Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find I(adsbygoogle = window.adsbygoogle || []).push({}); _{E}(the dc emitter current). For some reason, the book I'm using, when calculating V_{b}which is equivalent to V_{TH}took in consideration the resistances in the emitter which are in parallel with R_{2}(the given solution is also uploaded, check figure 2 please).

I know this makes sense because the current passing through R_{1}and R_{2}is not equal since some current goes through the base. However for the thevenin equivalent circuit shouldn't the part where there's R_{1}and R_{2}be open circuit?

If i were to calculate I_{E}myself I'd do it as follows:

V_{TH}= (15/25.1k)*5.1k = 3.0478V

R_{TH}= R1||R2 = 4064ohms

Considering the input loop of the first stage, V_{B}= V_{TH}= I_{B}R_{TH}+ V_{BE}+ I_{E}(R_{E1}+ R_{E2})

Using this i got I_{E}= 5.85mA which is quite close to what the book gave as an answer however I'd like to confirm my doubts.

Thanks in advance

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# Homework Help: Thevenin's Theorem

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