Tube Amp - Output Power Fuse Protection


The question is why do some manufactures of tube amps ,not have protection for the output stage of the amplifiers?
I have been auditioning some tube integrated amps that did not have any kind of fuse protection for the output stage of the amplifier.

I was in front of one amp when an EL34 tube flashed and knocked out the left channel. It blew out an output resistor , which has to be replaced to restore the amp. A simple fuse in the plate circuit would have protected the amp and with a new fuse , restored service. Instead , one would have to take the amp to a local repair shop , or have it shipped if not local.

Is this a common practice of manufactures to not have protection? Is it worth purchasing an amp that does not offer protection?

As we all know , it can be fun not to use protection , but sooner , or later it will cost you! One way or the other!
timo62

Showing 1 response by atmasphere

You need to know plate voltage and how the tube is biased to determine dissipation and how hard the tube is run. They won't tell you. So simply look at the watts per channel, and divide it by the number of tubes. 80 watts per channel out of two KT88's. That's 40 watts per KT88. The maximum dissipation of a KT88 is 42 watts. So that's toastier than you might want.

Kevin, in your example I quoted above the likelihood is that the tube is dissipating much less than the rating! In this case it might only be dissipating 20 watts at full power.

The dissipation is important- it gives you some idea of how much power you can dissipate in the tube. But if the tube is producing 40 watts of output power, there is a larger amount of power that is needed to cause it to do that. Typically it might be about 60 watts or so.

A better approximation of how much power is being dissipated in each tube is to look at the total power draw of the amp at full power. Subtract about 7% for the power transformers themselves. Then subtract about 10 watts for the driver circuit. Then subtract the output power. The remaining power is divided by the number of power tubes and will be pretty close to the dissipation of each tube at full power. Some variance will exist on account of the filaments and class of operation.