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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. Circuit

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  • \$\begingroup\$ I believe you can find the answer here. allaboutcircuits.com/textbook/semiconductors/chpt-4/…. \$\endgroup\$ – ASWIN VENU Aug 11 '20 at 3:46
  • \$\begingroup\$ 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) \$\endgroup\$ – ASWIN VENU Aug 11 '20 at 3:50
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The images you gave are IMO a kind of roundabout way of the following:

schematic

simulate this circuit – Schematic created using CircuitLab

This circuit is equivalent to the one shown in your figures. And calculating the Thevenin-equivalent of the boxed part will get you your answers:

$$e_T = \frac{R_{B1}}{R_{B1}+R_{B2}}\cdot V_{CC}$$ $$R_T = R_{B1}//R_{B2}$$

\$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, $$e_T > v_B > v_M=0V$$

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  • \$\begingroup\$ Can you explain why this circuit is equivelant to the original? \$\endgroup\$ – Tarmius Aug 11 '20 at 15:52
  • \$\begingroup\$ Splitting the voltage source in two separate voltage sources will still keep both nodes at the voltage "VCC". The only difference is that the current is split into two parts, but the author doesn't care about that. \$\endgroup\$ – Sven B Aug 12 '20 at 13:38

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