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Topic: Impedance matching between amplifier and speaker (Read 14277 times) previous topic - next topic
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Impedance matching between amplifier and speaker

Hello all. Although I have been wandering around for quite some time, this is my first post here.

One thing that occurred to me while reading about audio amplifiers is why impedance matching is not used to drive a speaker instead of impedance bridging. I know that when we are talking about connecting a pre-amp to an amp, we do not want power transfer but "signal transfer", in the way that the amplifier just cares about the waveform (voltage amplitude in time domain) and will draw power to amplify the input signal from its own source.

But when we are talking about a passive device, such as a speaker, wouldn't we want the maximum power transfer from the amplifier to the speaker so as to maximize system efficiency (from an energy-usage perspective)?

I understand that low impedance output from a voltage amplifier helps speaker damping, which in turn might translate in better sound (low THD and so on), but some loudspeakers, such as ones using full-range drivers, does not require that much damping as do those multi-drivers loudspeakers we are used to. Wouldn't that be the case for a voltage driver with output of impedance matched to the speaker?

Any ideas?

Best regards,
Bruno

Impedance matching between amplifier and speaker

Reply #1 –
Calling a speaker a passive device is highly deceptive. It is in fact a very complex electromechanical system with all kinds of resonances etc. Its impedance is basically dominated by a resistive element, the resistance of its windings, but if you drove it with anything other than a low-impedance voltage source you would be opening yourself up to all kinds of peculiarities in its frequency response.

Besides, low-impedance voltage-source amplifiers are a lot easier to design and cheaper to build than good quality speakers.

Impedance matching between amplifier and speaker

Reply #2 –
Quote from: Bruno Zoby de Moraes on
...One thing that occurred to me while reading about audio amplifiers is why impedance matching is not used to drive a speaker instead of impedance bridging....


And isn't that the way it is?! 

For example: I have my 8 ohm amplifier impedance output matched to my 8 ohm impedance load speakers.

Impedance matching between amplifier and speaker

Reply #3 –
Quote from: Light-Fire on
Quote from: Bruno Zoby de Moraes on

...One thing that occurred to me while reading about audio amplifiers is why impedance matching is not used to drive a speaker instead of impedance bridging....


And isn't that the way it is?! 

For example: I have my 8 ohm amplifier impedance output matched to my 8 ohm impedance load speakers.

Not at all. The amplifier's "8 ohm" spec is a recommendation for the the nominal impedance of speakers you should attach to it. The actual output impedance of the amp is bound to be less than 1 ohm, quite possibly under 0.1 ohm. And your "8 ohm" speakers' impedance will vary greatly over the frequency range: there will be dips and peaks, typically as low as 3 or 4 ohms and as high as 16.

Impedance matching between amplifier and speaker

Reply #4 –
Quote from: Light-Fire on
Quote from: Bruno Zoby de Moraes on

...One thing that occurred to me while reading about audio amplifiers is why impedance matching is not used to drive a speaker instead of impedance bridging....


And isn't that the way it is?! 

For example: I have my 8 ohm amplifier impedance output matched to my 8 ohm impedance load speakers.


When an amplifier is rated at "8 ohms" this simply means that the ratio of the voltage that it is capable of driving to the current that it is capable of driving (V/I) is approximately 8. This means that the maximum output power is achieved when driving an 8 ohm load. If the load is either higher impedance (limited by maximum voltage) or lower impedance (limited by maximum current) then the maximum output power will be reduced.

Impedance matching between amplifier and speaker

Reply #5 –
***sigh***

The impedence of most any power amp, other than specialty amps these days, is under .1 ohm.

For an active device with feedback, optimum power delivery != matched impedences, unless transmission lines come into play.

Most speakers, what's more, are made (these days) to be voltage-driven devices, not power or current-driven devices.

There's really no magic here, the speakers expect a constant-voltage device, and it's easiest to make an amp with feedback be pretty close to a constant-voltage device.

For RF or places where transmission lines come into play (please spare me from that 8 ohm impedence speaker coax, oh, please!), of course, the rules change massively.

Consider, using 50% of c for propagation (very slow), and 50kHz for a top end (very high), .1 wavelength (which is about when transmission lines come into play) is about 1000 feet, if I got my math right.  I just don't know too many speaker cables like that.  Those of you who get to use meters may adjust accordingly.
-----
J. D. (jj) Johnston

Impedance matching between amplifier and speaker

Reply #6 –
Actually I come up with 3000 meters, so I think you slipped a decimal place.

Edit: A friend of mine worked at Radio Shack many years ago. He actually told a customer that twin lead could not be used for speaker cable because of its 300 ohm impedance.

Impedance matching between amplifier and speaker

Reply #7 –
Quote from: Bruno Zoby de Moraes on
But when we are talking about a passive device, such as a speaker, wouldn't we want the maximum power transfer from the amplifier to the speaker so as to maximize system efficiency (from an energy-usage perspective)?

No, maximum power transfer is not the same as maximum efficiency. (A common misconception amongst first year Electrical engineering students).

Maximum power occurs when the source resistance is equal to the load resistance. Maximum efficiency occurs when the source resistance is as low as possible and the load resistance is as high as possible. The logical (and theoretically correct) conclusion is that 100% efficiency will occur when the source resistance is zero, and the load is an open circuit! This of course results in zero power transfer and is totally impractical. The real situation of a fraction of an ohm driving 8 ohm is a practical realization of a high efficiency circuit. (The reactance can be safely ignored in terms of what I'm saying).

Quote
When an amplifier is rated at "8 ohms" this simply means that the ratio of the voltage that it is capable of driving to the current that it is capable of driving (V/I) is approximately 8. This means that the maximum output power is achieved when driving an 8 ohm load. If the load is either higher impedance (limited by maximum voltage) or lower impedance (limited by maximum current) then the maximum output power will be reduced.

No (except in special cases), the lower the speaker impedance the higher the power. The rated impedance will correspond to the maximum power that can be safely supplied by the amplifier. If you use a lower impedance speaker than specified for an amplifier the power will certainly increase, although possibly for a short time...

(The special case would be an amplifier with output protection circuitry. Such circuits usually don't help the distortion figures though).
Cheers,
Alan

Impedance matching between amplifier and speaker

Reply #8 –
Quote from: alanofoz on
Quote
When an amplifier is rated at "8 ohms" this simply means that the ratio of the voltage that it is capable of driving to the current that it is capable of driving (V/I) is approximately 8. This means that the maximum output power is achieved when driving an 8 ohm load. If the load is either higher impedance (limited by maximum voltage) or lower impedance (limited by maximum current) then the maximum output power will be reduced.

No (except in special cases), the lower the speaker impedance the higher the power. The rated impedance will correspond to the maximum power that can be safely supplied by the amplifier. If you use a lower impedance speaker than specified for an amplifier the power will certainly increase, although possibly for a short time...

(The special case would be an amplifier with output protection circuitry. Such circuits usually don't help the distortion figures though).


By this logic one would conclude that the amplifier delivers its maximum power into a zero ohm load!

What we are really talking about is what is the maximum undistorted power into what impedance, and distortion can result from exceeding either the maximum output voltage or the maximum output current of the amplifier. If we operate at just below the maximum output voltage and the maximum output current then we are delivering the maximum undistorted power. The load impedance under these conditions is the ratio of the voltage to the current.

Of course, for nearly any modern solid state amplifier this occurs at less than 8 ohms, possibly even less than 4 ohms, but this has not always been the case. Early solid state amplifiers typically peaked at 6 to 8 ohms and often delivered less power into 4 ohms than 8 ohms. Vacuum tube amplifiers were always designed to give maximum power at their rated impedance, although they often had multiple taps for multiple impedances.

Impedance matching between amplifier and speaker

Reply #9 –
I was wondering when somebody would mention vacuum tubes (better known as valves on our side of the Atlantic).  I used to mess around with valve amplifiers.  They were mostly open loop designs, that is to say they had no feedback at all.  You could do that wiith a valve amplifier and get acceptable sound quality, at least to the average man in the street.  If you wanted hi-fi you used class A push-pull and a modest amount of feedback.

That brings me to the crux of the matter - feedback.  Solid state amplifiers typically have huge open loop gains and so their closed loop gain is set almost entirely by the feedback loop.  This gives them a very low output impedance which is good for driving speakers.  Their ability to deliver power to those speakers is determined by factors which have nothing to do with impedance.  As pdq points out ---

"When an amplifier is rated at "8 ohms" this simply means that the ratio of the voltage that it is capable of driving to the current that it is capable of driving (V/I) is approximately 8. This means that the maximum output power is achieved when driving an 8 ohm load. If the load is either higher impedance (limited by maximum voltage) or lower impedance (limited by maximum current) then the maximum output power will be reduced."

(I must learn how use the quote feature on this forum.)

Let's take an example.  An amplifier has a maximum linear voltage swing of ±20V (I know it because I built it) and a maximum safe current swing of ±2.5A.  I connect an eight ohm load and wind it up to its maximum undistorted (sinewave) output.  The current goes to 2.5A just as the voltage reaches 20V and so I have 25 watts rms power.  If I put a one ohm load on it the current will reach its safe maximum at just 2.5V and I will only get 3.125 watts (rms again).  For maximum power output you need voltage and current to reach their limits at the same time.

So where does that business about matched impedances come from?  A lot of it has to do with transmission lines.  These have capacitance and inductance distributed uniformly down their length which, working together, create a virtual resistance at an open end.  If you stuck an ohm meter across the end of an infinitely long, perfect transmission line (ie one with zero series resistance and zero leakage) you would measure a pure resistance; odd but true.  This is known as the characteristic impedance of the line.  It's there because you have a step waveform running down the line, charging the capacitance as it goes.  Current is flowing into your end of the line to supply this charge.  It also takes voltage to drive this waveform down the line because you are 'charging' the inductance.  The ratio of voltage to current is the square root of L/C where L and C are the inductance and capacitance per metre (or inch, or light year) respectively, and so your meter sees a resistor.

If you try this with a finite length you'll need a very fast meter because the waveform will soon hit the open end and bounce back.  When this reflection comes back to your meter the current will fall abruptly to zero and you'll see an open circuit.  (Very fast meters exist for the purpose of testing transmission lines.  It's called time domain reflectometry.)  If you connect a real resistive load to the far end of a finite line it should be equal to the characteristic impedance of the line.  If, and only if, you match the load in this way there will be no reflection and all the power you send down the line will go into the load.  Any mismatch will send power back to your end where it will either bounce again - and arrive at the load out of phase - or else it will be wasted.

It's also quite true that you will get maximum power out of a real resistive source if you match the load to it.  Example:  A 12 volt battery has an internal resistance of 0.1 ohms.  (OK, I know batteries are non-linear but this one is special.)  If you connect a 0.1 ohm load you will get a current of 60 amps.  That's because only 6V appears at the battery terminals; the rest is lost in the battery.  The resulting 360 watts is the maximum power you can draw from that battery but would you really want to do this?  As alanofoz points out, maximum power transfer does NOT equate to maximum efficiency.

Impedance matching between amplifier and speaker

Reply #10 –
Quote from: Johnny Neutron on
I was wondering when somebody would mention vacuum tubes (better known as valves on our side of the Atlantic).  I used to mess around with valve amplifiers.  They were mostly open loop designs, that is to say they had no feedback at all.  You could do that wiith a valve amplifier and get acceptable sound quality, at least to the average man in the street.  If you wanted hi-fi you used class A push-pull and a modest amount of feedback.


I would say that by the late sixties, when I was building both solid state and vacuum tube amplifiers from kits, all hi-fi amplifiers had feedback in their final stage. Ordinary radios etc. would have been a different story.

Also, a push-pull final stage would never be class A but either class B or (better) class AB.

Edit: Well maybe not NEVER never, but almost never.

Impedance matching between amplifier and speaker

Reply #11 –
Quote
Also, a push-pull final stage would never be class A but either class B or (better) class AB.

Edit: Well maybe not NEVER never, but almost never.


Pdq, you've got me thinking now.  I was messing about in those days and didn't do much 'proper' design.  More often than not I used suck-it-and-see - which probably explains why so much of my early stuff didn't work very well!  I vaguely remember an article in Practical Wireless which was all about the relative merits of amplifier classes A, AB, B, C, and even D.  I think it recommended class A as the best but I'm not sure now.  It was all about overlaying the transfer curves of the two valves so as to get the best straight line.  I still have those old magazines so I suppose I could try and find it again.

I'm not sure about the impedance matching either.  Pentodes have a very high anode impedance.  I'm thinking that the trick was to match to the maximum voltage and current swings available, just as we do today.  A valve with 300 volts on its anode and a maximum average anode current of 50 mA would do best if you loaded it with 6k, hence the output transformer.

Impedance matching between amplifier and speaker

Reply #12 –
Fond memories, eh?

The problem with class A amplifiers is that they draw just as much power at zero volume as at maximum volume. Great for low distortion, lousy for efficiency. Class B draws very little power at zero volume, but it is really hard to minimize crossover distortion. Class AB is a compromise. Some quiescent power consumption but easier to handle crossover distortion.

Impedance matching between amplifier and speaker

Reply #13 –
Quote from: pdq on
By this logic one would conclude that the amplifier delivers its maximum power into a zero ohm load!

You'll need to walk me through that logic!

You apparently haven't seen my reference to the maximum power transfer theorem which is quite clear: if a power amp has an output resistance of (say) 0.1 ohm, then maximum power would occur when the load is also 0.1 ohm. Of course the smoke will escape long before you could test it. But I did refer to delivering the power safely.

Quote from: pdq on
What we are really talking about is what is the maximum undistorted power into what impedance, and distortion can result from exceeding either the maximum output voltage or the maximum output current of the amplifier. If we operate at just below the maximum output voltage and the maximum output current then we are delivering the maximum undistorted power. The load impedance under these conditions is the ratio of the voltage to the current.


Of course we are talking about the maximum undistorted power - once we get into clipping the whole argument falls apart.

Now, what do you mean by maximum output current? The output current will simply follow ohm's law. If the voltage is just short of clipping, then the current is the ratio of the voltage to the load resistance. For the chosen resistance the current is the variable here.

The waters of this discussion are muddied by power supply regulation, and I assume you are taking this into account when you talk of maximum current, but this is not a simple relationship at all.

Quote from: pdq on
Of course, for nearly any modern solid state amplifier this occurs at less than 8 ohms, possibly even less than 4 ohms, but this has not always been the case. Early solid state amplifiers typically peaked at 6 to 8 ohms and often delivered less power into 4 ohms than 8 ohms. Vacuum tube amplifiers were always designed to give maximum power at their rated impedance, although they often had multiple taps for multiple impedances.


It's a good 40 years since I studied valve amps and I'll just keep it that way... In this case what you say is quite right though.

Probably your early solid state amplifiers had poor power supply regulation. In that case what you say may be true.

Having built the original Toby-Dinsdale quasi complementary-symmetry design in the early '60s, and then redesigned it for silicon a few years later, I can assure you that by no means all early solid state amplifiers peaked around 6-8 ohms. Driving a sine wave just short of clipping into a 4 ohm resistive load produced not quite double the power of an 8 ohm load.

Whichever way you look at it though, a (solid state) amplifier rated at 8 ohms is most unlikely to produce maximum power at 8 ohms, and will undoubtedly produce more power at lower speaker impedances, maximum ratings notwithstanding.
Cheers,
Alan

Impedance matching between amplifier and speaker

Reply #14 –
Quote from: Johnny Neutron on
Let's take an example.  An amplifier has a maximum linear voltage swing of ±20V (I know it because I built it) and a maximum safe current swing of ±2.5A.  I connect an eight ohm load and wind it up to its maximum undistorted (sinewave) output.  The current goes to 2.5A just as the voltage reaches 20V and so I have 25 watts rms power.  If I put a one ohm load on it the current will reach its safe maximum at just 2.5V and I will only get 3.125 watts (rms again).  For maximum power output you need voltage and current to reach their limits at the same time.

If you connect a 1 ohm load to your amp nothing (except prudence) stops you exceeding the 2.5A safe limit (presumably you have calculated this from the quoted power rating). Even if the power supply regulation cuts the available voltage swing to ±10V, that's 50W into 1 ohm.

If you read my posts you'll see that all along I've talked about what you can do safely, and I'm too prudent to actually exceed the ratings as I've just described, but it's still wrong to say that you can't get more power at reduced load impedances.
Cheers,
Alan

Impedance matching between amplifier and speaker

Reply #15 –
Quote from: alanofoz on
Quote from: pdq on

By this logic one would conclude that the amplifier delivers its maximum power into a zero ohm load!

You'll need to walk me through that logic!

You apparently haven't seen my reference to the maximum power transfer theorem which is quite clear: if a power amp has an output resistance of (say) 0.1 ohm, then maximum power would occur when the load is also 0.1 ohm. Of course the smoke will escape long before you could test it. But I did refer to delivering the power safely.



I see your problem now. You are obviously confusing output resistance with output impedance. Because of negative feedback an amplifier can have a very low output impedance, i.e. the ratio of a voltage perturbation to the current that produces it, but this is not the same as its output resistance, which is usually very much higher. If the amplifier had no negative feedback then the output impedance would be the same as the output resistance.

Impedance matching between amplifier and speaker

Reply #16 –
Quote from: pdq on
Quote from: alanofoz on

Quote from: pdq on

By this logic one would conclude that the amplifier delivers its maximum power into a zero ohm load!

You'll need to walk me through that logic!

You apparently haven't seen my reference to the maximum power transfer theorem which is quite clear: if a power amp has an output resistance of (say) 0.1 ohm, then maximum power would occur when the load is also 0.1 ohm. Of course the smoke will escape long before you could test it. But I did refer to delivering the power safely.



I see your problem now. You are obviously confusing output resistance with output impedance. Because of negative feedback an amplifier can have a very low output impedance, i.e. the ratio of a voltage perturbation to the current that produces it, but this is not the same as its output resistance, which is usually very much higher. If the amplifier had no negative feedback then the output impedance would be the same as the output resistance.


No confusion at all I'm afraid, I'm very familiar with the effect of negative feedback on output impedance. I chose the word "resistance" because I didn't want to complicate the discussion with thoughts of reactance which "impedance" could imply. While my "convenient" figure of 0.1 ohm is certainly on the high side, if I was confused I'd have probably used a figure a good deal higher still.

However, you still haven't explained the logical progression to "By this logic one would conclude that the amplifier delivers its maximum power into a zero ohm load!" from anything I stated.
Cheers,
Alan

Impedance matching between amplifier and speaker

Reply #17 –
You said: "No (except in special cases), the lower the speaker impedance the higher the power." Since the lowest possible speaker impedance is zero ohms, this must therefore correspond with the highest power.

The point that I was trying to make, with which you now seem to agree, is that as you decreaase the speaker impedance you pass through a point of maximum power, beyond which the power decreases. We seem to disagree only on what this speaker impedance is.

Impedance matching between amplifier and speaker

Reply #18 –
Quote from: alanofoz on
You apparently haven't seen my reference to the maximum power transfer theorem which is quite clear: if a power amp has an output resistance of (say) 0.1 ohm, then maximum power would occur when the load is also 0.1 ohm. Of course the smoke will escape long before you could test it. But I did refer to delivering the power safely.


Let's see if I understand you. Let's assume your output resistance of 0.1 ohm, which delivers maximum power into a 0.1 ohm load. If this amplifier is rated at, say, 100 watts RMS into 8 ohms (a very conservative value by today's standards) then the peak power is 144 watts and the peak voltage is approximatelt 34 volts.

Now let's deliver this 34 volts peak through a 0.1 ohm output stage into a 0.1 ohm load. This gives us 170 amps (34 volts / 0.2 ohms) at 17 volts to the speaker, or 2890 watts. The output stage is also dissipating 2890 watts, and the power supply is supplying 5780 watts. Oh, excuse me, this is at least two channels so the power supply is supplying 11560 watts for stereo, more if there are more channels.

I guess solid state amplifiers must have changed a lot in the 40 years since I last built one.

Impedance matching between amplifier and speaker

Reply #19 –
Quote from: pdq on
You said: "No (except in special cases), the lower the speaker impedance the higher the power."

Aha! Yes, I did indeed say that, and it is misleading. But to put it in context, that statement came just after I had discussed the max power transfer theorem, so a certain amount of understanding should be assumed.

We were also discussing speakers at this point, and I was thinking of real speakers, and as long as we keep real, my statement holds true.

Quote from: pdq on
Since the lowest possible speaker impedance is zero ohms, this must therefore correspond with the highest power.

No, it is not possible to have a speaker whose impedance is zero ohms. That's every bit as silly as max power into a short circuit!

Quote from: pdq on
The point that I was trying to make, with which you now seem to agree, is that as you decreaase the speaker impedance you pass through a point of maximum power, beyond which the power decreases. We seem to disagree only on what this speaker impedance is.

Since I started my discussion by referring to the max power transfer theorem, it's obvious that I always agreed with that first sentence. The second is also obviously true. 

My earlier discussion about a 0.1 ohm load should not be construed as a practical situation obviously. If there was any doubt, my reference to smoke should have dispelled it!

Now my career path took me away from the hands-on work years ago, but back in the days when I was a young engineer in a design lab, any power amp I tested produced more power as I reduced the load impedance. The practical limit of course is when the amp overheats to the point of destruction. Now I've already mentioned the effect of poor power supply regulation, and if this has been your experience then fair enough. But in my experience reducing the load impedance as much as I dared (and a couple of times more than I should have  ), always resulted in a higher power output.

And so I come back to my original assertion - while maximum safe power may very well occur with a load impedance above 4 ohms, the maximum possible power will usually occur at a lower load impedance. (Hopefully that's a clearer statement of my position.)
Cheers,
Alan

Impedance matching between amplifier and speaker

Reply #20 –
Fair enough.

Impedance matching between amplifier and speaker

Reply #21 –
Cheers,
Alan