[Patrick Turner]

In a recent post about heatsinks and fans the topic of
ME Techologies ME850 amp came up as a side issue.

Trevor Wilson, Sydney's ME "expert" could not seem to tell us exactly
how much class A
was produced in this SS amp, and expected me to inform him fully,
something I was unable to do
because I've never measured the bias current in an ME850 amp.

So lets forget about ME amps completely; If Trevor wants to talk about
the details of the ME 850, he's welcome,
but he better get his facts all straight and clear, before he makes a
monkey out of himself again.



Consider generic facts about SS amps configured with complementary pnp
and npn
output devices in series for the output stage.

Consider rails are +50V and -50V.

Assuming the peak load voltage travel is +/- 48V,
you can get 33Vrms, giving 136Watts into 8 ohms, and 272 watts into 4
ohms and 544 watts into 2 ohms
but ONLY IF the PSU is stiff enough and there are enough output devices
to ensure all remain
in their SOA.
But even with a 200 watt PSU, if the power supply caps have enough
capacitance,
and there are enough devices, power peaks with music can be handled up
to the theoretical
figure of 544 watts if the load is as low as 2 ohms. 48pk V into 2 ohms
is only 24 amps peak.

With only two output bjts or mosfets, T03 packaged, one would have
current limiting
to prevent more than say 100 watts of PO regardless of the load,
as well as have rail fuses, and some well placed thermal sensors mounted
on the TO3 hold down bolts.

All this power in most SS amps is mainly in class AB with very low bias
currents at idle of perhaps
typically 30mA per output device.

So say you had a total of 30mA rail to rail, idle Pd = 100V x 0.03A = 3
watts,
and the the amount of class A PO available depends on the idle current.

The maximum peak Iswing available in the load during class A operation =
2 x idle current.

So say Iq dc = 30mA, then max peak class A load current = +/- 60mA,
or 0.042Arms, and if the load is 8 ohms, class A PO = Irms squared x R =
0.042 x 0.042 x 8
= 14 milliwatts, which is hardly any class A at all so its why most SS
amps
are regarded as virtual class B amps.

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.
If you read Douglas Self, the high SS class A isn't worth the trouble
because the power supply
works hard and noise artifacts muddy the signal more than when there is
low bias AB.

The jury is still out of course on all this, personally
I prefer tubes.....

Douglas is very interesting on steam engines BTW....

Patrick Turner.

[Andre Jute]

On Feb 11, 4:05*pm, Patrick Turner wrote:

The jury is still out of course on all this, personally
I prefer tubes.....

Doesn't everyone?

Douglas is very interesting on steam engines BTW....

I like Douglas Self on Monowheels:
http://www.dself.dsl.pipex.com/MUSEU...l/motorwhl.htm

Andre Jute
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/
"wonderfully well written and reasoned information
for the tube audio constructor"
John Broskie TubeCAD & GlassWare
"an unbelievably comprehensive web site
containing vital gems of wisdom"
Stuart Perry Hi-Fi News & Record Review

[Trevor Wilson[_2_]]

"Patrick Turner" wrote in message
...
In a recent post about heatsinks and fans the topic of
ME Techologies ME850 amp came up as a side issue.

Trevor Wilson, Sydney's ME "expert" could not seem to tell us exactly
how much class A
was produced in this SS amp,

**Wrong. I have stated so many times. It is 16 Watts @ 8 Ohms.

and expected me to inform him fully,
something I was unable to do
because I've never measured the bias current in an ME850 amp.

**Wrong. _I_ explained to you that the amp delivered 16 Watts Class A into
an 8 Ohm load. I also informed you that the rail Voltage was +/- 50 Volts.
From these two facts, you would be able to work out the power dissipation in
each output stage. You erroneously stated that it was "32 Watts at idle".
Some simple arithmetic will show you that this was not correct.


So lets forget about ME amps completely; If Trevor wants to talk about
the details of the ME 850, he's welcome,
but he better get his facts all straight and clear, before he makes a
monkey out of himself again.

**Fine by me. It was you who decided to bring ME amplifiers into this and
the previous discussion. Not me. IOW: If you don't want to hear about ME
amplifiers, don't bring them into the discussion.




Consider generic facts about SS amps configured with complementary pnp
and npn
output devices in series for the output stage.

Consider rails are +50V and -50V.

Assuming the peak load voltage travel is +/- 48V,
you can get 33Vrms, giving 136Watts into 8 ohms, and 272 watts into 4
ohms and 544 watts into 2 ohms
but ONLY IF the PSU is stiff enough and there are enough output devices
to ensure all remain
in their SOA.
But even with a 200 watt PSU, if the power supply caps have enough
capacitance,
and there are enough devices, power peaks with music can be handled up
to the theoretical
figure of 544 watts if the load is as low as 2 ohms. 48pk V into 2 ohms
is only 24 amps peak.

With only two output bjts or mosfets, T03 packaged, one would have
current limiting
to prevent more than say 100 watts of PO regardless of the load,
as well as have rail fuses, and some well placed thermal sensors mounted
on the TO3 hold down bolts.

All this power in most SS amps is mainly in class AB with very low bias
currents at idle of perhaps
typically 30mA per output device.

So say you had a total of 30mA rail to rail, idle Pd = 100V x 0.03A = 3
watts,
and the the amount of class A PO available depends on the idle current.

The maximum peak Iswing available in the load during class A operation =
2 x idle current.

So say Iq dc = 30mA, then max peak class A load current = +/- 60mA,
or 0.042Arms, and if the load is 8 ohms, class A PO = Irms squared x R =
0.042 x 0.042 x 8
= 14 milliwatts, which is hardly any class A at all so its why most SS
amps
are regarded as virtual class B amps.

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

If you read Douglas Self, the high SS class A isn't worth the trouble
because the power supply
works hard and noise artifacts muddy the signal more than when there is
low bias AB.

The jury is still out of course on all this, personally
I prefer tubes.....

Douglas is very interesting on steam engines BTW....

**Indeed.

Trevor Wilson

[John Byrns]

In article ,
"Trevor Wilson" wrote:

"Patrick Turner" wrote in message
...

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

But Graham says 1 Amp idle current and 100 Watts of idle dissipation is
incorrect, so what is the correct answer, and who knows what they are
talking about?


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

Trevor Wilson wrote:

"Patrick Turner" wrote in message
...
In a recent post about heatsinks and fans the topic of
ME Techologies ME850 amp came up as a side issue.

Trevor Wilson, Sydney's ME "expert" could not seem to tell us exactly
how much class A
was produced in this SS amp,

**Wrong. I have stated so many times. It is 16 Watts @ 8 Ohms.

and expected me to inform him fully,
something I was unable to do
because I've never measured the bias current in an ME850 amp.

**Wrong. _I_ explained to you that the amp delivered 16 Watts Class A into
an 8 Ohm load. I also informed you that the rail Voltage was +/- 50 Volts.
From these two facts, you would be able to work out the power dissipation in
each output stage. You erroneously stated that it was "32 Watts at idle".
Some simple arithmetic will show you that this was not correct.


So lets forget about ME amps completely; If Trevor wants to talk about
the details of the ME 850, he's welcome,
but he better get his facts all straight and clear, before he makes a
monkey out of himself again.

**Fine by me. It was you who decided to bring ME amplifiers into this and
the previous discussion. Not me. IOW: If you don't want to hear about ME
amplifiers, don't bring them into the discussion.

I don't think anyone in this group wants to hear anymore about ME amps
at all,
unless its a way of educating the group. But you never educate anyone.

If you are going to insist you talk to us about them, please
***state all the facts at the outset***, and never let people guess
things
and do your ****ing work for free.

If you are not going to spell each and every single technical argument
out properly, then get off this news group.

And you have not done so yet, so I assume your knowledge is extremely
shallow.

You and Peter Stein, the ex CEO of ME technology are extremely lazy
people who unethically
try to convince the world they
know something but you have no website; nothing deep and meaningful is
avaliable online.
Basically, your'e a pair of hasbeens.

Personally I have nothing against either of you, but technically and
professionally,
your'e dumbclucks.

Get a decent website together to explain all about ME and you might just
gain more
respect than youse do.


Consider generic facts about SS amps configured with complementary pnp
and npn
output devices in series for the output stage.

Consider rails are +50V and -50V.

Assuming the peak load voltage travel is +/- 48V,
you can get 33Vrms, giving 136Watts into 8 ohms, and 272 watts into 4
ohms and 544 watts into 2 ohms
but ONLY IF the PSU is stiff enough and there are enough output devices
to ensure all remain
in their SOA.
But even with a 200 watt PSU, if the power supply caps have enough
capacitance,
and there are enough devices, power peaks with music can be handled up
to the theoretical
figure of 544 watts if the load is as low as 2 ohms. 48pk V into 2 ohms
is only 24 amps peak.

With only two output bjts or mosfets, T03 packaged, one would have
current limiting
to prevent more than say 100 watts of PO regardless of the load,
as well as have rail fuses, and some well placed thermal sensors mounted
on the TO3 hold down bolts.

All this power in most SS amps is mainly in class AB with very low bias
currents at idle of perhaps
typically 30mA per output device.

So say you had a total of 30mA rail to rail, idle Pd = 100V x 0.03A = 3
watts,
and the the amount of class A PO available depends on the idle current.

The maximum peak Iswing available in the load during class A operation =
2 x idle current.

So say Iq dc = 30mA, then max peak class A load current = +/- 60mA,
or 0.042Arms, and if the load is 8 ohms, class A PO = Irms squared x R =
0.042 x 0.042 x 8
= 14 milliwatts, which is hardly any class A at all so its why most SS
amps
are regarded as virtual class B amps.

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

But you can't work it ****ing out. Or else YOU would have composed a
post like this
which is truly informative.

I saw no evidence that the ME dissipates 100 watts per channel and has
an idle current of 1 amp dc
rail to rail, thus allowing 16 watts into 8 ohms of class A, per
channel.

If you did want 16 watts of class A, and you didn't want to have such a
high idle dissipation,
and only want about say 40 watts, then you'd have to have a much higher
load than 8 ohms.

In theory, to get 16 watts of class A then you'd need at least 32 watts
of disssipation,
but only if the on resistance of the bjts was zero, and the V swing was
perfect to 0V,
so hence my previous guess about the 32 watts needed.

But I guess only about the first 4 watts per channel is in class A in
the ME850.
I stated this guess in another recent post and you didn't disagree, and
that means only 16
watts per channel is dissipated, and its something that agrees with my
observations of the
temperature of the amp.

There is NO online resource ( website ) with guranteed full information
about ME amps.
No need to say to much about them here because nobody is ever going to
know if you are talking the truth.

If any SS amp dissipates 100 watts after turn on to heat up fast then
its right into class A 2 seconds after turn on, and the sound can't
change greatly
when the temp has risen 20C.
Class A means there is no switching in the output devices,
so there is very low THD/IMD, right from cold through to being hot.

This is the way my class A SS amps have worked.

Class A tube amps DO need a few minutes before nearly full cathode
emissions occur,
so a tube amp will sound a little harsh at 30 seconds after turn on.

Patrick Turner.



If you read Douglas Self, the high SS class A isn't worth the trouble
because the power supply
works hard and noise artifacts muddy the signal more than when there is
low bias AB.

The jury is still out of course on all this, personally
I prefer tubes.....

Douglas is very interesting on steam engines BTW....

[Patrick Turner]

John Byrns wrote:

In article ,
"Trevor Wilson" wrote:

"Patrick Turner" wrote in message
...

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

But Graham says 1 Amp idle current and 100 Watts of idle dissipation is
incorrect, so what is the correct answer, and who knows what they are
talking about?

Allow me to explain, since Trevor is so useless at correct explanations
he's as useless as tits on a bull.

You have +50V, and -50V rails, OK.

You have 1.0amp dc rail to rail, OK.

The middle junction of npn and pnp seriesed complementary BJT devices
are at 0V at idle.

Let us suppose they are arranged in emitter follower mode, so the base
signal voltage
is only slightly more than the emitter voltage.
The npn and pnp emitters are effectively tied together to make the
midpoint.

Load is from this midpoint to 0V.

Power input from PSU is 100 watts, because you have 100V x 1 amp.

Say we have 8 ohms connected as the load.

Say the driver circuitry swings the bases of both npn and pnp BJTs
positively.

A positive voltage appears at one end of the load and current flows
through the load to 0V.
More current flows in the npn BJT, less current flows in the pnp.
One device has increasing current, the other decreasing current.
Power supply input in watts does not change.

Class A action continues until one device cuts off.
So let us say the maximum positive V swing at the npn BJT emitter is
+16V.
The current in the npn BJT has doubled from 1 amp to 2 amps; 16V across
8 ohms = 2 amps.
Meanwhile the current in the pnp BJT has reduced to near zero, or to the
brink of switching right off.

So for +16V and -16V peak V swings on the 8 ohms, each BJT has control
of the load,
and power output of each BJT is single ended class A. Because in each
device there is a net +/- 1 amp dc change
in current, each BJT "sees" 16 ohms while in class A.
The npn and pnp BJTs have some similarity in their even order distortion
spectra so a lot of cancellation occurs
but no cancellation of odd orders occurs. But while no switching occurs,
the open loop distortion is
fairly low, and 2H, 3H and other H well down and very close to spectra
in a class A pentode/beam tetrode amp.
Except that with the emitter follower connection, the open loop
collector gain of about
perhaps 100 is reduced to less than 1.0, due to the large amount local
series negative feedback, about 40dB.
So the closed loop distortion in the class A region of working is VERY
low, 0.01% or less,
and this is before any global NFB is applied.

If the swing is beyond +/-16V peak, the amp begins to work in class AB,
and
the power supply has to increase its power to the output stage.

With 50V rails, 130W+ is available in class AB.

In this described typical example used in MOST class SS amps,
the pure class A peak load current cannot ever exceed twice the idle
current.

So, an amp with 50V rails with a 1 amp dc idle flow produces 16 watts
into 8 ohms of pure class A.
Above this po the class becomes AB, and having a load of say 23 ohms
would allow
class A of approximately slightly under 1/2 the idle dissipation of the
devices, say 46 watts,
and you'd only ever get 46 watts and no more before clipping commenced.

There are many SS amps with perhaps only 50mA of idle current in the
BJTs at idle.
If rails are 50V, they also make a max 130W approx at clipping.
But the max pure class A peak load current is only 100mA, and class A PO
is in milliwatts only.

Hence such "low bias current" SS amps are regarded as mainly class B
amps.

In practice, there are in fact many SS amps where very little sound
change occurs
when one alters the bias current between class C, with both devices
turned off at idle,
right up to heavy class A with an amp or two.

Next time you adjust the bias of a generic SS amp, have some music
going,
and adjust the bias pot across a wide range slowly. The vast amount of
NFB
will usually completely remove any sense of severe switching artifacts
even in class C at normal listening levels.
The wave form on the CRO does look a little strange.
But last time I did this with a budget model Cambridge amp,
there was no difference to its awful sound.

In most SS amps, you have a two BJT differential input pair
which has two inputs, one for input signal from a cd player etc,
and the other for the NFB signal from the output.
Typical gain of the input pair is 20x.
There is then often a single Voltage Amplifier Stage ( VAS ) with single
gain transistor
with a constant current collector load. Its gain is commonly in excess
of 2,000.
A buffer emitter follower stage between this driver single ended BJT
stage is often used
to increase the appallingly low input resistance of the output BJTs even
with their emitter follower arrangement.
Its called the Darlington Pair connection. Crown used a Darlington
Triple connection
to enable the driver stage to have huge gain and lower THD.
So, the driver THD without global NFB is typically 5% at full 130 watt
levels,
and very much like a pentode, ie, with lotsa ****ty sounding artifacts,
but nevertheless mainly 2H and 3H.
Total overall open loop gain = 20 x 2,000 x 1.0 = 40,000.
GNFB is used to typically reduce by a factor of 1/2,000 to 20, so for
30Vrms output, you need 0.67 vrms input.

The NFB gain reduction is 1/2,000, and THD of 4% is also reduced by
1/2,000 to 0.002%.

In a typical well made SS amp, 100 watts at 0.002% is easily possible
and this % declines about linearly with output voltage
so that at 3 Vrms its typically 0.0002%, or so darn low its difficult to
measure easily.

But some SS amps were not that well made, and THD can be 0.1% at 3Vrms,
a poorer
result than many of the poorer tube amps.

Usually it was because the output stage was configured in common emitter
rather than common collector
as described above.
And in many SS amps the driver amp is not well buffered before the
output stage base
input resistance which can widely vary in npn and pnp devices, thus
causing 25% THD in the open loop driver characteristic.

I suggest you read my pages on SS amps and also read Douglas Self on
amps.
Ben Duncan also put out a very informative book I read in 1996;
this has the history of the developments of SS circuit topologies from
the early days
when the invention of the fused junction silicon transistor became
so much better than the early germanium crap, and both germanium and
vacuum tubes
were declared obsolete by mainstream industry after 1960.

In my SS circuits I prefer power mosfets because their gate input
impedance
is the gate to source capacitance as reduced by the source follower
connection,
ie, less than 150pF with two mosfets in the output stage,
and the gate input resistance is as high as that of a vacuum tube.
No buffers are required. And mosfets don't latch up when turning off
at high F like BJTs, causing "cross conduction" rail to rail,
something not dealt with very well as D.Self's or anyone else'e website.

And BTW, you can have a "quasi complementary pair" of vacuum tubes,
with two tubes in series as used in many OTL amps.
The top acts in cathode follower, the bottom is arranged in anode
follower,
and so both are arranged to have closed gain below 1.0.

A similar arrangement was used in the first OTL transistor amps, with a
2,200 uF cap to the
speaker load. Despite the greater difficulty of driving such circuits,
the "top"
device is the same as the "bottom" device, so their PP action
is VERY complementary, and you get better even order H cancellations.
In such circuits, onlt one rail voltage was used, which made bean
counters smile.
The output coupling cap stopped dc from a fused top BJT reaching a
speaker and wrecking it.
I think Sugden class A and AB amps still use this early type of circuit.
But such amps still failed; a shorted speaker or lead caused the cap to
fry to a short,
BJTs would fry to a short, the speaker would cop the dc anyway, and a
few other bits would smoke,
and finally a fuse might blow......

The trouble with SS amps is mainly twofold.
1.The speaker lead terminals are usually way too close together on the
rear panel of most of 'em.
2.And there is NEVER any circuit which detects the presence of a load
that is too low, say 3 ohms,
and shuts down the amp. This would mean having some electronic circuit
which
divides output voltage by output current at all levels so that the load
is constantly
being measured.

I've never specialised in building SS amps, but if I did, I'd get some
guru
to design me up a board to make the load measurement and make an output
signal
suitable to turn a relay driven by an auxilliary psu to turn off the
mains
to the main amp.


Patrick Turner.










Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
"Trevor Wilson" wrote:

"Patrick Turner" wrote in message
...

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

But Graham says 1 Amp idle current and 100 Watts of idle dissipation is
incorrect, so what is the correct answer, and who knows what they are
talking about?

Allow me to explain, since Trevor is so useless at correct explanations
he's as useless as tits on a bull.

You have +50V, and -50V rails, OK.

You have 1.0amp dc rail to rail, OK.

Trevor is not the issue here, it's Graham who in the predecessor to this
thread repeatedly claimed that an idle current of 2A is required to get
16 Watts of class A power into 8 Ohms, resulting in an idle dissipation
of 200 Watts per channel for Trevors hypothetical amp with +/- 50 volt
rails.


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Eeyore]

John Byrns wrote:

Patrick Turner wrote:
John Byrns wrote:
"Trevor Wilson" wrote:
"Patrick Turner" wrote in message
...

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

But Graham says 1 Amp idle current and 100 Watts of idle dissipation is
incorrect, so what is the correct answer, and who knows what they are
talking about?

Allow me to explain, since Trevor is so useless at correct explanations
he's as useless as tits on a bull.

You have +50V, and -50V rails, OK.

You have 1.0amp dc rail to rail, OK.

Trevor is not the issue here, it's Graham who in the predecessor to this
thread repeatedly claimed that an idle current of 2A is required to get
16 Watts of class A power into 8 Ohms, resulting in an idle dissipation
of 200 Watts per channel for Trevors hypothetical amp with +/- 50 volt
rails.

If the bias is set at 1 amp, the power supply current can vary from 'near zero' to 2
amps without causing cut-off in either output device. This gives a current swing in
the load of ~ 2 amps.

However the requirement is a current swing of FOUR amps ( +/- 2 amps). Audio goes
negative as well as positive !

Graham

[John Byrns]

In article ,
Eeyore wrote:

John Byrns wrote:

Patrick Turner wrote:
John Byrns wrote:
"Trevor Wilson" wrote:
"Patrick Turner" wrote in message
...

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins
at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 =
16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max
of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850
dissipates
considerably more than 32 Watts per channel.

But Graham says 1 Amp idle current and 100 Watts of idle dissipation is
incorrect, so what is the correct answer, and who knows what they are
talking about?

Allow me to explain, since Trevor is so useless at correct explanations
he's as useless as tits on a bull.

You have +50V, and -50V rails, OK.

You have 1.0amp dc rail to rail, OK.

Trevor is not the issue here, it's Graham who in the predecessor to this
thread repeatedly claimed that an idle current of 2A is required to get
16 Watts of class A power into 8 Ohms, resulting in an idle dissipation
of 200 Watts per channel for Trevors hypothetical amp with +/- 50 volt
rails.

If the bias is set at 1 amp, the power supply current can vary from 'near
zero' to 2
amps without causing cut-off in either output device. This gives a current
swing in
the load of ~ 2 amps.

Yes, exactly, but note that is a peak current of 2 amps which yields a
class A sine wave power of 16 Watts into 8 Ohms.

However the requirement is a current swing of FOUR amps ( +/- 2 amps). Audio
goes
negative as well as positive !

Yes, 16 Watts sine wave power into 8 Ohms requires a "peak to peak"
current swing of 4 amps, but notice that we are no longer talking about
peak current and instead have switched to talking about "peak to peak"
current, which is twice the peak current. A class A amplifier requires
an idle current of only 1 amp to deliver a "peak to peak" current of 4
amps.


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/

[Phil Allison]

"flipper"


It would seem you're thinking 'mono polar' supply but it's stated to
be bipolar: +-50V. The current swing is 2A on both the positive and
negative rails, for +-2A, not just a 'single supply' of 2A.


** In the case of an amp operating from a single supply, there is no real
change.

Current in the DC supply still varies from 0 to 2 amps in sine wave fashion
(assuming sinewave drive) - only the voltage is now double the split supply
case.

The half rail coupling cap does the trick in supplying -2 amp peaks to the
speaker load.



....... Phil

[Patrick Turner]

Eeyore wrote:

John Byrns wrote:

Patrick Turner wrote:
John Byrns wrote:
"Trevor Wilson" wrote:
"Patrick Turner" wrote in message
...

But suppose we raise IQdc to 1.0 Amps, so Pdq = 100 watts.

Max load current for class A = +/- 2peak amps,
or 1.414Arms, and if the max V swing is 33Vrms, then the load for
maximum possible pure class A = 33 / 1.414 = 23.3 ohms.
One would get 46.7 watts max of pure class A PO,
and efficiency is 46.7%, since the power drawn from the PSU reamins at a
constant 100W
when the amp remains in the pure class A operating region.

But if we had 8 ohms, the PO is I squared x R = 1.414 x 1.414 x 8 = 16
watts of class A.
The output voltage could still reach 33vrms, and PO reaches a max of
136W but its mainly class AB.
With 4 ohms, only 8 watts of class A is made, with 100watts of idle
power.

**See? You can work it out. Now you understand why the ME850 dissipates
considerably more than 32 Watts per channel.

But Graham says 1 Amp idle current and 100 Watts of idle dissipation is
incorrect, so what is the correct answer, and who knows what they are
talking about?

Allow me to explain, since Trevor is so useless at correct explanations
he's as useless as tits on a bull.

You have +50V, and -50V rails, OK.

You have 1.0amp dc rail to rail, OK.

Trevor is not the issue here, it's Graham who in the predecessor to this
thread repeatedly claimed that an idle current of 2A is required to get
16 Watts of class A power into 8 Ohms, resulting in an idle dissipation
of 200 Watts per channel for Trevors hypothetical amp with +/- 50 volt
rails.

If the bias is set at 1 amp, the power supply current can vary from 'near zero' to 2
amps without causing cut-off in either output device. This gives a current swing in
the load of ~ 2 amps.

However the requirement is a current swing of FOUR amps ( +/- 2 amps). Audio goes
negative as well as positive !


4 amps pk to pk divided by 2.82 = 1.41Arms, and PO = I squared x R so
class A PO for any load where it is possible to get the 1.41Arms swing,
the maxima possible
for pure class A, then PO = 2 x RL, for where the idle current = 1
ampdc.

Thinking about 4 amps pk to peak isn't really helpful.

Better to simply think that for pure class A,
maximum peak load current = 2 x Idle Ampsdc, and load amps RMS is simply
1.41 x idle Amps DC.

Techs are expected to remember all this without consequent laborious
wokings out.

And you could have +/- 1,000V volt rails and get no more class A than
with +/-50V
using the same low load values for the +/-50V and 1 amp idle current.

But with +/- 1,000V rails, a swing of +/-950Vpk or 671Vrms is possible,
load current is simply 1.41Arms,
so the load would be 671 / 1.41 = 474 ohms, and class A PO = 949 watts
from an input of 2,000 watts from the
2,000V psu.

40 x 845 could be used to achieve this, with quasi complementary series
connection,
properly biased heaters, and floating transformer drive to the top
triodes, and
with an OPT cap coupled to the cathode-anode junction between top-bottom
triodes.
The efficiency would not be near the maximum of 50% possible, because
typical biasing of the
845 at Ea = 1,000V is 75mA for each of the 40 triodes, so power from psu
needs to be 3,000 watts
to get the 949 watts mentioned above.

It matters not whether you have series triode connection, or traditional
PP OPT coupled operation; for 949 watts in PURE class A,
you'd still need a 3,000 watt psu and 40 x 845.

Its a lot easier to settle for PP class AB2 and for 949 Watts you'd need
far less
psu power, and far fewer tubes, maybe 10 triodes would be plenty, Ea =
1,000V, and with a traditional
PP OPT.



Patrick Turner.

Graham