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Maximum Power Transfer Theory for Circuit Example

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ACharnley

Member
Hi,

According to maximum power theory, maximum transfer occurs when input impedence = output impedence. As represented in the image below, the combined impedence of LCR1 should equal R2. At this point X is the ideal voltage level.

The generator is 0.5A variable voltage (I may have used the wrong symbol).

I calculated LCR for various frequencies but for this example it's 50Hz. You can see that I computed 27ohm which is about 13V. This matches my expectations.

Screenshot_2019-03-20_11-50-50.png

I then created a spreadsheet to see how changing the input impedence would change. You can see that for this capacitor selection the lowest impedence is at 25Hz.

Screenshot_2019-03-20_11-51-44.png

Which suggests the highest power extraction occurs at 25Hz when R2 is lowered so that X is 2.32V. This is much lower than I was expecting, but could be right.

Can someone that knows what they are doing :) verify my calculations?

Regards,

Andrew
 

MikeMl

Well-Known Member
Most Helpful Member
Is the driving source a current-source, or is it supposed to be a voltage source?

If it is a current source, then the power delivered to R2 is dependent only the value of R2 (R2*I^2).

If it is a voltage of 0.5Vrms, then the maximum power transfer to R2 occurs when R2=4 Ohms at a frequency of 26.36 Hz.
 
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ACharnley

Member
I swore I wrote a line which said I'd used the wrong symbol. It's a current source, fixed 0.5A.

The delivered power also depends on the impedance of the source., which is what maximum power transfer calculation is about.
 

Ratchit

Well-Known Member
Hi,

According to maximum power theory, maximum transfer occurs when input impedence = output impedence. As represented in the image below, the combined impedence of LCR1 should equal R2. At this point X is the ideal voltage level.
Where did you learn that theorem? You have completely misstated it. If the source impedance cannot be changed, then the maximum power transfer occurs when the load impedance is the conjugate of the source impedance. If the source resistance can be lowered, then more power will be transferred, even if the load impedance is not changed from its previous value. The maximum power transfer occurs when the source resistance is zero.

The generator is 0.5A variable voltage (I may have used the wrong symbol).
Voltage is measured in volts (V), not amps (A). What kind of generator is it?

I calculated LCR for various frequencies but for this example it's 50Hz. You can see that I computed 27ohm which is about 13V. This matches my expectations.
You got two things going on here. The first is the current changes caused by the inductance and capacitance when the frequency changes. The second is the load distribution determined by the source and load impedances . A resonant frequency will enable the source generator to output the greatest power. Otherwise, the capacitor will inhibit the power at low frequencies and the inductance at high frequencies. If that is a current generator you have, it will have to be a monster to punch 0.5 A through the cap at low frequencies or the coil at high freqs.

View attachment 117257

I then created a spreadsheet to see how changing the input impedence would change. You can see that for this capacitor selection the lowest impedence is at 25Hz.
I don't see any figures for power transferred, so I guess it doesn't mean too much.

View attachment 117258

Which suggests the highest power extraction occurs at 25Hz when R2 is lowered so that X is 2.32V. This is much lower than I was expecting, but could be right.

Can someone that knows what they are doing :) verify my calculations?
Back to the drawing board for you.

Ratch



Regards,

Andrewhttps://www.youtube.com/watch?v=Hy-sVByUHqE[/QUOTE]
 

Ratchit

Well-Known Member
I swore I wrote a line which said I'd used the wrong symbol. It's a current source, fixed 0.5A.

The delivered power also depends on the impedance of the source., which is what maximum power transfer calculation is about.
Well, now that we know you have a 0.5 A current source driving things, we can deduce that there is no maximum or minimum transfer of power. The power transferred will be i^2 tunes R2 no matter what the the source impedance or frequency is. The voltage at "x" will be I times R2. That is because no matter what the source impedance is, 0.5 A will always be present through R2.

Ratch
 

ACharnley

Member
The voltage changes and is at it's maximum at point of impedance equality (complex conjugate). Power is 0.5A times this voltage.

https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem (see Reactive circuit).

So at some point there is a limit to how low R2 can go before the voltage begins to drop. If this point (maximum power transfer theory) is equal to input impedance, and knowing there is always 0.5A, I can calculate the voltage. For a digital circuit I can therefore monitor this voltage and alter a duty cycle to impersonate the resistive load value.

I think.
 

MikeMl

Well-Known Member
Most Helpful Member
You are missing a very important point. Since I1 is a current source, it forces 0.5A through the circuit regardless of what the impedances of L1 and C1 are!

Look at I(R2) and Power(R2) vs freq from 1Hz to 100Hz:
d1.png

Now look again at a fixed frequency of 20Hz but with R2 varying from 1 Ohm to 100 Ohms.
d1a.png
Notice that power in R2 is proportional to value(R2), so the max power in R2 occurs when R2 is infinite.

If the source is a 0.5V voltage source, I can tell you by inspection that the maximum power in R2 happens when it is 4 Ohms and when the frequency is such that the reactances of L1 and C1 are equal.
 
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ACharnley

Member
Agreed 0.5A is fixed, but you're missing another point - the voltage varies! If the input impedance is 10 ohm you will get less power out using a 5ohm or a 15ohm output resistor than a 10ohm resistor.

Maximum transfer is when the input impedance matches the output impedance. With an inductive source offsetting inductance with capacitance results in less reactance (aka phase difference).
 

ACharnley

Member
BTW I'm not saying my calculations are collect. The reactance is correct (I can see that by selecting different value capacitance). I suspect the voltage is wrong - yet if the resistances are equal it should be right. Still, if the lowest input impedance is at 25Hz (see the spreadsheet above) and that yields 2.32v then that x 0.5A is the lowest wattage. It's this I need clarity with.
 

crutschow

Well-Known Member
Most Helpful Member
Maximum transfer is when the input impedance matches the output impedance.
True.
But a current source has an infinite impedance so the output impedance would also need to be infinite for maximum power (which in this case would also be infinite).

You need to work with a voltage source, not a current source.
 

MikeMl

Well-Known Member
Most Helpful Member
True.
But a current source has an infinite impedance so the output impedance would also need to be infinite for maximum power (which in this case would also be infinite).
...
Which is exactly what the sim showed... (I just quit at 100 Ohms ;))
 

ACharnley

Member
It doesn't have an infinite impedance though - it changes it's impedance based on frequency and inductance, which I phase correct (as best as possible) using various capacitance values. The calculated figures are in the spreadsheet.
 

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MikeMl

Well-Known Member
Most Helpful Member
The equations used in the spread sheet do not reflect the reality of the circuit topology.

It is a series circuit fed by a current source. That means that the current in every element of the circuit is determined only by the the magnitude of the current delivered by the source.
 

crutschow

Well-Known Member
Most Helpful Member
It doesn't have an infinite impedance though
An ideal current source has an infinite impedance.
An ideal voltage source has zero impedance.
Take your pick.
 

Ratchit

Well-Known Member
It doesn't have an infinite impedance though - it changes it's impedance based on frequency and inductance, which I phase correct (as best as possible) using various capacitance values. The calculated figures are in the spreadsheet.
You cannot change the fact that 0.5 amp of current is present in R2 no matter what frequency you drive it with or how much reactance you put in series with the current source. If the current does not change, then neither does the power transferred. You defined it to be a constant value, so the power transferred must be constant also, unless you change the value of R2. Keep in mind that the power associated with the reactance voltage goes back and forth to/from the source generator and not to R2.

Ratch
 

ACharnley

Member
No dispute that the current going through R2 is 0.5A. Now, past and future. Fixed.

Power = current * voltage. Voltage changes, hence power changes.

I'm not dealing with a theoretical ideal current source. It has impedance, values on the spreadsheet.

Max power transfer is when the output impedance = input impedance, values on the spreadsheet. Too little a load or too much a load = less output.

Now assume the output load R2 is not a filament bulb or a plain resistor but a digital circuit which changes duty cycle. The means to determine the equivalent resistance is by measuring the voltage, since the current is known, and adjusting the duty cycle until the output impedance is close to the input impedance.

Got it? :)
 

Ratchit

Well-Known Member
No dispute that the current going through R2 is 0.5A. Now, past and future. Fixed.

Power = current * voltage. Voltage changes, hence power changes.
No, voltage across R2 does not change because the current is steady at 0.5 amps.

I'm not dealing with a theoretical ideal current source. It has impedance, values on the spreadsheet.
If the current source does not have a very high impedance, it is not a current source. If your load is going to match a current source, then we are not talking about a current source anymore. Instead it is a matter of voltage division.

Max power transfer is when the output impedance = input impedance, values on the spreadsheet. Too little a load or too much a load = less output.
Same answer as before. Not practical to match the load to a high impedance current source.

Now assume the output load R2 is not a filament bulb or a plain resistor but a digital circuit which changes duty cycle. The means to determine the equivalent resistance is by measuring the voltage, since the current is known, and adjusting the duty cycle until the output impedance is close to the input impedance.

Got it? :)
No, what does the duty cycle have to do with the input/output impedance?

Ratch
 

crutschow

Well-Known Member
Most Helpful Member
It has impedance, values on the spreadsheet.
That's the impedance of the network in series with the source, not the current source, which is infinite.
Max power transfer is when the output impedance = input impedance, values on the spreadsheet.
No.
You need to better understand what a current source is.
It generates a fixed current, independent of any other element in series with it, thus the value of the network impedance has no effect on the load power.

Got it? :rolleyes:
 

ACharnley

Member
Yes that is true enough though irrelevant for the course of computation. It is best to forget the current source and work with the LCR impedance.
 
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