Consider an npn transistor operating in the linear region at 450C. If VBE is increased by 15mV, by what factor will the collector current increase?

To determine the factor by which the collector current will increase when VBE is increased by 15mV in an npn transistor operating in the linear region at 450C, we need to consider the relationship between VBE and collector current (IC).

In an npn transistor, the collector current, IC, is related to the base-emitter voltage, VBE, through the exponential relationship given by the Ebers-Moll equation:

IC = IC0 * exp(VBE / (VT * n))

where IC0 is the collector current at the reference condition (VBE = 0), VT is the thermal voltage (which depends on the temperature), and n is the ideality factor.

Since the question mentions that the transistor is operating at 450C, we need to consider the impact of temperature on VT. The thermal voltage (VT) is given by:

VT = (k * T) / q

where k is the Boltzmann constant, T is the absolute temperature in Kelvin, and q is the electronic charge.

At room temperature (25C or 298K), the value of VT is approximately 25.85 mV.

Now, to find the factor by which the collector current will increase when VBE is increased by 15mV, we need to calculate the new IC with the increased VBE and compare it to the original IC.

Let's assume an initial IC, IC1, corresponding to the initial VBE, and a new IC, IC2, corresponding to the increased VBE.

To find the increase in IC, we can calculate the original IC by plugging in the initial VBE into the equation. Then, we can calculate the new IC by adding 15 mV to the initial VBE and plugging it into the equation. Finally, we can find the factor by which IC increased using the formula:

Factor = IC2 / IC1

Please provide the values of IC0 (the collector current at VBE = 0), n (the ideality factor), and the initial VBE to proceed with the calculation.