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In the following circuit:

enter image description here

Ib is given as:

enter image description here

I don't understand how they arrived at this. I thought it would be a voltage divider between Rsig and Rb||Rbase divided by Rbase:

enter image description here

Can anyone explain? Thanks.

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There is no emitter resistor to take current so this looks unanswerable to me. Where did the circuit come from? – Andy aka Aug 4 '14 at 10:57
@Andyaka Sedra/Smith Microelectronic Circuits – Mosho Aug 4 '14 at 11:09
I don't have that book but does it explain what this circuit is meant to be all about? – Andy aka Aug 4 '14 at 11:12
There's no way that circuit is in the Sedra Smith without some explanations. – Vladimir Cravero Aug 4 '14 at 11:32
@VladimirCravero I never said there were no explanations for the circuit, but the expression was given with one that wasn't sufficient for my understanding. – Mosho Aug 4 '14 at 11:37
up vote 4 down vote accepted

I assume there is a DC path for emitter current that isn't shown and, for some reason, doesn't significantly change the AC circuit.

The first equation is correct. Looking out of the base, there is Thevenin signal source with Thevenin voltage

$$V_{tb} = V_{sig}\frac{R_B}{R_B + R_{sig}}$$

and Thevenin reistance

$$R_{tb} = R_B||R_{sig}$$

The signal base current is then given by

$$I_b = \frac{V_{tb}}{R_{tb} + \left( \beta + 1\right)\left(r_e + \frac{1}{sC_E} \right)} = V_{sig}\frac{R_B}{R_B + R_{sig}}\frac{1}{R_B||R_{sig} + \left( \beta + 1\right)\left(r_e + \frac{1}{sC_E} \right) }$$

Your approach should give the same equation after some algebra.

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The shown circuit is a so called "small-signal equivalent ac circuit". That means: it contains only those element which contribute to the ac behaviour of the circuit. This also is the reason, that the collector path is connected to ground rather than to a positive supply voltage.

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