Question

An npn BJT having reverse saturation current I_S = 10^-15 A is biased in the forward active region with V_BE = 700 mV. The thermal voltage (V_T) is 25 mV and the current gain (β) may vary from 50 to 150 due to manufacturing variations. The maximum emitter current (in µA)  is  _______.

Answer 1

Concept:

Collector current (I_C) can be calculated as:

\({I_C} = {I_s}{e^{\frac{{{V_{BE}}}}{{{V_T}}}}}\)     …(1)

where,

I_S = Reverse Saturation Current

V_BE = Base-emitter biasing voltage

V_T = Thermal voltage

& Emitter current (I_E) is expressed as:

\({I_E} = \frac{{\left( {\beta + 1} \right)}}{\beta }{I_C}\)     …(2)

Where,

β = Current gain

I_C = Collector current

Calculation:

Given I_S = 10^-15 A, V_T = 25 mV and V_BE = 700 mV.

Using Equation (1),

I_C = 10^-15 × e ^(700/25) A

\(\;{I_E} = \left( {1 + \frac{1}{\beta }} \right){I_C}\)

I_E will be maximum when β is minimum.

So when β = 50,

\({\left( {{I_E}} \right)_{max}} = \left( {1 + \frac{1}{{50}}} \right) \times {10^{ - 15}} \times {e^{\left( {700/25} \right)\;}} \)  

\(= 1475 \times {10^{ - 6}}A = 1475\mu A\)

Hence it is the required emitter current.