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For example we have this circuit:

enter image description here

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.

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

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