Dear Patrick,
I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what kind
of wattage out can I expect and how much of that will be class A?
Also besides just giving me a figure, can you explain how you go about to
attain it? Thanks so much for all your wisdom & help.
Cordially,
west

west wrote:

Dear Patrick,
I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what kind
of wattage out can I expect and how much of that will be class A?
Also besides just giving me a figure, can you explain how you go about to
attain it? Thanks so much for all your wisdom & help.
Cordially,
west

You will find part of the answer at
http://frank.pocnet.net/sheets/111/8/845.pdf

But that results in a most dangerous piece of equipment.

For the safety of others & yourself a better alternative is possible with tubes
such as KT88, 6550 or EL34. And there are others.

Cheers, John Stewart

west wrote:

Dear Patrick,
I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what kind
of wattage out can I expect and how much of that will be class A?

Enough. No need to think you won't have enough.


Also besides just giving me a figure, can you explain how you go about to
attain it? Thanks so much for all your wisdom & help.
Cordially,
west

Just apply electronic theory, and out comes the OPT requirements.

300B only has low µ so its gain won't be high, but sure, its a nice driver.

You will need another stage in front of the 300b.

More applications of electronic theory, with ironic tendencies
to usin a lotta iron will get you across the line.

LOTTA volts needed for the whole caboodle.
But I assure you that with the best ideas of Lynn Olson you can have the
purdiest
lookin amp at nightime you ever laid eyes on.

Patrick Turner.

"John Stewart" wrote in message
...
west wrote:

Dear Patrick,
I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what
kind
of wattage out can I expect and how much of that will be class A?
Also besides just giving me a figure, can you explain how you go about
to
attain it? Thanks so much for all your wisdom & help.
Cordially,
west

You will find part of the answer at
http://frank.pocnet.net/sheets/111/8/845.pdf

John,
This is a great site. Lots of pertinent info, Thanks. BTW: how do access
other tube data? Site seems to be restricted.
west

But that results in a most dangerous piece of equipment.

Are you talking about 1kV B+?
west

For the safety of others & yourself a better alternative is possible with
tubes
such as KT88, 6550 or EL34. And there are others.

I have worked on CRT display ckts ~ 5KV or is this tube amp circuit more
dangerous? Any experience?
west

Cheers, John Stewart

west wrote:

"John Stewart" wrote in message
...
west wrote:

Dear Patrick,
I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what
kind
of wattage out can I expect and how much of that will be class A?
Also besides just giving me a figure, can you explain how you go about
to
attain it? Thanks so much for all your wisdom & help.
Cordially,
west

You will find part of the answer at
http://frank.pocnet.net/sheets/111/8/845.pdf

John,
This is a great site. Lots of pertinent info, Thanks. BTW: how do access
other tube data? Site seems to be restricted.
west

Home page is http://frank.pocnet.net/ from where you can
navigate to many tubes.

But that results in a most dangerous piece of equipment.

Are you talking about 1kV B+?
west

Yes

For the safety of others & yourself a better alternative is possible with
tubes
such as KT88, 6550 or EL34. And there are others.

I have worked on CRT display ckts ~ 5KV or is this tube amp circuit more
dangerous?

CRT's are happy with a few microamps PS. The 845's will need something like
1000X that much current. That kind of current is much more than enough to kill.

Any experience?

Yes, quite a bit. I worked in the HV test lab at Ferranti where transformers
were built for power stations. But we had lots of safety measures in place. Not
for amateurs at all.

Be careful, John Stewart

west

Cheers, John Stewart

"west" ha scritto nel messaggio
. ..
Dear Patrick,
I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what
kind
of wattage out can I expect

About 110W using full feeding voltage (1250V). Class A will be about
40-45W. A higher anode load will give less power but more class A.

and how much of that will be class A?
Also besides just giving me a figure, can you explain how you go about to
attain it?

Draw the load line on the anode curves chart. Locate the idle point and the
points corresponding to the maximum voltage swing. Get the relevant values
of I and V. Now You have all we need to calculate the amplitudes of two
sine waves, I and V. To get the power, we need to multiply the RMS value of
voltage times the RMS value of current.
RMS V at full power is: (V at Vg=0) - (V at idle) * 2 ('cos the other half
wave is provided by the other tube) * / (2*sqrt 2)
RMS I at full power is: (I at Vg=0) - (I at idle) * 2 ('cos the other half
wave is provided by the other tube) * / (2*sqrt 2)
To get the Class A power, we need to consider a smaller voltage/current
swing, ie, the one limited by the cut-off point (the one where the loadline
crosses the zero current axis) and its symmetrical one with reference to the
idle point.
RMS V class A power is: (V @ "symmetric point") - (V at idle) * 2 (2 tubes)
* / (2*sqrt 2)
RMS I class A power is: (I @ "symmetric point") - (zero) * 2 (2 tubes) * /
(2*sqrt 2)

If You print the anode curves and play a bit with pencil & ruler, it becomes
apparent.

Ciao

Fabio

Thanks so much for all your wisdom & help.
Cordially,
west

I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what
kind
of wattage out can I expect

About 110W using full feeding voltage (1250V). Class A will be about
40-45W. A higher anode load will give less power but more class A.
and how much of that will be class A?

Also besides just giving me a figure, can you explain how you go about
to
attain it?
Draw the load line on the anode curves chart. Locate the idle point and
the
points corresponding to the maximum voltage swing. Get the relevant
values
of I and V. Now You have all we need to calculate the amplitudes of two
sine waves, I and V. To get the power, we need to multiply the RMS value
of
voltage times the RMS value of current.
RMS V at full power is: (V at Vg=0) - (V at idle) * 2 ('cos the other half
wave is provided by the other tube) * / (2*sqrt 2)
RMS I at full power is: (I at Vg=0) - (I at idle) * 2 ('cos the other half
wave is provided by the other tube) * / (2*sqrt 2)
To get the Class A power, we need to consider a smaller voltage/current
swing, ie, the one limited by the cut-off point (the one where the
loadline
crosses the zero current axis) and its symmetrical one with reference to
the
idle point.
RMS V class A power is: (V @ "symmetric point") - (V at idle) * 2 (2
tubes)
* / (2*sqrt 2)
RMS I class A power is: (I @ "symmetric point") - (zero) * 2 (2 tubes) * /
(2*sqrt 2)
If You print the anode curves and play a bit with pencil & ruler, it
becomes
apparent.
Ciao
Fabio

Thanks Fabio, that helps me. You are a good friend and I wish you luck in
your project. It sounds ambitious.
west