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I am designing a Nixie tube clock and need to power some INS-1 neon bulbs. The bulbs have a striking voltage of 65 - 90 V and an operating current of 0.5 mA.

To power the bulbs should I divide 170 V to 68 V using a voltage divider (left) or should I step down 120 VAC to 48 V and rectify that to ~68 VDC (right)?

In the final design the circuit will be powering two blinking neon bulbs. Since I am dealing with mains and am fairly new to electronics, I thought I should probably ask here before testing these designs. Any suggestions are appreciated.

Example circuits

davidcary
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Warsp
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    Something to consider when powering a neon lamp. It has two electrodes in the glass bulb. If you power it off DC, only one electrode glows. Power it off AC and both glow. – Simon B Nov 17 '22 at 12:17
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    Per the UL White Book, 0.5 mA is the amount of current a 120/240V device is allowed to leak to safety ground, if it does not have access to the neutral wire (smart switch etc.) and certain other precautions are built-in. GFCIs trip at 5-6 mA. – Harper - Reinstate Monica Nov 17 '22 at 22:02

3 Answers3

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You don’t need or want to divide down the voltage. Neons will work fine from the full voltage, provided you adjust the resistor value to limit the current to the recommended range. They’re a bit like LEDs in that regard, except if they don’t get enough voltage to begin with they never light (and that voltage increases with age, and is higher in the dark).

So (170-68)/0.5mA is about 200k ohms.

Spehro Pefhany
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  • "except if they don’t get enough voltage to begin with they never light" same with LEDs really. Just a different value of enough! :-) – Dannie Nov 17 '22 at 09:14
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    @Dannie The neons have a strike voltage that is higher than the operating voltage, so a sort of memory due to the negative resistance character of the discharge. LEDs don't have that kind of characteristic. You don't need to connect the series resistor to 5V before they will operate on 3.3V. – Spehro Pefhany Nov 17 '22 at 10:01
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    one of the things that really got me interested in physics a million years ago when I was a kid was a nightlight that only lit during the day - it was of course just what you mention - photoelectrons ejected when light struck the bulb were needed to trigger breakdown. – uhoh Nov 18 '22 at 01:49
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    @uhoh Some of the bulbs had a whiff of radioactivity added in order to mitigate that issue. If they're decades old the radioactivity may have faded significantly due to short half-life of the materials used. – Spehro Pefhany Nov 18 '22 at 02:03
  • *Cool!* I never knew that. First [thorium in lantern mantles](https://en.wikipedia.org/wiki/Gas_mantle) made them slightly radioactive incidentally, now radioactive neon light bulbs. Next they'll be saying that my old TV makes X-rays and there's radon in the basement :-) – uhoh Nov 18 '22 at 11:08
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    @uhoh. Thats bananas! (0.1 microsieverts) – RedGrittyBrick Nov 18 '22 at 11:29
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As an alternative, consider a stack of ten 9V PP3 batteries in series. I made a replacement for a 90-V B126 valve radio battery this way: easily obtainable and has worked for years.

If nothing else, it's a good way to test your neons before you build the mains circuit.

jonathanjo
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I use a pair of 48V "wall warts" in series (with a 33kΩ resistor between them to limit the current to a very safe level) to power this thing:enter image description here

John Doty
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    I know you said there's a current-limiting resistor, but that just *looks* terrifying. – Hearth Nov 18 '22 at 02:27
  • @Hearth The current is so small, I can't even feel it . – John Doty Nov 18 '22 at 02:30
  • What are the black things in parallel with the neons? There really is something pleasing about the way [neons flash](https://electronics.stackexchange.com/a/479349/173919). – jonathanjo Nov 18 '22 at 07:56
  • @jonathanjo The "black things" are leaded insulators. Their purpose is mechanical support: the leads on the bulbs are flimsy. – John Doty Nov 18 '22 at 13:27