Products

## Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

## Holooly Tables

All the data tables that you may search for.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Help Desk

Need Help? We got you covered.

## 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.

## 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.