Hey Enzo.....

[daz]

Could you look at this post at another forum (by someone you'll know from the PV forum) on a simple way to run a c30 at 1/2 power and tell me if you see anything about it that may make it a bad idea? (other than the potential to insert the tubes wrong) And also, i use a 8 ohm speaker with the speaker jack instead of the internal wires. How will this translate? in other words, will i basically be running the speaker/OT at a load that is double what the tubes want to se? If so i guess i'm not going the get a good match. It's a bit confusing to me, but how would you do this besides going back to the stock 16 ohm speaker.

http://www.tdpri.com/forum/amp-centr...-watt-mod.html

[Rob Mercure]

While Enzo gets around to answering I think I would simply plug a pin 4 to 5 jumper into one "pulled" tube socket and a suitable value/power resistor between pins 4 and 5 on the other "pulled" socket. You're trying to drop 12.6 volts at 0.76 amps so you need 16.6 ohm 9.6 watt resistor using straight ohms law. But you'd want double the power rating and you could probaby get away with 15 ohm resistor - and 15 ohm power resistors could easily vary enough in value to get to 16-17 ohms so a 15 ohm 20 W resistor would work. And AES used to have little units consisting of a 9 pin socket and plug wired together to permanently plug into a bad 9 pin socket so that you wouldn't have to replace it. If they have them you could simply do the wiring on these plugs. Any of this seems simpler than cutting 7 pins off an EL34 - and I'm not sure that 2 pins hold the tube's weight when you shift the amp around.

But perhaps Enzo has other ideas.

Rob

[daz]

Thanks, but it would just be quick and easy with a tube and i have a whole drawer full of old 84's that i should throw away but never get around to it. So cuttin' a couple up is no problem at all. I guess my main concern is really the impedence question.

While Enzo gets around to answering I think I would simply plug a pin 4 to 5 jumper into one "pulled" tube socket and a suitable value/power resistor between pins 4 and 5 on the other "pulled" socket. You're trying to drop 12.6 volts at 0.76 amps so you need 16.6 ohm 9.6 watt resistor using straight ohms law. But you'd want double the power rating and you could probaby get away with 15 ohm resistor - and 15 ohm power resistors could easily vary enough in value to get to 16-17 ohms so a 15 ohm 20 W resistor would work. And AES used to have little units consisting of a 9 pin socket and plug wired together to permanently plug into a bad 9 pin socket so that you wouldn't have to replace it. If they have them you could simply do the wiring on these plugs. Any of this seems simpler than cutting 7 pins off an EL34 - and I'm not sure that 2 pins hold the tube's weight when you shift the amp around.

But perhaps Enzo has other ideas.

Rob

[Enzo]

Sure it would work. When I first saw that I thought Give the guy credit for thinking, but then for only that little reduction, is it worth the trouble.

The tube's weight is not held by the two legs, it is held by the spring clip that the C30 has on each tube. Not an issue. In fact, you can leave a couple more pins on the tube for stability. Pins 1 and 8 come to mind.

Pulling half the power tubes is a time honored trick - and you like the results or you don't, but that is a separate issue from whether it works. In the C30, though, the heaters of the four power tubes are wired in series. SO if you snip the legs off the tube saving the heater pins, then these dummy tubes keep the heater circuit intact.

Resistors would also work, but a couple old EL84s, snip snip, and your glass bottle resistors are ready to go - no calculating, no soldering, there you go. I can't think of anything easier, it can be done in seconds, literaly. So in its own way it is an elegant solution to the problem.

SInce the impedance changes you need to step the 16 ohm speaker down to the 8 ohm tap. So wire the speaker to a plug, plug it into the EXT jack which switches to 8 ohms, and tape off the internal speaker wires. If you want to use an 8 ohm speaker, there is no way for it to matvh, and it is up to you if you want to run it as a mis-match. Won't likely hurt it.

[daz]

Thats what i figured. (8 ohm speaker.....no way) but i tried it anyways. I'm not sure if it's the mismatch or the lower power but i can get a lot more drive at lower settings. But i think the mismatch makes for a little fuzzier tone. Not a bad little trick tho. Wish i could hear it with a 16 ohm speaker plugged into the jack as my 8 ohm is. I think it might be just the ticket for low volume gigs. But i use an EV and it's a critical part of my sound so it's staying. Thanks.

Sure it would work. When I first saw that I thought Give the guy credit for thinking, but then for only that little reduction, is it worth the trouble.

The tube's weight is not held by the two legs, it is held by the spring clip that the C30 has on each tube. Not an issue. In fact, you can leave a couple more pins on the tube for stability. Pins 1 and 8 come to mind.

Pulling half the power tubes is a time honored trick - and you like the results or you don't, but that is a separate issue from whether it works. In the C30, though, the heaters of the four power tubes are wired in series. SO if you snip the legs off the tube saving the heater pins, then these dummy tubes keep the heater circuit intact.

Resistors would also work, but a couple old EL84s, snip snip, and your glass bottle resistors are ready to go - no calculating, no soldering, there you go. I can't think of anything easier, it can be done in seconds, literaly. So in its own way it is an elegant solution to the problem.

SInce the impedance changes you need to step the 16 ohm speaker down to the 8 ohm tap. So wire the speaker to a plug, plug it into the EXT jack which switches to 8 ohms, and tape off the internal speaker wires. If you want to use an 8 ohm speaker, there is no way for it to matvh, and it is up to you if you want to run it as a mis-match. Won't likely hurt it.

[Satamax]

Sure it would work. When I first saw that I thought Give the guy credit for thinking, but then for only that little reduction, is it worth the trouble.

The tube's weight is not held by the two legs, it is held by the spring clip that the C30 has on each tube. Not an issue. In fact, you can leave a couple more pins on the tube for stability. Pins 1 and 8 come to mind.

Pulling half the power tubes is a time honored trick - and you like the results or you don't, but that is a separate issue from whether it works. In the C30, though, the heaters of the four power tubes are wired in series. SO if you snip the legs off the tube saving the heater pins, then these dummy tubes keep the heater circuit intact.
Resistors would also work, but a couple old EL84s, snip snip, and your glass bottle resistors are ready to go - no calculating, no soldering, there you go. I can't think of anything easier, it can be done in seconds, literaly. So in its own way it is an elegant solution to the problem.

SInce the impedance changes you need to step the 16 ohm speaker down to the 8 ohm tap. So wire the speaker to a plug, plug it into the EXT jack which switches to 8 ohms, and tape off the internal speaker wires. If you want to use an 8 ohm speaker, there is no way for it to matvh, and it is up to you if you want to run it as a mis-match. Won't likely hurt it.

Huh, daft idea, why not just cut the grid pin??? Ok, i see it now, it's fixed bias On a cathode biassed amp, that could be a way to keep the amp matched. I've been thinking along that line for a long while Here one could disconect the signal wire and make a separate bias supply so there's no coupling through bias . Ok, i'm going astray, i'm out

[Bruce / Mission Amps]

Not sure if you quite ment this, but if you cut the grid pin from a ground or bias source on either cathode or fixed bias with that tube still connected to B+, a filament supply and leave tube's cathode grounded in anyway, ... it will take off with no bias and draw max plate current, eventually failing.
To shut the tube nearly off, you could remove the B+ from the "other grid", the screen grid.
To keep current through the OT, how about something really different like a switching arrangement that takes the B+ off the plate of the unwanted power tube and steers that same plate B+ to a grounded, 25 watt 5K-10K resistor?
You'd have to figure out through Ohm's Law what resistance at the given B+ = the loaded power tube.

[Richard]

On amps that have a half power mode that turns off a couple of tubes, I believe they lift the ground on the cathode. These are fixed bias amps, If the tubes do not share a cathode bias resistor, I think this would work. You could merely cut the cathode pin. I believe this maintains the same impedance.

[Steve Conner]

The problem with pulling two tubes from a four tube amp is that the remaining two try to work twice as hard and wear out prematurely. Anything that prevents two of the tubes from pulling their weight will also overrun the remaining two: there is no combination of tube pins you can cut off to avoid this. (If there was, Randall Smith would have a patent on it.)

Moving the speaker impedance tap compensates for that by making the load twice as easy to drive, so now two tubes run happily, providing half the power.

If you wanted to pull tubes without moving the impedance tap and without overrunning the remaining tubes, the only thing I can think of that would work would be to reduce the B+ by quite a lot, and you would end up with nearer one quarter the power you had with four tubes.