On Jul 21, 11:48*am, "Bob Spooner" wrote:
"COLIN LAMB" wrote in message

m...



If the cathode resistor is the correct value, and the voltage across it is
correct, then the power drawn by the tube is correct. *The tubes are
running Class A and so over half of the power consumed is dissipated by
the tubes (divide by 2 of course). *You can increase the cathode
resistance or increase the screen resistor, to reduce power. *This could
cause a mismatch in the output transformer ratio, but being Class A, it
should still sound fine.

Generally, what I would expect you would be seeing (actually, feeling) is
a combination of small factors. *The filament voltage is probably above
6.5 and the plate and screen voltage may be a little above specified.
This is because we have a higher line voltage. *Often, you have to think
about incremental improvement, rather than night and day. *If you drop the
filament voltage about .5 volts, you will decrease heat from the tubes by
1 watt.

Also, adding a small muffin fan can do wonders to remove heat from those
old metl boxes.

73, *Colin *K7FM

Colin,
Do they actually run class A even though the output stage is push-pull?
73, Bob AD3K- Hide quoted text -

- Show quoted text -

Good question. Also if there is 18 volts across the cathode resistor
and it measures within limits (immediately after switching off i.e.
hot) that will allow calculation of the cathode current being drawn by
the tube or tubes.