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10-27-2026, 08:31 PM
Hi Guys
If you are lucky enough to have a schematic of the vintage amp with voltages marked then you could make an educated guess regarding how high to bump the Rk value.
Suppose the schematic indicates +25V at the tube cathode and 300V for B+ with 115Vac mains.
With the schematic voltages and Rk value, say 200R, we can calculate the total current through the tubes to be 25V / 200R = 125mA. Let's assume that on this schematic this is shared by two tubes, so 62.5mA per tube. The schematic indicates B+ as 300V, then 300V - 25V = 275V across each tube. The heat per tube is 17W.
We would expect that B+ would track the mains increase. The change from 115V to 120V is 4.4%. Is this enough to over-dissipate the tubes? Considering how self-biased amps are usually designed, the answer would be absolutely.
The increased mains should produce a B+ of 313V. If Vk stays in proportion, it only rises by 1V This seems like a small increase, but it is almost twice the increase in voltage.and total Ik is 130mA. There is now 287V across the tubes for a net dissipation of 37.3W, 18.7W each.
The take-away is that a small increase in mains voltage is enough to make a not so small increase in tube heat when considering cathode-biased amplifiers. The voltage increases. Current increases. The increase of heat is the product of those smaller increases.
One other "little detail" is that heater voltage has also risen by 4.4%. Does this put the new heater voltage out of spec for the tube?
Tube heater voltage is rated as +/-10%, meaning that a nominal 6.3V can be as low as 5.67V and as high as 6.93V. much depends on how the power transformer was specified and upon how well the PT regulates its voltages. A massive transformer has very good regulation where a small one does not. Copper losses can help to compensate against rising winding output voltages, but only so far.
Say we have a 6L6 with an ideal heater that is 900mA at 6.3V. The hot resistance is then 7R and the heat is 5.67W, which drops to 4.6W at the lowest allowed voltage, and rises to 6.9W at the highest. Again we see that the change in power is twice the percentage as the change in voltage. The heater resistance should go down a little at the low voltage and rise with the higher voltage, but we cannot know what is happening without monitoring the heater current.
If you are lucky enough to have a schematic of the vintage amp with voltages marked then you could make an educated guess regarding how high to bump the Rk value.
Suppose the schematic indicates +25V at the tube cathode and 300V for B+ with 115Vac mains.
With the schematic voltages and Rk value, say 200R, we can calculate the total current through the tubes to be 25V / 200R = 125mA. Let's assume that on this schematic this is shared by two tubes, so 62.5mA per tube. The schematic indicates B+ as 300V, then 300V - 25V = 275V across each tube. The heat per tube is 17W.
We would expect that B+ would track the mains increase. The change from 115V to 120V is 4.4%. Is this enough to over-dissipate the tubes? Considering how self-biased amps are usually designed, the answer would be absolutely.
The increased mains should produce a B+ of 313V. If Vk stays in proportion, it only rises by 1V This seems like a small increase, but it is almost twice the increase in voltage.and total Ik is 130mA. There is now 287V across the tubes for a net dissipation of 37.3W, 18.7W each.
The take-away is that a small increase in mains voltage is enough to make a not so small increase in tube heat when considering cathode-biased amplifiers. The voltage increases. Current increases. The increase of heat is the product of those smaller increases.
One other "little detail" is that heater voltage has also risen by 4.4%. Does this put the new heater voltage out of spec for the tube?
Tube heater voltage is rated as +/-10%, meaning that a nominal 6.3V can be as low as 5.67V and as high as 6.93V. much depends on how the power transformer was specified and upon how well the PT regulates its voltages. A massive transformer has very good regulation where a small one does not. Copper losses can help to compensate against rising winding output voltages, but only so far.
Say we have a 6L6 with an ideal heater that is 900mA at 6.3V. The hot resistance is then 7R and the heat is 5.67W, which drops to 4.6W at the lowest allowed voltage, and rises to 6.9W at the highest. Again we see that the change in power is twice the percentage as the change in voltage. The heater resistance should go down a little at the low voltage and rise with the higher voltage, but we cannot know what is happening without monitoring the heater current.
