Impedance matching theory.

[MichaelR]

Before anybody else steps in ,all subject to internal resistance of battery not changing with power being dissipated for DC conditions .

[anthonys radios]

Excellent, just so!
This can be achieved by making the duration of the test very short, before polarization and temperature changes occur.
Holds for that Car battery (kilowatts) or the mains(in theory (mega mega watts )

Any body checked for a 3.1 ohm or 2.9 ohm load resistor, for themselves ?

[anthonys radios]

Load 2.9 ohms.
R total =5.9 ohms
I (V/R 3/5.9)= 0.5084 amps.
I sq. 0.2585 amps.
I sq R (watts) =0.74978 watts.(Resultant power developed in the load R)

Load 3.1 ohms.
R total = 6.1 ohms
I (V/R 3/6.1) =0.4918 amps
I sq. 0.2418 amps.
I sq R (watts) = 0.74979 Watts.(Resultant power developed in the load R)

eaoe !

If a plot of various values for R load is made, the resultant curve is symetrical,about the maximum value, but with all calculated power level less than the maximum value which occurs when the load matches the internal resistance.
As RF engineers will confirm, the general idea is to match the load with the source impedance to maximize power transfer.
Impedance takes into acount the circuit resistance (always present, except for Cryogenics I suppose) and reactive (capacitance/inductive components)

[kalee20]

Anthon, again, your calculations are bang on. And this demonstrates the Maximum Power Transfer theorem.

If you do the same for a pentode (with very small grid drive), again, you'll find you'd get maximum power in the load when the load impedance equals the valve's ra. More than ra, or less than ra, and the power will drop.

The down side, is that when you turn up the volume control, you'll get distortion at a pretty low level.

When you return the volume control to your original level, you can imagine reducing the load impedance. Sure, the power will drop (because you've no longer got load impedance equals source impedance). But then you turn up the volume control (thus increasing grid drive), and you'll find you can get greater power before the distortion threshold (which is fairly well defined for pentodes).

That's why the maximum power theorem is inappropriate for output stage matching - because the source voltage is not constant, it's under our control.

[MichaelR]

Excellent thread

[anthonys radios]

Three cheers MichealR, it is an interesting topic , and everyones input is very important.

Some more to come yet I hope ?

[jimmc101]

For the Maximum Power Transfer Theorem to apply the system must be linear.
i.e. the voltage and current are directly proportional.

Consider approaching things a different way.

Taking the battery mentioned earlier

Measure open circuit voltage = 3V
Measure short circuit current = 1A
From which we can deduce that the internal resistance is 3ohm
And the maximum power can be extracted by using a 3ohm load

Take now our pentode with the load between anode and HT

Measure open circuit voltage = 250V
Measure short circuit current = 48mA
From which we can deduce that the internal resistance is 5200ohm
And the maximum power can be extracted by using a 5200ohm load

Nothing to do with ra (which is many times larger)!!!

Ok so where have I cheated?

The battery is a linear system the voltage and current are directly related.

For the pentode the voltage and current are not directly related
The maximum voltage is equal to the supply, not (Ia x ra) as in a linear system.
The maximum current is limited by the bias conditions , not (Va / ra) as in a linear system.
(Ignoring saturation voltage, allowable distortion etc 5200ohm would be the correct load to obtain maximum power from this valve.
(see EL84 data sheet http://www.drtube.com/datasheets/el84-philips1969.pdf )

As kaylee20 pointed out for small signals where neither the signal voltage or current are limited by supply voltage or bias current then matching ra gives maximum power.

For large signals maximum power is obtained when supply voltage and bias current are matched.

If any one wants a to see how it's done for RF power amplifiers see (and very good luck to you!)
http://www.freescale.com/files/rf_if...ote/AN282A.pdf
and
http://www.freescale.com/files/rf_if...note/AN267.pdf


Jim

[smiler411]

The maximum voltage in a transformer coupled system is not limited by the supply and can be much greater.

The anode impedance at any time must be equal to Va/Ia.

As can be seen from the characteristic curves of the valve where Va is plotted against Ia this is not a constant but depends upon Ia and Va at any point. That is Ra varies depending upon how the valve is biased. Ra is given by the gradient of the curves.

Above a certain anode voltage for any bias condition however the Ra becomes very high and virtually constant.

Ideally for a pentode the Ra should be infinite. i.e. Anode voltage has no effect on anode current at all.

The pentode should be operated in the high impedance region. Therefore the ideal load would be a load line where the line passes through the knee in the curve for 0V grid bias. Generally the 0V grid bias gives the lowest anode voltage for the transition to high impedance.

This would ensure that for all sensible grid bias (grid always negative) the valve will be in the constant high impedance region.

If the load was higher the valve would enter the lower impedance region at the less negative grid bias and would be very non linear and cause distortion.

Ra would be much lower at this time

This would be less of a problem at lower loads than ideal but would mean that the volatge gain is reduced and a larger input would be required to provide the current to drive the load.

There is a limit to how large the input could be as you will start to drive the grid positive with very undesirable results and possibly exceed maximum cathode current.

The ideal load line would go through the 0V knee in the curve and be at twice supply volatge at no current. The DC quiescent conditions should be arranged to sit on this line.

[anthonys radios]

The maximum voltage in a transformer coupled system is not limited by the supply and can be much greater.

I agree,in a transformer coupled class A amplifier, I recall, in fact anode voltage can rise to twice supply.

Yet to read it fully, but I see no reason to disagree with what smiler411 says.

[anthonys radios]

Above a certain anode voltage for any bias condition however the Ra becomes very high and virtually constant.

In an ideal case would be nice, but not I fear in practice.

[jimmc101]

Originally Posted by smiler411

The anode impedance at any time must be equal to Va/Ia.

As can be seen from the characteristic curves of the valve where Va is plotted against Ia this is not a constant but depends upon Ia and Va at any point. That is Ra varies depending upon how the valve is biased. Ra is given by the gradient of the curves.

Sorry, I think this is confusing.
The first statement refers to the 'large signal impedance', the second to the 'small signal impedance'.

These two are not equal for all points unless the curve is a straight line passing through the origin i.e. linear.

For the EL84 I referred to earlier Va/Ia = 250v/48mA = 5.2k
Whereas the gradient is given as 38k at this point

Jim

[jimmc101]

Originally Posted by anthonys radios

The maximum voltage in a transformer coupled system is not limited by the supply and can be much greater.

I agree,in a transformer coupled class A amplifier, I recall, in fact anode voltage can rise to twice supply.

I agree too, but remember that this is the maximum peak to peak voltage, (assuming saturation voltage is zero)
Also that peak to peak current is twice quiescent current (assuming the valve's emission is up to it)

Which means using peak to peak values, impedance for maximum power = (2 x supply) / (2 x quiescent current) which gives exactly the same impedance as I gave using peak values.

Jim

[PJL]

As others have said, a valve is a non-linear device and a pentode is close to a constant current source. Ra is the effective plate impedance under normal operating conditions that appears across the anode load not a series resistance. If you look at the characteristics of a pentode, it rises quickly and then runs almost as a flat line with Ia changing very little with Va. You can estimate Ra by drawing a straight line on this graph and calculating Ra=dV/dI. The high Ra of a pentode means it can provide more voltage gain and reduced damping in tuned anode circuits.

The equivalent circuit for a valve can not be transposed into a fixed voltage with a series fixed impedance so the power transfer theorem does not apply.

[MichaelR]

Originally Posted by jimmc101

Sorry, I think this is confusing.
The first statement refers to the 'large signal impedance', the second to the 'small signal impedance'.

These two are not equal for all points unless the curve is a straight line passing through the origin i.e. linear.

For the EL84 I referred to earlier Va/Ia = 250v/48mA = 5.2k
Whereas the gradient is given as 38k at this point

Jim

This reply from Jim I think summarises very neatly what all of the contributors have been saying in their own way some with their own"subject to" some without. I would like to just clarity on one possible point though and that the word impedance is banded about a lot and sometimes often in the context of DC conditions not signal conditions.

Mike

[anthonys radios]

Are we agreed upon the following in the current context?
Maximum power transfer takes place if the source impedance matches the load.

Anode impedance is related/changes according to DC conditions.
Specifically Anode impedance varies with changes in anode voltage in practice.
Distortion can increases as signal amplitude increase.
Valves in general are not linear devices.
It is not practical to match the Anode Impedance of a Pentode output valve to the loudspeaker.
Valve makers suggest the Optimum Anode Resistance that should be used for impedance matching calculations, being the best compromise between power output and distortion.
It is possible we may all be saying the same thing, in different ways

[jimmc101]

No, sorry I disagree.

For a small signal device operating in its linear region where there is no need to consider any limits imposed by the device bias point the maximum power transfer theorem applies.

For a large signal amplifier where the voltage and current limits are not defined by the signal but by the device bias point, the load resistance for maximum power is simply determined by maximum voltage /maximum current.


Jim

[MichaelR]

Jim, the key thing is conditions of Linearity between Volts and Amps, then you cannot get it wrong.

I think you will confirm this

Mike

[jimmc101]

Originally Posted by MichaelR

Jim, the key thing is conditions of Linearity between Volts and Amps, then you cannot get it wrong.

I think you will confirm this

Mike

Absolutely, as I've tried to explain the moment anything interferes with this relationship the maximum power transfer theorem does not apply.

Jim

[Skywave]

Just a quick contribution from myself, if I may . . .

Picking up on ppppengion's earlier Post:

Originally Posted by ppppenguin

Maximum power transfer will always happen when the resistance of the source equals the resistance of the load but a few caveats:

(a) It assumes a linear system. If either R changes with current then it doesn't really apply.

(b) If the resistance is actually an impedance then the best match is at the complex conjugate impedance.

Where the source impedance is of the form Rs + jXs & the load is of the form RL + jXL, where the subscripts denote source and load respectively, a 'conjugate match' requires that XS = -XL and that RS = RL. In effect, this establishes resonance, since the reactances are mutually cancelling. If only the resistive parts are equated, this achieves a 'resistive match'.

This fundamental aspect of 'matching' is what causes so much trouble in the arena of matching a transmitter O/P to its aerial, via a transmission line, for maximum power transfer from the Tx. to the aerial proper.

Al. / Skywave.

[G8HQP Dave]

For linear conditions (e.g. small signal) the theorem applies, but it may be irrelevant because usually one wants to obtain some output/distortion compromise at higher power (where 'impedance' varies over the AC cycle).

It gets harder for RF output stages, as the 'impedance' varies even more (e.g. Class C - even when smoothed by the flywheel effect of the output tank circuit). Then another spanner gets thrown into the works, as sometimes we want maximum efficiency rather than maximum power. For some designs, such as Class E, these two maxima occur at different tunings but not everyone realises this.

I think the maximum power transfer theorem should always be taught with a health warning: in a given situation this may be true and helpful, true but misleading, or not applicable!