Question

An npn BJT with η=1.2 and α=0.96, when measured at room temperature has VBE=0.67 V while having an emitter current of IE=5.2 mA. If the VBE becomes 0.75 V, what would be the new emitter current (express the result in mA units)?

Answer 1

To find the new emitter current with a VBE of 0.75V in an npn BJT, use the formula I(net) = I(o) * (e^(VBE/(kBT)) - 1), where I(net) is the net current, I(o) is the current with no applied voltage, VBE is the base-emitter voltage, k is the Boltzmann constant, B is the base current, and T is the temperature. Substituting the given values and calculating the expression, the new emitter current is approximately 6.57 mA.

Step-by-step explanation:

To find the new emitter current with a VBE of 0.75V, we will use the formula for the net current in a BJT:

I(net) = I(o) * (e^(VBE/(kBT)) - 1), where I(net) is the net current, I(o) is the current with no applied voltage, VBE is the base-emitter voltage, k is the Boltzmann constant, B is the base current, and T is the temperature.

Substituting the given values, we have: I(net) = 5.2 mA * (e^(0.75V/(1.2 * 8.617333262145 * 10^-5 eV/K)) - 1). Calculating this expression, we find that the new emitter current is approximately 6.57 mA.