A symmetrical p^+-n-p^+ bipolar transistor has the following properties: Emitter Base A = 10^-4 cm² Na = 10^17 Nd = 10^15 cm^-3 Wb = 1 μm τn = 0.1 μs τp = 10 μs μp = 200 μn = 1300 cm² V-s μn = 700 μp = 450 cm² /V-s (a) Calculate the saturation current IES = ICS. (b) With VEB = 0.3V and

Question

A symmetrical p[latex]^+[/latex]-n-p[latex]^+[/latex] bipolar transistor has the following properties:

Emitter Base

[latex]\ \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 [/latex]

(a) Calculate the saturation current [latex]I_{ES} = I_{CS}[/latex].
(b) With [latex]V_{EB} = 0.3 V and V_{CB} = -40 V[/latex], calculate the base current [latex]I_B[/latex], 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 [latex]L_n[/latex].

Accepted Answer

In the base,

[latex]p_n=n_i^2/n_n=(1.5 imes 10^{10})^2/10^{15}=2.25 imes 10^5 \ \space \ D_p=450(0.0259)=11.66, \space L_p=(11.66 imes 10^{-5})^{1/2}=1.08 imes 10^{-2} \ \space \ W_b/L_p=10^{-4}/1.08 imes 10^{-2}=9.26 imes 10^{-3} \ \space \I_{ES}=I_{CS}=qA(D_p/L_p)p_nctnh(W_b/L_p) \ \space \ =(1.6 imes 10^{-19})(10^{-4})(11.66/1.08 imes 10^{-2})(2.25 imes 10^5)ctnh \space 9.26 imes 10^{-3} \ \space \ =4.2 imes 10^{-13}A \ \space \ \Delta p_E=p_ne^{qV_{EB}/kT}, \Delta p_C\approx 0 \ \space \ \Delta p_E=2.25 imes 10^5 imes e^{(0.3/0.0259)}=2.4 imes 10^{10} \ \space \ I_B=qA(D_p/L_p)\Delta p_E anh(W_b/2L_p) [/latex]

or

[latex]I_B=\frac{Q_b}{ au_p}=qAW_b\Delta p_E/2 au_p=1.9 imes 10^{-12}A[/latex]

In the emitter,

[latex]D_n=700(0.0259)=18.13\ \space \ L_n=(18.13 imes 10^{-7})^{1/2}=1.35 imes 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} anh\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 imes 1.08 imes 10^{-2} imes 10^{15}}{11.66 imes 1.35 imes 10^{-3} imes 10^{17}} anh9.26 imes 10^{-3}\Big]^{-1}=0.99885 \ \space \ B=sech\frac{W_b}{L_p}=sech9.26 imes 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[/latex]

Question 7.4: A symmetrical p^+-n-p^+ bipolar transistor has the following...

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