# The maximum power transfer theorem

#### sober

Joined May 5, 2023
4
The maximum power transfer theorem is a polular one. It matches impedances between source and load. In ac circuits, under what condition do we use abs(Zs)=abs(ZL) since matching the source impedance Zs = Rs+Xs to the conjugate of the load impedance ZL = RL-jXL does not give desired results?

#### Papabravo

Joined Feb 24, 2006
20,399
The maximum power transfer theorem is a polular one. It matches impedances between source and load. In ac circuits, under what condition do we use abs(Zs)=abs(ZL) since matching the source impedance Zs = Rs+Xs to the conjugate of the load impedance ZL = RL-jXL does not give desired results?
Why do you think that? When dealing with complex impedances we aim for a conjugate match. The transfer of power, at 50%, is maximized when the real parts are equal, and the imaginary parts cancel each other.

#### sober

Joined May 5, 2023
4
Why do you think that? When dealing with complex impedances we aim for a conjugate match. The transfer of power, at 50%, is maximized when the real parts are equal, and the imaginary parts cancel each other.
Thanks Papabravo, matching both sides creastes very untenable conditions (conditions that conflict and cannot be practically achieved) when the conjugate match is used. The real parts gives a conditon, the imaginary part gives another condition. Both conditions do not agree in any way as the load is a series R-L type.
But what do you mean by "at 50%"?

#### sober

Joined May 5, 2023
4

#### Papabravo

Joined Feb 24, 2006
20,399
Thanks Papabravo, matching both sides creastes very untenable conditions (conditions that conflict and cannot be practically achieved) when the conjugate match is used. The real parts gives a conditon, the imaginary part gives another condition. Both conditions do not agree in any way as the load is a series R-L type.
But what do you mean by "at 50%"?
If source and load impedances are equal, then 50% of the available power is dissipated in the source and 50% is dissipated in the load. We do this all the time with RF power amplifiers and antennas. So don't tell me it can't be done. Is it difficult to achieve at power line frequencies? Absolutely.

#### crutschow

Joined Mar 14, 2008
32,926
Of course that theorem applies when varying the load impedance for a fixed source impedance.
You never increase the source impedance, for a fixed load impedance as that reduces the efficiency and power to the load.

That's why you never use matching for power line loads.

#### sober

Joined May 5, 2023
4
Of course that theorem applies when varying the load impedance for a fixed source impedance.
You never increase the source impedance, for a fixed load impedance as that reduces the efficiency and power to the load.

That's why you never use matching for power line loads.
Sure Crutschow, the source impedance is fixed, it is the load that is varying

#### nsaspook

Joined Aug 27, 2009
11,794
Maximum Power Doesn’t Mean Maximum Efficiency
The Maximum Power Transfer Theorem does not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to the load impedance.

Similar to AC power distribution, high fidelity audio amplifiers are designed for a relatively low output impedance and a relatively high speaker load impedance. As a ratio, “output impedance”: “load impedance” is known as damping factor, typically in the range of 100 to 1000.

Maximum power transfer does not coincide with the goal of the lowest noise. For example, the low-level radio frequency amplifier between the antenna and a radio receiver is often designed for the lowest possible noise. This often requires a mismatch of the amplifier input impedance to the antenna as compared with that dictated by the maximum power transfer theorem.
In most circuits The Maximum Power Transfer Theorem is irreverent to the circuit solution because we want efficiency.

#### Papabravo

Joined Feb 24, 2006
20,399
In RF work we want the conjugate match to avoid having any or minimal reflections back to the source (power amplifier) that might have the potential to destroy the output stage. Efficiency in not exactly a top concern.

#### nsaspook

Joined Aug 27, 2009
11,794
In RF work we want the conjugate match to avoid having any or minimal reflections back to the source (power amplifier) that might have the potential to destroy the output stage. Efficiency in not exactly a top concern.
It's always a concern when you deal with scores of 30kw rf systems in acceleration linacs every day. In RF the match is to the transmission line and load. If they are matched there won't be reflections. Efficiency is critical in the output stage so it's designed with that as a critical requirement.

#### Papabravo

Joined Feb 24, 2006
20,399
It's always a concern when you deal with scores of 30kw rf systems in acceleration linacs every day. In RF the match is to the transmission line and load. If they are matched there won't be reflections. Efficiency is critical in the output stage so it's designed with that as a critical requirement.
You are of course correct. I was thinking of efficiency in terms of output power with respect to input power, rather than radiated power to output power. The input power comes from the utility company and is available in whatever amount you can pay for.

#### nsaspook

Joined Aug 27, 2009
11,794
The input power from the utility is usually a DC resistive load to the RF output stage so designing for MPTT at that point (RF power generation) would be extremely wasteful of power that's not free from the utility.
Conjugate and reflectionless impedance matching power transfers become identical when all the components in a circuit are purely resistive. My point was not about the need for conjugate and reflectionless impedance matching in RF transmission lines and complex loads where the MPTT condition is not dissipative because it's a transform of energy in those conditions.

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