DC Analysis of BJT Amplifier Circuits

Using the common-emitter amplifier circuit shown in the figure as an example, the use of equivalent circuits assists with analyzing circuits. DC analysis of a common-emitter amplifier circuit begins with determining the dc bias values and then removing coupling and bypass capacitors, the load resistor, and the signal source to produce a dc equivalent circuit by applying Thevenin’s theorem and Kirchoff’s voltage law.

AC analysis of a common-emitter amplifier circuit begins by recognizing the capacitive reactance (XC) remains very low at the signal frequency. By considering XC as equal to zero, reducing the circuit to an ac equivalent circuit requires replacing the three capacitors in the circuit with effective shorts. Then, the analysis continues by replacing the dc source with ground. From the perspective of ac analysis, a dc voltage source has an internal resistance of zero ohms.

Since no ac voltage can develop across the dc source, it serves as an ac ground. Electrically, the ac ground and actual ground exist at the same point. All this reduces the equivalent circuit to three resistors and the transistor. Connecting an ac voltage source to the input of the circuit. Because the AC source voltage has an internal resistance of zero ohms, the source voltage appears at the base of the transistor.

Finding the ac signal voltage at the transistor base requires combining the source resistance (RS), the bias resistance, and the ac input resistance at the base to produce the total input resistance (Rin(tot) ) seen by the ac source connected to the input. Using the voltage divider formula, the signal voltage at the base of the transistor (Vb) equals:

         V_b = (R_in(tot) / R_S + R_in(tot) ) V_S

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