11. Maximum Power Transfer
Every practical signal source has an internal resistance — a battery has its internal ohms, a signal generator has its output impedance, a solar panel has a resistance that changes with sunlight. Connecting a load to that source always splits the available energy between the internal resistance and the load. The question every engineer and designer 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 (sensors, measurement instruments, audio preamps) where extracting every available microwatt matters more than efficiency. In high-power applications (power amplifiers, motor drives, battery systems), 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 — A 600 Ω audio line source
Maximum power (matched, RL = 600 Ω):
Pmax = Vth²/(4 × 600) = 4/2400 = 1.667 mW
Case A — Perfect match: RL = 600 Ω:
PL = 1.667 mW Mismatch loss = 0 dB ✓
Case B — 2:1 mismatch: RL = 1200 Ω:
I = 2.0/(600 + 1200) = 1.111 mA
PL = (0.001111)² × 1200 = 1.481 mW Mismatch loss = −10 log(1.481/1.667) = 0.51 dB
Case C — 5:1 mismatch: RL = 3000 Ω:
I = 2.0/(600 + 3000) = 0.556 mA
PL = (0.000556)² × 3000 = 0.926 mW Mismatch loss = −10 log(0.926/1.667) = 2.55 dB
Case D — Short circuit: RL = 0:
PL = 0 W (all power in Rth).
Summary: Mismatch loss grows as the ratio moves away from 1:1, but slowly at first: a 2:1 mismatch costs only about half a decibel. The same numbers apply to any source and load with the same ratio.
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 signal source driving an ultrasonic transducer has Zth = 50 + j30 Ω (50 Ω resistive, +30 Ω inductive reactance). The transducer at its operating frequency presents ZL = 73 − j40 Ω (73 Ω resistance, −40 Ω capacitive reactance).
Is the transducer directly matched to the source?
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 ZL = 73 − j40 Ω into 50 − j30 Ω (the conjugate of Zth). An L-network or pi-network of an inductor and capacitor does this. After matching, the source sees a load that is the conjugate of its own output impedance, and maximum power is transferred to the transducer.
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 Practice
Signal sources and standard impedances
Signal generators, test equipment and many communication systems use standard impedances — 50 Ω for high-frequency test equipment and cables, 75 Ω for video and TV, 600 Ω for traditional audio lines. When source, cable and load all share the same impedance, every link transfers power at or near maximum with no reflections, and readings taken with one instrument can be compared with another. A signal generator's output level is usually calibrated on the assumption that it drives a matched load.
Solar panels and MPPT
A solar panel behaves like a source whose internal resistance changes with light and temperature. For each condition there is one load that extracts the most power — the maximum power point. An MPPT (maximum power point tracking) charge controller continually adjusts the load it presents to the panel to stay on that point, delivering typically 10–30% more energy than a simple controller that connects the panel straight to the battery.
Audio amplifiers and loudspeakers: deliberately not matched
A modern audio amplifier has an output impedance of a small fraction of an ohm, while the loudspeaker it drives is 4–8 Ω. This is deliberate. The amplifier is designed as a voltage source, not to transfer maximum power from its internal resistance: a low output impedance wastes almost no power, keeps the amplifier efficient, and damps the speaker cone so that it stops moving when the signal stops (the “damping factor”). The speaker's rated impedance tells you the minimum load the amplifier can safely drive, not a value it must be matched to.
Power systems: efficiency first
Batteries, mains supplies and motor drives also avoid matching. A battery with 0.05 Ω internal resistance matched to a 0.05 Ω load would waste half its energy as heat and quickly overheat. Power systems are designed with a source resistance far below the load resistance, so that almost all the energy reaches the load. Maximum power transfer is for signals; maximum efficiency is for power.
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 (sensors, 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 (power amplifiers, 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.
Why aren't audio amplifiers matched to their loudspeakers?
Because matching would waste half the power as heat inside the amplifier and ruin its control of the speaker. Amplifiers are built as voltage sources with a very low output impedance (often under 0.1 Ω). The speaker draws the power it needs from that voltage, the amplifier stays efficient, and the low source impedance damps unwanted cone movement. The impedance rating on the amplifier is the lowest load it can drive safely.
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 practice, a matching network of inductors and capacitors cancels the load'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-resistance cable, a better sensor, a lower-noise 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 of this lesson. Every answer can be found in the lesson above.