Electronic Devices Test 4

Concept:

ICEO is the reverse leakage current in the common-emitter configuration of BJT when the base is open.

ICBO is the reverse leakage current in the common-base configuration of BJT when the emitter is open.

Also, ICEO > ICBO

And they are related by the relation:

ICEO = (1 + β) ICBO

Total collector current is given by:

IC = βIB + ICBO(β + 1)

Calculation:

I_CEO = 410 μA, I_CBO = 5 μA, I_B = 30 μA

I_C­ = βI_B + (1 + β) I_CBO

I_CEO = (1 + β) I_CBO

410 = (1 + β)5

β = 81

I_C = 81 × 30 + 82 × 5

I_C = 2.84 mA

Concept:

The built-in potential is defined as

\({V_{bi}} = \frac{{kT}}{q}\;In\;\left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)      ---(1)

Also, the carrier majority concentration is related to the Fermi-level as:

\({N_A} = {n_i}\;{e^{\frac{{\left( {{E_i} - {E_F}} \right)}}{{kT}}\;}}\)

\({N_p} = {n_i}{e^{\frac{{\left( {{E_F} - {E_i}} \right)}}{{kT}}}}\)

Ei = Intrinsic Fermi level position

EF = Fermi Level of the doped semiconductor

The above two equations can be rearranged as:

\({\left( {{E_i} - {E_F}} \right)_{p - side}} = kT\;In\;\left( {\frac{{{N_A}}}{{{n_i}}}} \right)\)

\({\left( {{E_F} - {E_i}} \right)_{n - side}} = kT\;In\;\left( {\frac{{{N_D}}}{{{n_i}}}} \right)\)

So, the built-in potential from Equation (1) can also now be defined as:

\({V_{bi}} = \frac{1}{q}\left( {{{\left( {{E_i} - {E_F}} \right)}_{p - side}} + {{\left( {{E_F} - {E_i}} \right)}_{n - side}}} \right)\)   ---(1)

Application:

The built-in potential barrier of the collector base will be:

\({V_{bi\left( {CB} \right)}} = \frac{{kT}}{q}\;ln\left( {\frac{{{N_C}}}{{{n_i}}}} \right)\)

Similarly, for the emitter-base junction, the built-in potential will be:

\({V_{bi\left( {EB} \right)}} = \frac{{kT}}{q}\;In\left( {\frac{{{N_E}}}{{{n_i}}}} \right)\)

Since the base is the common region between the two, the potential difference between the collector and emitter is obtained using Equation (1) as:

\(\left| {{\rm{\Delta }}{V_{CB}}} \right| = \frac{1}{q}\left[ {{{\left( {{E_i} - {E_f}} \right)}_{emitter}} - {{\left( {{E_i} - {E_f}} \right)}_{collector}}} \right]\)

\(= \frac{{kT}}{q}\left[ {ln\left( {\frac{{{N_{AE}}}}{{{n_i}}}} \right) - ln\left( {\frac{{{N_{AC}}}}{{{n_i}}}} \right)} \right]\)

\(= \frac{{kT}}{q}\;ln\;\left( {\frac{{{N_{AE}}}}{{{N_{AC}}}}} \right)\)

\(\left| {{\rm{\Delta }}{V_{CB}}} \right| = 0.026 \times \;ln\;\left( {\frac{{5 \times {{10}^{17}}}}{{{{10}^{14}}}}} \right)\)