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230v Outlet Tester

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Keep the horizontal ones 68K - assuming the previous example worked OK on 110V? (and I presume it did?) then all you're doing is duplicating it with slightly higher resistors (but the same ratio).


If I did that, and given a reverse polarity outlet. Wouldn't L1 remain unlit?
(R2 / (R1 + R2)) * V1 = V2
(68k / (150k + 68k)) * 230v = 72v

It was stated earlier that neons require 90v to strike.

With 150k resistors...
(R2 / (R1 + R2)) * V1 = V2
(150k / (150k + 150k)) * 230v = 115v
This would enable the neon bulb to strike, thus signifying a reverse polarity?

It's quite the balancing act it seems
 
If I did that, and given a reverse polarity outlet. Wouldn't L1 remain unlit?
(R2 / (R1 + R2)) * V1 = V2
(68k / (150k + 68k)) * 230v = 72v

It was stated earlier that neons require 90v to strike.

With 150k resistors...
(R2 / (R1 + R2)) * V1 = V2
(150k / (150k + 150k)) * 230v = 115v
This would enable the neon bulb to strike, thus signifying a reverse polarity?

It's quite the balancing act it seems

Not really - if you substitute 68K (as post #14) then you get the exact same value - although I don't know why you're calculating it (two identical resistors in a potential divider simply give you half).

So as I said, assuming the #14 circuit works at 110V, then you need to increase the division ration (150k for the top would be a decent starting point).

Hoping the neons don't strike until 90v is rather hopeful, and obviously NOT the case in #14
 
Ok, here is another...
Tester3.png

Now if I understand this correctly (which I haven't so far)
Normal wiring would light L1 and L3
Reverse polarity would make a potential divider of R4~L1~R1 effectively dropping the voltage across L1 to 72v. This may or may not strike the neon (depending on the bulb type as some linked in the mouser catalog strike at 65v and others at 90v)

An open ground would make a potential divider of (R6+R5)~L2~R2 effectively dropping the voltage across L2 to 43v, which would prevent the neon from striking.
In this scenario, what would the voltage across L3 be? Would it strike?


Furthermore, in the event of an open neutral, what is preventing L1 and L2 from striking?
 
It's quite the balancing act it seems
Indeed it is. So far we've only discussed RMS voltages. But its peak voltage which is the critical value. Assuming R4,5,6 = 150k then if R1,2,3 = 68k a 90V neon will only strike very near the peak of a 230V RMS mains, so any sag in mains voltage would not strike the neon. Making R1,2,3 = 100k gives more reliable striking for 90V neons, but the peak voltage across L2 or L3 if ground is open would then be 65V which, as you found, could be enough to strike some neons. You will probably need to experiment using something between 68k and 100k for R1,2,3.
 
I'm also in Europe looking to make a mains wiring tester. I have found some for sale but they are a minimum of around 20 Euros. Considering there are so few components involved, it seems like a waste of money. In a previous post, it was mentioned that they are available on eBay for 99p. If anybody knows where they are sold at anywhere near that price, please provide a link. As for the schematics, there seems to be an issue with the neons lighting up when they should be dark. Aren't you guys misreading the datasheet on the ne-2? From what I understand, they're supposed to have a maximum breakdown voltage of 65V AC or 90V DC. Since we're discussing AC circuits here, shouldn't we be adjusting our resistor values for a 65V AC drop or less if we don't want to be pushing the bulbs to the maximum? https://www.mouser.com/ds/2/423/Neon Indicator Lamps_7___8-552457.pdf
 
Generally, when the voltage is high enough 90 VDC or 65 VAC*1.414 = 90, the lamp will light. When the lamp lights you can basically assume 0V across the lamp. Turn-on and turn-off voltages are significantly different for a neon lamp.

Brightness isn't a killer her because it's only used for a short time.
 
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