(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given, β=100, Is = 6E-16

See attachment.

a) Find min value of Rb such that the BJT is in active region

b) For the Rb found, what is the base-collector voltage if β=200

2. Relevant equations

3. The attempt at a solution

Assuming, VCE min for active region operation as 0.75V, I got Ic = 0.87mA

since β=100, Ib = 0.0087mA

Vbe = Vt ln (Ic/Is) = 0.728V

using Ib = (Vth-Vbe)/Rth {thevenin equivalent}, I got Rbmin as 7K. The voltage divider voltage is 0.75v and Rth is 2.1k.

But now, when I plug in Ib = Vth-Vbe/Rth, I get 0.00104mA. Which Ib is correct?

I can't figure out how to calculate the beta change. Where do I use the beta?

Ib is 0.0087mA, with new beta Ic will be 1.74mA. which will make IcRc = 3.48v, which doesn't make sense.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: BJT voltage divider bias beta variation

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