How To Draw Ac Equivalent Circuit

2 Answers 2

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In order to find the operating point for your circuit you have to perform the steps 1-5 given above. The operating point is equivalent to the DC analysis of the circuit. Inductors and capacitors are therefore removed by a short or open circuit, respectively.

For this DC solution the circuit is linearized. Your transistor is replaced by a simple transconductance (voltage-controlled current source). The AC components are of course required for the AC model of the circuit and need to be included again.

To answer your questions:

C1 is left in the circuit because we need the AC model of the circuit
The transistor is now a voltage-controlled current source, with a transconductance gm
Vpi is simple the controlling voltage (the base-emitter voltage) of the transistor
Re is the emitter resistor which is given as 1kOhm

answered Mar 13 '16 at 11:21

MarioMario

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\$\begingroup\$ Is Ro always in parallel with the transconductance and calculated using beta/gm? Second, if this was PNP, then the current source direction would just be flipped right? \$\endgroup\$

Mar 13 '16 at 20:14
\$\begingroup\$ Yes, Ro always is in parallel with the transconductance. For a PNP the model would be same. The base-emitter voltage is reversed and the current direction as well, therefore the same model results. If the current source was flipped the direction of Vpi would have to be flipped as well. \$\endgroup\$

Mar 13 '16 at 20:26

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(1) C1 is not necessary because the circuit is an "equivalent diagram" that is valid for small signals only (no dc values within the circuit).

(2) The product (gm * Vpi) is identical to the product (beta * Ib) because of beta/gm=Vpi/Ib=r,pi

(3) In the transconductance model the controlling voltage is the base-emitter voltage Vbe=Vpi.

(4) The emitter resistor is given and fulfills Ohms law: Re=Ve/Ie (DC values!).

answered Mar 13 '16 at 16:02

LvWLvW

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