Maximum power transfer question !

[Bloodninja]

Hello all,

while studying i had this odd question. we know that the maximum power transfer theorm indicates that impedence matching is necessery to achieve max power transfer to the load, but when it comes to amplifier networks, and studying the effect of load resistance, we say that the load resistance better be large to achieve higher gain, so how do these two facts get along ?

i'm saying that creating higher load for amplifier networks establishes some change in current gain to keep power to its max maybe ?

 

[simonbramble]

Power is a combination of current and voltage. To get maximum voltage transfer you need minimum load. To get maximum current transfer you need maximum load. To get maximum power transfer, you need a balance. Without going into the equations, if the output impedance is equal to the load (or the complex conjugate of the load) this results in maximum power transfer

 

[Bloodninja]

hmmm i believe you got it reversed... isn't max voltage transfer need larger load so the voltage drop over it becomes larger ? and maximum current transfer needs least load to make it easier to pass ?
max power is achieved when both voltage and current are in half

i guess i was confused a bit but its sorted now

thanks anyway and sorry for wasting space

 

[simonbramble]

Just on a point of contention, the maximum voltage transfer is when the load is least (highest resistance). Any internal resistance of the voltage source (battery etc) will develop a voltage drop across it when loaded. If you have an open circuit, there is no load hence no voltage drop, hence maximum voltage transfer.

You can work out the voltage at the junction of the load and series resistance of the battery. You can also work out the current into the load. Multiply the 2 together to get power in the load then differentiate with respect to the variable (Rload) and put equal to zero to get the maximum.

You should find that max power transfer occurs when Rload^2 = Rout^2.

It is interesting to note that because the equation above involves squares, this also accounts for complex loads (max power transfer occurs when the load is the complex conjugate of the output resistance)

 

[Ratchit]

If the load impedance is fixed, then more power will be transferred if the source impedance is smaller than if the source impedance equaled the load impedance. Of course it is usually not possible to change the source impedance.

Ratch

 

[Nigel Goodwin]

If the load impedance is fixed, then more power will be transferred if the source impedance is smaller than if the source impedance equaled the load impedance. Of course it is usually not possible to change the source impedance.

Ratch

No it won't, do the sums

It's only ohms law.

Maximum power transfer (which is only 50%) occurs when load impedance equals source impedance. But in electronics this is only VERY rarely a good thing.

 

[audioguru]

A modern audio amplifier has an output impedance of 0.04 ohms or less so it can damp the resonances of an 8 ohm speaker.
The impedances are not matched.
Years ago the very high output impedance of a vacuum tube amplifier fed an output transformer that matched the impedance of a speaker. Then the speakers had internal damping of their resonances or else they sounded boomy.

 

[Ratchit]

Nigel Goodwin,

No it won't, do the sums

I have, and have known this to be true for years.

It's only ohms law.

Maximum power transfer (which is only 50%) occurs when load impedance equals source impedance. But in electronics this is only VERY rarely a good thing.

It is very easy to prove what I said. It is true that if you can change only the load impedance, then the max power theorem holds true. But if the source impedance can be changed, and the load cannot be changed, then you want the source impedance to be as small as possible. Of course, as the audio guru pointed out, other factors might be in play to limit what you can do with the source impedance.

OK, take a 16 volt source with a 8 ohm resistance driving a load with a 8 ohms resistance. The current is 1 amp and the power dissipated in the load is 1²*8 watts. Now drop the source resistance to 2 ohms. Now the the current is 1.6 amps and the power dissipated in the 8 ohm load is 1.6²*8 watts, an increase of over twice as much power delivered to the load. The source now dissipates 1.6²*2 = 5.12 watts, down considerably from its previous 8 watts. So lowering the impedance of the source makes for a more efficient power transfer and a higher amount of power to the load.

Ratch

 

[Nigel Goodwin]

It is very easy to prove what I said. It is true that if you can change only the load impedance, then the max power theorem holds true. But if the source impedance can be changed, and the load cannot be changed, then you want the source impedance to be as small as possible. Of course, as the audio guru pointed out, other factors might be in play to limit what you can do with the source impedance.

OK, take a 16 volt source with a 8 ohm resistance driving a load with a 8 ohms resistance. The current is 1 amp and the power dissipated in the load is 1²*8 watts. Now drop the source resistance to 2 ohms. Now the the current is 1.6 amps and the power dissipated in the 8 ohm load is 1.6²*8 watts, an increase of over twice as much power delivered to the load. The source now dissipates 1.6²*2 = 5.12 watts, down considerably from its previous 8 watts. So lowering the impedance of the source makes for a more efficient power transfer and a higher amount of power to the load.

Yes, it's very easy to prove what you said - but your proof is that you're wrong

You're describing voltage transfer (which is far more efficient), we're talking about 'maximum power transfer' which occurs when the load and source are matched, and is only 50% efficient.

Like I said - do the simple maths.

 

[dougy83]

No it won't, do the sums

- but your proof is that you're wrong ...Like I said - do the simple maths.

Nigel, why don't you put your money where your foot is and do the sums.

Ratchit said that the power delivered to a fixed load is maximised when the source impedance is minimised. That is correct.
Where the source impedance is fixed and the load is variable, the maximum power delivered to the load is when the load and source impedances are equal.

 

[dougy83]

when it comes to amplifier networks, and studying the effect of load resistance, we say that the load resistance better be large to achieve higher gain,

Is this a [common emitter] transistor amplifier you're talking about? Increasing the collector resistance increases the output voltage swing by virtue of the fact that for a given input voltage change we have a given output current; by ohms law we find we have a larger output voltage for a larger resistance.. e.g. if an [input] base voltage change of 0.05V gives an [output] collector current change of 1mA, then the output voltage change will be 1V for a 1k resistor and 5V for a 5k resistor.

 

[Ratchit]

Nigel Goodwin,

Yes, it's very easy to prove what you said - but your proof is that you're wrong

You're describing voltage transfer (which is far more efficient), we're talking about 'maximum power transfer' which occurs when the load and source are matched, and is only 50% efficient.

Like I said - do the simple maths.

Well, I did the simple maths and posted them. I calculated power before and after the source impedance change. I did not calculate or describe any voltage changes. Did you even look at my derivation?

If the concept is hard to grasp, think of it this way. Reducing the impedance of the source will increase the series current. That can only increase the power dissipated in the load because its impedance has not changed while its current has increased.

Ratch

 

[Nigel Goodwin]

Nigel Goodwin,



Well, I did the simple maths and posted them. I calculated power before and after the source impedance change. I did not calculate or describe any voltage changes. Did you even look at my derivation?

If the concept is hard to grasp, think of it this way. Reducing the impedance of the source will increase the series current. That can only increase the power dissipated in the load because its impedance has not changed while its current has increased.

Ratch

Again - that isn't the point - maximum power transfer occurs when source and load impedances equal each other. If you vary the source impedance, you need to vary the load impedance to maintain maximum power transfer.

If the load is higher than the source, you don't get maximum power, if the load if lower than the source you don't get maximum power either.

Here's a simple example:

Now 10V source, 1 ohm impedance, 1 ohm load (matched).
Current = 5A, power in load 25W

Now 10V source, 1 ohm impedance, 2 ohm load (not matched).
Current = 3.333A, power in load 22.217W

Now 10V source, 2 ohm impedance, 1 ohm load (not matched).

Current = 3.333A, power in load 11.109W

Notice that the matched one is highest power.

If you plot all the points, you will get a curve on the graph, which shows a single point of maximum power transfer, which is when source and load are equal.

I'm fully aware that this isn't usually what you want, a low source feeding a high load is much more normal, but it's not the maximum power point.

 

[Ratchit]

Nigel Goodwin,

Again - that isn't the point - maximum power transfer occurs when source and load impedances equal each other. If you vary the source impedance, you need to vary the load impedance to maintain maximum power transfer.

I am referring to the case where the load is fixed and cannot be changed. In that case, you want the source impedance to be the lowest possible to maximize the power transfer to the load. The lower the source impedance, the more power will be transferred to the nonchangable load, and it will be transferred more efficiently.

If the load is higher than the source, you don't get maximum power, if the load if lower than the source you don't get maximum power either.

I keep saying that the conditions for my statement are that the load cannot be changed.

If you plot all the points, you will get a curve on the graph, which shows a single point of maximum power transfer, which is when source and load are equal.

In your example, you are juggling both the load and source impedance. That is not what my premise is about.

I'm fully aware that this isn't usually what you want, a low source feeding a high load is much more normal, but it's not the maximum power point.

If the load impedance is fixed, you will transfer more power to the load if the source impedance is less than the load impedance. The ultimate impossible best is for the source to be zero impedance. Then all the power is transferred to the load.

Ratch

 

[dougy83]

Here's a simple example:

Why don't you show the example for RL = 1 ohm, RS @ {0.5, 1, 2} ohms?

 

[indulis]

Nigel is correct; this is very simple… an ideal voltage source connected to a resistive divider, where the “upper resistor” is the source resistance. The maximum power that can be extracted from the source and be dissipated in the “lower load resistance” is when the resistors are equal.

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

 

[Nigel Goodwin]

I keep saying that the conditions for my statement are that the load cannot be changed.


In your example, you are juggling both the load and source impedance. That is not what my premise is about.

But your 'premise' isn't anything to do with the case in point, which is MAXIMUMM POWER TRANSFER - in case you haven't noticed, it's the title of this thread.

You might try a little googling, there's lot's out there about it, including this:

https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

 

[Grossel]

Would just say that the math behind can under certain circumstances fool the understanding about max transfer. What is very easy to forget is that in most cases output impedance isn't possible to change.

When doing the math, one can easilly be fooled to think about output impedance as a variable.

 

[Ratchit]

indulis,

Nigel is correct; this is very simple… an ideal voltage source connected to a resistive divider, where the “upper resistor” is the source resistance. The maximum power that can be extracted from the source and be dissipated in the “lower load resistance” is when the resistors are equal.

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

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

 

[Ratchit]

Nigel Goodwin,

But your 'premise' isn't anything to do with the case in point, which is MAXIMUMM POWER TRANSFER - in case you haven't noticed, it's the title of this thread.

You might try a little googling, there's lot's out there about it, including this:

My premise is pertinent to the point in reference to the maximum power transfer for a fixed load. More power at a higher efficiency is delivered when the source impedance is the smallest it can be.

I saw nothing in the link you gave that contradicts what I said. If fact, it 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. It does not say how to choose the source resistance, once the load resistance is given. Given a certain load resistance, the source resistance that maximizes power transfer is always zero, regardless of the value of the load resistance."

You cannot dispute the mathematics.

Ratch