Hi!
Sometimes the simplest things can stump me.
Here's one that I can't find any two matching descriptions for, so I thought I'd pass it by you all before I wreck a neon indicator. I saw some good neon posts on this forum, but I'm still not sure I have the resistor placement & sizing correct.

I have a spark discharge machine I'd like to get operating (& probably put on the shelf forever, knowing me).
On the control panel is a neon indicator (GE BA1 65 volt) which is stated on the machine label to show the following:
Bright if there is no electrode contact
Off if the electrode is shorted OR
Blinking if the electrode is working properly (which is intermittent contact with the work piece at 60 cycles per second).

The existing machine wiring sends 120 volts AC to a 6.9k Ω resistor which is in series with the indicator bulb.

I have no way of knowing if the machine, as found, is wired correctly or has been 'customized'. One clue is that the neon is burned out. (I have a spare.)
The manufacturer has no details of this machine (quite old).

Since the bulb is rated 65 volts at 1.2 ma- my brain tells me to do this:
Peak minus lamp voltage divided by milliamps equals resistor size
120 volts peaks at 169.7 volts.
Therefore: 169.7 - 65 = 104.7 volts to cross a resistor for 65 at the bulb.
104.7 volts / 1.2ma = 87.25k resistor in series with the lamp.

But, I just read a couple of places using the 120 volts & 65 volt lamp saying to put a 100k Ω parallel to the lamp. One place wants 500k Ω, also parallel. One said 120k Ω in series

Are my math & placement correct, or did I goof this up?

Thanks for ideas!
Paul

 

Don't subtract the 65 volts. Once the gas breaks over, its voltage collapses.
I'd do 170V/.001A because you don't have to run all lamps at their maximum rated ability.

 

220kΩ resistor in series with a small neon lamp is not unusual.

 

Thanks for replying Mr. Chips & #12
I appreciate your advice. I'll put the properly sized resistor in series.

I didn't even think about the gas breaking over. Makes sense- similar to a fluorescent tube.

Thanks Again for helping. I appreciate you guys sharing knowledge.
Paul

 

Using old familiar NE2H, strike V 95V , operating 65 V, series resistor 33k , 1.9 mA. V only falls to 65V when current is limited.

 

Just a comment, the collapsing voltage of the ionized gas is used to make a relaxation oscillator. It's as simple as 3 parts: a series resistor, a capacitor to ground, and the neon bulb in parallel with the capacitor. In 1977, I used (15) 9 volt batteries, a 10 meg in series with a .1 capacitor and an NE-2 bulb to make a blinker that lasted for over a year. I left it in my mother's car and it was mistaken for a car alarm!

As my own variation, I added a 22k resistor in series with the neon bulb to "stretch" the "on" time and reduce the peak current (which tends to blacken the glass). It was just a fun thing to do.

If I wanted to do that today, I'd use the 65 volt neon bulb and the 22 volt batteries that look like 9 volt batteries. Four to six batteries would last quite a while pushing current through a 10 meg resistor.

 

Thank you Bernard for the mention of NE-2H. I got a little confused by the resistor size of 33k, so I looked it up & found it was shown as 105-125 volts. Does it already have an internal resistor to let it operate with the smaller 33k?
Just curious.

I like your neon bulb oscillator project, #12.
I know I'll end up making one just for fun. Oops, I meant to say "to learn".

I remember being a kid in the 60's and taking apart a junk-picked two-way radio I dragged home tied to my sled. I wanted to give my dad the tubes. He asked for the 'little flat top bulbs' too.
I wonder if those were neon & being used as oscillators in the radio? I forgot all about it until #12 mentioned his oscillator project.

Thanks again for the advice on my neon project!
Paul

 

The NE-2H can survive higher current and has a higher break-over voltage. That's why it can be used with 33K. It does not have an internal resistor.

 

Thank you #12 for explaining the difference.
I like reading this forum. I learn lots every day, thanks to all of you!
Paul