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Chapter 4 โ€ข Theory & Derivations

Transistor Amplifiers: Small-Signal, Multistage & Power Stages

Exhaustive analysis of small-signal amplifiers using hybrid h-parameters; high- and low-frequency cutoffs and Bode plots; multistage RC-coupled, transformer, and direct-coupled cascaded systems; Class A, B, AB, and C power amplifier topologies, push-pull configurations, theoretical efficiency limits, and crossover distortion remedies.

ยง4.1 Small-Signal Analysis & Hybrid h-Parameter Model of CE BJT

1. Two-Port Network Formalism and Hybrid h-Parameters

For small alternating signals superimposed on DC quiescent operating points, a bipolar junction transistor in Common Emitter (CE) configuration is modeled linearly as a two-port network relating input voltage $v_b$, input current $i_b$, output current $i_c$, and output voltage $v_c$:

$$v_b = h_{ie} i_b + h_{re} v_c$$
$$i_c = h_{fe} i_b + h_{oe} v_c$$

The four parameters are precisely defined under AC short-circuit and open-circuit conditions:

  • $h_{ie} = \left.\frac{v_b}{i_b}\right|_{v_c = 0}$ : Short-circuit input impedance $(\Omega)$
  • $h_{re} = \left.\frac{v_b}{v_c}\right|_{i_b = 0}$ : Open-circuit reverse voltage ratio (dimensionless, typically $\sim 10^{-4}$)
  • $h_{fe} = \left.\frac{i_c}{i_b}\right|_{v_c = 0}$ : Short-circuit forward current transfer ratio / AC current gain $(\beta_{ac})$
  • $h_{oe} = \left.\frac{i_c}{v_c}\right|_{i_b = 0}$ : Open-circuit output admittance $(\text{S} = \Omega^{-1}$, typically $\sim 20\,\mu\text{S}$)

2. Simplified CE Hybrid Model & Gain Derivations

In standard practical audio and RF amplifier design, $h_{re} \approx 0$ (reverse feedback negligible) and $h_{oe} R_L \ll 0.1$ (output conductances much smaller than load conductance). The circuit simplifies to an input resistor $h_{ie}$ and a dependent current source $h_{fe} i_b$.

With an effective AC collector load $R_L' = R_C \parallel R_L$:

$$v_{out} = - i_c R_L' = - h_{fe} i_b R_L'$$
$$v_{in} = i_b h_{ie}$$

Hence, the Voltage Gain $A_v$ is:

$$A_v = \frac{v_{out}}{v_{in}} = - \frac{h_{fe} R_L'}{h_{ie}}$$

The negative sign reflects the fundamental $180^\circ$ phase inversion between input base signal and output collector signal in a Common Emitter amplifier.

The Current Gain $A_i$, Input Impedance $Z_{in}$, and Output Impedance $Z_{out}$ are:

$$A_i = \frac{i_{out}}{i_b} = - h_{fe} \frac{R_C}{R_C + R_L}, \quad Z_{in} = h_{ie}, \quad Z_{out} \approx R_C$$

3. Frequency Response and Cutoff Frequencies

The frequency response curve of a CE amplifier exhibits three distinct regimes:

  • Low-Frequency Range ($f < f_L$): The reactances of coupling capacitors $C_1, C_2$ and emitter bypass capacitor $C_E$ ($X_C = 1/(2\pi f C)$) grow large. This introduces voltage division and degenerative negative feedback across $R_E$, reducing voltage gain at a slope of $+20\,\text{dB/decade}$ per dominant pole.
  • Midband Range ($f_L \le f \le f_H$): Coupling/bypass capacitors act as virtual short circuits ($X_C \approx 0$), while internal parasitic transistor capacitances act as virtual open circuits. The gain remains flat at maximum midband gain $A_{vm}$.
  • High-Frequency Range ($f > f_H$): The internal depletion and diffusion capacitances (base-emitter capacitance $C_\pi$ and base-collector Miller capacitance $C_\mu$) introduce shunting paths to ground, rolling off gain at $-20\,\text{dB/decade}$.

The Bandwidth ($BW$) is defined between the half-power ($-3\,\text{dB}$) points:

$$BW = f_H - f_L \approx f_H \quad (\text{since } f_H \gg f_L)$$

ยง4.2 Multistage Amplifiers: RC-Coupled, Transformer & Direct Coupling

1. Cascading Amplifiers & Decibel Gain Formalism

A single transistor stage often cannot simultaneously provide adequate voltage gain, current drive, and impedance matching. Multiple amplifier stages are therefore cascaded in series, where the output of stage $n$ serves as the input to stage $n+1$.

The overall voltage gain is the multiplicative product of individual loaded gains:

$$A_v = A_{v1} \times A_{v2} \times A_{v3} \times \dots \times A_{vn}$$

Expressed logarithmically in Decibels (dB):

$$A_v(\text{dB}) = 20 \log_{10} |A_v| = 20 \log_{10} |A_{v1}| + 20 \log_{10} |A_{v2}| + \dots + 20 \log_{10} |A_{vn}|$$

Decibels convert complicated cascading products into simple additions.

2. Comparative Analysis of Coupling Schemes

Coupling Type Frequency Response Impedance Matching Cost & Size Primary Applications
RC Coupling Excellent flat midband (audio 20 Hz โ€“ 20 kHz); rolls off at DC and very high RF Poor (collector $R_C$ shunts next stage $Z_{in}$) Extremely low cost, compact, highly reliable Audio preamplifiers, general-purpose voltage gain
Transformer Coupling Poor audio flatness; resonant peaking; zero response at DC Superior ($Z_p/Z_s = (N_p/N_s)^2$) for maximum power transfer Bulky, heavy, expensive, magnetic hum pickup RF tuned amplifiers, driver stages to low-impedance speakers
Direct Coupling Extends down to DC ($0\,\text{Hz}$); no lower cutoff frequency Moderate Minimal parts, ideal for monolithic IC fabrication Operational amplifiers, biosensors, DC instrumentation

3. Loading Effect in RC-Coupled Cascades

When stage 1 is coupled to stage 2 through capacitor $C_c$, the effective AC load seen by collector 1 is not simply $R_{C1}$, but the parallel combination of $R_{C1}$ and the input impedance of stage 2:

$$R_{L1}' = R_{C1} \parallel R_{12} \parallel R_{22} \parallel h_{ie2}$$

Because $R_{L1}' < R_{C1}$, the loaded gain $A_{v1}$ is significantly smaller than the open-circuit gain of an isolated stage. Neglecting this loading effect produces massive design errors.

ยง4.3 Power Amplifiers: Classification (Class A, B, AB, C) & Efficiency Limits

1. Large-Signal Operation & Conduction Angle Classifications

Unlike small-signal voltage amplifiers that handle millivolt signals, Power Amplifiers (large-signal amplifiers) deliver significant power (watts to kilowatts) into low-impedance loads (speakers, antennas) with high efficiency while keeping device dissipation within thermal tolerances ($T_j < T_{j,max}$).

Amplifier classes are categorized strictly by the conduction angle $\theta_c$ of the collector current during one complete sinusoidal input cycle ($360^\circ$):

  • Class A ($\theta_c = 360^\circ$ or $2\pi$ radians): The transistor conducts continuously for the entire input cycle. The Q-point is biased in the center of the active linear region. Distortions are minimal, but static DC power is continually wasted even with zero signal.
  • Class B ($\theta_c = 180^\circ$ or $\pi$ radians): The transistor is biased exactly at cutoff ($I_{CQ} = 0$). Conduction occurs for exactly one half-cycle ($180^\circ$). Zero static DC power at idle.
  • Class AB ($180^\circ < \theta_c < 360^\circ$): Biased slightly above cutoff with a small quiescent current ($I_{CQ} > 0$). Conduction occurs for slightly more than half a cycle, eliminating Class B crossover distortion.
  • Class C ($\theta_c < 180^\circ$): Biased well beyond cutoff. Conduction occurs in brief pulses ($80^\circ - 120^\circ$). High harmonic distortion; used exclusively with tuned resonant LC tanks in high-efficiency RF transmitters.

2. Mathematical Derivation of Maximum Collector Efficiency

Collector Efficiency $\eta$ is defined as the ratio of average AC output power delivered to the load to the average DC power drawn from the power supply:

$$\eta = \frac{P_{ac,out}}{P_{dc,in}} \times 100\%$$
(a) Series-Fed Class A Amplifier

For a series-fed Class A amplifier with supply $V_{CC}$ and collector resistor $R_C$:

$$V_{CQ} = \frac{V_{CC}}{2}, \quad I_{CQ} = \frac{V_{CC}}{2 R_C}$$
$$P_{dc,in} = V_{CC} I_{CQ} = \frac{V_{CC}^2}{2 R_C}$$

Under maximum unclipped sinusoidal swing, $V_m = V_{CC}/2$ and $I_m = I_{CQ} = V_{CC}/(2 R_C)$:

$$P_{ac,max} = \frac{V_m I_m}{2} = \frac{(V_{CC}/2)(V_{CC}/(2 R_C))}{2} = \frac{V_{CC}^2}{8 R_C}$$
$$\eta_{max, \text{Series-Fed Class A}} = \frac{P_{ac,max}}{P_{dc,in}} = \frac{V_{CC}^2 / (8 R_C)}{V_{CC}^2 / (2 R_C)} = \frac{2}{8} = 25\%$$

For a transformer-coupled Class A amplifier, the DC drop across the primary is near zero, allowing peak-to-peak voltage swing up to $2 V_{CC}$, doubling the theoretical limit to $\eta_{max} = 50\%$.

(b) Push-Pull Class B Amplifier

In a complementary-symmetry push-pull Class B configuration operating from a supply $V_{CC}$ (or dual $\pm V_{CC}$):

For peak load voltage $V_m$ across load $R_L$, peak collector current is $I_m = V_m / R_L$. The average current drawn from the DC supply during half-wave pulses by each rail is $I_{dc} = \frac{2}{\pi} I_m$. Thus:

$$P_{dc,in} = V_{CC} \left( \frac{2}{\pi} I_m \right) = \frac{2}{\pi} \frac{V_{CC} V_m}{R_L}$$

The AC power delivered to the load is:

$$P_{ac,out} = \frac{V_m^2}{2 R_L}$$

Collector efficiency as a function of output swing is:

$$\eta = \frac{P_{ac,out}}{P_{dc,in}} = \frac{V_m^2 / (2 R_L)}{2 V_{CC} V_m / (\pi R_L)} = \frac{\pi}{4} \left( \frac{V_m}{V_{CC}} \right)$$

Under maximum theoretical output voltage swing where $V_m = V_{CC}$:

$$\eta_{max, \text{Class B}} = \frac{\pi}{4} \approx 0.7854 = 78.54\%$$

ยง4.4 Push-Pull Configurations, Crossover Distortion & Class AB Biasing

1. Push-Pull Operation & Harmonic Distortion Cancellation

A push-pull amplifier utilizes two matched transistors ($Q_1$ and $Q_2$) operating in anti-phase: $Q_1$ conducts during the positive half-cycle of the input signal, pushing current into the load, while $Q_2$ conducts during the negative half-cycle, pulling current from the load.

Mathematically, expanding the nonlinear collector current transfer characteristics in a Taylor series:

$$i_{c1} = I_Q + a_1 v_{in} + a_2 v_{in}^2 + a_3 v_{in}^3 + \dots$$
$$i_{c2} = I_Q + a_1 (-v_{in}) + a_2 (-v_{in})^2 + a_3 (-v_{in})^3 + \dots = I_Q - a_1 v_{in} + a_2 v_{in}^2 - a_3 v_{in}^3 + \dots$$

In a balanced push-pull output transformer, net load current is proportional to the difference $i_{c1} - i_{c2}$:

$$i_{load} = k (i_{c1} - i_{c2}) = 2 k \left( a_1 v_{in} + a_3 v_{in}^3 + a_5 v_{in}^5 + \dots \right)$$

Crucial Result: All even-order harmonic distortion terms ($a_2 v_{in}^2, a_4 v_{in}^4$) completely cancel out! Furthermore, DC core saturation in output transformers is eliminated because quiescent currents produce opposing magnetic fluxes.

2. Crossover Distortion in Pure Class B

In a pure Class B push-pull stage, transistors are biased at $V_{BE} = 0$. However, real silicon bipolar transistors require a threshold forward voltage $V_{BE} \approx 0.6\text{ to }0.7\,\text{V}$ before base-emitter conduction begins.

Consequently, whenever the input signal passes through the zero-crossing within the deadband $-0.7\,\text{V} < v_{in} < +0.7\,\text{V}$, neither transistor conducts ($i_{c1} = i_{c2} = 0$). The output waveform flattens to zero near every crossing, generating severe high-order harmonic distortion known as Crossover Distortion.

3. Class AB Biasing and Diode Thermal Tracking

To eliminate crossover distortion, the transistors are biased into Class AB by applying a slight forward bias ($V_{bias} \approx 2 V_D \approx 1.4\,\text{V}$) across the base terminals using two series silicon diodes ($D_1, D_2$) or an active $V_{BE}$-multiplier circuit.

This maintains both transistors barely conducting at an idle quiescent current $I_{CQ} \approx 10\text{ to }50\,\text{mA}$. When the input swings through zero, the conduction smoothly transfers from $Q_1$ to $Q_2$ with zero deadband. Mounting the biasing diodes on the same physical heatsink as the power output transistors ensures thermal tracking: as junction temperature rises, diode voltage drops at $-2\,\text{mV}/^\circ\text{C}$, matching the transistor $V_{BE}$ drop and preventing thermal runaway.

Solved Problem Example 4.1: Small-Signal CE Amplifier Analysis via Hybrid h-Parameters

A Common Emitter BJT amplifier operates with a load resistance $R_L = 10\text{ k}\Omega$ and collector bias resistor $R_C = 4.7\text{ k}\Omega$. Transistor $h$-parameters are $h_{ie} = 2.0\text{ k}\Omega$, $h_{fe} = 100$, $h_{re} = 2.5 \times 10^{-4}$, and $h_{oe} = 25\ \mu\text{S}$. (a) Calculate the exact current gain $A_i = i_L / i_b$. (b) Calculate the input impedance $R_{\text{in}}$ seen at the base terminal. (c) Determine the overall voltage gain $A_v = v_o / v_{\text{in}}$ and output impedance $R_{\text{out}}$.

Step 1: Calculate Effective AC Collector Load $R_L'$

The AC collector resistance is the parallel combination of $R_C$ and external load $R_L$:

$$R_L' = R_C \parallel R_L = \frac{4.7\,\text{k}\Omega \times 10\,\text{k}\Omega}{4.7\,\text{k}\Omega + 10\,\text{k}\Omega} = \frac{47}{14.7} \approx 3.197\,\text{k}\Omega$$

Step 2: Check Simplified Model Validity

$$h_{oe} R_L' = (25 \times 10^{-6}\,\text{S}) \times (3.197 \times 10^3\,\Omega) = 0.0799 < 0.1$$

Since $h_{oe} R_L' < 0.1$, the simplified model yields high accuracy within $5\%$.

Step 3: Calculate Voltage Gain $A_v$

Using the simplified formula:

$$A_v = - \frac{h_{fe} R_L'}{h_{ie}} = - \frac{120 \times 3.197\,\text{k}\Omega}{1.2\,\text{k}\Omega} = - \frac{383.64}{1.2} \approx -319.7$$

Using the exact formula including $h_{re}$ and $h_{oe}$:

$$A_v = \frac{- h_{fe} R_L'}{h_{ie} + (h_{ie} h_{oe} - h_{fe} h_{re}) R_L'}$$
$$\Delta h = h_{ie} h_{oe} - h_{fe} h_{re} = (1200 \times 2.5 \times 10^{-5}) - (120 \times 2.5 \times 10^{-4}) = 0.030 - 0.030 = 0$$

Remarkably, $\Delta h \approx 0$ here, so the exact gain equals precisely the simplified gain: $A_v = -319.7$.

Step 4: Calculate Current Gain $A_i$ and Input Impedance $Z_{in}$

$$A_i = \frac{- h_{fe}}{1 + h_{oe} R_L'} = \frac{-120}{1 + 0.0799} = \frac{-120}{1.0799} \approx -111.1$$
$$Z_{in} = h_{ie} - \frac{h_{re} h_{fe} R_L'}{1 + h_{oe} R_L'} = 1200 - \frac{(2.5 \times 10^{-4})(120)(3197)}{1.0799} = 1200 - 88.8 = 1111.2\,\Omega \approx 1.11\,\text{k}\Omega$$
Solved Problem Example 4.2: Two-Stage RC-Coupled Amplifier Cascaded Analysis

A two-stage RC-coupled BJT amplifier consists of identical Common Emitter stages. For each transistor, $h_{ie} = 1.5\text{ k}\Omega$, $h_{fe} = 80$, and $h_{oe} \approx 0$. The collector resistors are $R_{C1} = R_{C2} = 3.3\text{ k}\Omega$, biasing resistors are $R_1 = 47\text{ k}\Omega$, $R_2 = 10\text{ k}\Omega$, and the load is $R_L = 4.7\text{ k}\Omega$. (a) Determine the effective AC load of the first stage $R_{L1}'$. (b) Calculate the individual voltage gains $A_{v1}$ and $A_{v2}$. (c) Compute the total overall voltage gain $A_v = A_{v1} \times A_{v2}$ in decibels.

Step 1: Calculate Input Impedance of Stage 2 ($Z_{in2}$)

The input impedance of Stage 2 includes the bias resistors in parallel with the transistor base input:

$$Z_{in2} = (R_1 \parallel R_2) \parallel h_{ie2} = 15\,\text{k}\Omega \parallel 1.5\,\text{k}\Omega = \frac{15 \times 1.5}{15 + 1.5} = \frac{22.5}{16.5} \approx 1.364\,\text{k}\Omega$$

Step 2: Calculate Loaded Voltage Gain of Stage 2 ($A_{v2}$)

The effective AC load on Stage 2 is $R_{L2}' = R_{C2} \parallel R_L$:

$$R_{L2}' = 3.3\,\text{k}\Omega \parallel 4.7\,\text{k}\Omega = \frac{3.3 \times 4.7}{3.3 + 4.7} = \frac{15.51}{8.0} \approx 1.939\,\text{k}\Omega$$
$$A_{v2} = - \frac{h_{fe} R_{L2}'}{h_{ie2}} = - \frac{100 \times 1.939\,\text{k}\Omega}{1.5\,\text{k}\Omega} \approx -129.27$$

Step 3: Calculate Loaded Voltage Gain of Stage 1 ($A_{v1}$)

Stage 1 is loaded by its own collector resistor $R_{C1}$ in parallel with the entire input impedance $Z_{in2}$ of Stage 2:

$$R_{L1}' = R_{C1} \parallel Z_{in2} = 3.3\,\text{k}\Omega \parallel 1.364\,\text{k}\Omega = \frac{3.3 \times 1.364}{3.3 + 1.364} = \frac{4.501}{4.664} \approx 0.965\,\text{k}\Omega$$
$$A_{v1} = - \frac{h_{fe} R_{L1}'}{h_{ie1}} = - \frac{100 \times 0.965\,\text{k}\Omega}{1.5\,\text{k}\Omega} \approx -64.33$$

Step 4: Calculate Overall Voltage Gain $A_v$ and Decibel Gain

$$A_v = A_{v1} \times A_{v2} = (-64.33) \times (-129.27) \approx +8315.9$$

Note that the double phase inversion produces an overall positive gain (in-phase output).

$$A_v(\text{dB}) = 20 \log_{10}(8315.9) = 20 \times 3.9199 \approx 78.4\,\text{dB}$$
Solved Problem Example 4.3: Class B Push-Pull Power Amplifier Performance & Thermal Dissipation

A complementary-symmetry Class B push-pull amplifier operates from dual power supplies of $\pm V_{CC} = \pm 18\text{ V}$ and drives an $8.0\ \Omega$ loudspeaker load. (a) Calculate the maximum unclipped output signal swing and maximum AC output power $P_{L,\text{max}}$ (assuming ideal zero saturation voltage). (b) Determine the DC power supplied $P_{\text{dc}}$ under maximum output conditions. (c) Calculate the conversion efficiency $\eta$ and the maximum thermal power dissipation per transistor $P_{D,\text{max}}$.

Step 1: Calculate Maximum AC Power Delivered to Load $P_{ac,max}$

With peak output voltage $V_m = V_{CC} = 18\,\text{V}$ and load $R_L = 8\,\Omega$:

$$P_{ac,max} = \frac{V_m^2}{2 R_L} = \frac{(18\,\text{V})^2}{2 \times 8\,\Omega} = \frac{324}{16} = 20.25\,\text{W}$$

Step 2: Calculate DC Power Drawn from Supply $P_{dc}$

Peak output current is $I_m = V_m / R_L = 18 / 8 = 2.25\,\text{A}$. Average supply current is $I_{dc} = \frac{2}{\pi} I_m$:

$$I_{dc} = \frac{2}{\pi} \times 2.25\,\text{A} = \frac{4.5}{\pi} \approx 1.432\,\text{A}$$
$$P_{dc} = V_{CC} I_{dc} = 18\,\text{V} \times 1.4324\,\text{A} \approx 25.78\,\text{W}$$

Step 3: Collector Efficiency $\eta$ at Maximum Swing

$$\eta = \frac{P_{ac,max}}{P_{dc}} \times 100\% = \frac{20.25\,\text{W}}{25.78\,\text{W}} \times 100\% \approx 78.54\%$$

Step 4: Power Dissipated by Transistors at Maximum Swing

The total heat power dissipated across both transistors is:

$$P_{D,total} = P_{dc} - P_{ac,max} = 25.78\,\text{W} - 20.25\,\text{W} = 5.53\,\text{W}$$

For each individual transistor:

$$P_{D1} = \frac{P_{D,total}}{2} = \frac{5.53\,\text{W}}{2} \approx 2.765\,\text{W}$$

Step 5: Worst-Case Transistor Thermal Dissipation Peak

In Class B amplifiers, maximum transistor dissipation does NOT occur at maximum signal swing! Total dissipation is $P_D(V_m) = \frac{2 V_{CC} V_m}{\pi R_L} - \frac{V_m^2}{2 R_L}$. Differentiating with respect to $V_m$ and setting to zero:

$$\frac{d P_D}{d V_m} = \frac{2 V_{CC}}{\pi R_L} - \frac{V_m}{R_L} = 0 \implies V_m = \frac{2}{\pi} V_{CC} \approx 0.6366 V_{CC}$$
$$V_m^* = 0.6366 \times 18\,\text{V} \approx 11.46\,\text{V}$$

At this specific output voltage swing, the worst-case transistor dissipation is:

$$P_{D,max}(\text{per transistor}) = \frac{V_{CC}^2}{\pi^2 R_L} = \frac{(18\,\text{V})^2}{\pi^2 \times 8\,\Omega} = \frac{324}{78.957} \approx 4.10\,\text{W}$$

Heatsink sizing must be designed to withstand this $4.10\,\text{W}$ peak dissipation rather than the $2.765\,\text{W}$ full-power condition.

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