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06 · Intermediate

Norton equivalence and maximum power transfer

The dual current-source view, and why maximum power is not the same as maximum efficiency.

75–100 minIntermediate
In this lesson
  • Convert exactly between Thévenin and Norton forms.
  • Compute load current naturally from the Norton model.
  • Derive the maximum-power condition for a resistive source and load.
  • Explain the 50% efficiency consequence at the maximum-power point.
Starting question

If Thévenin and Norton are exactly equivalent at the terminals, why keep both forms — and why does a load equal to the source resistance maximize load power but only give 50% efficiency?

This chapter answers both questions from the terminal law, not from memorized source-transformation rules. The goal is to know which equivalent is useful, what “maximum power” actually means, and when matching is or is not the engineering objective.

same port lawdifferent convenient coordinates

06.0Prerequisite recap — one terminal line, two source forms

V = VTH − RTHI
I = IN − V/RN

Both equations describe the same straight line in the V–I plane. Matching coefficients immediately gives

RN = RTH
IN = VTH/RTH
VTH = INRN

06.1Complete equivalence check with one load

Take VTH = 5 V and RTH = 100 Ω. The Norton current is 50 mA. Connect RL = 900 Ω.

Figure 6.1a — Thévenin form

5 V in series with 100 Ω drives the same 900 Ω load.

equivalent topology verified
Thévenin equivalent with load+−VTH5 VRTH100 ΩRL900 ΩVOUTGND

Figure 6.1b — Norton form

50 mA in parallel with 100 Ω preserves the same terminal V–I law.

equivalent topology verified
Norton equivalent with loadIN50 mARN100 ΩRL900 ΩVOUTGND
Thévenin:  IL = 5/(100 + 900) = 5.00 mA
VL = 5.00 mA · 900 Ω = 4.50 V

In Norton form, the source current divides between RN and RL.

IL = IN · RN/(RN + RL)
= 50 mA · 100/(100 + 900) = 5.00 mA

The same load current and voltage appear because the two source forms are the same one-port model.

06.2Derive the maximum-power condition

For a positive resistive source RTH driving a resistive load RL, the load current is

I = VTH/(RTH + RL)
PL = I2RL = VTH2RL/(RTH + RL)2

Differentiate with respect to the load resistance while keeping the source model fixed.

dPL/dRL = VTH2(RTH − RL)/(RTH + RL)3
dPL/dRL = 0   ⇒   RL = RTH

The derivative is positive below the match and negative above it, so this stationary point is a maximum.

PL,max = VTH2/(4RTH)

06.3Worked example — maximum power is not the same as high output voltage

For the 5 V / 100 Ω source, maximum power occurs at RL = 100 Ω.

I = 5/(100 + 100) = 25 mA
VL = 2.50 V
PL = 2.502/100 = 62.5 mW

Compare that with RL = 900 Ω from the earlier example.

VL = 4.50 V
PL = 4.502/900 = 22.5 mW

The 900 Ω load receives a much larger fraction of the open-circuit voltage, but less power. Voltage transfer and power transfer optimize different quantities.

06.4Why the matched resistive case is only 50% efficient

If RTH represents a real dissipative series resistance, the same current flows through source resistance and load.

PRTH = I2RTH
PL = I2RL
η = PL/(PL + PRTH) = RL/(RTH + RL)

At RL = RTH, exactly half of the power is dissipated in each resistance:

η = 1/2 = 50%

06.5See the trade-off with one dimensionless ratio

Define r = RL/RTH. Then three useful normalized quantities are

VL/VTH = r/(1 + r)
PL/PL,max = 4r/(1 + r)2
η = r/(1 + r)   when the source resistance is a real loss
rVL/VTHPL/PL,max
0.10.0910.331
10.5001.000
100.9090.331
1000.9900.039

Large load resistance gives excellent voltage transfer and, for a passive series source resistance, high efficiency — but it extracts little of the source’s maximum available power. Matching gives maximum load power, not maximum voltage and not maximum efficiency.

06.6When should you actually use a power match?

GoalTypical preferenceReason
Measure a voltageInput impedance ≫ source impedanceMinimize loading.
Deliver DC power efficientlySource loss ≪ loadA 50% matched-loss condition is usually undesirable.
Extract available small-signal/RF powerImpedance matchPower transfer and reflection control become central objectives.

06.7AC extension — conjugate matching

In sinusoidal steady state, the real waveform remains lowercase, for example v(t). AoN writes the associated complex phasor as an underlined lowercase quantity, for example vTH(ω).

ZTH = RTH + jXTH
ZL = RL + jXL

Average load power is

PL = |vTH(ω)|2RL/|ZTH + ZL|2

First remove unnecessary reactive current by cancelling the source reactance, then apply the resistive optimum:

XL = −XTH
RL = RTH
therefore  ZL = ZTH*

For example, if ZTH = 50 + j20 Ω, the conjugate-matched load is 50 − j20 Ω.

06.8Independent practice

Problem A — transform and verify

A 12 V Thévenin source has RTH = 600 Ω. Find the Norton form and the load current for RL = 1.8 kΩ.

Solution. IN = 20 mA, RN = 600 Ω, and IL = 20 mA × 600/(600 + 1800) = 5 mA.
Problem B — power versus efficiency

For a passive resistive source, choose RL so that efficiency is 80%. What fraction of maximum load power is then obtained?

Solution. r/(1+r)=0.8 ⇒ r=4. Then PL/PL,max = 4r/(1+r)² = 16/25 = 0.64. You get 80% efficiency but only 64% of the maximum available load power.
Problem C — AC match

If ZTH = 35 + j15 Ω, what load maximizes average power under the stated linear AC assumptions?

Answer. ZL = 35 − j15 Ω.

06.9Mastery checklist

  • Derive Norton parameters from the terminal law, not only from a memorized transformation.
  • Choose Thévenin or Norton according to the surrounding topology.
  • Derive RL = RTH for maximum resistive load power.
  • Explain why matched passive resistance gives 50% efficiency.
  • Distinguish voltage transfer, maximum power and efficiency as different objectives.
  • State and use the conjugate-match condition in AC.
Continue to chapter 07 problems
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