Impedance matching theory.

[anthonys radios]

Split from this thread:-

https://www.vintage-radio.net/forum/...ad.php?t=43333

Graham, Station X, Forum Moderator.
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There should be damage in a valve amplifier by using a lower impedance speaker, output valves are said to have an "optimum output impedance", this like many things is a compromise between power output versus distortion.
Valve amplifiers object to having no speaker (if there is an input sigal)as large back emf occur, even causing arcing across valve bases and transformer.
For maximum power transfer the load impedance should match the source impedance.
You cannot however measure impedances with a simple multimeter, this only measure the DC resistance.
Speaker impedances vary with frequency, the manufacturer often quote speaker impedance as measured at some predefined frequency, often 1Khz.

[anthonys radios]

If the O/P finds the book explanations complicated, then remember for maximum power transfer from a to b (whatever a or b actually is), the load impedance must always match the source impedance. This applies even if the source was a dry cell, and the load a torch bulb.
However in this particular case the load is a Loudspeaker, and the source is an output valve.
So far as a valve amplifier goes, the source impedance is generally higher than the load impedance, this then requires a Matching step down transformer.
(AC Input voltage high, AC input current low, AC output volts low, AC output current high)

The output impedance of a typical pentode is very high, but for a Power Output Vave the impedance actualy used in matching calculations is the value thats gives the best compromise between output power and low distortion, and this "optimum" value is usually much lower than the normal pentode valve output impedance.

Valve amps are not normally harmed by low impedance, or short circuit loads.
Transistor amps are the other way round, they can be damaged by loads of lower impedance than normal.

[PJL]

I am not sure the power transmission efficiency rule applies here. A pentode has a higher output impedance simply because it behaves almost like a constant current source which is a distinct advantage when considering gain or damping of tuned circuits. Kalee20 though does make a good point on limiting voltage swing in this case.

A simple way to look at it the effect of load impedance is by looking at the limiting conditions and remembering that power = volts x current:

1) If the impedance is too high the voltage swing extends outside the linear operating range of the valve (eventually hitting the rail) but the current swing is small.
2) If the impedance is too low the voltage swing is small and the output power is limited by the current that the valve can deliver.

There is an optimal load for each valve type which will deliver maximum power with acceptable distortion. This load impedance is normally specified in the manufacturers datasheet.

The load impedance in this case is the output transformer. A transformer does exactly what it says and transforms the voltage swing at the primary to a voltage swing at the secondary. The primary impedance is therefore a 'reflection' of the load on the secondary and will change in direct proportion to the change of the secondary load (your speaker).

Lowering the load impedance as you have done is going to lower the load impedance on the valve and reduce the available output power but will probably reduce the distortion as the reduced voltage swing will keep the operating conditions in a more linear operating area.

[MichaelR]

Originally Posted by PJL

I am not sure the power transmission efficiency rule applies here. A pentode has a higher output impedance simply because it behaves almost like a constant current source which is a distinct advantage when considering gain or damping of tuned circuits. Kalee20 though does make a good point on limiting voltage swing in this case.

.

Surely maximum power transfer theory always applies, for maximum power transfer. I do accept what you are detailing is correct in terms of mismatch may improve linearity however it will be at the expense of power.

Mike

[anthonys radios]

As newby I would not wish to disturb the equilibrium with established contributors of course,
Perhaps on this occasion this discussion can be laid to rest in the best interests of the group?

[MichaelR]

I do not think there is any argument developing with the discussion, purely from my point of view getting a clear understanding of fact.

Mike

[PaulR]

Absolutely! This is fascinating.

[ppppenguin]

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

• It's not always appropriate - just try doing it with mains or a car battery
• Assumes a linear system. If either R changes with current then it doesn't really apply
• If the resistance is actually an impedance then the best match is at the complex conjugate impedance. I think

Modern audio amps, and some older designs such as the Quad II which have a lot of negative feedback, have a very low output impedance so max power transfer isn't appropriate. It's just a question of how much current they will supply before overheating.

As others have explained, in a standard valve amp it's a matter of optimising the transformer ratio for best compromise between power and distortion. In most radios this is not at all critical, the worst that will happen is less volume or more distortion.

[Ray Cooper]

The danger with reducing the apparent anode load value, is that you may end up driving the output valve into over-dissipation on the anode. See the attached sketch...

The red curve shown is the maximum anode dissipation for the valve: it's a hyperbolic curve, since you get it from plotting Va*Ia = max dissipation (constant). When the set manufacturers design the output stage, they clearly want to get as much oomph as possible from the valve, so they bias it as close as possible to the max dissipation curve - the black blob in the sketch. The anode load is then represented by the sloping green line, and this shows the range that the output valve can be driven through. Note that at all times, the green line is 'inside' the max dissipation curve - so the anode dissipation never exceeds it's maximum permitted value.

But if you reduce the anode load resistance, you steepen the curve: yet the valve bias (and hence standing anode current) remains the same. With a lower value of anode load, the curve may end up as shown by the orange line: note that the maximum linear excursion is reduced (so you get less output) but the load-line now cuts and passes beyond the curve of maximum dissipation. So the mean power that the anode is dissipating will probably exceed the max value for the valve, which will do it no good in the long-term. You might be able to improve matters by reducing the standing bias (moves the load-line further left) which will reduce dissipation and slightly increase output power, but may introduce further (linearity) problems.


Get yourself a decent 15-ohm speaker...

[jimmc101]

I think some of the confusion her is caused by missing a part of the maximum power transfer theorem (as Jeff has suggested).

For an ohmic source, maximum power transfer occurs when the load resistance is equal to the source resistance.
Ohmic source means that voltage drop is proportional to current.

Consider a theoretical pentode with Ia = 100mA at Va = 200v
and Ia = 90mA at 100V (Vg constant).

The anode resistance is 10k ((200v-100v)/(100mA-90mA) but the current is not defined by this resistance and thus the maximum power transfer theorum does not apply.


Jim

[MichaelR]

Jim, you are right but just to be absolutely pedantic as pppenguin stated the maximum power transfer theory still applies to impedance as long as the reactive components of the source and load are conjugate i.e cancel each other out. When this occurs you are just dealing with the resistive componenets of the impedance of dource and load and as long as they are equal maximum power will be transferred.

For transmission line engineers rf or low frequency they aim for conjugate matching to achieve maximum power transfer whenever there is a discontinuity or connection in their transmission run.

If the source and the load are purely reactive i.e no resistve element then power is transferred to and from the reactive components during an AC cycle

Regards
Mike

[paulsherwin]

Originally Posted by anthonys radios

As newby I would not wish to disturb the equilibrium with established contributors of course,
Perhaps on this occasion this discussion can be laid to rest in the best interests of the group?

Don't be concerned about this Anthony. We're all friends here and often discuss technical issues like this. Your basic question has been answered - you can use the 4 ohm speaker without causing any damage - but the thread has moved on to a theoretical discussion of impedence matching in valve equipment. This is on topic for the subject and the thread will be allowed to run while matters of general interest are still being discussed.

Paul

[G8HQP Dave]

Maximum power transfer theorem assumes not only that the output is linear (ohmic, or perhaps reactive) but that there is no current limit i.e. pure impedance. Most of the arguments which arise about this theorem do so because people try to apply it to non-linear situations such as audio or RF power output stages.

[smiler411]

Originally Posted by Ray Cooper

The danger with reducing the apparent anode load value, is that you may end up driving the output valve into over-dissipation on the anode. See the attached sketch...

I agree with Ray. The main reason for the transformer is to set the correct load line for the valve based on its characteristics.

Reducing the speaker impedance will mean that the load line will be steeper and the anode power disipation may well exceed its maximum causing long term damage. See Ray's graph it explains it well

The idea of the output transformer is to set this load line not to match the anode impedance of the valve which would be much higher

[PJL]

Sorry guys, I will throw another curve ball on that theory. We are talking about an AC signal on top of the DC anode current. Dissipation in a class A output stays pretty much constant. Whatever goes up the load line must come down again...

In class A the anode voltage on an (ideal) output valve could swng from 0 to HT*2 and the current from 0 to (bias condition) *2.

[kalee20]

The reason that the maximum power transfer theorem is inappropriate, is that it assumes a constant source voltage. In the very rare cases where the signal source is limited (say a 1-valve radio trying to drive a loudspeaker) then yes, you will get maximum volume if the load impedance equals the source impedance. And that carries the proviso that the drive signal is not large enough to either cut off the valve, drive it into grid current, 'bottom' it (so the anode voltage swings to near zero), or exceed its ratings.

In real world cases, the source voltage is not fixed - you drive your output valve with as much signal as it needs, by adding previous stages as necessary. And in this situation, you examine which of the limiting conditions above will apply first as you increase the signal, and calculate the power at this limiting point (which is either the onset of distortion or the onset of valve abuse). You do this, with load impedamce as an independent variable. And the impedance which gives the maximum power, is is usually chosen as the optimum load.

For an 'ideal' pentode operating into an 'ideal' transformer, the optimum load impedance is equal to quiescent anode voltage divided by quiescent anode current (and this figure is much less than ra). Real pentodes operate at slightly less than this. 'Ideal' triodes, rather bizarrely, give maximum power when the load impedance is equal to twice ra (some conditions); or quiescent anode voltage divided by anode current minus twice ra (other conditions); or another calculation again.

[anthonys radios]

OK then, to return with fingers in my ears!

I accept that there are elements of truth in many of the postings, but there are a number of misconceptions as well, two many to deal with in one reply

An ideal class A amplifier (max 50% efficiency) should show no change in DC anode current with changes in signal input level, if it does then distortion will be occuring.

The principle function of an output transformer is impedance matching between the valve and the speaker.
The impedance ratio is the square root of the turns ratio.

Pentodes aproximate to constant current devices, and have high dynamic impedance, (anode resistance is high)

Valve manufactures give values of optimum loads for output valves which are much lower than the valve anode impedance, and are a compromise between distortion and power handing.

There are two sets of load lines for a class A amplifier.
The DC load line, set by the DC condition of the valve.
(The slope of which should not change with signal changes)
The AC load line which is concerned with AC signal excursions.

[kalee20]

Anthony, you are just about right on all counts in your summary!

Just don't worry about triodes - the optimum load is harder to pin down (fortunately, it isn't very critical in practice...)

[anthonys radios]

Refering to maximum power transfer, which is important for RF and for that matter DC.
Maximum power transfer occurs when the load impedance matches, (equals) source impedance.

Consider a dry cell as a perfect generator, in series with the internal resistance of the battery.
Make V = 3 volts , and the internal resistance as 3 ohms (for conveinience).
Maximum power will be transfered to any load connected providing this has a resistance of 3 ohms also.


Take the condition when the load is infinite (open circuit).
No current will flow, Power (watts) is given by IsqR=0.(no power dissipated in the load)

Take the condition when the load is zero (s/c) as the load resistance is zero, IsqR is still 0 (no power dissipated in the load)

Maximum power disipation in the load will occur somewhere between zero and infinite load resistance.

In the case of the short circuit load, the entire battery voltage will be developed across the internal battery resistance and will be, from ohms law, I=V/R = 3/3 = 1Amp.(3 watts disipated)
Thus there can never be more than 1 amp in any load resistance, what ever it's value.

With a load resistance of 3 ohms, the toal circuit resistance becomes 6 ohms. I will be 0.5 amps.
Power dissipated in the load will be (from W=Isq R) 0.75 watts.

It is easy to calculate that increasing the resistance of the load would reduce the load current, and reducing the load resistance would reduce the voltage across the load,
both conditions would result in less power being dissipated in the load.Try it yourself!


Done in a hurry, hope there are no silly mistakes !

[anthonys radios]

Well thats a releif, thanks for the message, I appreciate