Maximum power transfer question !

[indulis]

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.

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.

 

[Ratchit]

indulis ,

The “ideal voltage source” doesn’t exist, so there will always be a source resistance.

That's true, but how does it pertain to the argument? I said in post #14 "The ultimate impossible best (max power transfer for a fixed load) is for the source to be zero impedance." That implies a limit, not a hard value.

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.

I never said that lowering the source impedance had anything to do with the max power theorem. In fact, the quote I posted from Wikipedia says "The theorem states how to choose (so as to maximize power transfer) the load resistance, once the source resistance is given, not the opposite" In post #16 you said

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

Well, lowering the source impedance will send more power to the load at a higher efficiency. That assumes you cannot change the load impedance. My math example proves that.

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.

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.

FiOk... lets do the math and kick it up a notch.

OK, sounds great. I already did one example, so it's your turn. Make sure you don't change the load impedance. Also show me how the maximum power transfer to the load occurs when the source impedance is equal to the load impedance (it doesn't).

Ratch

 

[simonbramble]

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

Simon

 

[simonbramble]

... 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]

simonbramble,

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

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. Your are treating the load impedance as a variable, which you would do if you are trying to adjust the load to the source. But since I specified the load as fixed, it is a constant, and you cannot differentiate with respect to a constant as you did. So the current I = V/(Rs+Rl), and the power is V²*Rl/(Rs+Rl)² . As you can see from the result, V and Rl do not change, so the maximum load power occurs when the denominator is the smallest. That happens when Rs is the smallest, or Rs becomes the closest to zero as it can be.

If you had checked my math example, and not found anything wrong with it, you would not have made that wrong assumption.

... 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

Yes, but changing the load (antenna) to match the source (transmitter) is not what we are talking about.

Ratch

 

[BrownOut]

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.

That's simply not true, and was not proved by the example. Maximum power transfer occurs when load impedance is matched to source impedance. Nothing to the contrary has been proved on this thread or anywhere else.

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.

simonbramble is right, and you are wrong.

 

[Grossel]

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.

 

[Ratchit]

Grossel,

I know this is quoted before, so I shouldn't repeat.

There is nothing wrong with repeating a quote if you want to draw attention to some aspect of it.

However, I have mentioned in an earlier post. In what situations are you able to adjust output impedance?

There are countless situations where the output impedance can be changed.

Also your explanation is based on a very popular misunderstanding.

What "misunderstanding" is that?

The consept is that output impedance is fixed and load impedance can vary.

Aren't those two impedances one and the same? Neither one is the source impedance.

The load effect is then highest when load impedance == source impedance. Disprove that.

What is the load effect? I already proved that if the load impedance if fixed, then the maximum power transfer occurs when the source impedance is the lowest possible value. See my example in post #8 of this thread for the proof.

Ratch

 

[Grossel]

Hi.

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).

 

[Ratchit]

Grossel,

If assumed that output impedance is possible to vary, then the situation flips.

Why are you even considering that situation? I said many times that the assumption is the load impedance is fixed.

If this is tha case, a lower output impedance would cause total effect to increase. Or - I should use the term resistance.

Don't change the output impedance. It is assumed to be fixed.

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.

Check what I said using resistors only. Then you don't have to worry about angles.

Misunderstanding - the teorem is based on a fixed output impedance. You assume variable output impedance.

No, I said many times the assumption is that the output impedance is fixed.

Load impedance - I meant to say impedance of the load (resistor).

Tube, transistor, resistor and or anything else that can conduct a current, it does not matter.

Ratch

 

[dougy83]

Wow, this is still going.

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.

 

[BrownOut]

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.

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. This can easily be proved by induction. Assuming RL=0 would result in P=0. Likewise, RL=∞ also resluts in P=0. Thus, RS=RL is a maximum by the maximum value theroem.

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.

That's irrelevant, since we are not talking about RL being fixed. We're discussing maximum power transfer, which always occurs when RL=RS.

 

[dougy83]

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.

I never said that. Try reading.

That's irrelevant, since we are not talking about RL being fixed.

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.

 

[BrownOut]

I never said that. Try reading.

I did read. Maybe you should read you own comments:

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.

Yeah I know, you were possessed.

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.

Rachet's point is equally irrelevant. The subject is maximum power transfer, not whatever trival, off-topic blather Ratchit is going on about.

 

[dougy83]

I did read.

Then try understanding.

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.

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.

Rachet's point is equally irrelevant. The subject is maximum power transfer, not whatever trival, off-topic blather Ratchit is going on about.

That "trivial, off-topic blather" has been stated as being completely wrong by a number of people - but it's correct, irrelevant or otherwise.

 

[BrownOut]

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.

Wrong again. It does give us all the answers we want. As I already proved, the result is a maximum at RS=RL per the maximum value theorem. Also,

SimonBramble's proof relied on RS^2 = RL^2. That doesn't mean RS=RL; RS=-RL is also a solution.

As I stated above, RS=-RL being a solution does not prove that RS=RL isn't the solution. You could use some understanding.

That "trivial, off-topic blather" has been stated as being completely wrong by a number of people - but it's correct, irrelevant or otherwise.

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]

As I stated above, RS=-RL being a solution does not prove that RS=RL isn't the solution.

I never stated that it did.

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.

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.

 

[BrownOut]

I never stated that it did.

You did.

SimonBramble's proof relied on RS^2 = RL^2. That doesn't mean RS=RL; RS=-RL is also a solution.

And the bolded part is wrong.

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.

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.

 

[dougy83]

The 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.

 

[BrownOut]

Since when is "V^2/20*10= V^2/10" correct? V^2/20*10= V^2/2

Should'a checked my edits before you posted.

There's not much point in even responding to the rest of your post.

Becuase you know I'm right.