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How does an RF power amplifier work?

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An RF power amplifier (PA) converts DC supply power into RF output power at the antenna. Its design centers on the load line: the transistor’s optimum load resistance is approximately Ropt = (VDD − Vknee)/(Imax/2) for a class-A stage biased at Imax/2, i.e. peak voltage swing over peak RF current, which is usually much less than 50 Ω and must be transformed by the output network. Key metrics are output power, efficiency (drain efficiency or PAE), gain, linearity and ruggedness.[1]

Why isn't a PA conjugately matched?

Conjugate matching maximizes small-signal gain, but maximum output power is limited by voltage and current swing. Cripps’ load-line approach chooses Ropt so that voltage and current both reach their limits simultaneously; the result can deliver 1–2 dB more power than a conjugate match.[1] Load-pull measures the optimum empirically.

What determines output power?

For a class-A or class-B stage with sinusoidal output, Pout = (VDD − Vknee)²/(2RL). Higher supply voltage and breakdown allow higher Ropt, easing matching; this is one reason GaN (high breakdown) and LDMOS dominate high-power PAs. See RF transistor technologies.

What other concerns apply?

Frequently asked questions

What is a typical PA load impedance?

A few ohms for watt-level handset PAs at low voltage; a 1 W, 3.4 V PA needs roughly 5 Ω.

Why do PAs need harmonic filtering?

Nonlinear operation generates harmonics that must meet regulatory spurious limits.

Is PA efficiency the same as battery life?

PA efficiency is a major factor in transmit current and therefore battery life.

References

Online sources accessed October 21, 2026; printed works cited by edition. See all sources.

  1. “RF Power Amplifiers for Wireless Communications, 2nd ed.”. S. C. Cripps, Artech House. 2006.