# Maximum Power Transfer Theorem and Transformers

Discussion in 'General Electronics Chat' started by Spacerat, Jul 15, 2016.

1. ### Spacerat Thread Starter New Member

Aug 3, 2015
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0
Hello everyone,
I just studies the maximum power transfer theorem which states that the max power that can be transferred from a source to a load is 50% of the total power generated by the source. This max transfer only occurs in the case of of conjugate impedance matching between the source and the load.

That said, an ideal transformer is a device that changes (increases or decreases) voltage with a consequential change in the current while the input power remains equal to the output power. How is that possible, in light of the max power transf. theorem, to transfer power in its totality and not only 50% of it?
What am I missing?

thanks.

2. ### AnalogKid Distinguished Member

Aug 1, 2013
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1,297
You cannot apply a theorem based on real world impedances to an ideal device with no impedance.

ak

ps. The MPTT is an excellent way to analyze and optimize the fun you have playing poker.

3. ### Spacerat Thread Starter New Member

Aug 3, 2015
19
0
Ok, so let's consider a real transformer (with losses) that sits between a source and a load. For things to operate properly, I think we need

a) impedance matching between the source and the transformer
b) impedance matching between the transformer and the load

But at the source/transformer connection, even in the case of perfect impedance matching, do we lose 50% of the incoming power in virtue of the MPTT?

4. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
No. You cannot consider the primary circuit and the secondary circuit independently. A real transformer will transform voltage, current, and impeadance. When you analyse the primary side you have to include the secondary impedance reflected into the primary side and vice versa. Read the following chapter of the eBook and come back to us.

5. ### AnalogKid Distinguished Member

Aug 1, 2013
4,689
1,297
No. The transformer primary draws current from the source based on its own equivalent impedance and the impedance of the source. In a normal case, the transformer has a much higher input impedance (ohms) than the AC power line equivalent output impedance (milliohms). This is why a transformer that needs 1 amp can be connected to a 200 A domestic feed and a) not burn up; b) not short out the local electric grid. And the power being transferred is nowhere near the maximum available. But nothing is "lost", and the system efficiency is very high. In a power station where high voltage transmission lines are stepped down for local distribution feeds, things *do* go for maximum power transfer, and the impedances are matched for this.

At the transformer secondary, MPTT can apply if you are trying to wring out every possible watt and keep thermal losses to a minimum. Again, if you have a 100 W power supply running a 10 W load, then things will not be operating near MPTT, but who cares?

Back to your use of the word "lose" - Something to keep in mind is that efficiency is the ratio of total power in the load to total power into the system, NOT total power *available* to the system. Without getting into how the conversion efficiency of a power supply changes based on its load conditions, a source with an ESR of 1 ohm driving a load of 9 ohms is 90% efficient, and the power loss is 10% of a low number. Sure, maximum power transfer happens when the load is 1 ohm, but the efficiency drops to 50% and it's 50% of a much larger number. Load power went up, but system power and system power losses went up by much more.

ak

6. ### crutschow Expert

Mar 14, 2008
13,492
3,372
The problem is, they they taught you about the MPTT with teaching where it is normally used.
The MPTT is only used where we want the maximum power with no concern for efficiency and the source has a relatively high impedance (such as a solar panel or an RF amplifier).
This does not apply to typical power usage, such as power from the mains, or a battery, or an audio amplifier.
In those cases you generally want high efficiency, so the source resistance is much lower (ideally zero) than the load resistance.
Note that for those types of sources, the maximum power available would be much higher than those sources could safely deliver.

7. ### Spacerat Thread Starter New Member

Aug 3, 2015
19
0
Hello everyone and thank your help. I am still a little confused and going through the information at

That said, I can see different types of transfer taking place. We could talk about the transfer of maximum power (energy per unit time), or of maximum voltage, of maximum current or of maximum energy.

Is that correct?

8. ### crutschow Expert

Mar 14, 2008
13,492
3,372
Yes, you are basically correct.
Of course, 100% efficiency means there is no internal resistance, not that the internal resistance doesn't dissipate power (which would be true only at a zero output current).

9. ### anhnha Active Member

Apr 19, 2012
776
48
By the way, I have just read an old thread here. Below is part of post #4. Is it wrong?
I checked some cases and it is wrong. However, I would like to be confirmed from members here.

10. ### Spacerat Thread Starter New Member

Aug 3, 2015
19
0
Thanks crutschow.

In the case of power transmission through the powerline, the power company generates electric energy at some location (where the power plant is, for example). The total amount of energy generated per second (total power) must then be transferred not to a single customer but to the many customers who have electronic devices needing electric energy to function. This scenario seems to require that the least amount of energy (not power) is dissipated during the transfer from the power plants to the customer premises. I know that AC current and transformers are used to limit Joules "power" losses" by increasing the voltage and decreasing the current so that most of the generated power is transferred to the customers, correct? But I know that the MPTT does not apply in this case.

As mentioned, all electronic devices need electric energy to function. The needed power P is always specified and specific to the device. Power is P=IV. For example, USB devices must always be provided a 5V voltage. A voltage transformer is used if the available voltage is not 5V. Once provide the 5V, the devices will draw the correct current it needs if that current can be supplied. What is a situation in which the necessary current would not be supplied?

When charging USB devices, the voltage must be 5V but the device can function with different amounts of current (the larger the current the faster the charging, up to a limit). Devices that contain LEDs and coin batteries (3V) must have small chips that keep the current passing through the LED constant even when the energy of the battery is decreasing, correct? For example, if the available power decreases because the battery is discharging, I guess the chip will try to keep the current at fixed value while the supplied voltage decreases.

11. ### crutschow Expert

Mar 14, 2008
13,492
3,372
That statement is not wrong, it is correct.
If you reduce the source impedance with a given load resistance, the load power will increase.
What kind of cases did you "check" that show otherwise?

12. ### anhnha Active Member

Apr 19, 2012
776
48
Thank you. You are right. I made a mistake. I kept source resistance constant and changed load resistance.
If source resistance can not be reduced then the maximum power transfer will be when load and source resistances are matched.

13. ### AnalogKid Distinguished Member

Aug 1, 2013
4,689
1,297
No. First, it would not be at a slow rate. Second, you changed the conditions in the middle of the comparison.

MPTT is best understood with a power source with a constant output impedance (Zout) driving one of three possible loads. For example, a 10 V source with a 10 ohm output impedance, and loads of 1, 10, and 100 ohms (or any other spread that has 10 ohms in there somewhere). When you calculate the power dissipated in each of the loads, you see that the maximum power is in the load that matches Zout. A finite and constant Zout is a condition that is explicitly stated in the theorem.

As mentioned above, MPTT usually is applied in signal systems rather than in power systems, partly because signal systems frequently have relatively high circuit impedances (25 to 300 ohms) and relatively low signal power. In these situations, the power dissipated in an output impedance usually is not high enough to cause heating and reliability effects.

ak

Last edited: Jul 23, 2016
14. ### Spacerat Thread Starter New Member

Aug 3, 2015
19
0
Hello AnalogKid, thank you.

Papabravo mentioned that "...No. You cannot consider the primary circuit and the secondary circuit independently. A real transformer will transform voltage, current, and impeadance. When you analyse the primary side you have to include the secondary impedance reflected into the primary side and vice versa. Read the following chapter of the eBook and come back to us....."

Does that mean that a transformer does not have a specific, rated input impedance and output impedance? In very basic terms, impedance is the ration V/I. I see how the transformed changes the incoming voltage (and current). There are devices, however, whose impedance is specified (antennas, cables, etc.)

15. ### crutschow Expert

Mar 14, 2008
13,492
3,372
A transformer does have some impedance due to the resistance of the windings and the leakage inductance, but that's not related to the transformers that have a specified impedance rating.
The impedance rating means the transformer was designed to operate with those values for the input source and the output load.
If its rated input impedance is different than its rated output impedance, that just means the turns ratio are such that the source input impedance of the input value will appear as the output value at the load.
The actual transformer impedance is typically much less then its rated "impedance" since you want the transformer to be transparent (ideal) to the transforming process.

Power transformers are designed to have as low an impedance as possible (ideally zero).
Large power transformers come close to this ideal with efficiencies over 95%.

16. ### Spacerat Thread Starter New Member

Aug 3, 2015
19
0
AnalogKid and crutschow,

please be patient and let me start from the very beginning. I am still not grasping the fundamental ideas. Let me share with you what I know.

A generic electronic device can be characterized by its impedance X which is a function of frequency omega: X(w)=R(w)+j*I(w).
I have studies the maximum power transfer in the context of a source connected to a load (for ex. antenna) via a transmission line. the goal is to send the most power (ideally 100%) from the source to the load. the source, the t-line and the load have their own impedance.
To do this, the MPTT is applied and the conjugate impedance matching is performed at the following interfaces:

Source/Transmission line

This way there are no energy reflections at the the interfaces and all the energy flows to the load.

Back to the transformer which is supposed to transfer power from one end to the other end while changing the voltage. Could you try to explain to me again how to connect the concepts I know to how the transformer works and the impedance matching idea for max power transfer?

17. ### AnalogKid Distinguished Member

Aug 1, 2013
4,689
1,297
By its nature, a transformer has a finite output impedance. If the transmission line and load both match this impedance, then there are no reflections at either of the two interfaces and all of the energy leaving the transformer reaches the load (minus some power loss in the transmission line). With respect to the energy entering the primary of the transformer, some of it is lost in the core. 50% of what is left is dissipated in the transformer output impedance and 50% reaches the load.

ak

18. ### Spacerat Thread Starter New Member

Aug 3, 2015
19
0
ok, thanks. So, even in the case of a transformer, perfect impedance matching does not prevent the dissipation of 50% of the incoming power.
That still happens even if there are no energy reflections at any of the interfaces...

19. ### crutschow Expert

Mar 14, 2008
13,492
3,372
That's not correct.
A transformer impedance is normally very low compared to the load impedance (due mainly to leakage inductance and winding resistance).
It transfers the load impedance back to the source by the square of the turns ratio, so you can match the load impedance to the source impedance by adjusting the turns ratio.
This gives the maximum power transfer between the source and load (50% to each).
Very little of that power (perhaps a few percent) is lost in a well designed transformer.
You never match the transformer impedance to either the load or source impedance, you just match the load impedance to the source impedance.

Note that when a transformer is designated with an impedance (such as 600Ω) that's not the actual impedance of the transformer, it's the impedance it's designed to work with. The measured transformer impedance would be much less than that.