8
\$\begingroup\$

I'm trying to get an old soviet era cathode ray tube working.

On the data sheet it says the heating element needs between 5.7V and 6.9V, and between 0.54A and 0.66A, but I've measured the resistance of the heating element and it reads as 1.5 ohms. If I apply 6V then I would get 4A.

Am I missing something here? Is it possible to limit the current but keep the voltage at 6V?

I really don't want to burn out the heater as it will be impossible to replace.

\$\endgroup\$
1
  • \$\begingroup\$ When posting questions about physical items, it can help to add photos. Partially to help flesh out the question with unstated details, and also because things are nice to look at. Welcome to the site. \$\endgroup\$
    – Criggie
    Commented Oct 19, 2022 at 0:05

1 Answer 1

Reset to default
34
\$\begingroup\$

Many tubes are designed to run on a nominal 6.3V for the heater.

The reason you are reading a very low resistance is that the heater is cold. As it heats up, the resistance increases.

If in doubt, try running it off a bench power supply with a current limit. But allow for a current surge when it's first turned on.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Instead of saying "current surge", I think it may be better to observe the relationship between power, voltage, and current for hot and cold filaments. The filament is designed to dissipate between 3 and 4.5 watts when hot. If one were to drive it with 0.66 amps when cold, it would only dissipate about 0.66 watts, which might not be enough to heat it up usefully. If one sets e.g. a 1.5 amp limit, then initial power would be about 3.4 watts, which should be enough to heat the tube but far below the 24 watts the tube would initially dissipate if fed 6 volts without a current limit. \$\endgroup\$
    – supercat
    Commented Oct 19, 2022 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.