What's the trick to get the most efficiency out of an SS AB solid state amp?

[ELS122]

What's the trick to get the highest efficiency out of a solid state push pull output? the 73 or whatever % of max efficiency on paper. And also couldn't this be even higher if you shift the output into pure stage B as the amplitude goes up, I think some Sony amp did this... can't recall the name of it.
So what's the reason an amp would not reach this theoretical max efficiency? I can only think of the reason being that both output transistors are on at the same time leading to wasted power, so would a class B output stage ALWAYS have it's max efficiency?
Is there some compensation circuitry that improves the linearity when the transistors are barely on, so that I could have a class B-ish output stage with minimal negative feedback and still have acceptable THD specs?

Also, would BJT class AB amps all be considered as class AB2 since there would always be base current? Or does the 1/2 type class only apply to tubes, or fets?

[Doctorandus_P]

So what's the reason an amp would not reach this theoretical max efficiency? I can only think of the reason being that both output transistors are on at the same time leading to wasted power, ...

There are several reasons, and the one you mention is a very minor one.
Class AB normally has a fairly low quiescent current, just enough to get rid of cross over distortion. As a result, when the amplifier delivers some significant power, then one of the transistors (alternating of course) is off nearly all the time, and the time that both transistors conduct, the current is small.

A significant part of the power loss is headroom between the power rails and the maximum output voltage. With BJT's you have a minimum of 600mV, but the power transistors are usually powered from a current source, and that also needs a minimum voltage drop, so in practice there is some 2 or 3 volt difference on each of the power rails.

Another mayor cause is the output ripple of the power supply. Under a load the output voltage of the power supply both sags and develops a ripple because of the electrolytic capacitors charge / discharge cycles, and you have to add this voltage drop to the minimum headroom.

If you are really interested in the design of audio amplifiers, then I highly recommend the book: Audio power design Handbook by Douglass Self. It discusses many aspects of designing audio amplifiers in depth and in an easy to understand way. It assumes very little pre-existing knowledge about audio amplifiers, but it does assume general electronics knowledge. And of course it also has several schematics of discrete class AB amplifiers you can build and do measurements on. Even if your knowledge about transistors and stuff is still a bit limited you will learn during the study of that book.

[floobydust]

A second excellent book is Designing Audio Power Amplifiers by Bob Cordell.

"Audiophiles usually care most about maximum dissipation as opposed to overall efficiency. These are two very different things. Audiophiles don’t care as much about power drawn from the outlet. They care about how big they must make their heat sinks in order to achieve a given output power and sound quality. This is why hybrid class D may be attractive for some audiophiles."

[Kim Christensen]

What's the trick to get the highest efficiency out of a solid state push pull output? the 73 or whatever % of max efficiency on paper.

Play it loud. 
Just before clipping is where it's going to be most efficient (Hard clipping is even more efficient). Remember that the "73%" would be calculated from the DC input power to the output stage of the amp, and not the AC input feeding the transformer-rectifier-filter & driver circuitry. Of course it's not going to hit the theoretical max because the output swing is never going to reach the supply rails, because there'll always be a bit of voltage drop across the output transistors to waste additional power.

[ELS122]

What's the trick to get the highest efficiency out of a solid state push pull output? the 73 or whatever % of max efficiency on paper.

Play it loud. 
Just before clipping is where it's going to be most efficient (Hard clipping is even more efficient). Remember that the "73%" would be calculated from the DC input power to the output stage of the amp, and not the AC input feeding the transformer-rectifier-filter & driver circuitry. Of course it's not going to hit the theoretical max because the output swing is never going to reach the supply rails, because there'll always be a bit of voltage drop across the output transistors to waste additional power.

Are there amps that use 2 power rail voltages then? if you use a lower power rail voltage for the lower part of the waveform and switch to a higher power rail voltage for the peak of the waveform, there would be less power dissipation in the output transistors and theoretically the peak efficiency should be at max drive instead of like ~80% drive, if done right, right?

[floobydust]

See Class G and Class H, discussed here: https://sound-au.com/articles/class-g.htm

[macboy]

What's the trick to get the highest efficiency out of a solid state push pull output? the 73 or whatever % of max efficiency on paper.

Play it loud. 
Just before clipping is where it's going to be most efficient (Hard clipping is even more efficient). Remember that the "73%" would be calculated from the DC input power to the output stage of the amp, and not the AC input feeding the transformer-rectifier-filter & driver circuitry. Of course it's not going to hit the theoretical max because the output swing is never going to reach the supply rails, because there'll always be a bit of voltage drop across the output transistors to waste additional power.

Are there amps that use 2 power rail voltages then? if you use a lower power rail voltage for the lower part of the waveform and switch to a higher power rail voltage for the peak of the waveform, there would be less power dissipation in the output transistors and theoretically the peak efficiency should be at max drive instead of like ~80% drive, if done right, right?

Look up class H. At least that's what some manufacturers called it.

[Kim Christensen]

I have seen some that have 2 power rail voltages and switch between them depending on signal level. Only really worth it for very powerful amplifiers.
Yamaha B-6 does that kind of thing:

[macboy]

A company called Sunfire made (maybe still does?) very powerful amplifiers which were class AB technically, but with variable voltage supply rails. They used a very fast switch mode power supply which tracked just a few volts above the input signal at all times. We are talking hundreds of watts of power per channel in multi channel amplifiers or way over 1000W for subwoofers.  These were so efficient that in some models they didn't even fit heatsinks to the amplifier output transistors (or so I had read). Sunfire was founded by Bob Carver, who had engineered some really amazing audio gear over the years. There are some white papers about the technology on these amps. Worth the read.

Maybe that's the answer to the title question. Of course the real trick then is designing an SMPS fast enough to power the amp.

[ELS122]

Well I looked at some of the designs for class H and class G amps. But I was thinking more like having 2 sets of output transistors, when the one for low voltage started going into saturation you would start turning the other set on that had a higher power rail voltage.

Class G imo is just a wacky class D amp, similar to the magnetic amps in that way - a class D amp with added wackyness and then audiophiles get crazy over it 


Nevermind, I misunderstood how class G works.
Also half the articles say class H is the stepped one and class G is continuously variable one, and half say the other way around.

[magic]

Well I looked at some of the designs for class H and class G amps. But I was thinking more like having 2 sets of output transistors, when the one for low voltage started going into saturation you would start turning the other set on that had a higher power rail voltage.

Self calls this "class G-shunt" and says that it actually is a thing in commercial amps with large power output.

The stupid problem of having two emitter followers in series is avoided and efficiency should be higher.
I wonder how much glitching occurs at the transition between those output stages...

[langwadt]

Well I looked at some of the designs for class H and class G amps. But I was thinking more like having 2 sets of output transistors, when the one for low voltage started going into saturation you would start turning the other set on that had a higher power rail voltage.

Self calls this "class G-shunt" and says that it actually is a thing in commercial amps with large power output.

The stupid problem of having two emitter followers in series is avoided and efficiency should be higher.
I wonder how much glitching occurs at the transition between those output stages...

afair Douglas Self claims the switching is not really an issue when using Schottky diodes

[langwadt]

I have seen some that have 2 power rail voltages and switch between them depending on signal level. Only really worth it for very powerful amplifiers.
Yamaha B-6 does that kind of thing:

I believe there were some that used a doubler to generate the higher voltage, music is usually much higher crest factor than sinewaves so you only need the higher voltage to avoid clipping the peaks to be as loud as higher power amp with music

[magic]

afair Douglas Self claims the switching is not really an issue when using Schottky diodes

Nope, I'm talking about this kind of output stage. The high voltage part turns on when Tr1 base voltage is high enough with respect to the output to forward bias D1. The low voltage part turns off when Tr1 saturates. There is zero reason why these events should occur simultaneously, as necessary for clean transition.

And transition seems to be driven by the magnitude of load current rather than output voltage. It looks so dodgy that I wonder if this circuit is really used or if Self made some mistake here.

Possibly a better switching method may exist, but it's not immediately obvious what it would be.

[TimFox]

There is an old rule of thumb that if you use a "fancy" power supply (e.g., feedback-regulated linear supply or later higher-tech stuff) to power anything other than class-A (which has constant high current draw), it has to be at least as "good" as the amplifier itself, lest it degrade the performance.

[langwadt]

afair Douglas Self claims the switching is not really an issue when using Schottky diodes

Nope, I'm talking about this kind of output stage. The high voltage part turns on when Tr1 base voltage is high enough with respect to the output to forward bias D1. The low voltage part turns off when Tr1 saturates. There is zero reason why these events should occur simultaneously, as necessary for clean transition.

And transition seems to be driven by the magnitude of load current rather than output voltage. It looks so dodgy that I wonder if this circuit is really used or if Self made some mistake here.

Possibly a better switching method may exist, but it's not immediately obvious what it would be.

http://www.utn-eaplicada.com.ar/DownLoads/Audio_Power_Amplifier_Handbook__3rd_edition.pdf

page 299 and forward

[magic]

... or page 36.

While page 37 is the circuit that I am talking about.
You may notice that it has better output swing (by not connecting EF's in series) and that it sucks for reasons previously described.

[MrAl]

What's the trick to get the highest efficiency out of a solid state push pull output? the 73 or whatever % of max efficiency on paper. And also couldn't this be even higher if you shift the output into pure stage B as the amplitude goes up, I think some Sony amp did this... can't recall the name of it.
So what's the reason an amp would not reach this theoretical max efficiency? I can only think of the reason being that both output transistors are on at the same time leading to wasted power, so would a class B output stage ALWAYS have it's max efficiency?
Is there some compensation circuitry that improves the linearity when the transistors are barely on, so that I could have a class B-ish output stage with minimal negative feedback and still have acceptable THD specs?

Also, would BJT class AB amps all be considered as class AB2 since there would always be base current? Or does the 1/2 type class only apply to tubes, or fets?

Hello there,

To start with, the maximum efficiency you can get is 0.785398 (which is pi/4) and that is when the output sine peak equals the Vcc voltage (you may have plus and minus Vcc though).

The expression for the efficiency when the sine output peak is LESS than (or equal to) the Vcc voltage is:
eff=(pi/4)*A/Vcc   [Note A is the peak voltage of the output sine wave as in: A*sin(w*t)]

and that expression is valid only when A is equal to or less than Vcc.  Of course that also implies the transistor can reach 0 Ohms when turned fully on, and there are no other losses such as in the drive circuitry.

Judging from that expression, we can see we reach the max pi/4 only when A=Vcc.  This implies that to get maximum efficiency ALL of the time, we would have to vary Vcc as needed based on the volume setting.
For example, if we set the volume such that we get 10v peak output then we would have to set Vcc=10 volts DC (ideal case again).
If we set the volume such that we get 5v peak output then we would have to set Vcc=5 volts DC.
This would mean the volume knob would also have to control the DC supply voltage, and in order to keep that part at the highest efficiency we would have to use a buck circuit to lower the DC voltage when needed.  We cannot use a linear regulator.

Note that in the above if we kept Vcc=10v and Voutpeak=5v, the efficiency would drop to 1/2 of what it was when they were equal.

Also note that a lower efficiency does not necessarily mean we use up MORE power, it just means that we don't get as much bang for the buck.  That's because the total input power also goes down as we reduce the output voltage using a constant voltage DC buss.

It might not be that hard to vary the buss voltage using a buck circuit, and this is about the same thing that high efficiency power supplies do.  They have the switching regulator front end put out a lower voltage when the user turns the output voltage down.  The front-end switcher 'tracks' the output voltage.  You should be able to set up the same thing just that the output will be a combination of sine waves so you'd have to use a circuit to find the peak.  You'd have to work out how fast you want it to respond too and see what works best without too much added distortion between widely varying output levels.

[CaptDon]

Langwadt, Thanks for the handbook link. The electronic edition is more easily searchable than the printed version!! Trying to recall my stack of Carver / Clair Brothers CBA-1000 amplifiers I think they had 30vdc / 50vdc / 85vdc / 125vdc rails using the transition method of higher output voltage. These amps were industrialized versions of the PM-2.0T with 7 large capacitors for each of the 125vdc rails. The caps were rated at 150vdc and known to blow up violently if the line power for the 120vac models rose above 135vac as can happen at local fairs with poor power regulation and huge motor loads. Also the audio section could be destroyed by sustained high power output at high frequencies when the amps were used in certain testing applications not related to audio service. It seems they suffered some sort of lock-up condition and went bang in a smokey sort of way.

[Kleinstein]

Well I looked at some of the designs for class H and class G amps. But I was thinking more like having 2 sets of output transistors, when the one for low voltage started going into saturation you would start turning the other set on that had a higher power rail voltage.

Self calls this "class G-shunt" and says that it actually is a thing in commercial amps with large power output.

The stupid problem of having two emitter followers in series is avoided and efficiency should be higher.
I wonder how much glitching occurs at the transition between those output stages...

The glitches would only happen at already rather large volume. As the "switching" happens at the transistor collector side (for a BJT based solution) there should not much reach the output. In the usual form it is not really switching the supply, but going from a constant lower supply to the collector voltage going up with the needs, so more like a fixed CE voltage. It would be not so much a glitch, but a break in the curve. A true glitch can come from diode reverse recovery, but this should not be that bad with fast diodes, as slew rate is usually not that bad with audio (much of the volume is from the low frequencies) and there is still a bit reserve for the output driver to work with.

For high power amplifiers the class G/H makes sense, alone from reducing the voltage that the transistors see and this way reducing the SOA needs for the transistors. While in the simple form one has 4 instead of 2 power transistors, that actual number of transistors needed may go down. 2 in series may replace 3 or 4 in parallel when it comes to the SOA with higher voltage.

[Smokey]

A second excellent book is Designing Audio Power Amplifiers by Bob Cordell.
...

+1

[xavier60]

afair Douglas Self claims the switching is not really an issue when using Schottky diodes

Nope, I'm talking about this kind of output stage. The high voltage part turns on when Tr1 base voltage is high enough with respect to the output to forward bias D1. The low voltage part turns off when Tr1 saturates. There is zero reason why these events should occur simultaneously, as necessary for clean transition.

And transition seems to be driven by the magnitude of load current rather than output voltage. It looks so dodgy that I wonder if this circuit is really used or if Self made some mistake here.

Possibly a better switching method may exist, but it's not immediately obvious what it would be.

That doesn't seem right, does it? It would work with the Emitters of the high voltage transistors connected to the Collectors of the low voltage transistors. The low voltage rail would then be the reference for the point at which the high voltage transistors begin to be driven, just before the low voltage transistors saturate.

[magic]

That's how class G usually works, but it has the disadvantage that two emitter followers appear in series, requiring higher rails for the same peak output swing and pointlessly burning extra power.

The solution that Self calls "class G-shunt" or "commutating amplifier" switches between a few parallel output stages, each powered by a different rail.
The switching seems not entirely trivial to get right.

[ELS122]

What about the Proton "Dynamic Power on Demand" system, apparently "Was way ahead of it's time (1980s)"... Delivering like 5x higher burst power than normal power.
"the power amp section has about 6dB of headroom. This receiver is rated at 40 watts, but during instantaneous peaks (up to 200ms), it can deliver up to about 150-160 watts." -about the Proton D940
The D275 car amp can deliver the 'up to 5x higher power' mentioned. But there doesnt seem to be any schematics for that one.

Looking at the schematic for the D940 it just looks to me like a typical class G amp, (the continous type). How does it get so much burst power? is it just because the power supply has a ton of headroom? it uses 10'000uF capacitors for each 74V rail. The big caps are proprietary caps manufactured by Nichicon specifically for Proton.
For the D940 the filter caps for the 34V rails are 2*4700uF for each rail
The D275 uses 10'000uF @ 80V caps for the HV rails, and 2200uF @ 50V caps for the LV rails

Also would you run into beta droop problems at such high currents with BJT output?
   Speaking of which, I also learned about secondary breadkdown in BJT's, I presume an advantage of class G is that the risk of this is greatly reduces with class G since the voltage across the conducting output transistor would be much lower than with a typical AB amp circuit, right?

Apparently this technology was patented but I can't find a patent for this.


Here's some articles:
https://worldradiohistory.com/hd2/IDX-Audio/Archive-Stereo-Review-IDX/IDX/80s/HiFi-Stereo-Review-1985-11-OCR-Page-0041.pdf
https://worldradiohistory.com/hd2/IDX-Audio/Archive-Audio-IDX/IDX/80s/Audio-1987-05-OCR-Page-0051.pdf
https://youtu.be/eYDHFD5ITa4?t=80
He shows the internals a bit in the video.


It doesn't seem like anything special, other than a class G amp.
But I thought output power in amplifiers was limited by the output transistors not being able to pass the current. So is it instead limited by the power supply instead? Would then putting 10'000uF caps on the power rails in a typical power amp make it's burst power skyrocket to like 5x more than normal power?
So the emitter resistor sense protection circuit in typical amps is there to keep the transistors from exceeding their max rated current, and if you put transistors that are the same, except they can pass a lot more current (other specs being the same for example sake), with the protection circuitry removed, if would get super high dynamic burst power? Then the power would be limited by the max voltage swing instead?
And then all that's limiting continous power will be the power supply, and thermals ofc.

Why do many amps have the protection circuitry sensing only the top output transistor? is it for simplicity and cost saving? how reliable would that be?