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Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find I_{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} = R1R2 = 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
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} = R1R2 = 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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