TexasRF@aol.com wrote:
>
> I have been experimenting with some NTC thermistors to use in a filament
> transformer 120vac primary to limit the secondary (tube filament) current
> flow.
> The unit showing the most promise, model CL90, is rated at 120 ohms cold,
> dropping to about 1.2 ohms hot.
Is a resistor not good enough for you? If the transformer is not huge in
comparison to that needed, then it will limit the inrush current. A
simple resistor will do a bit more.
> Can anyone here help with the method to calculate the resistance change vs.
> time in one of these devices?
To do it properly, it would be horribly complex. You have a set of
non-linear effects going on.
1) The transformer is not going to be a perfect voltage source with zero
impedance.
In fact, that is good, as it limits inrush current, but it does make
things more complex to work out.
So you would have to measure the properties of the transformer.
2) The thermistor is a non linear device, whose temperature and
resistance depend not only on the current through it now, but what it
has been previously.
3) The heater is non linear, with the same issues as the thermistor.
> At turn on, the tube filament resistance increases about 10 X with time as
> the thermistor decreases about 100 X with time. What I really want to
> determine
> is the peak filament current while all the resistance changes are taking
> place.
You also need to know if the heater will *ever* get hot. You may well
find that the NTC thermistor will limit the current too much to allow
the heater to draw sufficient current to heat up the NTC thermistor.
> As a first step, I have inserted a 1 ohm resistor in series with the primary
> and added a diode and filter C that charges to the peak voltage across the
> resistor.
In the primary? That is not a good place to put any resistor. In any
case, the diode drop would be significant, but it has a more fundamental
flaw.
> If Ohms law works here, at turn on, it seems that the peak current
> in Amps would equal the charged voltage in Volts.
No.
If you assume an ideal transformer, the primary current will be less
than the secondary current by a factor of the turns ratio or voltage
ratio. If the primary is 120 V, the secondary 6 V, then the voltage is
decreased by a factor of 20. So the current in the primary will be
1/20th of that in the secondary.
To measure the peak current, put a very low resistor and measure the
voltage across it with a scope, not a volt meter. Unless you have a
decent bench volt meter that can do 100s or 1000's of readings per
second and give you the peak, or allow you to read them over a
RS-232/GPIB bus.
> It would be reassuring if the measured data agreed with some calculations.
I think you will be hard pressed to do it accurately. There are too many
non-linear effects going on. Solving non-linear equations is not
trivial, but its not helped here the fact you will not know what numbers
to use to try to solve those equations.
> Anyone been there/done that?
Not been there, not done it and will not attempt it.
I'd just use a sensible (not overrated transformer) and/or a resistor.
> Thanks/73,
> Gerald K5GW
--
David Kirkby,
G8WRB
Please check out http://www.g8wrb.org/
of if you live in Essex http://www.southminster-branch-line.org.uk/
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|