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Chapter 6

Q. 6.4

Ebers-Moll Equations

Calculate the reverse-bias voltage present on the base-emitter junction of an npn BJT when the emitter is open-circuited and a reverse bias is placed on the base-collector junction. Assume \alpha_ F = 0.98, \alpha_R = 0.70, I_{CS} = 1 \times 10^{-13} A, I_{ES} = 7.14 \times 10^{-14} A.

Step-by-Step

Verified Solution

For this bias condition, the collector current is I_{CB0} as given by Equation 6.4.15. Because the emitter current is zero, Equation 6.4.10a establishes that

I_{C B 0}=\left.I_{C}\right|_{I_{E}=0}=I_{C S}\left(1-\alpha_{F} \alpha_{R}\right)                                        (6.4.15)

I_E=-I_F+\alpha _RI_R                                            (6.4.10a)

I_F=\alpha _RI_R\approx -\alpha _RI_{CS}=I_{ES}\left(\exp \frac{qV_{BE}}{kT} -1 \right)

where we have used I_R\approx -I_{CS} .

Using the reciprocity relationship (Equation 6.4.8), we solve for V_{BE}

\alpha _FI_{ES}\equiv \alpha_RI_{CS}\equiv I_S                                            (6.4.8)

V_{BE}=\frac{kT}{q} \ln(1-\alpha_F)=-0.10 \ V

at T = 300 K.

The reverse bias on the base-emitter junction depends only on \alpha_F for this case of an open-circuited emitter. This dependence occurs because the bias is established by balancing the linking current with the current returned through the back-biased, base-emitter junction diode so that the emitter current is zero.