Comments on Thévenins Theorem for Transistor Circuit
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Thévenins Theorem for Transistor Circuit
I am calculating the base current of transistor Q1. To simplify the circuit equations, I applied the Thévenin's theorem twice and converted V1, R1 and R2 into Vth and Rth, as shown in the image.
\[ \begin{align} V_{th} &= V_1 \times (R_2/R_1+R_2) && & R_{th} &= R_1 \parallel R_2 \\[1 em] V_{th2} &= V_{th} \times (R_7/R_{th}+R_3+R_7) && & R_{th2} &= R_7 \parallel (R_{th}+R_3) \\[1 em] V_a &= \pu{0.7V} + I_b \times (h_{fe}+1) \times \pu{200\Omega} \end{align} \]
I can accurately calculate $I_b = \pu{2.09 mA}$ by using Kirchoff's voltage law, but I'm getting an inaccurate value ($I_b= \pu{213 \mu A}$) by applying Thévenin's theorem.
Is Thévenin's theorem being applied incorrectly for the circuit?
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Is the Thevenin's Theorem being applied incorrectly for the circuit ?
It's being applied inaccurately. The final Thevenin voltage should be 2.782 volts (rather than your calculated value of 2.61 volts). I calculate the Thevenin resistance to be 759 ohms (near enough to yours so that it doesn't matter). Double check with simulator for the Thevenin voltage: -
I can accurately calculate the Ib = 2.09mA by using Kirchoff's Voltage law
Then you have miscalculated because, if 2 mA were flowing into the base then, due to hFE (~40), the emitter current would be about 80 mA and, if 80 mA were flowing through the emitter resistor of 200 Ω then there would be an emitter voltage of 16 volts and clearly, that has to be nonsense.
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