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Lunat1c
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Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find IE (the dc emitter current). For some reason, the book I'm using, when calculating Vb which is equivalent to VTH took in consideration the resistances in the emitter which are in parallel with R2 (the given solution is also uploaded, check figure 2 please).
I know this makes sense because the current passing through R1 and R2 is not equal since some current goes through the base. However for the thevenin equivalent circuit shouldn't the part where there's R1 and R2 be open circuit?
If i were to calculate IE myself I'd do it as follows:
VTH = (15/25.1k)*5.1k = 3.0478V
RTH = R1||R2 = 4064ohms
Considering the input loop of the first stage, VB = VTH= IBRTH + VBE + IE(RE1 + RE2)
Using this i got IE = 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 R1 and R2 is not equal since some current goes through the base. However for the thevenin equivalent circuit shouldn't the part where there's R1 and R2 be open circuit?
If i were to calculate IE myself I'd do it as follows:
VTH = (15/25.1k)*5.1k = 3.0478V
RTH = R1||R2 = 4064ohms
Considering the input loop of the first stage, VB = VTH= IBRTH + VBE + IE(RE1 + RE2)
Using this i got IE = 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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