Series resistor.

[klem kedidelhopper]

I think I know this but I'd like to be sure. I need to affix a small
neon indicator on to a 240V line. Would 80K be an appropriate resistor
to use? Thanks, Lenny

 

[N_Cook]

I think I know this but I'd like to be sure. I need to affix a small
neon indicator on to a 240V line. Would 80K be an appropriate resistor
to use? Thanks, Lenny

usually a bit more , 150 K perhaps

 

[[email protected]]

I think I know this but I'd like to be sure. I need to affix a small
neon indicator on to a 240V line. Would 80K be an appropriate resistor
to use? Thanks, Lenny

82K is correct if the neon indicator has an internal resistor. If
it's a 'bare' NE-2 lamp, use 130K.

PlainBill

 

[Phil Allison]

"klem kedidelhopper"

I think I know this but I'd like to be sure. I need to affix a small
neon indicator on to a 240V line. Would 80K be an appropriate resistor
to use?

** The value of 80k is fine, but needs to be rated at 1 watt to have a long
life.

The usual values found inside 240V neon bezels and illuminated switches is
150k to 220 k ohms - chosen mainly because the resistor needs to be of
small physical size it and this dictates one of 0.5 or 0.25 watt rating.

The voltage across the resistor is about 200V so 150k dissipates 0.27 watts.


..... Phil

 

[Robert Macy]

I think I know this but I'd like to be sure. I need to affix a small
neon indicator on to a 240V line. Would 80K be an appropriate resistor
to use? Thanks, Lenny

bare bulb?

assume neon turns on at 90V and extinguishes around 60V.

Sinewave are 'so fast' just assume the neon is like a bidirectional
zener of 60V:

First what power idssipates if there were NO neon bulb:
to find R try 240*240/R=Power
therefore, R=240*240/Power
rule of thumb for resistors is use half their rating, so change
formula to
R=2*240*240/Power
so a 1/4W means R = 460k, close value 470k 5%
and 1/2W means R = 230k, close value 220k 5%

Now let's 'clip' the voltage that goes to the neon:
as a simplistic estimate, 240-60=190, and try again
R=2*190*190/Power
1/4W R = 289k
1/2W R = 144k

use 1/2W 220k, looks like won't hurt anything.

just as a check, how much power is going into the neon?
current is (240-60)/220k = 0.86 mA
power into the neon bulb is 60*0.86mA = approx 50 mW, so neon is not
likely to burn up.

Plus, dopn't those GFI outlets limit AC current to something like less
than 1 mA, so even if you get your fingers in there you're not likely
to be killed.

 

[Phil Allison]

"Robert Macy"

use 1/2W 220k, looks like won't hurt anything.


** Correct.


Plus, don't those GFI outlets limit AC current to something like less
than 1 mA,


* Trip current is normally 10mA and time to trip is 30mS for units that can
be plugged in.

This is for a 240V country.



..... Phil