GB> RE: Tubes in Series was (Re: [Heathkit] Question on 12AU7) LONG

[N2EY at aol.com]

In a message dated 2/3/08 12:02:12 AM Eastern Standard Time, 
ckepus at comcast.net writes:


> Based on the prior posts that recommended not using the 5814A in place of
> the 12AU7 led me to think that even a small percent delta in current
> requirements was to be avoided.  With the math calcs you and the others have
> done, I still remain unsure of why a +/- tolerance of less than 10% would be
> a problem. 

The problem is that it's an error that is being designed-in. Sure, it will 
probably "work", but why stress the tubes unnecessarily? All it takes is a 
resistor around the lower-current tubes to put everything right at design-center.

 What I found looking for an answer on the web before I posted my
> 
> question was that any *voltage excess* provided by a transformer (or AC line
> voltage) to the series string total voltage requirement was typically dealt
> with by a ballast resistor to get it closer with a ballast resistor
> (probably using resistors with no better than +/- 10% accuracy).  Most of
> the articles I found were silent on the current issue.  I think that was
> because all the tubes in the strings described had the same current draw
> requirement. 
> 

Exactly. This is why the "All American Five" type tubes were developed.


> I have never seen a specification in the tube data sheets that claimed an
> accuracy of regard to the current draw stipulated. Is the actual current
> draw likely to be within +/- 5% of the rating?  
> 

I've never seen one either. Point is, whatever the specification, they should 
all be the same or have equalizing resistors.


> At this point, knowing that I don't know the answer to your question, my
> brain overrides my thinking that +/- 10% is OK....so I would err on the side
> of caution and say that in a series filament string that kind of imbalance
> would not be a good thing and I would avoid constructing the condition.
> 
> Is that the correct answer?? :-)


Yep. Particularly when you're building new for today's environment, where 
good bottles aren't inexpensive.

Heck let's do the math.

Say you want to use a tube with 12.6 volt 0.175A heater in series with one 
that has a 12.6 volt 0.15A heater.

The difference is 0.025 A, which should go through a resistor in parallel 
with the tube that has the *lower* current heater.

12.6/0.025 = 504 ohms 
12.6 x 0.025 = 0.315 watts

So a 510 ohm 1 watt (or greater) resistor will do the job. Or a combination 
of other resistors, like two 1000 ohm half-watters in parallel. More watts 
can't hurt.

73 de Jim, N2EY





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