Maximum Power Transfer
Every practical signal source has an internal resistance — a battery has its internal ohms, a transmitter has its output impedance, an antenna has its radiation and loss resistance. Connecting a load to that source always splits the available energy between the internal resistance and the load. The question every engineer and ham operator faces is: what load resistance extracts the most power? The answer — one of the most elegant results in circuit theory — is that maximum power is transferred when the load resistance equals the source resistance. For AC circuits with reactive components, the condition extends to conjugate impedance matching: the load reactance must cancel the source reactance. Both DC and AC cases are covered in this lesson.
Power delivered to a load peaks at RL = Rth. On both sides of the optimum, delivered power falls — steeply for low RL (voltage lost across Rth), gradually for high RL (little current flows).
View LargerThe Maximum Power Transfer Theorem
Pmax = Vth² / (4 Rth)
The Thevenin equivalent is the key: any linear network driving a load can be reduced to Vth with series Rth. Once you have those two values, the maximum power transfer condition is simply RL = Rth.
At that condition, the circuit is two equal resistances in series, driven by Vth. The current is I = Vth / (2 Rth). Power in the load:
At RL = Rth: Pmax = [Vth / (2Rth)]² × Rth = Vth² / (4 Rth)
Derivation and the 50% Efficiency Rule
To find the RL that maximises PL, take the derivative with respect to RL and set it to zero. The power in the load is:
PL = Vth² × RL / (Rth + RL)²
Differentiating with the quotient rule:
dPL/dRL = Vth² × [(Rth + RL)² − 2RL(Rth + RL)] / (Rth + RL)⁴
Setting the numerator to zero:
(Rth + RL) − 2RL = 0 → Rth − RL = 0 → RL = Rth
The 50% efficiency consequence
At the matched condition (RL = Rth), the source delivers I = Vth/(2Rth) through two equal resistors. The total power from the source is Psource = Vth² / (2Rth). The load receives Pmax = Vth² / (4Rth). Efficiency:
η = Pload / Psource = [Vth² / (4Rth)] / [Vth² / (2Rth)] = 50%
This is a fundamental consequence: when power to the load is maximum, exactly 50% of the total generated power is wasted as heat in Rth. This is why maximum power transfer matters for signal applications (receivers, measurement instruments, audio preamps) where extracting every available microwatt matters more than efficiency. In high-power applications (transmitter PA stages, motor drives), 50% efficiency is unacceptable — those systems deliberately mismatch to achieve 70–90% efficiency, accepting less delivered power in exchange for far less wasted heat.
Power vs Load Resistance: The Parabolic Curve
| RL / Rth ratio | PL / Pmax | What is limiting power? |
|---|---|---|
| 0 (short circuit) | 0% | All voltage drops across Rth; load voltage = 0 |
| 0.25 (RL = Rth/4) | 64% | Low load voltage; most voltage still across Rth |
| 0.5 (RL = Rth/2) | 89% | Near peak; rising steeply |
| 1.0 (RL = Rth) | 100% (peak) | Optimum — equal split of voltage and current |
| 2.0 (RL = 2 Rth) | 89% | Near peak; falling gradually |
| 4.0 (RL = 4 Rth) | 64% | Current very low; load voltage high but little current |
| ∞ (open circuit) | 0% | No current flows; load voltage = Vth but I = 0 |
On a linear RL axis the curve is asymmetric — it falls steeply below Rth, gradually above. On a logarithmic RL axis it is symmetric, which explains why ratios of 0.5:1 and 2:1 both give 89%. A practical implication: a 2:1 mismatch either way delivers 89% of maximum power — only 0.5 dB loss. A 10:1 mismatch delivers only 33% — a 4.8 dB loss. This is why matching to within VSWR ≤ 2:1 is often considered acceptable in practice.
Worked Examples
Example 1 — Signal generator to load
(a) Optimum RL: RL = Rth = 50 Ω
(b) Maximum power:
Pmax = Vth² / (4 Rth) = (2)² / (4 × 50) = 4/200 = 20 mW
(c) Power at RL = 150 Ω:
I = 2/(50 + 150) = 10 mA
PL = (0.010)² × 150 = 15 mW
(d) Mismatch loss:
PL/Pmax = 15/20 = 0.75
Loss = −10 log₁₀(0.75) = 1.25 dB
(3:1 mismatch ratio → 75% of peak power, 1.25 dB loss)
Example 2 — HF transmitter and antenna system
Thevenin source for 100 W into 50 Ω:
Pmax = Vth²/(4 × 50) = 100 W → Vth² = 400 × 50 = 20,000 → Vth = 141.4 V
(This is the open-circuit peak voltage; at 50 Ω matched load: I = 141.4/(2×50) = 1.414 A, P = 1.414² × 50 = 100 W ✓)
Case A — Perfect match: RL = 50 Ω:
PL = 100 W VSWR = 1.0:1 Mismatch loss = 0 dB ✓
Case B — 2:1 mismatch: RL = 100 Ω:
I = 141.4/(50 + 100) = 0.943 A
PL = 0.943² × 100 = 88.9 W Mismatch loss = −10 log(88.9/100) = 0.51 dB
VSWR = 100/50 = 2.0:1 — only 0.51 dB loss, usually acceptable.
Case C — 5:1 mismatch: RL = 250 Ω:
I = 141.4/(50 + 250) = 0.471 A
PL = 0.471² × 250 = 55.5 W Mismatch loss = −10 log(55.5/100) = 2.55 dB
VSWR = 250/50 = 5.0:1 — over half a dB wasted, antenna tuner strongly recommended.
Case D — Short-circuit fault: RL = 0:
PL = 0 W (all power in Rth); transmitter protection circuits will fold back power. Real transmitters shut down or reduce drive into a short.
Summary: Mismatch loss increases dramatically as the ratio moves away from 1:1. The transmitter's output network and protection circuits are designed around the 50 Ω condition — large mismatches stress the output transistors and reduce lifespan.
Conjugate Matching for AC Circuits
In DC circuits, maximum power transfer requires RL = Rth. In AC circuits, the source may have a complex Thevenin impedance Zth = Rth + jXth, where Xth is the source reactance (positive for inductive, negative for capacitive). The load impedance is ZL = RL + jXL.
For maximum power delivered to ZL from a source Zth = Rth + jXth:
RL = Rth (equal resistive parts)
XL = −Xth (opposite reactive parts — the load reactance cancels the source reactance)
ZL = Zth* (the complex conjugate of Zth)
Pmax = |Vth|² / (4 Rth) (same formula as DC, using the magnitude of Vth)
Conjugate matching: the load reactance XL = −Xth cancels the source reactance, making the total series impedance purely resistive. Power transfer is then maximised by setting RL = Rth.
View LargerThe reason for cancelling the reactance: any reactance in the circuit reduces the current for a given source voltage (reactance increases impedance magnitude without doing useful work). By choosing XL = −Xth, the total series reactance is zero — the circuit is resonant at the operating frequency — and only the resistive parts determine the power. Then the DC condition RL = Rth applies to these resistive parts.
A transmitter output has Zth = 50 + j30 Ω (50 Ω resistive, +30 Ω inductive reactance). The antenna feedpoint has Zantenna = 73 − j40 Ω (73 Ω radiation resistance, −40 Ω capacitive reactance).
Is the antenna directly matched to the transmitter?
Resistive parts: 50 vs 73 Ω — not equal.
Reactive parts: +30 vs −40 Ω — opposite signs, but different magnitudes. These do not cancel.
Total impedance: Ztotal = (50 + j30) + (73 − j40) = 123 − j10 Ω.
This is not a conjugate match and power transfer is not maximum.
Matching network requirement:
The matching network must transform Zantenna = 73 − j40 Ω into ZL = 50 − j30 Ω (the conjugate of Zth). An L-network, pi-network, or antenna tuner is needed. After matching, the transmitter sees a load that is the conjugate of its own output impedance, and maximum power is transferred from transmitter to antenna.
Maximum Power Transfer Calculator
Maximum Power Transfer Calculator
Enter the Thevenin equivalent of the source (Vth and Rth). Optionally enter a specific load resistance RL to compare its delivered power against the theoretical maximum and compute the mismatch loss in dB.
⚖ Experiment: Finding the Optimum Load
Vary the load resistance across a fixed source and measure the power delivered at each value. Identify the load that gives maximum power and confirm it equals the source internal resistance.
- 9 V battery (source)
- 100 Ω fixed resistor (represents Rth — wire in series with the battery)
- A set of load resistors: 10, 22, 47, 100, 150, 220, 330, 470 Ω
- Digital multimeter
- Breadboard and jumper wires
- Wire the 100 Ω fixed resistor in series with the battery terminals. This models a Thevenin source with Vth ≈ 9 V, Rth = 100 Ω.
- Connect your first load resistor (start with 10 Ω) across the output terminals (the far side of the series 100 Ω).
- Measure the voltage across the load. Calculate PL = V²/RL. Record RL and PL.
- Repeat for each load value (22, 47, 100, 150, 220, 330, 470 Ω).
- Plot or tabulate your results. Find which RL gave the highest power reading.
- Compare your measured optimum RL to the predicted Rth = 100 Ω. Calculate the mismatch loss for the 47 Ω and 220 Ω points using the calculator above and compare to your measurements.
Peak power occurs at or very near RL = 100 Ω. For a 9 V source with 100 Ω series resistance: VL at matched load = 9/2 = 4.5 V; Pmax = 4.5²/100 = 202.5 mW. At 47 Ω and 220 Ω you should see roughly 89% of that peak (about 180 mW), consistent with the 2:1 mismatch result in the table. At 10 Ω and 470 Ω power drops significantly. Measure VL at each load and compute P = V²/R — the parabolic power curve should be clearly visible in your data.
Maximum Power Transfer in Ham Radio
The 50 Ω standard — why this number?
The universal RF impedance standard of 50 Ω is not arbitrary. Coaxial cable has two competing properties: minimum loss (attenuation) occurs at 77 Ω (optimal inner/outer diameter ratio for air-dielectric cable), while maximum power handling (before dielectric breakdown) occurs near 30 Ω. The 50 Ω standard is a deliberate compromise between these two optima, established by military and commercial radio engineers in the 1940s. Connecting a 50 Ω source to a 50 Ω load — the transmitter, the coax, the connector, the antenna feedpoint — ensures that every link in the RF chain transfers power at or near maximum with minimum reflections, without the engineering complexity of handling non-standard impedances.
Antenna impedance matching
A dipole antenna at resonance has a feedpoint impedance of approximately 73 Ω in free space, dropping to 50–65 Ω near ground. Ham operators often connect such antennas directly to 50 Ω coax without a matching network because the mismatch ratio is small (73/50 ≈ 1.46:1 VSWR) and the power loss is less than 0.5 dB. When antennas are used off-resonance — an 80 m dipole on 40 m, a shortened mobile whip at a non-resonant length — the feedpoint impedance may be very high or very low, producing serious mismatch losses and damaging reflections. A matching network (L-network, pi-network, or antenna tuner) transforms the antenna's complex impedance to 50 Ω as seen from the transmitter, satisfying both the resistive-match and the conjugate-match conditions.
Transmitter output and load matching
A solid-state transmitter's power amplifier is designed to work into 50 Ω. The output transistor is actually a current source with an optimum load impedance Ropt set by the supply voltage and desired output power (not simply Rth). The output matching network transforms 50 Ω to Ropt for the transistor, while presenting 50 Ω to the antenna connector. When the antenna system presents a different impedance, the transistor's output network is incorrectly loaded, output power drops, and the transistor experiences higher stress — increasing the risk of failure. Modern transmitters include VSWR fold-back protection that reduces drive power when the VSWR rises above 2:1 or 3:1.
Receiver antenna terminals and noise figure
In receive mode, the antenna and transmission line deliver a tiny signal to the receiver's input. The receiver's input impedance is designed to match the antenna system for maximum signal power transfer. For weak-signal work — meteor scatter, moonbounce (EME), satellite communication, or low-band DX — every tenth of a dB of mismatch loss is significant because the signal is at or near the noise floor. The preamplifier and matching network are designed for minimum noise figure into the antenna's source impedance, which is essentially a minimum-noise conjugate match that simultaneously maximises signal power transfer.
Frequently Asked Questions
Why is maximum power transfer efficiency only 50%? Isn't that wasteful?
Yes — and that is why maximum power transfer is used selectively. The 50% efficiency is a mathematical consequence of splitting voltage and current equally between Rth and RL. For signal-level applications (receivers, measurement instruments, audio preamps) the absolute power is so small that wasting 50% in heat is irrelevant — what matters is squeezing out every available microwatt from a weak source. For high-power applications (transmitter PA stages, motor drives, power distribution) 50% efficiency is unacceptable. Those systems deliberately mismatch in favour of higher efficiency, accepting some reduction in delivered power as the price of keeping the wasted heat manageable.
What is VSWR and how does it relate to impedance mismatch?
VSWR (Voltage Standing Wave Ratio) is the ratio of maximum to minimum voltage amplitude on a transmission line and is a direct measure of mismatch. A perfect match gives VSWR = 1:1. A mismatch of RL/Rth = k (where k ≥ 1) gives VSWR = k. The reflection coefficient is Γ = (RL − Rth) / (RL + Rth), and the fraction of power reflected is Γ². A VSWR of 2:1 corresponds to Γ = 1/3, reflected power = 11%, and mismatch loss ≈ 0.5 dB. The ham radio convention is that VSWR ≤ 2:1 is acceptable; above 3:1 or 4:1 most transmitters begin to fold back power to protect their output transistors.
Does maximum power transfer apply to AC and RF circuits?
Yes, with an extension: for AC circuits the source has a complex impedance Zth = Rth + jXth, and maximum power is transferred when ZL = Zth* (the complex conjugate): RL = Rth and XL = −Xth. The reactive parts cancel each other, leaving a purely resistive circuit at the operating frequency, and the resistive condition RL = Rth applies to those resistive parts. The maximum power formula remains Pmax = |Vth|² / (4 Rth). In practical RF matching, the matching network cancels the antenna's reactance and transforms the resistive parts to equality.
If maximum power transfer wastes 50%, why not just use a lower source resistance?
Reducing Rth does increase both efficiency and the power delivered to a fixed load. But Rth is usually set by the physics of the source — a battery's internal resistance is determined by its chemistry, a transistor's output impedance by its geometry and bias point. You cannot always choose Rth freely. When you can reduce source impedance (by using a lower-loss feedline, a better antenna, a lower-noise-figure preamplifier), you should — that always improves power transfer more than matching alone. Maximum power transfer tells you what to do given a fixed Rth; it does not prevent you from improving Rth itself when that is physically possible.
What is the difference between maximum power transfer and maximum efficiency?
They are fundamentally different conditions. Maximum power transfer (RL = Rth) delivers the most power to the load, but wastes 50% in Rth — efficiency is fixed at 50%. Maximum efficiency requires RL → ∞ (open circuit), which gives 100% efficiency but zero delivered power (no current flows). In practice, power system engineers choose RL much larger than Rth to prioritise efficiency, while RF engineers match RL = Rth to prioritise signal transfer. Class A PA stages run at modest efficiency (~25–50%) to preserve linearity; Class D and E stages use switching to achieve high efficiency (>90%) but are non-linear.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.