# Trouble analyzing BJT circuits when we change β from very high to 100

For example we have this circuit:

Originally we assumed β to be very high and we assume VBE = 0.7. So for this, we have IB = 0 and therefore V1 = -0.7V and V2 = 3.4V.

Now that we have β at 100, IV1 is still 2*10-3 but now IV2 = αIV1 where α = 0.99 so we get IV2 = 1.98*10-3.

We also have now instead of IB = 0, IB = IV2/β = 1.98*10-5 and therefore VB = IB *47k = 0.9306

So then, V1 = VB - VBE = 0.2306V. I don't have any other of what the answers should be but I do know that V1 should be -1.63V and I'm not entirely sure where I've gone wrong.

• Vb is negative so vb-vbe is -0.93 - 0.7 – Dorian Nov 2 '18 at 12:42

You forget about $$\R_B = 47\textrm{k}\Omega\$$ resistor and the base current.
The base current is $$\I_B = \frac{2\textrm{mA}}{101} = 19.8\mu\textrm{A}\$$ therefore the voltage drop across $$\R_B\$$ resistor is $$\VR_B = 0V - 19.8\mu\textrm{A}\cdot47\textrm{k}\Omega = -0.930\textrm{V}\$$
and the voltage at the emitter is $$\-0.930\textrm{V} - 0.7\textrm{V} = -1.63\textrm{V}\$$