+\frac{D_n^E/L_n^En_p^E}{D_p^B/L_p^Bn_n^B}\tanh\frac{W_b}{L_p}\Big]^{-1} \quad \quad \Big(use \space \frac{n_p^E}{p_n^B}=\frac{n_n^B}{p_p^E}\Big) \\ \space \\ =\Big[1+\frac{18.13\times 1.08\times 10^{-2}\times 10^{15}}{11.66\times 1.35\times 10^{-3}\times 10^{17}}\tanh9.26\times 10^{-3}\Big]^{-1}=0.99885 \\ \space \\ B=sech\frac{W_b}{L_p}=sech9.26\times 10^{-3}=0.99996 \\ \space \\ \alpha=B\gamma =(0.99885)(0.99996)=0.9988 \\ \space \\ \beta=\frac{\alpha}{1-\alpha}=\frac{0.9988}{0.0012}=832