A long time ago I understood this, but right now it totally escapes me.

Taking an audio amplifier for example. Say it drives a speaker which is 8 ohms at all frequencies. Taking the amplifier as a Thevenin equivalent. If the Thevenin resistance is zero ohms, that would drive the speaker at the highest voltage, hence highest current and power?


I think your confusion is shared by many after they first encounter the maximum power transfer theorem. In essence, many forget the question the theorem answers.

The question the theorem answers is: For a given source impedance, what load impedance results in maximum power transfer from source to load.

Unfortunately, I've seen trained engineers reverse the question and then try to apply the same theorem, i.e.: For a given load impedance, what source impedance results in maximum power transfer from source to load

Many who should know better will answer "match the source to the load". But that's just plain wrong. A zero source impedance delivers maximum power to any load because there is no power loss in the zero source impedance.

  • 1
    \$\begingroup\$ Professor Moritz von Jacobi of St. Petersburg (1801-1874) stated the maximum power theorem better in mid-1800's. "Maximum power is transferred when the internal resistance of the source equals the resistance of the load AND When the external resistance can be varied, and the internal resistance is constant." BUT "When the load resistance is fixed, maximum power transfer occurs for zero internal resistance." mysite.du.edu/~jcalvert/tech/jacobi.htm \$\endgroup\$ Jul 14 '12 at 13:37
  • \$\begingroup\$ It is worth noting, that zero Thevenin circuits do not exist in reality. All sources had a finite Z (or as I prefer non-zero ESR. Speakers are also never flat impedance since it is controlled by eddy current losses in air pressure, not simply winding resistance which adds other resonant effects as well. They are not even flat over 1 decade span, let alone >3 decades \$\endgroup\$ Jul 14 '12 at 13:44
  • \$\begingroup\$ It is also important to remember than Maximal Power Theorem criteria applies to output power only and does not care about losses or efficiency. It is most important for broadcast, radio and Output considerations only. The key point being refections lower output from mismatch. Otherwise for Efficiency, Zero internal transmission line or zero source ESR motor winding impedance is best for transmission lines & generators. !! Edison failed that question on motors. \$\endgroup\$ Jul 14 '12 at 13:55
  • \$\begingroup\$ Another important factor is sound quality is not best when maximum power is transferred with a source impedance matched to speaker as dampening factors on eddy current air pressure dictate a ratio of Rl/Rs of at least 50:1 for good dampening and better with 500:1 So zero source impedance is ideal for drivers and jumbo oxygenated copper wire \$\endgroup\$ Jul 14 '12 at 14:03
  • \$\begingroup\$ +1 for clarifying which way round things go, it looks like this is where the OP may be have been getting confused. \$\endgroup\$
    – Oli Glaser
    Jul 14 '12 at 14:29

Let's say the amplifier's output is 10 V RMS. Then with zero output resistance you have this 10 V across the speaker's 8 Ω, which gives you 1.25 A RMS and 12.5 W RMS.

Now if there were a 1 Ω output resistance this would form a voltage divider with the speaker's 8 Ω, and only 8/9 of the 10 V would get across the speaker, that's 8.89 V RMS. Current will also be reduced to 1.11 A RMS, and power to 9.9 W. The higher the output resistance the lower voltage and current, and therefore the lower the speaker power.

enter image description here

So the highest output power is attained when the amplifier's output resistance is zero, any internal resistance will lower the speaker's power. More or less what we could expect.

But if you have an internal impedance, like for instance an HF amplifier with 50 Ω impedance, then it's completely different. Then you're not looking for the best matching output impedance for a given load, but for the best load for a given output impedance.

enter image description here

The graph shows output power for a 10 V signal with a 50 Ω impedance, for a load varying from 10 Ω to 100 Ω. You can see that we have the maximum output power when the load's impedance matches the input impedance.

That there's a maximum can be intuitively explained: if the load's impedance would decrease the voltage caused by the divider would decrease, and therefore also the power. If the impedance would increase, then the current would decrease, and therefore also the power. That the optimum is reached when both impedances are equal is a matter of mathematics:

\$ P_{OUT} = \dfrac{\left(V \dfrac{R_L}{R_i + R_L} \right)^2} {R_L} = \dfrac{V^2 R_L}{(R_i + R_L)^2} \$

To find an extremum we have to find a zero for the derivative to \$R_L\$:

\$ \dfrac{d P_{OUT}}{d R_L} = \dfrac{(R_i -R_L) V^2}{(R_i + R_L)^3} = 0 \$

from which it's clear that \$R_i = R_L\$.

No matter what your supply voltage and output impedance is you'll always get the same kind of graph. Let's have a look at that audio amplifier again, with that bad 1 Ω impedance. If the output level is 10 V RMS then we get again a maximum output power when the speaker's impedance matches the 1 Ω:

enter image description here

Note the 9.9 W we got at an 8 Ω load. We have a lower current (the greenish curve) because of the much higher total impedance of 9 Ω instead of 2 Ω. The increase of the voltage (purple curve) isn't enough to compensate the current decrease, so power will decrease too.

  • \$\begingroup\$ Thanks, but then when does impedance matching increase power throughput in some situations? \$\endgroup\$
    – CL22
    Jul 14 '12 at 12:08
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    \$\begingroup\$ @Jodes: Impedance matching as you are referring to it means matching the load impedance to the source. See the second paragraph of my answer. That's fine when the source is what it is and you have some range for the load. For any given load though, a 0 Ohm resistance Thevenin source is still best for transferring the most power. \$\endgroup\$ Jul 14 '12 at 12:24
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    \$\begingroup\$ Nicely explained, as usual.. \$\endgroup\$ Jul 14 '12 at 20:34

I think I see what you want to know:

Lets forget about the unreal 0 ohms source scenario (which will always provide the most power for a given load) and take a "normal" example - if we have a source with a 10 ohm impedance, how do we get the most power out of it?

The answer is to use a load with 10 ohms impedance. If we go higher or lower the power dissipated by the load drops.

To visualise this I threw together this simulation:


Simulation of above, stepping Rload from 0 to 50 ohms:

MPT sim

The blue line is the voltage across RLoad, the red line is the current through Rload, and the green line is the power dissipated by Rload. The X axis is Rloads resistance.
It's easy to see that the power reaches a peak when Rload is at 10 ohms, when current and voltage are halfway between their respective max/min values.

Why? Because as you increase RLoad the voltage increases but the current decreases, and if you decrease Rload the current increases but the voltage decreases.

Lets take a couple of points just above and below 10 ohms:

Rload = 12 ohms:

V(Rload) = 1V * (Rload / (Rload + 10)) = 1V * (12 / (12 + 10)) = 0.545V
I(Rload) = 1V / (Rload + 10) = 1V / (12 + 10) = 0.0454A
Power dissipated in Rload = 0.545 * 0.0454 = 24.7mW

Rload = 8 ohms:

V(Rload) = 1V * (Rload / (Rload + 10)) = 1V * (8 / (8 + 10)) = 0.444V
I(Rload) = 1V / (Rload + 10) = 1V / (8 + 10) = 0.0556A
Power dissipated in Rload = 0.444 * 0.0556 = 24.7mW

We see either side of the maximum power point (0.5V * 50mA = 25mW) it drops off.

Reference: Maximum Power Transfer theorem

  • \$\begingroup\$ So that was the maximum power transfer theorem? Wow. I hate how the terminology confuses and frightens me. \$\endgroup\$ Jul 14 '12 at 20:33
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    \$\begingroup\$ @abdullah - yep, sometimes the terminology can make things sound far more complex than they really are (and vice versa) \$\endgroup\$
    – Oli Glaser
    Jul 15 '12 at 16:31

Yes, a 0 Ohm Thevenin source will always provide the most current and voltage to a load, compared to ones with positive resistance.

This is not what you asked, but it is interesting to note. For a Thevenin source with some resistance, the load resistance for the most load power is equal to the Thevenin resistance. In other words, if you had a power supply with 10 Ω effective resistance, it can deliver the most power to a 10 Ω resistor. Higher resistors don't have as much current, which becomes 0 in the limit as the resistance approaches infinity. Lower resistances don't have as much voltage, which becomes 0 at 0 resistance.

  • \$\begingroup\$ Yes for positive resistance but negative impedance slope for telephony voice POTS linear amplifiers they use Thevenin source and Norton receiver with negative impedance drivers and balance the gain for Bode stability. So a negative ESR gives voltage gain on driver and receiver is current controlled for a shared 2 wire POTS line. Separation is done with voltage and current audio transformers. \$\endgroup\$ Jul 14 '12 at 13:48
  • \$\begingroup\$ @TonyStewart and olin, play nice. Tony, Abdullah is a non-native speaker and was asking why he has a hard time understanding your writing, this is a common discussion point, it is often hard to follow your style of writing, especially for non-native speakers. \$\endgroup\$
    – Kortuk
    Jul 15 '12 at 0:26
  • \$\begingroup\$ Sometimes I wonder if people prefer to read what they can understand rather than what they do not know. Learning is about connecting the dots of assumptions, proof, logic and references, which often take more space than permitted. \$\endgroup\$ Jul 15 '12 at 3:02
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    \$\begingroup\$ @Tony: If people can't understand what they are reading, then it has no value. Learning also needs to be done incrementally starting from a person's existing base of knowledge. This is why going too far off field often has the opposite effect of overwhelming and thereby preventing any learning. True teaching is way more than simply dumping information out there. Like all useful information transfer, it is most effective if done within the recipients context. \$\endgroup\$ Jul 15 '12 at 12:45
  • \$\begingroup\$ I have to agree with Olin here, I think a good teacher tries to see things from the pupils point of view, and on feedback adapts their instruction as necessary. You can't make something out of nothing, so you need to use existing knowledge to form new knowledge. \$\endgroup\$
    – Oli Glaser
    Jul 15 '12 at 16:38

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