Ac voltage regulator

[Yamamaya42]

A 1k load on 6.3v is infinitesimal, smaller than most tubes, if by any chance the power resistor gets warm, try 5k or larger, I just pulled it at random, all that matters is there is something to divide 0-6.3 one end to the other on it, seems to me 1k should do fine.

[bandersen]

The heater is 0.55 volts at 0.06A or 60mA. That is 0.03 watts.

You would need a 100 ohm resistor in series with the CRT filament to run it from a 6.3 VAC transformer. The resistor would put out 1/3 watt. Nothing should be getting warm at all.

Did you put a resistor in parallel with the filament instead of in series?

[timmy]

Originally Posted by bandersen

The heater is 0.55 volts at 0.06A or 60mA. That is 0.03 watts.

You would need a 100 ohm resistor in series with the CRT filament to run it from a 6.3 VAC transformer. The resistor would put out 1/3 watt. Nothing should be getting warm at all.

Did you put a resistor in parallel with the filament instead of in series?

A 100 ohm resistor would bring the voltage down to around .55v the wattage sounds good. Would that be a carbon or power resistor? The full voltage passes freely I have to be able to measure the voltage befor putting the heater on it otherwise it will burn the filament out instantly. I did put a resistor in parallel was the only way to get the correct voltage nothing in series would work.

[timmy]

The 6.3v transformer no load is around 8.50v with a 6.3v heater load it’s around 7v. But the ohms load of the .55v tube heater is 3.5 ohms

[bandersen]

No wonder the transformer got so hot - you were basically shorting it out.

Saying 6.3 volt heater load is meaningless. What is the current or resistance?

Since this load is so light, better to adjust calculations for 8.5 volts.

(8.5 - 0.55)V / 0.06A = about 130 ohms. Wattage = (0.06 * 0.06) * 130 = 0.5.

Use a 1 watt resistor to be safe. Carbon comp, metal film, wirewound. Makes no difference.

[timmy]

Originally Posted by bandersen

No wonder the transformer got so hot - you were basically shorting it out.

Saying 6.3 volt heater load is meaningless. What is the current or resistance?

Since this load is so light, better to adjust calculations for 8.5 volts.

(8.5 - 0.55)V / 0.06A = about 130 ohms. Wattage = (0.06 * 0.06) * 130 = 0.5.

Use a 1 watt resistor to be safe. Carbon comp, metal film, wirewound. Makes no difference.

So if I use a 130 ohm resistor in series and then check the voltage it will show the full voltage at the resistor but putting it on the heater if it drops to where it should be what happens with the initial surge to the heater it could open ?

[timmy]

Well I tried it with 140 ohms and it brings it down but to 150mv the target voltage is .55v but before a load is placed it’s got full voltage but at the split second it’s put to the heater it could open the heater I believe.

[timmy]

The heater resistance on this tube is 3.4 ohms so I would use a 3.3ohm resistor to get a quick reference of the voltage befor plugging the tube in so the 150mv is to low 550mv would equal.55v but still full voltage for that split second I fear it wit take out the heater I only have 2 of these tubes that are good and I already burned the 3rd one I had.

[Yamamaya42]

Originally Posted by timmy

The heater resistance on this tube is 3.4 ohms so I would use a 3.3ohm resistor to get a quick reference of the voltage befor plugging the tube in so the 150mv is to low 550mv would equal.55v but still full voltage for that split second I fear it wit take out the heater I only have 2 of these tubes that are good and I already burned the 3rd one I had.

That's cold I'm sure. as it heats up, the resistance in the heater will go up.

[timmy]

Originally Posted by Yamamaya42

That's cold I'm sure. as it heats up, the resistance in the heater will go up.

Yes I’m sure it will but I need to know the initial surge voltage if any so the heater don’t open

[bandersen]

Add a thermistor in series. They have the opposite property. High resistance when cold.

[timmy]

Does all this apply to dc as well ? Same result with the thermistor?

[Yamamaya42]

The way that Bandersen describing is one way to do this, the adjustable power resistor is another.

Using a resistor such as, https://www.digikey.com/en/products/...WA210/11644319

Very cheap at under $7, and 1k won't put any real load on the filament transformer at all.
To be wired up like so, http://suzaku.live-evil.org/CRT-POWER.jpg
And starting with the resistor slider set to the lowest extreme, (0 volts) just slowly raise it up until you get the 0.55 volts you need on the heater for the tube.
This is how I'd do it anyway.

[timmy]

Originally Posted by Yamamaya42

The way that Bandersen describing is one way to do this, the adjustable power resistor is another.

Using a resistor such as, https://www.digikey.com/en/products/...WA210/11644319

Very cheap at under $7, and 1k won't put any real load on the filament transformer at all.
To be wired up like so, http://suzaku.live-evil.org/CRT-POWER.jpg
And starting with the resistor slider set to the lowest extreme, (0 volts) just slowly raise it up until you get the 0.55 volts you need on the heater for the tube.
This is how I'd do it anyway.

L
So using the 1k adjustable resistor there may not be enough current and the question is would there be a large surge upon turning it on for a nano second which don’t take much to open a heater filament like this one @ only.55v ideally I would like to measure voltage & .830v then plug the tube in and see .50v or alittle more but I can’t do this because no load shows full voltage.

[Yamamaya42]

1k on the filament transformer places no noticeable load on it at all, especially with the heater being 3.4 ohms, with the CRT unhooked, the transformer and resistor should hardly get warm at all, and all power would go the crt when hooked up and adjusted right.