Available Power
Understanding Available Power
Every RF source, whether a signal generator, an antenna, or an amplifier output port, has an internal impedance. The maximum power transfer theorem states that maximum power flows from the source to the load when the load impedance is the complex conjugate of the source impedance. This maximum is the available power. It depends only on the source's open-circuit voltage (or Norton current) and its internal impedance. Changing the load impedance can never increase delivered power beyond Pavs.
In a standard 50-ohm measurement system, signal generators are designed to present a 50 + j0 ohm source impedance. When connected to a perfect 50-ohm load, the delivered power equals the available power. But real loads are never perfect. A VSWR of 1.5:1 corresponds to a return loss of 14 dB and a mismatch loss of 0.18 dB, meaning 4% of the available power reflects back to the source.
Thevenin Derivation
Pavs = |Vs|² / (8 · Re{Zs})
Wave Parameter Form:
Pavs = |bs|² / (1 − |Γs|²)
Delivered Power Under Mismatch:
Pdel = Pavs · (1 − |ΓL|²)
Mismatch Loss (dB):
ML = −10 log(1 − |ΓL|²)
Where Vs is the open-circuit source voltage, Zs is the source impedance, bs is the source wave amplitude, Γs is the source reflection coefficient, and ΓL is the load reflection coefficient.
Mismatch Loss Reference
| VSWR | Return Loss (dB) | |Γ| | Mismatch Loss (dB) | Power Delivered (%) |
|---|---|---|---|---|
| 1.0:1 | ∞ | 0.000 | 0.00 | 100.0% |
| 1.2:1 | 20.8 | 0.091 | 0.04 | 99.2% |
| 1.5:1 | 14.0 | 0.200 | 0.18 | 96.0% |
| 2.0:1 | 9.5 | 0.333 | 0.51 | 88.9% |
| 3.0:1 | 6.0 | 0.500 | 1.25 | 75.0% |
| 5.0:1 | 3.5 | 0.667 | 2.55 | 55.6% |
Frequently Asked Questions
What is the difference between available power and delivered power?
Available power is the theoretical maximum a source can deliver, achieved only when the load impedance equals the complex conjugate of the source impedance. Delivered power is what the load actually receives. The relationship is Pdel = Pavs × (1 − |ΓL|²). With a VSWR of 2:1, mismatch loss is 0.51 dB, meaning only 89% of the available power reaches the load. In a perfectly matched 50-ohm system, the two values are identical.
Why is available power used as the reference for noise figure?
Noise figure is defined as F = SNRin / SNRout, where SNRin uses the available signal power from the source and the available noise power from a matched resistor at 290 K. Using available power standardizes the reference so that noise figure is a property of the device under test, independent of the measurement system's match quality. This convention is specified in IEEE Standard 1139 and used universally for amplifier and receiver characterization.
How do you calculate available power from S-parameters?
In scattering parameter notation, Pavs = |bs|² / (1 − |Γs|²). When the source is perfectly matched (Γs = 0), this simplifies to Pavs = |bs|². For a signal generator with 20 dB source return loss (Γs = 0.1), the available power is approximately 0.04 dB higher than the power delivered into a perfect 50-ohm load.