Re: Maximum power from 6V6 in push-pull

[Gregg]

Hi The Cheat,

Strong Bad here.....just kidding ;-)

I have measured 15W out of a pair of 6V6 in PP pure pentode before
distortion was objectionable. The transformer was 8,800 ohms measured
reflected load from an 8 ohm resistive dummy load.

Va was 330V.

--
Gregg t3h g33k
"Ratings are for transistors....tubes have guidelines"
http://geek.scorpiorising.ca

[Ned Carlson]

On Wed, 03 Nov 2004 08:49:57 +0000, The Cheat wrote:
> The usual method to estimate
> power is to draw the composite load line for one tube from the zero grid
> Volts curve to the maximum plate voltage on the plate voltage axis. The
> calculation is based on converting a peak sinusoid to rms by dividing by
> root 2. The composite load line with slope 4K (what each tubes sees) is
> shown below.

Each tube is seeing 1/2 the primary, thus each tube sees
1/4 the plate-to-plate load, right? Not 1/2?

Also, while I am no engineer, it seems to me the way you want
to calculate this would only work for A1. The tube is cut off
for part of the cycle thus the most negative voltage would be
off the end of the curve chart.

--
Ned Carlson Triode Electronics Chicago,IL USA
www.triodeelectronics.com

[Tim Williams]

"Ned Carlson" > wrote in message
.. .
> Each tube is seeing 1/2 the primary, thus each tube sees
> 1/4 the plate-to-plate load, right? Not 1/2?

But since both act in parallel (inverse being on opposite sides of the OPT),
each plate sees 1/2 the a-a impedance.

The fun stuff happens at cutoff: in class B, each tube drives the OPT for
its respective half cycle; neither is on at the same time (theoretically).
Thus they cannot act in parallel and each sees 1/4 the load, as the turns
ratio suggests.

Now, run class AB. It goes from 1/2 to 1/4 as the opposite tube cuts off -
curved loadline!

I missed the OP (probably a good thing, with a geigh name like that >:D ) so
I don't know what the question was.

Tim

--
"I've got more trophies than Wayne Gretsky and the Pope combined!"
- Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

[John Byrns]

In article >, "The Cheat"
> wrote:

> I am a retired electronics engineering professor and I am always finding new
> things that I don't understand. Here is the latest problem. The tube manuals
> give the maximum power output (with 5% or so distortion) for a pair of 6V6
> running in class AB push-pull with an 8K load as 14 W. This is with 285 V
> plate-to-cathode and screen-to-cathode and a grid bias of -19 V. Using self
> bias requires a cathode resistor of 250 Ohms. This is a classic
> configuration used in countless power amps over the years. However, from
> load line analysis of the 6V6 characteristics (attached), it doesn't seem
> possible to get this much power. The usual method to estimate power is to
> draw the composite load line for one tube from the zero grid Volts curve to
> the maximum plate voltage on the plate voltage axis. The calculation is
> based on converting a peak sinusoid to rms by dividing by root 2. The
> composite load line with slope 4K (what each tubes sees) is shown below. The
> actual operating conditions at zero signal are shown as point Q. The maximum
> power calculation is shown, which yields 9.3 W, not 14 W. When I test this
> output stage (with screens tied to the center-tap of the output
> transformer), this is about what I get when the output begins to clip on the
> scope. The output voltage is measured with a true rms voltmenter with an 8
> Ohm resistive load. So I am satisfied that theory gives a good estimate of
> what I measure but where does the 14 W value in the tube manuals come from?
> Any comments from the experts out there would be appreciated.


Hi Lloyd,

A few comments. First with respect to your load line analysis, the anode
characteristics you posted are for a screen voltage of 250 volts, while
you said you are operating your test circuit with a screen voltage of 285
volts, this would result in a lower power output from the analysis. I
haven't had a chance to study your analysis in any detail at all yet, but
I did notice that you have drawn the 4k load line on the anode curves as
if you were analyzing a class B1 amplifier, instead of a class AB1
amplifier as in the tube manual operating conditions. In class B the load
line with an 8k transformer would be 2k resulting in a higher power
output, and in class AB1 the 4k load line would occupy a different
position on the anode curves, again affecting the calculated power. I
will try to figure out exactly how you analyzed the circuit later.

As far as your test circuit goes, you state that you are measuring the
power into an "8 Ohm resistive load" which implies that you have an output
transformer in the circuit. The tube manual characteristics are measured
without an output transformer, so to get a comparable measurement you must
take into account the losses in your output transformer, which may be as
much as 2 or 3 Watts, or even more, depending on the quality of your
output transformer. Also you must be sure that the primary impedance,
including transformer losses is actually 8k as you are assuming. The DC
voltage at the plates, under test conditions, must also be 285 volts,
allowing for the DC resistance of the transformer primary this would mean
the DC voltage at the transformer center tap must be greater than 285
volts, and the screen must then not be feed from this higher source. The
operating conditions must also be precisely as given in the tube manual.
I notice that you mentioned cathode bias while the tube manual specifies a
fixed bias of -19 volts. If your test circuit does use cathode bias you
must be sure that the drop across the cathode is 19 volts under the test
conditions with 14 Watts output, and of course that would mean the plate
and screen voltages should be 304 volts, not considering the DC loss in
the transformer primary. Also are you sure you are measured it at 3.5%
distortion as specified in the tube manual? If you are measuring at the
onset of clipping you may not be pushing it quite hard enough. When I was
designing audio amplifiers professionally, I could set the distortion
precisely to 5% by eye, but I wouldn't trust my eye today.

Just a few ideas to check on as far as the analysis and test circuit go.
It's hard to get the full power specified by the tube manual operating
conditions into a load on the secondary of an output transformer, the tube
manual conditions are for ideal loss less operating conditions.


Regards,

John Byrns


Surf my web pages at, http://users.rcn.com/jbyrns/

[Patrick Turner]

John Byrns wrote:

> In article >, "The Cheat"
> > wrote:
>
> > I am a retired electronics engineering professor and I am always finding new
> > things that I don't understand. Here is the latest problem. The tube manuals
> > give the maximum power output (with 5% or so distortion) for a pair of 6V6
> > running in class AB push-pull with an 8K load as 14 W. This is with 285 V
> > plate-to-cathode and screen-to-cathode and a grid bias of -19 V. Using self
> > bias requires a cathode resistor of 250 Ohms. This is a classic
> > configuration used in countless power amps over the years. However, from
> > load line analysis of the 6V6 characteristics (attached), it doesn't seem
> > possible to get this much power. The usual method to estimate power is to
> > draw the composite load line for one tube from the zero grid Volts curve to
> > the maximum plate voltage on the plate voltage axis. The calculation is
> > based on converting a peak sinusoid to rms by dividing by root 2. The
> > composite load line with slope 4K (what each tubes sees) is shown below. The
> > actual operating conditions at zero signal are shown as point Q. The maximum
> > power calculation is shown, which yields 9.3 W, not 14 W. When I test this
> > output stage (with screens tied to the center-tap of the output
> > transformer), this is about what I get when the output begins to clip on the
> > scope. The output voltage is measured with a true rms voltmenter with an 8
> > Ohm resistive load. So I am satisfied that theory gives a good estimate of
> > what I measure but where does the 14 W value in the tube manuals come from?
> > Any comments from the experts out there would be appreciated.
>
> Hi Lloyd,
>
> A few comments. First with respect to your load line analysis, the anode
> characteristics you posted are for a screen voltage of 250 volts, while
> you said you are operating your test circuit with a screen voltage of 285
> volts, this would result in a lower power output from the analysis. I
> haven't had a chance to study your analysis in any detail at all yet, but
> I did notice that you have drawn the 4k load line on the anode curves as
> if you were analyzing a class B1 amplifier, instead of a class AB1
> amplifier as in the tube manual operating conditions. In class B the load
> line with an 8k transformer would be 2k resulting in a higher power
> output, and in class AB1 the 4k load line would occupy a different
> position on the anode curves, again affecting the calculated power. I
> will try to figure out exactly how you analyzed the circuit later.
>
> As far as your test circuit goes, you state that you are measuring the
> power into an "8 Ohm resistive load" which implies that you have an output
> transformer in the circuit. The tube manual characteristics are measured
> without an output transformer, so to get a comparable measurement you must
> take into account the losses in your output transformer, which may be as
> much as 2 or 3 Watts, or even more, depending on the quality of your
> output transformer. Also you must be sure that the primary impedance,
> including transformer losses is actually 8k as you are assuming. The DC
> voltage at the plates, under test conditions, must also be 285 volts,
> allowing for the DC resistance of the transformer primary this would mean
> the DC voltage at the transformer center tap must be greater than 285
> volts, and the screen must then not be feed from this higher source. The
> operating conditions must also be precisely as given in the tube manual.
> I notice that you mentioned cathode bias while the tube manual specifies a
> fixed bias of -19 volts. If your test circuit does use cathode bias you
> must be sure that the drop across the cathode is 19 volts under the test
> conditions with 14 Watts output, and of course that would mean the plate
> and screen voltages should be 304 volts, not considering the DC loss in
> the transformer primary. Also are you sure you are measured it at 3.5%
> distortion as specified in the tube manual? If you are measuring at the
> onset of clipping you may not be pushing it quite hard enough. When I was
> designing audio amplifiers professionally, I could set the distortion
> precisely to 5% by eye, but I wouldn't trust my eye today.
>
> Just a few ideas to check on as far as the analysis and test circuit go.
> It's hard to get the full power specified by the tube manual operating
> conditions into a load on the secondary of an output transformer, the tube
> manual conditions are for ideal loss less operating conditions.
>
> Regards,
>
> John Byrns
>
> Surf my web pages at, http://users.rcn.com/jbyrns/

To get the 14 watts into 8k, one needs to have 334 vrms across the OPT primary.

This means a peak voltage swing of 236 v occurs at each end of the primary.

When considering one sine wave, this swing is produced in a 6V6 amp which
an anode supply greater than 300v as John suggests above,
and the swing is produced by class AB action, and to get the full swing
one tube is cut off, and has no RL connected, while the other
sees a load of 2k. So what Tim said about the load changing is correct,
and the load line is a curved line on a graph, or rather a bent line,
because in class A each tube sees 1/2 RLa-a, ie, 4 k, then the load becomes 2k for
one tube
after cut off in the other.

It is easy to plot the action going on in one tube, rather than draw composite Ra
lines
across which one can plot a PP load.

With a copy of the Ra tube data lines in front of you,
draw a line from the value of Ea, say +320v, (the idle voltage between anode and
cathode), with a slope
RLa-a/4 ( 2k )towards the Ia axis.
So 320/2,000 = 0.16, so the the 2k load line is from 160mA on the vert axis
down to 320v on the horizontal axis. This is the load line for each tube
in a pure class B amp.
This load line intersects the curved data line of Ra where Eg1= 0.0v on a 6V6.
The position of the intersection will occur when Ia = 118ma, and
the Ea minimum value during the negative going swing is 320v - 236v = 84v.

The anode voltage cannot swing more than the boundary outlined by
the Ra line for Eg=0.0, unless you go to class AB2, and push the grid positive,
and then you'd get an extra55 approx volts of swing, and then you'd get 21 watts
fromthe pair of 6V6.

But all you need to have to calculate the po is the Ra at Eg=1.0v and 1/4 of RLa-a

to be able to work out the max po for any class AB1 amp, regardless of the
bias current at idle.

To fill in the drawing of the single 6V6 condition to show the class A load
portion, the
chosen bias point might be where Ia = 30 mA, plotted vertically above the Ea =
320v,
and through this point you draw a load line with a value of 1/2 RLa-a, or 4k,
and this will intersect the horizontal Ea axis at 440v, and the vertical Ia axis
at 110 mA.
You will also see that the 4 k class A load line intersects the 2k class B load
line at 200v.

From this, you can see that each 6V6 cuts off when its anode voltage rises to
440v,
and ends its class A action when the swing is down to 200v, and the
class A swing is 120v peak. This translates to 240 peak v anode to anode,
which is 170vrms, and this gives 3.6 watts of class A power.

So from the load line analysis with *one* tube, we can deduce what happens for the
two in a PP circuit.

You should see that if the class A line of 4k is raised to pass through a higher
current bias point,
the amount of class A v-swing is increased, ( and the thd is reduced ).
But then the bias current point is limited by the Pd max of say 12 watts for the
tube.
To get a greater % of class A in the pair of 6V6, the load value must be perhaps
12k a-a, and the graph re-plotted, and then you may find the class A amount
may extend to where the class B and class A load line intersect on the Ra line
where Eg1 = 0.0v.
Its impossible to get class A po = more than about 45% of the summed Pd of the
tubes,
so if you have 24 watts of Pd from 2 tubes, the class A po max will be 10.8 watts.

The actual load line seen by the single tube isn't really represented
by two straight intersecting lines, but is a tangent off the class A line to a
tangent
off the class B line, since the Ia / Eg cut off behaviour of any tube used in
class AB, especially
triodes, isn't sharp, but gradual, so you get a curved load line.

To work out power before you measure it, it is thus very simple for any tube.

For the measurements to comply to the graphical analysis,
the curves used for Ra must be the correct ones for the
value of screen voltage used.
The Ra line for Eg = 0.0v is located higher on the page for when Eg2 is higher.

Also, you must use fixed bias in your amp to get a correct swing figure when
measuring
full po with a class AB amp, because the flow of DC varies in the tubes during
class AB,
and this raises the cathode voltages in class AB action, so Ea becomes less,
Eg2 effectively may become less, and you won't get the anode swing.

And finally, all the power outputs mentioned above are based on having a perfect
OPT.

No such animal exists, and in many el-cheapo OPTs the power losses are 15%
in the OPT, due to the power lost in the resistances of the windings,
since P = I squared x R.

So with many 6V6 amps, one is lucky to get 9 watts of class A.

I'd probably prefer 9 watts of class A from a single KT88 in UL......


Patrick Turner.

[John Stewart]

Gregg wrote:

> Hi The Cheat,
>
> Strong Bad here.....just kidding ;-)
>
> I have measured 15W out of a pair of 6V6 in PP pure pentode before
> distortion was objectionable. The transformer was 8,800 ohms measured
> reflected load from an 8 ohm resistive dummy load.
>
> Va was 330V.
>
> --
> Gregg t3h g33k
> "Ratings are for transistors....tubes have guidelines"
> http://geek.scorpiorising.ca

I got 26 watts into an R load out of a pair of 6V6GT's after losses
in the OP Transformer. I will leave it to the whizes in this thread to
figure out how I did that!!

Cheers, John Stewart

[Nothing40]

Gregg > wrote in message news:<vrgid.49648$E93.42612@clgrps12>...
> Hi The Cheat,
>
> Strong Bad here.....just kidding ;-)
>

The Cheat,we put that light swich in for you to turn the light on,and
off,not so you could throw lightswitch raves! (*funky music starts*)

(sorry,couldn't resist...I love that one...) ;-)



> I have measured 15W out of a pair of 6V6 in PP pure pentode before
> distortion was objectionable. The transformer was 8,800 ohms measured
> reflected load from an 8 ohm resistive dummy load.
>
> Va was 330V.

15W? really?..hmm.. not bad. :-)
I get ~8wpc from my stereo PP 6V6 amp,both channels driven before any
signs of distortion... maybe ~10-11W one channel at a time. B+ in this
amp is only about 310V tho,and the OPT's are sorta small. (But it
sounds darn nice!)

[Gregg]

Behold, Nothing40 signalled from keyed 4-1000A filament:

> Gregg > wrote in message
> news:<vrgid.49648$E93.42612@clgrps12>...
>> Hi The Cheat,
>>
>> Strong Bad here.....just kidding ;-)
>>
>>
> The Cheat,we put that light swich in for you to turn the light on,and
> off,not so you could throw lightswitch raves! (*funky music starts*)
>
> (sorry,couldn't resist...I love that one...) ;-)

LIGHTSWITCH RAVE! Huzzuh! :-)))))


>> I have measured 15W out of a pair of 6V6 in PP pure pentode before
>> distortion was objectionable. The transformer was 8,800 ohms measured
>> reflected load from an 8 ohm resistive dummy load.
>>
>> Va was 330V.
>
> 15W? really?..hmm.. not bad. :-)
> I get ~8wpc from my stereo PP 6V6 amp,both channels driven before any
> signs of distortion... maybe ~10-11W one channel at a time. B+ in this
> amp is only about 310V tho,and the OPT's are sorta small. (But it sounds
> darn nice!)

I used the Russian 6V6's and I did push them beyond their limits. No plate
glow though.

<accent>
Russkie toob tuff, no glow, no problem!
</accent>

;-)

--
Gregg t3h g33k
"Ratings are for transistors....tubes have guidelines"
http://geek.scorpiorising.ca

[Patrick Turner]

John Stewart wrote:

> Gregg wrote:
>
> > Hi The Cheat,
> >
> > Strong Bad here.....just kidding ;-)
> >
> > I have measured 15W out of a pair of 6V6 in PP pure pentode before
> > distortion was objectionable. The transformer was 8,800 ohms measured
> > reflected load from an 8 ohm resistive dummy load.
> >
> > Va was 330V.
> >
> > --
> > Gregg t3h g33k
> > "Ratings are for transistors....tubes have guidelines"
> > http://geek.scorpiorising.ca
>
> I got 26 watts into an R load out of a pair of 6V6GT's after losses
> in the OP Transformer. I will leave it to the whizes in this thread to
> figure out how I did that!!
>
> Cheers, John Stewart

I have seen 36 watts from a couple of 6BQ5, Ea = 700v, Eg2 = 350v, and
RL = 14 ka-a.
It was in a Musical Reference amp.

Patrick Turner.

[Iain M Churches]

"Gregg" > wrote in message
news:1NFid.77627$df2.19231@edtnps89...
> Behold, Nothing40 signalled from keyed 4-1000A filament:
>
>> Gregg > wrote in message
>> news:<vrgid.49648$E93.42612@clgrps12>...
>>> Hi The Cheat,
>>>
>>> Strong Bad here.....just kidding ;-)
>>>
>>>
>> The Cheat,we put that light swich in for you to turn the light on,and
>> off,not so you could throw lightswitch raves! (*funky music starts*)
>>
>> (sorry,couldn't resist...I love that one...) ;-)
>
> LIGHTSWITCH RAVE! Huzzuh! :-)))))
>
>
>>> I have measured 15W out of a pair of 6V6 in PP pure pentode before
>>> distortion was objectionable. The transformer was 8,800 ohms measured
>>> reflected load from an 8 ohm resistive dummy load.
>>>
>>> Va was 330V.
>>
>> 15W? really?..hmm.. not bad. :-)
>> I get ~8wpc from my stereo PP 6V6 amp,both channels driven before any
>> signs of distortion... maybe ~10-11W one channel at a time. B+ in this
>> amp is only about 310V tho,and the OPT's are sorta small. (But it sounds
>> darn nice!)
>
> I used the Russian 6V6's and I did push them beyond their limits. No plate
> glow though.
>
> <accent>
> Russkie toob tuff, no glow, no problem!
> </accent>


The Ruskies call them "lamps" in Russian. Makes it even
better:-)

Iain

[Wbittle]

Throw out your formulas and use a scope, volt meter and a signal
generator. Nothing is perfect in the mathematical world of tubes. I have
been servicing as well as designing and building tube amps for many
years. I use load lines and calculated specifications as a starting
point. Then I 'tweak' by trial and error. Some things end up making no
sense on paper but the amp is right there in front of me producing what
they say can't be done. Tubes are just not a perfect science.
The 6V6, for most hi-fi applications and circuits, seems to be pretty
well topped out at around 10 watts per pair for very low distortion. But
with any amp, your full rated power is determined by the types of
tubes, the circuit, AND how much distortion you are willing to accept.
My Bell 3D Binaural claims 10 watts RMS per channel using two 6V6's per
channel. It can deliver more, but not at below 1% distortion. At 5%, it
can probably deliver close to 15 watts. The Mullard 5-20 is rated at 20
watts RMS at something like .1% distortion. Mullard calls it a 20 watt
amp, but it can crank out close to 35 watts at 3 - 5% distortion.
You can get far more power out of an amp with a regulated grid supply
and fixed bias then using a simple non regulated power supply and
cathode bias. Also, a class AB2 will deliver far more power then an A or
AB1.
Bill B.

The Cheat wrote:

> I am a retired electronics engineering professor and I am always finding new
> things that I don't understand. Here is the latest problem. The tube manuals
> give the maximum power output (with 5% or so distortion) for a pair of 6V6
> running in class AB push-pull with an 8K load as 14 W. This is with 285 V
> plate-to-cathode and screen-to-cathode and a grid bias of -19 V. Using self
> bias requires a cathode resistor of 250 Ohms. This is a classic
> configuration used in countless power amps over the years. However, from
> load line analysis of the 6V6 characteristics (attached), it doesn't seem
> possible to get this much power. The usual method to estimate power is to
> draw the composite load line for one tube from the zero grid Volts curve to
> the maximum plate voltage on the plate voltage axis. The calculation is
> based on converting a peak sinusoid to rms by dividing by root 2. The
> composite load line with slope 4K (what each tubes sees) is shown below. The
> actual operating conditions at zero signal are shown as point Q. The maximum
> power calculation is shown, which yields 9.3 W, not 14 W. When I test this
> output stage (with screens tied to the center-tap of the output
> transformer), this is about what I get when the output begins to clip on the
> scope. The output voltage is measured with a true rms voltmenter with an 8
> Ohm resistive load. So I am satisfied that theory gives a good estimate of
> what I measure but where does the 14 W value in the tube manuals come from?
> Any comments from the experts out there would be appreciated.
>
> Cheers
>
> Lloyd
>
>
>
> --
> Lloyd Peppard
> Mapletree Audio Design
> R. R. 1, Seeleys Bay, ON
> Canada K0H 2N0
> 613-387-3830
>
> http://www.mapletreeaudio.com
>
>
>