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Push-Pull Amplifiers

A single-transistor common-emitter amplifier has a fundamental limitation: one transistor handles both the positive and negative halves of the output signal cycle. For Class A operation, it must be biased to conduct a large quiescent current so that it can sink or source enough current for the full output swing. This wastes DC power as heat whenever no signal is being amplified. For Class B or AB operation, a single transistor can only amplify one polarity of signal — you need a second transistor to handle the other half of the cycle.

The push-pull amplifier solves both problems elegantly. Two transistors are connected so that one handles the positive half-cycle of the signal (pushing current through the load) while the other handles the negative half-cycle (pulling current through the load). The two halves are combined at the output. This configuration delivers twice the output power of a single stage from the same supply voltage, cancels even-order harmonic distortion by symmetry, and permits efficient Class AB operation that eliminates crossover distortion while maintaining acceptable linearity.

Push-pull design is the dominant topology for virtually every power amplifier stage in ham radio — from the final stage of an HF SSB transceiver to the audio power amplifier driving the loudspeaker, from VHF solid-state power amplifiers to the modulator in an AM broadcast transmitter. Understanding push-pull operation is essential for anyone who wants to understand or troubleshoot RF power equipment.

What you will learn: How push-pull operation works; transformer-coupled and transformerless (complementary-symmetry) topologies; even-order harmonic cancellation by symmetry; Class AB biasing for push-pull stages; output power and efficiency; impedance transformation with output transformers; balun transformers; and practical push-pull amplifier design for ham radio.
Push-pull amplifier circuit diagrams: transformer-coupled BJT push-pull on the left, complementary-symmetry NPN/PNP push-pull on the right
Two push-pull topologies. Left: transformer-coupled using two matched NPN transistors — the standard for HF RF power amplifiers. Right: complementary-symmetry using an NPN and a PNP pair — the standard for audio amplifiers.

The Push-Pull Principle

In a push-pull amplifier, the input signal is split into two equal and opposite signals — one drives transistor Q1, the other drives transistor Q2. Q1 amplifies the positive half of the cycle; Q2 amplifies the negative half. The outputs of Q1 and Q2 are combined in a way that reconstructs the full output waveform.

The key insight is symmetry. Because the two transistors handle equal and opposite signals:

  • Odd-order distortion products (fundamental, 3rd harmonic, 5th harmonic…) add constructively in the output — they are preserved and amplified.
  • Even-order distortion products (2nd harmonic, 4th harmonic…) cancel because they are in phase in both transistors and the output circuit subtracts one half from the other. This cancellation is automatic — it requires no tuning or adjustment.

The cancellation of even-order harmonics is a major advantage of push-pull design. A single-ended (one transistor) Class A stage produces significant 2nd harmonic distortion. The same transistors in push-pull produce no 2nd harmonic — their 2nd harmonic components cancel, leaving only 3rd, 5th, and higher odd-order distortion, which are typically much smaller in amplitude. This makes push-pull amplifiers substantially cleaner than their single-ended counterparts for the same transistors and bias conditions.

Transformer-Coupled Push-Pull

The classic implementation uses transformers at both input and output. The input transformer (driver transformer) has a centre-tapped secondary winding. When the signal goes positive, the top half of the secondary drives Q1's base positive and the bottom half drives Q2's base negative — giving Q1 and Q2 exactly opposite drive signals. The output transformer has a centre-tapped primary. Q1's collector current flows through one half of the primary winding; Q2's collector current flows through the other half. The two currents produce magnetic flux in the transformer core that adds to drive the secondary winding and the load.

Circuit components:

  • Q1, Q2: matched NPN transistors (or matched MOSFETs in higher-power designs)
  • T1: input/driver transformer with centre-tapped secondary
  • T2: output transformer with centre-tapped primary, turns ratio chosen for impedance matching
  • Bias resistors/diodes: set Class AB idle current for both transistors
  • Emitter resistors (small value, 0.1–1 Ω): force current sharing between Q1 and Q2

Transformer-coupled push-pull is used in virtually all HF solid-state power amplifiers producing more than a few watts. The transformers provide galvanic isolation, allow easy impedance transformation, and the centre-tap arrangement ensures precise 180° phase splitting. Ferrite core transformers (wound on toroidal or binocular ferrite cores) are the standard choice for HF work — they achieve low loss and wide bandwidth from 1.8 MHz to 30 MHz in a single winding.

Complementary-Symmetry Push-Pull

The complementary-symmetry topology replaces the output transformer with a matched NPN/PNP transistor pair. Q1 (NPN) handles positive half-cycles; Q2 (PNP) handles negative half-cycles. Because Q2 is a PNP transistor, it is connected so that a positive base signal turns it off and a negative base signal turns it on — the opposite of Q1. This provides the necessary push-pull action without a transformer.

Both transistors receive the same input signal — no phase-splitting transformer is required at the input. When the signal goes positive, Q1 (NPN) turns on and drives current through the load to the negative supply rail. When the signal goes negative, Q2 (PNP) turns on and drives current from the positive supply rail through the load. The load sees a full, bidirectional current waveform.

Complementary-symmetry is the standard topology for audio power amplifiers in transceivers (for speaker drive) and for low-power RF stages where the bandwidth requirements can be met without transformers. Its advantages are fewer components, no transformer bandwidth limitation, and no core saturation risk. Its disadvantages are the difficulty of finding perfectly matched NPN/PNP pairs and the need for careful biasing to avoid crossover distortion.

Even-Order Harmonic Cancellation

The cancellation of even-order harmonics can be understood mathematically. Suppose each transistor has a non-linear transfer function and produces output with a second-harmonic component a₂:

Q1 output: IC1 = a₁·v + a₂·v² + a₃·v³ + …
Q2 receives −v (inverted signal): IC2 = a₁·(−v) + a₂·(−v)² + a₃·(−v)³ + …
= −a₁·v + a₂·v² − a₃·v³ + …

Output combines Q1 pushing and Q2 pulling:
Iout = IC1 − IC2
= (a₁·v + a₂·v² + a₃·v³) − (−a₁·v + a₂·v² − a₃·v³)
= 2a₁·v + 2a₃·v³ + …

The a₂·v² (second harmonic) terms cancel completely. Only odd-order terms remain.

This is why push-pull HF amplifiers have no 2nd harmonic output (or negligible 2nd harmonic in practice, if the circuit is not perfectly balanced). The 2nd harmonic of a 7 MHz transmitter would fall at 14 MHz — right in the 20-metre band. A well-balanced push-pull stage can achieve 50–60 dB of 2nd harmonic suppression, dramatically simplifying the harmonic filter requirements of the transmitter.

Class AB Biasing for Push-Pull Stages

The most critical design element of a push-pull audio or linear RF amplifier is the idle (quiescent) current bias that places both transistors in Class AB operation. Without idle current, the stage operates in Class B and exhibits crossover distortion. With too much idle current, efficiency drops and the transistors overheat.

Diode Bias

The simplest approach for complementary-symmetry audio stages uses two silicon diodes (one for each transistor's VBE) in series between the bases. The diodes are thermally coupled to the transistors — mounted on the same heatsink. As the transistors heat up, VBE decreases; the diodes, also heating up, produce a lower voltage drop across the bias network, tracking the transistors' VBE reduction and keeping the idle current approximately constant. This is simple, cheap, and effective for audio power stages.

VBE Multiplier (Rubber Zener)

For more precise idle current control, a VBE multiplier circuit is used. This consists of a single transistor (Qbias) with a resistor divider from collector to base. It produces an output voltage that is a multiple of VBE, adjustable by the resistor ratio. The multiplied VBE voltage sets the idle current of the output pair. VBE multiplier circuits are thermally coupled to the output transistors for temperature tracking. They allow idle current to be set precisely by adjusting a single trimmer resistor.

Setting Idle Current

Idle current for push-pull HF RF power amplifiers is typically set at 50–200 mA total (25–100 mA per transistor). For audio amplifiers, 20–100 mA total is common. Too low gives crossover distortion and poor IMD. Too high wastes power and can cause thermal runaway in poorly heatsunk designs. The correct idle current is found experimentally by measuring 3rd-order IMD with a two-tone test while adjusting bias, or by looking at the output waveform at low signal levels and adjusting until the crossover region disappears.

Spectrum display showing a single-ended amplifier output with strong second harmonic and a push-pull amplifier output with the second harmonic cancelled by symmetry
Even-order harmonic cancellation in push-pull operation. The single-ended stage (left) shows a prominent 2nd harmonic. The balanced push-pull stage (right) has the 2nd harmonic suppressed by 40 dB or more through symmetrical cancellation.

Output Power and Efficiency

For a Class AB push-pull stage with supply voltage VCC and a load resistance RL (reflected to the primary through the transformer):

Maximum output power (Class AB, approximately Class B):
Pout(max) = VCC² / (2 × RL)

where RL is the load impedance as seen by one transistor (collector to collector / 4 for a centre-tapped primary, or the collector load of a complementary pair).

Maximum efficiency (Class B limit): ηmax = π/4 ≈ 78.5%

Typical Class AB efficiency: 60–70% at full output power
Worked Example: Push-Pull HF Power Amplifier

A 100 W HF push-pull amplifier uses two transistors with VCC = 28 V. What collector load resistance must each transistor see?

Pout = VCC² / (2 × RL(each))
100 = 28² / (2 × RL)
100 = 784 / (2 × RL)
RL = 784 / 200 = 3.92 Ω

Each transistor should see approximately 4 Ω from collector to the centre-tap (the transformer provides this from a 50 Ω antenna load).

DC input power: Assuming 65% efficiency:
PDC = Pout / η = 100 / 0.65 = 154 W

Heat dissipated: Pheat = PDC − Pout = 154 − 100 = 54 W
Each transistor dissipates approximately 27 W — hence the need for heatsinking in any push-pull power stage above a few watts.

Output Transformer Impedance Matching

The output transformer in a push-pull RF amplifier has two functions: it reconstructs the full output waveform from the two half-cycle contributions of Q1 and Q2, and it transforms the impedance from the transistors' optimum load to the antenna system's 50 Ω.

The transformer turns ratio required for impedance matching:

n = Nprimary / Nsecondary = √(Rprimary / Rsecondary)

For a centre-tapped primary, the effective primary impedance is Rcollector-to-collector = 4 × RL(each transistor)

Example: Rcc = 4 × 4 Ω = 16 Ω, Rsecondary = 50 Ω
n = √(16/50) = √0.32 = 0.566 (step-up transformer)

In practice, ferrite-core wideband transformers (wound on toroidal ferrite rings or binocular (BN) cores) are used instead of tuned transformers for HF broadband amplifiers. Wideband transformers can cover the full 1.8–30 MHz HF range without retuning, making them essential in solid-state HF transceivers where the operator switches bands without mechanical changes. The transformer is wound with bifilar or trifilar wire so that each winding has equal coupling to the core.

Balun Transformers in Push-Pull RF Stages

A balun (balanced-to-unbalanced) transformer is used at the output of a push-pull amplifier when the load is unbalanced (e.g., a coaxial cable with one grounded conductor). The push-pull stage has a balanced output — equal and opposite voltages on the two collectors. The antenna feedline is unbalanced — one conductor is at ground potential. The balun converts between these two forms without disturbing the symmetry of the push-pull stage or introducing common-mode current on the feedline.

Most broadband HF power amplifiers use a ferrite-core balun transformer (sometimes called a choke balun) at the output. A well-designed balun has insertion loss < 0.5 dB across the HF band and provides 20–40 dB of common-mode rejection.

Push-Pull Output Power Calculator

Push-Pull Amplifier Output Power Calculator

Calculate maximum output power, DC input power, dissipation, efficiency, and transformer turns ratio for a push-pull RF or audio power amplifier stage.

Results will appear here.

Experiment: Observe Even-Order Harmonic Cancellation

Purpose: Demonstrate that a balanced push-pull amplifier produces no 2nd harmonic, while an equivalent single-ended stage does.

You will need: Audio function generator (produces sine waves at 1 kHz), two transistors (BC547 or 2N3904, matched beta if possible), a centre-tapped audio transformer (such as those used for 600 Ω audio matching), a multimeter or audio analyser/spectrum app on a smartphone, 12 V supply, resistors per circuit below.

Part A — Single-ended stage:

  • Build a simple common-emitter amplifier with VCC = 12 V, RC = 2.7 kΩ, RE = 470 Ω (bypassed), voltage-divider bias for IC ≈ 2 mA.
  • Apply a 1 kHz, 200 mVpp sine wave to the input.
  • Connect a spectrum analyser app to the output and observe the fundamental (1 kHz) and the 2nd harmonic (2 kHz). Record the 2nd harmonic level relative to the fundamental.

Part B — Transformer-coupled push-pull:

  • Connect two identical transistors in push-pull using a centre-tapped input transformer (or a phase-splitter circuit) and a centre-tapped output transformer.
  • Apply the same 1 kHz, 200 mVpp input.
  • Connect the spectrum analyser app to the output and observe the 1 kHz fundamental and the 2 kHz second harmonic.

Expected results:

  • Single-ended: visible 2nd harmonic, typically −30 to −40 dBc depending on signal level and transistor characteristics.
  • Push-pull: 2nd harmonic reduced by 20–40 dB compared to the single-ended stage. The circuit symmetry should give at least −50 dBc for the 2nd harmonic.

To explore further: Deliberately unbalance the push-pull stage by changing one transistor's bias slightly — observe the 2nd harmonic reappearing. This demonstrates that perfect cancellation requires good balance, which is why HF amplifiers use matched transistors and care is taken with transformer winding symmetry.

Frequently Asked Questions

Why do push-pull amplifiers cancel even-order harmonics but not odd-order?

Even-order harmonics (2nd, 4th…) are produced by the squared term (a₂·vin²) in the non-linear transfer function. When one transistor sees +vin and the other sees −vin, squaring gives (+vin)² = (−vin)² = vin² — the same positive value for both transistors. These equal components cancel when the output transformer subtracts one transistor's output from the other. Odd-order harmonics (1st, 3rd, 5th…) come from terms like a₁·vin and a₃·vin³: cubing a negative gives a negative (−vin)³ = −vin³, so these terms are opposite in the two transistors and add constructively at the output. Hence the fundamental and all odd harmonics are amplified; even harmonics cancel.

Why must transistors be matched in a push-pull amplifier?

Even-order harmonic cancellation depends on the two halves of the push-pull stage being identical — same gain, same VBE, same non-linearity coefficients. If one transistor has significantly different characteristics than the other, the cancellation is incomplete and even-order harmonics appear in the output. For audio amplifiers, transistors from the same production batch are usually matched closely enough. For high-power HF RF amplifiers, transistors are sometimes characterised individually and paired by matching their gain and saturation characteristics. Even a 5–10% gain mismatch between Q1 and Q2 degrades 2nd harmonic cancellation from 60 dB to perhaps 30–40 dB.

What happens if the idle current in a Class AB push-pull stage is too low?

Insufficient idle current places the stage closer to Class B operation. Near the zero-crossing of the output signal, neither transistor is fully conducting and the output passes through the non-linear crossover region, producing crossover distortion. This appears on a spectrum analyser as increased odd-order harmonics (3rd, 5th) and on a two-tone test as increased IMD. In a transceiver's audio output stage, crossover distortion sounds harsh and grating. In an SSB power amplifier, it causes excessive IMD and splatter. The fix is to increase idle current by adjusting the VBE multiplier or diode bias network — usually via a trimmer resistor on the bias potentiometer.

What is a ferrite core toroidal transformer and why is it used in HF push-pull amplifiers?

A toroidal transformer is wound on a doughnut-shaped ferrite core. The closed toroidal path contains the magnetic flux efficiently, resulting in low leakage inductance (important for wideband frequency response) and minimal electromagnetic interference radiation. Ferrite cores are chosen for their permeability and loss characteristics at HF frequencies — a Type 43 ferrite works well from 1–10 MHz; Type 61 ferrite is better from 10–200 MHz. By winding bifilar (two wires twisted together) on a ferrite toroid, wideband 4:1 or 9:1 impedance transformers can be made that cover 1.8–30 MHz with less than 0.5 dB insertion loss — essential for the broadband push-pull finals in modern solid-state HF transceivers.

Check Your Understanding

Answer each question and click Check to see the explanation.

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