# Applying Thevenin's theorem between 3 points

I'm studying about transistor polarisation and got confused by this application of Thevenin's theorem, can someone please explain why eT is between B and M and why Vcc stayed even though it's not removed when finding eT? I'm not familiar with using Thevenin's theorem to a charge that's connected to the circuit via 3 points.

• I believe you can find the answer here. allaboutcircuits.com/textbook/semiconductors/chpt-4/…. – ASWIN VENU Aug 11 '20 at 3:46
• As to why eT is between B and M, because biasing a bjt calls for a forward biased Base-Emitter junction.Vcc stays the same for biasing the Collector-Base junction in reverse bias. (Vcc>eT) – ASWIN VENU Aug 11 '20 at 3:50

$$e_T = \frac{R_{B1}}{R_{B1}+R_{B2}}\cdot V_{CC}$$ $$R_T = R_{B1}//R_{B2}$$
$$\e_T\$$ is the output voltage of the boxed part if Q1 is not connected, or in that case $$\v_B = e_T\$$. After connecting Q1, the voltage $$\v_B\$$ will be partially pulled to GND via $$\B\$$ -> Base-Emitter-diode -> RE -> GND. Hence, $$e_T > v_B > v_M=0V$$