Pilot light

S

Steve Wolf

Guest
I have taken some pilot lights out of a broken stove. These lamps indicated whether the element was on or the element was hot. I think they are what you call Neon lamps eg. Small glass bulb with thick Metal bars in the bulb.

I want to use one of them as a pilot light for a 120vac light telling me that the switch is on . (120vac.)

The light from the stove is in a nice little plastic package, that when I took a part it looks like this. Very simple.



+--------------.
| (0)Lamp
---R120k--+--R120k-------+
-------------------------+

It is NOT an LED and says on it 250v. Clearly it is meant for 120-250 volts.

To test it I just connected 120 volts to the two terminals. However it did not light.

Is the above this some sort of shunt? Do I need a load to make it work.

eg

---+--------+
MyLamp Load
---+--------+

thank you
 
In article <0ffb5984-5446-459f-b11e-9826298a1322@googlegroups.com>,
stevwolf58@gmail.com says...
I have taken some pilot lights out of a broken stove. These lamps indicated whether the element was on or the element was hot. I think they are what you call Neon lamps eg. Small glass bulb with thick Metal bars in the bulb.

I want to use one of them as a pilot light for a 120vac light telling me that the switch is on . (120vac.)

The light from the stove is in a nice little plastic package, that when I took a part it looks like this. Very simple.



+--------------.
| (0)Lamp
---R120k--+--R120k-------+
-------------------------+

It is NOT an LED and says on it 250v. Clearly it is meant for 120-250 volts.

To test it I just connected 120 volts to the two terminals. However it did not light.

Is the above this some sort of shunt? Do I need a load to make it work.

eg

---+--------+
MyLamp Load
---+--------+

thank you
Click to expand...

It is difficult to tell by your drawing how it actually is.
It looks like you have two resistors in series with the neon bulb across
one of them.

Most neon bulbs need about 90 volts to fire off (light) then around 70
volts to keep them lit. A series resistor is needed to keep the current
to a save level.

If you have the two resistors in series and the bulb across one of them
you only have about 1/2 of the 120 volts or 60 volts which is not enough
to fire off the bulb. Even allowing for the 1.414 peak voltage of a
sine wave over the rms value you just barely do not have enough voltage
to fire off the bulb.

Try unhooking the resistor in parallel with the bulb, but leave the
resistor that is in series with the bulb in place to limit the current
when you hook it to 120 volts.
 
You are right. The first resistor is in series and the other I would call in parallel I think? Second resistor runs from hot to neutral "bridging" them. Sadly my diagram when saved on web got screwed. If you copy and paste into notepad it may show up more clearly. However your conclusion may be right. IF this light used to have 240 running an element in a stove through it then maybe removing one resistor will give it enough to start the lamp.

I will try and get rid of second Resistor and get back with results.
Thanks.
 
On Tue, 7 Jan 2020 10:45:16 -0800 (PST), Steve Wolf
<stevwolf58@gmail.com> wrote:

You are right. The first resistor is in series and the other I would call in parallel I think? Second resistor runs from hot to neutral "bridging" them. Sadly my diagram when saved on web got screwed. If you copy and paste into notepad it may show up more clearly. However your conclusion may be right. IF this light used to have 240 running an element in a stove through it then maybe removing one resistor will give it enough to start the lamp.

I will try and get rid of second Resistor and get back with results.
Thanks.
Click to expand...

One hundred K resistor in series with one leg of the lamp is about
right for 120VAC.

see:
http://www.bristolwatch.com/ele/neon.htm
 
In article <666b29a7-38cf-494d-979c-bb7262470b06@googlegroups.com>,
stevwolf58@gmail.com says...
You are right. The first resistor is in series and the other I would call in parallel I think? Second resistor runs from hot to neutral "bridging" them. Sadly my diagram when saved on web got screwed. If you copy and paste into notepad it may show up more clearly. However your conclusion may be right. IF this light used to
have 240 running an element in a stove through it then maybe removing one resistor will give it enough to start the lamp.

I will try and get rid of second Resistor and get back with results.
Thanks.
Click to expand...

If you remove one resistor, make sure the wiring is from hot wire,
resistor, bulb, neutral. That puts it in series.

If you remove the resistor that is across the bulb , the bulb will
probably explode in one bright flash as there is nothing to limit the
courrent to a safe value.
 
On 1/7/2020 4:59 PM, Ralph Mowery wrote:
If you remove one resistor, make sure the wiring is from hot wire,
resistor, bulb, neutral. That puts it in series.
Click to expand...

Right
If you remove the resistor that is across the bulb , the bulb will
probably explode in one bright flash as there is nothing to limit the
courrent to a safe value.
Click to expand...

Wrong - backwards. Not what you said above. Do remove the one across
(parallel to) the bulb. Do not remove the one in series with the bulb.
 
I have removed the resistor in parallel so it is now simply a lamp with one resistor in series. Works fine. I might try a 100k R as one person suggested to get a little brighter light. It's fine but flickers a little.

Thanks for all your help.
Regards
 
In article <qv3a3q01gop@news1.newsguy.com>, BobEngelhardt@comcast.net
says...
If you remove the resistor that is across the bulb , the bulb will
probably explode in one bright flash as there is nothing to limit the
courrent to a safe value.

Wrong - backwards. Not what you said above. Do remove the one across
(parallel to) the bulb. Do not remove the one in series with the bulb.
Click to expand...

Glad you caught that. It should have been do not remove the series
resistor or the bulb will probably explode.

The bulb probably takes about 90 volts to fire off. When used on the
240 volt stove the bulb is like an open circuit to start with. So the
resistors form a simple voltage devider and puts 120 volts on the bulb.
It fires off and then drops to around 70 volts. The series resistor
limits the current to a save value for the bulb and the parallel
resistor takes a small ammount of current, but can probably be ignored.

If hooked across the 120 volt circuit you only have 60 volts across the
bulb. Not enough to fire it off . If the parallel resistor is removed,
then there is not a voltage devider, so the bulb gets the full 120
volts for a fraction of a second then the bulb fires and has 70 volts
across it and the resistor had the 120-70 or 50 volts.
 
Thanks for the physics. Always amazes me how complex a simple 2 or 3 component circuit is.
Regards.
 
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