Could someone explain this Hafler/Strickland output topology to me?

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Hi Paul

This is the way I visualize it. I may be wrong.

Imagine the +OUTPUT swinging upwards and downwards. When N-fet draws current, +OUTPUT is pulled upwards towards + rails. Hence, current at +OUTPUT flows through speaker to 0V.

When N-fet shorts, full 20,000uF discharges through mosfet to 0V. +OUTPUT hits rail instantaneously. Fuse blows from high discharge. 20,000uF is disconnected from shorted mosfet. +OUTPUT then settles back to mid-point of +/- supply rails.

Cheers
Mike
 
Hi Mike,

You're missing one piece, though. There is no current path from 0V ground to the "other" side of the +V supply except through the speaker load.

Easiest way to see that is to look at C16. To short C16 and blow the fuse instantly as you describe, you'd have to short both the N-Channels and the connection from OUT- to OUT+.

Cheers,
Paul
 
When the N-channel FET:s conduct more the positive output will go _"down"_, not up! They pull down on the positive supply, making the center tap connected to positive output go "down" too.

In a fault condition it won't be much different from a conventional amp. A shorted N-channel will make the positive output try to go negative. The feedback will try to correct this and the p-channels will turn on harder (which is true in a conventional topology amp too) and one of the fuses is probably going to blow.

If the feedback is successful not much current will flow through the speaker yet.

Trouble is that the fuses are esentially in series so the negative one might blow as likely as the positive one. If the fuse on the defective polarity blows, output won't go DC. If the fuse on the non-defective side blows, output _will_ go DC though.

On the other hand, those output transistors are very hard to destroy - there seems to be two reasons: Strongly decreasing transconductance with increasing temperature and some internal gate-source structure becoming leaky and latching up like a thyristor at high temperature, effectively shorting out gate drive and turning the transistor off. I'd guess it's the gate protection zeners that are responsible for this and in a test I did it happened at estimated around 200 degrees C junction temperature.
 
When the N-channel FET:s conduct more the positive output will go _"down"_, not up! They pull down on the positive supply, making the center tap connected to positive output go "down" too.

Hi megajocke

I don't think so. When N-Fet conducts, upper 20,000uF discharges. +OUTPUT moves upwards due to discharge. Lower cap voltage increases with +OUTPUT point.

For this kind of configuration, supply caps have to be double the voltage because each cap must be able to swing rail to rail. In other words, if supply rails are +/- 50V, supply caps will have to be rated at least 100V, preferably 120V.
 
Tome it is mainly leading to this question :

Why use high quality polypropylene caps like any audiophie would do and then end the audio chain by using such a design where the signal goes through very high value electrolytics, very unstable over time and aging quickly because of high DC voltage across them...

Still, these amplifiers had very good reviews so why not...

But it's just not really audiophile like to me.

Regards
 
Michael Chua said:


Hi megajocke

I don't think so. When N-Fet conducts, upper 20,000uF discharges. +OUTPUT moves upwards due to discharge. Lower cap voltage increases with +OUTPUT point.

Why? The power supply is connected across those capacitors and will keep voltage over both at whatever the supply voltage is. The only difference to the conventional type of output stage is where the ground reference is located and how the gate drive is arranged.

The output stage is identical to the commonly used topology, but instead of using the negative output/center tap of supply as ground reference they chose the positive because it gives them the previously mentioned advantages.

The output stage functions in exactly the same way as earlier, it's just the way it's driven that is different.


For this kind of configuration, supply caps have to be double the voltage because each cap must be able to swing rail to rail. In other words, if supply rails are +/- 50V, supply caps will have to be rated at least 100V, preferably 120V.

No, that's not true. It makes circuit operation easier to visualize if you think of those capacitors as two batteries of rail voltage like someone mentioned earlier.

If disregarding how the gates are driven and where ground reference is chosen the circuit is exactly the same as the conventionally used EF output - remember that the power supply is floating and it's output has no connection to ground.
 
darian said:
Tome it is mainly leading to this question :

Why use high quality polypropylene caps like any audiophie would do and then end the audio chain by using such a design where the signal goes through very high value electrolytics, very unstable over time and aging quickly because of high DC voltage across them...

Still, these amplifiers had very good reviews so why not...

But it's just not really audiophile like to me.

Regards

Those capactiors are just as much or as little in the "signal path" as the power supply capactitors in a conventional amplifier - the power circuit is identical, only the ground reference is different.
 
An externally hosted image should be here but it was not working when we last tested it.


Hi megajocke

This is a conventional PSU that we are all so familiar with. Center of PSU caps is referenced to 0V.

When positive cycle, upper transistor conducts.

Current from upper filter cap flows through transistor, through speaker to 0V.

Because center of PSU caps is fixed at 0V, upper filter cap voltage drops when current is drawn.


Let's take a closer look at the Hafler PSU below.

An externally hosted image should be here but it was not working when we last tested it.


Our Fixed 0V is now relocated to the emitters of the transistors.

Center of PSU caps is not referenced to our Fixed 0V anymore.

When positive cycle, upper transistor conducts. Because the center point of the PSU caps is not referenced to Fixed 0V, top of upper filter cap will remain at 50V but the mid-point now will swing upwards, going positive.

When that happens, you have a positive potential at the +OUTPUT. Current then flows from +OUTPUT, through speaker to 0V.

That is why I said earlier that the filter caps need to be rated for -ve to +pos rails as a minimum because they will be charging from 50V to 100V instead of discharging from 50V downwards as in a conventional PSU.

I may be wrong but I think that's the way it works. Very clever.

Mike
 
I you want your eyes to really go batty take a good look at what Crown does with their MacroTech series, where things are swinging like monkeys in trees! :_)

The have some nice name for the topology that at the moment I can't recall... doubtless someone else will.

I remember seeing a schematic where the power supply is unipolar (like 0,+50V), but since the whole PS is modulated towards ground, the amp output can produce +/-50V with unipolar +50 supply
 
Michael Chua said:
An externally hosted image should be here but it was not working when we last tested it.


Hi megajocke

This is a conventional PSU that we are all so familiar with. Center of PSU caps is referenced to 0V.

When positive cycle, upper transistor conducts.

Current from upper filter cap flows through transistor, through speaker to 0V.

Because center of PSU caps is fixed at 0V, upper filter cap voltage drops when current is drawn.


Let's take a closer look at the Hafler PSU below.

An externally hosted image should be here but it was not working when we last tested it.


Our Fixed 0V is now relocated to the emitters of the transistors.

Center of PSU caps is not referenced to our Fixed 0V anymore.

When positive cycle, upper transistor conducts. Because the center point of the PSU caps is not referenced to Fixed 0V, top of upper filter cap will remain at 50V but the mid-point now will swing upwards, going positive.

When that happens, you have a positive potential at the +OUTPUT. Current then flows from +OUTPUT, through speaker to 0V.

That is why I said earlier that the filter caps need to be rated for -ve to +pos rails as a minimum because they will be charging from 50V to 100V instead of discharging from 50V downwards as in a conventional PSU.

I may be wrong but I think that's the way it works. Very clever.

Mike

HaflerOutput.JPG


Hi Michael,

Attached is the equivalent circuit from Stickland's patent. Please explain to me why the N-Channel (upper) transistor will cause the voltages output by the two batteries (or the two rails of the power supply in the real amp) to change, as you show in your diagram.

My understanding, which agrees with Megajocke's, is that a 63V power supply is a 63V power supply, regardless of whether it's under load or not; it's not going to suddenly change into a 30V power supply under load unless something is wrong with it.

If you look at the equivalent, you'll see that, when the upper transistors are conducting, they will pull the +OUT NEGATIVE relative to the -OUT. If the N-Channel were short, and all fuses were good, the speaker would "feel" -63V between its terminals. Another way to see this is, if you connect the + of PS1 to ground through Q3, then E will be at minus whatever voltage PS1 puts out relative to ground.

Cheers,
Paul
 
bear said:
I you want your eyes to really go batty :xeye: take a good look at what Crown does with their MacroTech series, where things are swinging like monkeys in trees! ;)

The have some nice name for the topology that at the moment I can't recall... doubtless someone else will.

_-_-bear


"Grounded Bridge". It's a bridge amp with a "normal" topology driving the high side of the speaker, and one of these twisted sons of.... driving the floating ground. Only a single rail voltage with one filter cap is needed for a DC-coupled amp.
 
If you look at the equivalent, you'll see that, when the upper transistors are conducting, they will pull the +OUT NEGATIVE relative to the -OUT.

Hi Paul

I'm sorry, somehow I'm not seeing that. Maybe I'm getting old. Gray matter becomes sluggish.

What I'm seeing is the +OUT swings like the output of a normal amplifier.

"lumanauw" puts it succinctly

the whole PS is modulated towards ground

Cheers
Mike
 
Michael,

If PS1 in the diagram were 9V and Q3 were a closed switch (equivalent of a saturated transistor) then the + of the 9V battery (PS1) would be connected to ground.

If you then put the black probe of your meter to ground, and the red probe to E, what would your meter read?

-9V.

Paul
 
If PS1 in the diagram were 9V and Q3 were a closed switch (equivalent of a saturated transistor) then the + of the 9V battery (PS1) would be connected to ground

Hi Paul

Yes, PS1 will be connected to ground in a short.

If you then put the black probe of your meter to ground, and the red probe to E, what would your meter read? -9V

No, reading will be +9V


I've drawn out another diagram with Haflers' patent.

The one on the right is with Q3 injected with a 9V square wave.

E swings to + 9V rails. PS2 is now -9V to +9V, making it +18V.

But when looking at E from Ground, it would be +9V. E will have the same waveform as the injected signal.

An externally hosted image should be here but it was not working when we last tested it.


Cheers
Mike
 
So is this a form of sliding bias where neither O/P device is allowed to turn off fully?

Has the topology been used in a DIY implementation or any listening impressions of it?

Edit: Is this the same operational concept as is used in Susan Parkers Zeus amplifier (although this uses an O/P transformer)?
 
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