Filament Voltage Too High

[dbishopbliss]

My filament voltage is about 7V (AC) (just over, but close enough).  I have two tubes that filaments draw 350mA each.

Assuming I want to drop the voltage by 0.7V to achieve 6.3V do I simply put a 1 Ohm 1 watt resistor in series with both filament wires?

0.7V / .7A = 1 Ohm.
0.7V * 0.7A = 1 Watt.

Is that right?

[JC]

A couple of questions:  Are we to presume that the heaters are AC?  And, are we to presume that the two of them are wired in parallel?

[Grainger49]

There is some non linearity to the current, voltage, resistance relationship in heaters.  They change resistance with different current/voltage.  But to get much closer to 6.3V a 1 ohm resistor will get you very close.

[JC]

Actually, the way the arithmetic works out for me, the resistor in the equation would be dissipating .49W.  I think that by indicating a 1W resistor, he was already figuring in a safety factor of ~2.

My questions are with regard to the "in series with both filament wires" part of the post.

Perhaps we should also discuss some nomenclature while we are on the subject:  It has always been my understanding that "filaments" is most generally used to describe the heating portion of a directly heated tube.  If it is an indirectly heated tube, I have most often seen "heater" used for that bit.

So, I should probably add to my two questions above a third question:  Are the tubes in question directly heated or indirectly heated?

Not that it is likely to matter much in this discussion, but I find that this format of communications on a topic such as this often benefits from a general agreement on terms and usage of language.

[dbishopbliss]

A couple of questions:  Are we to presume that the heaters are AC?  And, are we to presume that the two of them are wired in parallel?

The heaters are AC.  The transformer secondaries are rated as 6.3VCT@2A.  When I measure across the leads, I get 7.1V (connected to the tubes).  When I measure from CT to one of the leads I get 3.54V (which makes sense).  I have the CT connected to a potential divider to so it is raised about 70V. 

I can try the 1 ohm resistor, but perhaps what I really need is 1/2 ohm resistor on each lead.  Could I run into problems with unmatched resistors?  Seems like things have to be pretty precise when you get this low of a value.

Another option I was considering is using two diodes strapped together in opposite directions.  I should be able to find a diode that drops 1/2 volt.  What do you think?

[Grainger49]

An unloaded transformer "floats" high in voltage.  You are drawing less than the transformer can deliver.  The 1 ohm and 2 - 1/2 ohm resistors will have exactly the same effect.

The diodes might inject some noise into the heaters.  If it were me I would buy several 1 ohm resistors and parallel them till you get the voltage you want.

[JC]

A single 1 Ohm resistor in series with the paralleled heaters would do it.  You may not get precisely the Voltage drop you're calculating, but you should get your heater Voltage to well within +or- 10% of nominal.

As you lower the heater Voltage, you may also lower emission from the cathode a tiny bit.  Thus, your operating point may shift slightly.  But, this is by no means a linear relationship in my experience.  You may not measure any shift in operating point at all.

[Jim R.]

David,

In my SE el84 amp, the filaments were also a bit high -- also around 7v, and now have a pair of high power schottky  diodes, back to back and in series with one leg of the heater circuit.  This brought the filament voltage down to where it needs to be and is absolutely quiet on my 100 dB cornwalls.

-- Jim