Analog Electronics I · 120 h · Topic 4 of 8

Power amplifiers

The last stage, the one that drives the loudspeaker. Here voltage gain no longer matters; what matters are the watts delivered, the efficiency and the heat: a badly designed power amplifier does not sound bad, it burns out.

Amplifiers Class A / B / AB Push-pull Efficiency Heat sinks

01How this differs from the small signal

In a small-signal amplifier, the swing is millivolts and the transistor always works on a tiny linear stretch. In power amplifiers, the signal travels along the whole load line and three new problems appear:

Distortion. The transistor's curves are not perfectly linear over the whole range.
Dissipation. The transistor may have to dissipate several watts, and heat sinks have to be calculated.
Efficiency. What matters is how much of the energy taken from the supply ends up in the load and how much is lost as heat.

η = \frac{P_{l o a d ​}}{P_{s u p p l y}} \times 100 %

P_{l o a d ​} = \frac{V_{p}^{2}}{2 R_{L}} = \frac{V_{rm s}^{2}}{R_{L}}

Power delivered to the load with a sinusoidal signal. V_p is the peak amplitude across the loudspeaker.

The power that is dissipated is the power that did not reach the load

Everything the supply delivers and the load does not receive is turned into heat in the transistors. That is why efficiency is not just an energy-saving figure: it determines the size of the heat sink and whether the amplifier survives or not.

02Class A

The transistor conducts during the full 360° of the cycle: it never cuts off. The Q point is in the middle of the load line.

Figure 1. Conduction of classes A, B and AB, animated. In class B and AB the contribution of the other transistor is also drawn: between the two they cover the complete cycle.

Advantages

The signal never leaves the active region: minimum distortion.
Simple circuit, a single transistor.
No crossover distortion.

Disadvantages

Theoretical maximum efficiency: 25% with RC coupling, 50% with a transformer. In practice, much less.
It draws current and heats up even with no signal, because the quiescent current is at its maximum.
For 1 W of audio, 3 W or more must be dissipated.

That is why class A is restricted to preliminary stages (where the power is negligible) and to high-end audio equipment, where a unit that draws 200 W to deliver 20 W is accepted in exchange for exceptional linearity.

03Class B and complementary symmetry

The idea that solves the efficiency problem: use two transistors, one for each half-cycle. Each one conducts for 180° and rests for the other 180°. At rest no current flows, so with no signal there is no consumption.

The simplest way to do it is complementary symmetry: an NPN for the positive half-cycle and a PNP for the negative one, both in an emitter-follower configuration.

Figure 2. Complementary symmetry, animated: Q1 pushes in the positive half-cycle and Q2 pulls in the negative one. The two emitters join at the loudspeaker, and the diodes provide the class AB bias.

η_{\max} = \frac{π}{4} = 78.5 %

P_{D} \approx 0.2 \cdot P_{\max}

Maximum efficiency of class B, with the signal using the whole available swing. The dissipation per transistor is at its maximum not with the maximum signal, but with the signal at 63% of it.

A fact that often surprises

The dissipation of a class B amplifier is not at its maximum when it plays loudest. With the signal at maximum, most of the energy goes to the loudspeaker; with the signal at zero, there is no consumption. The worst case is in the middle, at around 63% of the maximum swing. An amplifier that withstands full volume can burn out at medium volume if the heat sink was calculated badly.

04Crossover distortion and class AB

Class B has a serious flaw. Neither transistor conducts until the signal exceeds the 0.7 V needed to overcome its V_BE. In the stretch between −0.7 V and +0.7 V the output is zero: a step appears at every zero crossing.

Figure 3. Crossover distortion. It is proportionally worse the smaller the signal, which is why it is especially noticeable at low volume.

The solution is class AB: the bases are pre-biased with a voltage of approximately 1.4 V (two junctions), so that both transistors are barely conducting at rest. When one starts to turn off the other is already on, and the transition is continuous.

Pre-biasing method	How it works	Remarks
Two diodes in series	Each diode contributes 0.7 V between the bases.	Simple. If the diodes are mounted on the heat sink, they compensate the thermal drift of the transistors.
V_BE multiplier	A transistor with a trimmer between base and collector generates an adjustable voltage from 1.2 to 2.5 V.	It is what commercial equipment uses. It allows the quiescent current to be adjusted precisely.
Resistor with trimmer	Adjustable divider between the bases.	Does not compensate temperature. Obsolete.

Thermal runaway

It is the most dangerous failure of class AB. As the transistors heat up, their V_BE drops by about 2 mV per degree, so the quiescent current rises, and that heats them up even more. The loop feeds on itself and ends with the transistors destroyed within seconds. It is avoided with two measures, and both must be used:

Emitter resistors of 0.22 to 0.47 Ω in series with each output transistor. They are the negative feedback that holds the current back.
Thermal coupling of the diodes (or of the V_BE multiplier transistor) to the same heat sink as the output devices, so that they sense the same temperature.

05Split supply or single supply

Split supply (±V_CC)

The output at rest is at 0 V, so the loudspeaker goes directly to ground.

No output capacitor: better bass response.
More power for the same total voltage.
Needs a center-tapped transformer and dual rectification.
Risk: if a transistor fails, DC goes to the loudspeaker and destroys it. Commercial equipment includes an output protection relay.

Single supply (+V_CC)

The output at rest sits at V_CC/2, so a large output capacitor (1000 to 2200 µF) is needed to block that DC.

Simpler and cheaper power supply.
The output capacitor limits the bass response and adds distortion.
The available power is approximately one quarter of that with a split supply of the same total voltage.
It protects the loudspeaker by itself: the capacitor does not let DC through.

f_{c} = \frac{1}{2 π R_{L} C}

Lower cutoff frequency imposed by the output capacitor. With R_L = 8 Ω and C = 2200 µF it gives 9 Hz: acceptable. With 470 µF it would be 42 Hz, and the bass would be lost.

06Complete calculation of a power stage

Worked example · 20 W into 8 Ω

Design a class AB stage in complementary symmetry that delivers 20 W RMS into an 8 Ω loudspeaker, with a split supply.

1 · Required peak voltage. From P = V_p²/(2R_L):

V_{p} = \sqrt{2 \cdot P \cdot R_{L}} = \sqrt{2 \times 20 \times 8} = 17.9 V

2 · Supply voltage. The drop across the output transistor and the emitter resistors has to be added, plus some margin: ±22 V. That is, a 15+15 V RMS transformer (15 × 1.414 = 21.2 V peak per branch).

3 · Peak current. I_p = 17.9 / 8 = 2.24 A. The transistors must withstand at least 3 A and a V_CEO of 44 V (the voltage between rails). A TIP41C/TIP42C (6 A, 100 V) complies with room to spare.

4 · Dissipation per transistor. The worst case in class B:

P_{D} = \frac{V_{C C}^{2}}{4 π^{2} R_{L}} = \frac{22^{2}}{4 \times 9.87 \times 8} = 1.53 W

With real music and peaks, size for 4 to 5 W per transistor to leave a margin.

5 · Supply consumption. With η = 70%: P_supply = 20 / 0.7 = 28.6 W. The supply must deliver at least 1.5 A of DC per branch.

Heat sink calculation

Heat travels from the transistor junction to the air through three thermal resistances in series, measured in °C/W:

T_{j} = T_{a} + P_{D} \cdot (R_{j c} + R_{c s} + R_{s a})

T_j = junction temperature (typical limit 150 °C, designed not to exceed 100 °C), T_a = ambient, R_jc = junction to case (datasheet value), R_cs = case to heat sink (0.5 °C/W with mica and thermal grease), R_sa = heat sink to air (heat sink specification).

Example · What heat sink is needed?

P_D = 5 W, T_a = 40 °C, maximum design T_j of 100 °C, TIP41C with R_jc = 1.92 °C/W, mounting with mica R_cs = 0.5 °C/W.

R_{s a} = \frac{T_{j} - T_{a}}{P_{D}} - R_{j c} - R_{c s} = \frac{60}{5} - 1.92 - 0.5 = 9.6 °C/W

A heat sink of 9.6 °C/W or less is needed: a finned aluminum plate of about 5 × 5 cm with a profile. The lower the value in °C/W, the better the heat sink.

07Integrated power amplifiers

Except in high fidelity, a power stage is no longer built from discrete transistors today: an integrated circuit is used that already includes the push-pull, the pre-biasing, the current limiting and the thermal protection.

IC	Power	Supply	Load	Notes
LM386	0.5 W	4 to 12 V	8 Ω	DIP-8, gain 20 to 200. The classic of school projects.
TDA2003	6 W	8 to 18 V	4 Ω	5 pins, made for car stereos. Needs a heat sink.
TDA2030	14 W	±6 to ±18 V	4 or 8 Ω	Single or split supply. Widely used.
TDA2050	32 W	±4.5 to ±25 V	4 or 8 Ω	Requires a large heat sink.
TDA7297	2 × 15 W	6 to 18 V	8 Ω	Stereo on a single chip, with mute and standby.

What must always be added to them

Supply decoupling: 100 nF ceramic + 100 µF electrolytic, as close as possible to the pins.
Zobel network at the output: 4.7 Ω in series with 100 nF, to ground. It stabilizes the amplifier against the inductive load that a loudspeaker presents at high frequency. Without it, many ICs oscillate at hundreds of kHz — you cannot hear it, but it heats things up and distorts.
Star ground: all grounds to the same physical point, next to the filter capacitor. It is what prevents the 50 Hz hum.

08The other classes

Class	Conduction	Typical η	Use
A	360°	20-25%	Preliminary stages, high-end audio.
B	180°	70-78%	Almost never used pure, because of crossover distortion.
AB	200-220°	50-70%	The audio standard.
C	< 180°	> 85%	Radio frequency with a tuned tank circuit. Useless for audio: it distorts enormously.
D	switching	90-95%	The transistors work as switches with PWM and an LC filter reconstructs the signal. It is what every modern portable device uses.

Class D is not “another way of amplifying linearly”: it is a paradigm shift. The transistor is only ever in cutoff or saturation, where it barely dissipates (see the transistor as a switch), and the information is carried in the width of the pulses. It is covered in Industrial Electronics, in Year 6.

09In the lab

Lab 1 · Seeing crossover distortion

Build a complementary stage without pre-biasing (pure class B) with a TIP31/TIP32, a ±9 V split supply and an 8 Ω load (or a 10 Ω / 5 W resistor, which does not bother the neighbors). With the oscilloscope the step at the zero crossing is clearly visible. Then add the two diodes between the bases and watch it disappear.

Lab 2 · Measuring the efficiency

With a resistive load and a 1 kHz sine wave:

Measure V_p across the load and calculate P_out = V_p²/(2R).
Measure the DC current drawn from each branch of the supply (ammeter in series) and calculate P_supply = V_CC × I_total.
Calculate η. It should come out between 50% and 70%.
Repeat at low volume and check that the efficiency drops.

Lab 3 · Amplifier with the LM386

Minimal LM386 setup: input through a 10 kΩ potentiometer, a 250 µF capacitor at the output, an 8 Ω loudspeaker, a Zobel network and decoupling. With pins 1 and 8 open the gain is 20; bridged with a 10 µF capacitor, it rises to 200. It is the most accessible audio project in Year 4.

Precautions

Power stages handle amperes: an accidental short circuit melts the traces.
Heat sinks burn. Many TO-220 packages have the tab connected to the collector: if the heat sink is exposed, there is voltage exposed.
Never run the stage without a load or with the output short-circuited.
Check the DC at the output (it should be ≈ 0 V with a split supply) before connecting the loudspeaker. If there is DC, the loudspeaker is destroyed.

10Common mistakes

Symptom	Usual cause
Harsh sound, especially at low volume	Crossover distortion: the class AB pre-biasing is missing.
The output transistors heat up with no signal	Excessive quiescent current: the pre-biasing was set too high. Adjust the V_BE multiplier.
The transistors burn out after a few minutes	Thermal runaway: the emitter resistors are missing or the diodes are not coupled to the heat sink.
The loudspeaker “crackles” and the voice coil heats up	There is DC at the output. With a split supply it indicates a shorted transistor: disconnect immediately.
50 or 100 Hz hum	Insufficient supply filtering, or poor grounding (not star-connected).
It clips before reaching the calculated power	The supply sags under load: transformer or filter capacitor undersized.
The IC shuts off every so often and comes back	The internal thermal protection is kicking in. Heat sink missing.
High-frequency oscillation (it heats up, visible on the oscilloscope)	The Zobel network or the supply decoupling is missing.

11Self-assessment

What is the theoretical maximum efficiency of class A with RC coupling, and why is it so low?

25%. Because the transistor conducts quiescent current throughout the cycle, even with no signal, and that current flows through the collector resistor, dissipating heat continuously.

Why does crossover distortion appear in class B?

Because no transistor conducts until the signal exceeds its V_BE of 0.7 V. Between −0.7 and +0.7 V the output is zero, and that generates a step at every zero crossing.

Calculate the power delivered to a 4 Ω loudspeaker with 12 V peak.

P = V_p²/(2R_L) = 144 / 8 = 18 W.

What are the 0.22 Ω resistors in the output emitters for?

They introduce negative current feedback: if the quiescent current tends to rise with temperature, the drop across those resistors reduces the effective V_BE and holds it back. They are the main protection against thermal runaway.

Why are the pre-biasing diodes mounted on the heat sink?

So that their voltage drop (which also falls by about 2 mV/°C) follows the same thermal drift as the V_BE of the output transistors. That way the quiescent current stays stable as the equipment warms up.

What concrete advantage does the split supply have over the single supply?

It needs no output capacitor, so the bass response is flat down to DC and there is less distortion. It also delivers about 4 times more power for the same total voltage. In exchange, it requires a more complex supply and a loudspeaker protection system.

A transistor dissipates 8 W, the ambient is at 35 °C, R_jc = 1.5 °C/W and R_cs = 0.5 °C/W. If T_j must not exceed 110 °C, what R_sa is needed?

R_sa = (110 − 35)/8 − 1.5 − 0.5 = 9.375 − 2 = 7.4 °C/W. A heat sink of 7.4 °C/W or less is needed (lower values = larger and better heat sink).

At what signal condition is the dissipation of a class B amplifier at its maximum?

Not at maximum power or at rest, but at approximately 63% of the maximum swing. That is the case to use for sizing the heat sink.

What is a Zobel network and why is it used?

A 4.7 Ω resistor in series with a 100 nF capacitor, connected from the output to ground. It presents the amplifier with a resistive load at high frequency, where the loudspeaker becomes inductive. Without it many amplifiers oscillate at hundreds of kHz.

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