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Electronics · Ch 3 — Transistor Amplifiers

DC and AC equivalent circuit of amplifiers

3.4

DC and AC equivalent circuit of amplifiers

In a working transistor amplifier, direct current from the biasing supply and alternating current from the signal being amplified flow through the same components at the same time. To study such a circuit cleanly, its analysis is split into two independent parts — a DC analysis that fixes the transistor's operating point, and an AC analysis that gives its signal (small-signal) performance. Mastering both equivalent circuits is essential for the Karnataka 2nd PUC Electronics board exam, so work through each one carefully.

DC analysis and the DC equivalent circuit

DC analysis considers only the DC sources and finds the steady voltages and currents that set the operating point. The DC equivalent circuit is obtained in two steps:

  1. Reduce every AC source to zero.
  2. Open-circuit every capacitor, because a capacitor blocks DC.

Applying this to the voltage-divider-biased CE amplifier of Figure 3.4.1 gives the DC equivalent of Figure 3.4.2. The supply divides across R1 and R2, and the voltage V2V_2 developed across R2 forward-biases the base–emitter junction. Writing Kirchhoff's voltage law (KVL) around the base–emitter loop gives the emitter current IEI_E, and KVL around the collector–emitter loop relates the supply to VCEV_{CE}. From these relations every DC voltage and current in the amplifier can be calculated.

Note

The textbook prints the base–emitter loop as V2=VBE−IEREV_2 = V_{BE} - I_E R_E. For its own quoted result IE=(V2−VBE)/REI_E = (V_2 - V_{BE})/R_E to follow, the loop must actually read V2=VBE+IEREV_2 = V_{BE} + I_E R_E — the divider voltage across R2 equals the base–emitter drop plus the drop across the emitter resistor. Treat the printed minus sign as a sign misprint and use the "+" form (as captured in the formula card below) when solving numerical problems.

AC analysis and the AC equivalent circuit (the re' model)

AC analysis finds how the amplifier responds to the signal alone. Its equivalent circuit is drawn in two steps:

  1. Replace every DC source by a short circuit.
  2. Replace every capacitor by a short circuit (at signal frequencies the coupling and bypass capacitors behave as shorts).

The first step gives Figure 3.4.3, in which R1 and R2 appear in parallel across the input and RC is the collector load. The second step replaces the transistor by its small-signal model, giving Figure 3.4.4, the re' model: the emitter–base junction is replaced by an AC resistance and the collector–base junction by a constant-current source. For a small AC signal the emitter diode does not rectify — it simply offers an AC resistance — while the collector diode behaves as a current source.

From the re' model, the section derives five performance quantities of the CE amplifier:

  • Input impedance ZinZ_{in} — the ratio of input voltage to input current. It is the parallel combination of the two bias resistors and the impedance looking into the base, and the impedance looking into the base works out to βre′\beta r_e', where re′r_e' is the AC emitter-junction resistance. …
Figure 1Circuit diagram of a voltage-divider-biased common-emitter (CE) transistor amplifier, the reference circuit for both DC and AC analysis.
Fig. 1 — Circuit diagram of a voltage-divider-biased common-emitter (CE) transistor amplifier, the reference circuit for both DC and AC analysis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 3.4.1 — the complete CE amplifier: supply +Vcc, base voltage-divider R1–R2, collector resistor RC, emitter resistor RE with bypass capacitor CE, input coupling capacitor C1 and output coupling capacitor C2, driven by the AC source Vin. This is the starting circuit from which the DC equivalent (Figure 3.4.2) and …

Figure 2DC equivalent circuit of the CE amplifier, with all capacitors opened and the AC source removed, showing the bias voltages and emitter current.
Fig. 2 — DC equivalent circuit of the CE amplifier, with all capacitors opened and the AC source removed, showing the bias voltages and emitter current.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 3.4.2 — with every capacitor opened and the AC source set to zero, only the DC bias path survives: the supply Vcc divides across R1 (drop V1) and R2 (drop V2), VBE is the base–emitter drop, and the emitter current IE flows through RE to ground. This circuit sets the transistor's operating poi …

Figure 3First step of the AC equivalent circuit of the CE amplifier, with the DC supply and all capacitors replaced by short circuits.
Fig. 3 — First step of the AC equivalent circuit of the CE amplifier, with the DC supply and all capacitors replaced by short circuits.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 3.4.3 — shorting the DC supply and all capacitors places R1 and R2 in parallel across the input and grounds the emitter for AC, leaving RC as the collector load. This is the intermediate step, drawn before the transistor itse …

Figure 4Small-signal re' model AC equivalent circuit of the CE amplifier, showing the base resistance beta-re' and the collector constant-current source beta-ib.
Fig. 4 — Small-signal re' model AC equivalent circuit of the CE amplifier, showing the base resistance beta-re' and the collector constant-current source beta-ib.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 3.4.4 — the re' model. The emitter–base junction is replaced by the AC resistance βre′\beta r_e', into which the base current ib flows, and the collector–base junction by a constant-current source βib\beta i_b feeding the collector load RC. The input impedance Zi and the impedance looking into the base are marked. Every gain and impedan …

Formula 5DC bias equations of the CE amplifier

Voltage across R2 (forward-biases the base–emitter junction):

V2=VCCR1+R2 R2V_2 = \frac{V_{CC}}{R_1 + R_2}\,R_2

Emitter current from the base–emitter loop (using the corrected V2=VBE+IEREV_2 = V_{BE} + I_E R_E):

IE=V2−VBEREI_E = \frac{V_2 - V_{BE}}{R_E} …

Formula 6Input and output impedance of the CE amplifier (re' model)

Input impedance (ratio of input voltage to input current):

Zin=viniin=R1 ∥ R2 ∥ Zin(base)Z_{in} = \frac{v_{in}}{i_{in}} = R_1 \,\|\, R_2 \,\|\, Z_{in(base)}

Impedance looking into the base:

Zin(base)=VT βIC=βre′(strictly (1+β)re′≈βre′)Z_{in(base)} = \frac{V_T\,\beta}{I_C} = \beta r_e' \quad (\text{strictly } (1+\beta)r_e' \approx \beta r_e')

AC emitter-junction resistance, with VT=26 mVV_T = 26\ \text{mV} (silicon) and 52 mV52\ \text{mV} (germanium):

re′=VTIEr_e' = \frac{V_T}{I_E}

Output impedance: …

Formula 7Voltage, current and power gain of the CE amplifier (re' model)

Voltage gain (the negative sign is the 180° phase reversal):

Av=−Zore′=−RCre′;Av=−RC ∥ RLre′ (with load)A_v = -\frac{Z_o}{r_e'} = -\frac{R_C}{r_e'} \quad ;\quad A_v = -\frac{R_C \,\|\, R_L}{r_e'} \ \text{(with load)}

Current gain: …

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