Question

Consider two BJTs biased at the same collector current with area A₁ = 0.2 μm × 0.2 μm and A₂ = 300 μm × 300 μm. Assuming that all other device parameters are identical, kT/q = 26 mV, the intrinsic carrier concentration is 1 × 10¹⁰ cm^-3, and q = 1.6 × 10^-19 C, the difference between the base-emitter voltages (in mV) of the two BJTs (i.e., V_BE1 – V_BE2) is _____. 

Answer 1

Concept:

Relation between collector current and base current is given by I_C = βI_B

Where, \({I_B} \propto A\;exp^\left( {\frac{{e{V_{BE}}}}{{KT}}} \right)\)   also  \( \frac{{{I_{{C_1}}}}}{{{I_{{C_2}}}}} = \frac{{{A_1}e^\left( {\frac{{e{V_{B{E_1}}}}}{{KT}}} \right)}}{{{A_2}e^\left( {\frac{{e{V_{B{E_2}}}}}{{KT}}} \right)}}\)

Application:

Assuming all other parameter is same for both of the BJT.

Now,

\( \frac{{{I_{{C_1}}}}}{{{I_{{C_2}}}}} = \frac{{{A_1}e^\left( {\frac{{e{V_{B{E_1}}}}}{{KT}}} \right)}}{{{A_2}e^\left( {\frac{{e{V_{B{E_2}}}}}{{KT}}} \right)}}\)

\(1 = \frac{{{{\left( {0.2 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {300 \times {{10}^{ - 6}}} \right)}^2}}}.e^\left[ {\frac{e}{{KT}}\left( {{V_{B{E_1}}} - {V_{B{E_2}}}} \right)} \right]\)

\(2.25 \times {10^6} = e^\left[ {\frac{{{V_{B{E_1}}} - {V_{B{E_2}}}}}{{{V_T}}}} \right]\)

\(ln\left( {2.25 \times {{10}^6}} \right) = \frac{{{V_{B{E_1}}} - {V_{B{E_2}}}}}{{{V_T}}}\)

\(14.626 \times {V_T} = {V_{B{E_1}}} - {V_{B{E_2}}}\)

\(\therefore{V_{B{E_1}}} - {V_{B{E_2}}} = 14.626 \times 26\;mV \)

= 380 mV