Is it theoretically possible for power tube plate dissipation to drop to zero
with large signals?

I'm not making this up... I read it on the web.

"wsc" wrote in message
...
Is it theoretically possible for power tube plate dissipation to drop
to zero with large signals?

No actually, strictly speaking, since a tube will never fully cut off. In
practical terms, yes it often does.

Tim

--
"I have misplaced my pants." - Homer Simpson | Electronics,
- - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting
and Games: http://webpages.charter.net/dawill/tmoranwms

wsc wrote:

Is it theoretically possible for power tube plate dissipation to drop to zero
with large signals?

I'm not making this up... I read it on the web.

It *is* possible for a power tube to sit merilly in its socket,
with 500 v applied to its anode, and no current flow,
and no signal applied, so voltage x current = zero watts.
This occurs in a class C amp at idle.
A class B amp at idle also has zero dissipation at idle,
but it only exists theoretically, since the tiniest signal causes
plate current to flow, and in fact tubes have a gradual cut off
so most "class B" amps do have some tiny bias current at idle.

On the other hand, it is very ununusual for a tube to have a current flow,
but with no voltage between the anode and cathode.
This also is a case of zero dissipation.
However in solid state mosfet devices for switching, the 'on' dissipation can be
very low
even with 10 amps in the device because the drain to source voltage is very low.


Patrick Turner.

Duncan Munro wrote:

On 14 Apr 2004 19:22:35 GMT, wsc wrote:

Is it theoretically possible for power tube plate dissipation to drop to zero
with large signals?

Yes, in some cases.

Class B and AB tubes do not draw any power at all for part of the cycle.

Yes, this is correct.
During part of the cycle, the tube current is cut off, even though there is a
high voltage rise at the anode.

But at idle, all B and AB amps draw a current, and Ea x Ia = the idle power
dissipation.

Patrick Turner.



--
Duncan Munro
http://www.duncanamps.com/

No actually, strictly speaking, since a tube will never fully cut off. In
practical terms, yes it often does.

Tim

To clarify, what I was talking about is was a tube with a large signal where
the bias is irrelevant - i.e. a square signal where the tube current goes from
zero to the saturation current. If the dissipation were zero in this case, the
plate efficiency would naturally be infinite.

I think it may boil down to how you compute dissipation. To me its the average
plate current times the average plate voltage. But to the guy who's making the
claim of zero dissipation, he does the multiplication before the average.
Here's the link:

http://www.aikenamps.com/Why70percent.html

"wsc" wrote

To clarify, what I was talking about is was a tube with a large signal
where
the bias is irrelevant - i.e. a square signal where the tube current goes
from
zero to the saturation current. If the dissipation were zero in this
case, the
plate efficiency would naturally be infinite.


** It can never be zero since tubes are not perfect switches.


I think it may boil down to how you compute dissipation.


** There is only one correct answer - but various to derive it.


To me its the average
plate current times the average plate voltage.


** Completely wrong.

The product of plate voltage and current has to be computed *at each
instant* for a given waveform and a dissipation waveform derived - the
average value of this waveform is then the average dissipation.


But to the guy who's making the
claim of zero dissipation,


** He makes no such claim - he merely gives the result for an imaginary
tube that acts like a perfect switch.




.......... Phil

"wsc" wrote in message
...
To clarify, what I was talking about is was a tube with a large signal
where the bias is irrelevant - i.e. a square signal where the tube
current goes from zero to the saturation current. If the dissipation
were zero in this case, the plate efficiency would naturally be
infinite.

Ah, something like a horizontal output (sweep amp). In this case, the
dissipation is in series with the load on the current-drawing half of the
cycle.

At Pat said, transistors can manage very good class D (either fully on
(saturated; by the plate-curve knee) or cutoff) because of their typical
Vsat of under 4V for the higher-power devices (equivalent to say, a 4X150's
Vsat around 300V). Most stuff is below 2V, with small signal devices
getting less than a diode drop (.7V) in some cases. Likewise such a tube
would be a 6AU6 with Vsat down below 50V. MOSFETs experience the same
characteristic knee curve, but at still lower voltages. Indeed, an IRFZ44
can do what, .02 ohm at 20A? All of 8W peak dissipation!

Tim

--
"I have misplaced my pants." - Homer Simpson | Electronics,
- - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting
and Games: http://webpages.charter.net/dawill/tmoranwms

wsc wrote:
No actually, strictly speaking, since a tube will never fully cut off. In

practical terms, yes it often does.

Tim


To clarify, what I was talking about is was a tube with a large signal where
the bias is irrelevant - i.e. a square signal where the tube current goes from
zero to the saturation current. If the dissipation were zero in this case, the
plate efficiency would naturally be infinite.

You're essentially describing a switching situation, at which tubes
don't typically do nearly as well as transistors (especially power
MOSFETs). No problem at the cutoff end - current is essentially zero, so
dissipation is also zero. However, at the saturation end, there will
typically be a considerable voltage drop across the tube. Look at a set
of plate curves to see this. For the 6L6GC, for instance, with a grid
voltage of zero and a resulting plate current of 200 mA, the plate
voltage will be some 50 volts -- a 10 watt dissipation (not counting the
substantial screen power dissipated under these conditions).

I think it may boil down to how you compute dissipation. To me its the average
plate current times the average plate voltage. But to the guy who's making the
claim of zero dissipation, he does the multiplication before the average.

No, it's not the order of the multiplication. Remember that
instantaneous plate dissipation equals *instaneous* plate voltage times
*instantaneous* plate current. As plate current goes up, voltage goes
down -- but not by the same factor. That explains the curves in the link
you provided.

If you simply multiply average plate current by average plate voltage,
you'll get a wrong result. How wrong? Depends largely on the waveshape
of the signal. For a "right" result, you have to take each point of the
signal, multiply the instantaneous voltage by the instantaneous current,
add them up all those instantaneous powers (integrate them), and only
then divide by time to get the true average power. In principle, one
could derive the power curves for a sine wave signal in conjunction with
an approximation of the characteristic plate curves; in practise,
however, even this would be quite cumbersome. With "real-life" signal
sources - forget it. That's why we tend to use "rules of thumb" based on
experience and a safety margin.

A practical approach for the home experimenter is to remember that power
into the system must equal power out of the system. So, connect a known
dummy load to the speaker output, then at various amplitudes, measure
the DC input power (average plate current times plate voltage), and
subtract the measured output power plus estimated losses in the OPT. The
difference obviously has to go somewhere -- that's your plate
dissipation at that output power.

If you repeat the experiment for a square-wave signal, you'll find the
results quite different!

Clear as mud?

Cheers,
Fred
--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects: http://dogstar.dantimax.dk |
+--------------------------------------------+

Remember that
instantaneous plate dissipation equals *instaneous* plate voltage times
*instantaneous* plate current.

Fred,

Thanks for responding. I happened across your cool web site earlier this year
and was fascinated by dogzilla.

I regularly do the measurements you suggest and the dissipation never goes
anywhere near zero. When I've driven tubes way past clipping things always end
up in a situation where the input power is roughly evenly divided between the
load and the tube (minus some little bits due to heating of the transformer et
cetera.) Here's a typical example, where I measured everything except the
dissipation -- and figured that out by subtraction. This is just as you said,
but I got it from a suggestion in RDH4 on p578. By the time you get to the
right side of the graph the output is essentially square. I have a bunch more
like this for different bias methods et cetera.

http://members.aol.com/wsjcrane/SE_EL84.gif

This is mainly why I find the idea of zero dissipation such a conundrum.

"wsc" wrote in message
...
This is mainly why I find the idea of zero dissipation such a
conundrum.

Zero *instantaneous* dissipation. In a square wave, half on, half off, the
on half is dissipating twice the average power if the off half dissipates
none. Simple enough, just follow the watts.

Tim

--
"I have misplaced my pants." - Homer Simpson | Electronics,
- - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting
and Games: http://webpages.charter.net/dawill/tmoranwms

"wsc" wrote in message
...
Remember that
instantaneous plate dissipation equals *instaneous* plate voltage times
*instantaneous* plate current.

Fred,

Thanks for responding. I happened across your cool web site earlier this
year
and was fascinated by dogzilla.

I regularly do the measurements you suggest and the dissipation never goes
anywhere near zero. When I've driven tubes way past clipping things
always end
up in a situation where the input power is roughly evenly divided between
the
load and the tube (minus some little bits due to heating of the
transformer et
cetera.) Here's a typical example, where I measured everything except the
dissipation -- and figured that out by subtraction. This is just as you
said,
but I got it from a suggestion in RDH4 on p578. By the time you get to
the
right side of the graph the output is essentially square. I have a bunch
more
like this for different bias methods et cetera.

http://members.aol.com/wsjcrane/SE_EL84.gif

This is mainly why I find the idea of zero dissipation such a conundrum.



** There was no claim of zero dissipation in any real situation.

You really need to go figure what was meant by the phrase "taken to the
extreme... "

Plus read all the posts in reply to yours.




.................. Phil

Zero *instantaneous* dissipation. In a square wave, half on, half off, the
on half is dissipating twice the average power if the off half dissipates
none. Simple enough, just follow the watts.

Tim

Tim, Correct me if I'm misreading this, but I think Mr. Aiken is saying that
with a large square wave the average dissipation approaches zero -- his
reasoning being that when the current is high the voltage is low and when the
voltage is high the current is low, and that the switching part can be ignored?
I think his graph for push-pull EL34s shows the average dissipation.

wsc wrote:
Zero *instantaneous* dissipation. In a square wave, half on, half off, the
on half is dissipating twice the average power if the off half dissipates
none. Simple enough, just follow the watts.

Tim


Tim, Correct me if I'm misreading this, but I think Mr. Aiken is saying that
with a large square wave the average dissipation approaches zero -- his
reasoning being that when the current is high the voltage is low and when the
voltage is high the current is low, and that the switching part can be ignored?
I think his graph for push-pull EL34s shows the average dissipation.

I think that the key word here is "approaches". This is true for any
switching situation, an ideal switch will never dissipate any power;
either the current is zero, or the voltage drop across it is zero -
either eventuality producing zero dissipated power.

Some switches are more ideal than others; vacuum tubes tend to be pretty
close to ideal in the off-state (zero current), but not great in the
on-state (maximum current) because of the substantial voltage drop that
still exists across them under this condition. In the example I gave in
a previous post, a 6L6GC in full saturation still dissipates about 10
watts. Averaged out over a 50% duty cycle square wave, this would still
be a dissipation around 5 watts -- and whilst that "approaches zero"
it's still not even "approximately" zero compared to many solid-state
devices under similar situations.

Cheers,
Fred
--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects: http://dogstar.dantimax.dk |
+--------------------------------------------+

I think that the key word here is "approaches".

Fred,

It took a while, but I finally convinced myself you guys are right about the
power calculation.

I guess the reason I've never seen anything remotely close to zero dissipation
does have to do with the word "approaches."

The biggest hurdle, as you said, would seem to be getting the voltage to
approach zero. This is a rather monumental task involving a robust driver
circuit. In such a case the plate dissipation in the power tube is just a
small fraction of the power losses which now include grid dissipation in the
power tube as well as plate dissipation in the driver tube. I think a
meaningful statement about dissipation would probably have to include that
stuff?

"wsc" wrote in message
...
In such a case the plate dissipation in the power tube is just a
small fraction of the power losses which now include grid dissipation
in the power tube as well as plate dissipation in the driver tube.
I think a meaningful statement about dissipation would probably have
to include that stuff?

Absolutely! Unless you're trying to market of course, in which case you'll
include only plate efficiency. Grid ratings be damned!

Also the other reason why MOSFETs are so superior to just about anything.
Until you break the grid insulator. #$@^.

Tim

--
"I have misplaced my pants." - Homer Simpson | Electronics,
- - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting
and Games: http://webpages.charter.net/dawill/tmoranwms