You would do well to look at the mathematical example I gave which proves what I averred. If the load impedance is fixed, lessening the source resistance increases both the power delivered to the load and the efficiency. As I said before, at a zero source resistance, all the power of the source if transferred to the load at a 100% efficiency. When the source impedance is equal to the load, the efficiency is only 50%.
Ratch
The “ideal voltage source” doesn’t exist, so there will always be a source resistance.
If you re-read my post, I actually agreed with you in terms of delivering max power to a fixed load, but that has nothing to do with the maximum power theorem.
When it’s a fixed load, obviously to maximize I^2*R in the load, the source resistance needs to be as small as possible, but that is not the same as extracting the maximum power to the load
As for the math, in a simple “real” source/resistive load circuit, to find the maximum power in the load, you differentiate the power in the load with respect to the load resistance. The maximum occurs when the derivative is zero, or when the load resistance equals the source resistance.
FiOk... lets do the math and kick it up a notch.
Here is the maths. Nigel is right - you need to put the load and source equal (or equal to the complex conjugate) for max power transfer. Attached is the maths. Hope this helps
... and maximum power transfer is used in RF systems where you have to match the antenna to the output impedance of the transmitter to get maximum power (not voltage or current) into the antenna
Ratchit said:Instead of a mathematical derivation, I used a simple example. I showed that if the load impedance is fixed, then maximum load power does not occur when the source impedance is equal to the load impedance. It occurs when the source impedance is as low as possible.
Ratchit said:Your math is correct, but your application of it is faulty. Thereby, your result is wrong. And everyone else who believes your conclusion is wrong.
I know this is quoted before, so I shouldn't repeat.Instead of a mathematical derivation, I used a simple example. I showed that if the load impedance is fixed, then maximum load power does not occur when the source impedance is equal to the load impedance. It occurs when the source impedance is as low as possible.
I know this is quoted before, so I shouldn't repeat.
However, I have mentioned in an earlier post. In what situations are you able to adjust output impedance?
Also your explanation is based on a very popular misunderstanding.
The consept is that output impedance is fixed and load impedance can vary.
The load effect is then highest when load impedance == source impedance. Disprove that.
If assumed that output impedance is possible to vary, then the situation flips.
If this is tha case, a lower output impedance would cause total effect to increase. Or - I should use the term resistance.
Magnitude of impedance could match, but still be at different angle, then the statement may not be true any more. Lets keep the angle values the same.
Misunderstanding - the teorem is based on a fixed output impedance. You assume variable output impedance.
Load impedance - I meant to say impedance of the load (resistor).
dougy83 said:SimonBramble's proof relied on RS^2 = RL^2. That doesn't mean RS=RL; RS=-RL is also a solution. A negative resistance is not valid, so for all realisable resistances, i.e. R >= ~0, the derivative=0 method won't give us all the answers we want, as it doesn't take into account that resistance may not be negative.
Here's the graphs for the power through a fixed load; one includes a negative resistance range just for viewing pleasure, the other only positive resistances.
For a fixed source resistance, RS, you can see maximum power is at RS=RL=10 ohms.
For a fixed load resistance, you can clearly see it's maximum at RS=0 ohms.
I never said that. Try reading.You are incorrect. The existance of a solution for negative resistance does not disqualify the result RS=RL as a solution to the maxumum power transfer equation.
It's relevant to the point made by Rachit. There seems to be a bunch of blind followers around; what Rachit was saying (at least in the first post, I haven't really been paying much attention) has been true.That's irrelevant, since we are not talking about RL being fixed.
dougy83 said:I never said that. Try reading.
dougy83 said:the derivative=0 method won't give us all the answers we want, as it doesn't take into account that resistance may not be negative.
It's relevant to the point made by Rachit. There seems to be a bunch of blind followers around; what Rachit was saying (at least in the first post, I haven't really been paying much attention) has been true.
Then try understanding.I did read.
the derivative=0 method won't give us all the answers we want, as it doesn't take into account that resistance may not be negative.
That "trivial, off-topic blather" has been stated as being completely wrong by a number of people - but it's correct, irrelevant or otherwise.Rachet's point is equally irrelevant. The subject is maximum power transfer, not whatever trival, off-topic blather Ratchit is going on about.
dougy83 said:I marked it in bold for you to see. I said it won't give all the answers. It gives an answer, but without the constraints mentioned previously.
dougy83 said:SimonBramble's proof relied on RS^2 = RL^2. That doesn't mean RS=RL; RS=-RL is also a solution.
That "trivial, off-topic blather" has been stated as being completely wrong by a number of people - but it's correct, irrelevant or otherwise.
I never stated that it did.As I stated above, RS=-RL being a solution does not prove that RS=RL isn't the solution.
Ok, so what is the source resistance of a supply that will result in the maximum power being transferred to a 10 ohm load? Please easily prove that.Actually no it's not correct. It only considers one resistance changing. It can easily be proved that his simple example does not give maximum power.
dougy83 said:I never stated that it did.
dougy83 said:SimonBramble's proof relied on RS^2 = RL^2. That doesn't mean RS=RL; RS=-RL is also a solution.
The answer is 10 ohms. The power transfered is: V^2/20^2*10= V^2/40 Watts. If you change the source resistance, to maintain maximum power transfer, you must change the load resistance to match.Ok, so what is the source resistance of a supply that will result in the maximum power being transferred to a 10 ohm load? Please easily prove that.
Since when is "V^2/20*10= V^2/10" correct? V^2/20*10= V^2/2The answer is 10 ohms. The power transfered is: V^2/20*10= V^2/10 Watts.
Since when is "V^2/20*10= V^2/10" correct? V^2/20*10= V^2/2
There's not much point in even responding to the rest of your post.
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