Understanding maximum power transfer theorem

[PainterGuy]

Hi :)


I'm trying to understand 'background' working of the maximum power transfer which says that power transfer takes place across R_L when it is equal to R_th (R_L is load resistance and R_th is Thevenin resistance). It is easily proved by doing some math.


P=VI, P=I^2.R, P=V^2/R, V=IR


Let's focus on P=V^2/R_L. To have maximum power transfer to R_L, V should be as large as power (V is volts dropped across R_L). V across R_L could be found using voltage divider rule: (R_L x E)/(R_L + R_th). Further R_L should be as little as possible because it is denominator. But also note that making R_L smaller would reduce the volts dropped across R_L.


But now focus on P=I^2.R_L. Compare it with the previous analysis of P=V^2/R_L. In (I^2.R_L), R_L should be as large as possible which is in contrast with the previous analysis which required R_L to be minimum. So some compromise is needed. But there is another point to note. R_L and R_th are in series so if R_L is made too big then the value of I would drop.


I hope you could see where I'm coming from (or, rather trying to come from! :) ). Could you please help me to really understand the working of maximum power transfer theorem? Many thanks.


Cheers

 

[KingNothing]

You are absolutely right that it is a "compromise". Not surprisingly, this comes at a point of great symmetry. Let's imagine that we have a 5V power supply with 1-ohm V_th. If the load resistor has a value of x (so we can vary it), what is the expression for power across the load resistor?

Wolfram Alpha Expression

 

[Jiggy-Ninja]

Power is a combination of voltage and current. If you have 0 load resistance, you have no voltage dropped across it. But if you have infinite load resistance, you will have no current.

I used King's formula to make a better plot with Wolfram. It shows the load power for a 5V 50 ohm source relative to the load resistance. You can clearly see that the plot peaks at the 50 ohm mark.

 

[PainterGuy]

Many thanks, King, Ninja, for the help.

Cheers

 

[artikk]

There's a better way of proving it by using calculus, if you have the background. My professor in a circuit analysis course did this. Find the equation of power for the load resistor and then take it's derivative with respect to RL. Make Vg(source voltage) and Rth(source impedance) constants, though, to simplify the differentiation. The load resistor value for which the derivative of the power is 0 is the value that will achieve maximum power transfer.
http://iamsuhasm.wordpress.com/2009/04/27/proof-for-the-maximum-power-transfer-theorem/

 

[PainterGuy]

Hi again,

Just curious to know if there is also a minimum power transfer theorem of some kind.

When R_L is not equal to R_Th, then the power dissipated by the R_L won't be maximum. Is this because of the voltage drop across R_Th?

Please help me out. Thank you

Cheers