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I need a high voltage (240 vac) (led) lamp which can be mounted on a
panel. It is to show the presence or absence of the voltage. Does
anyone know where I can find one?
** Folk normally use neon lamps for that job.
That would be fine. Can you indicate where I can get one? Remember,
it has to be 240 VAC, not 120 VAC.
it has to be 240 VAC, not 120 VAC.** Folk normally use neon lamps for that job.
That would be fine. Can you indicate where I can get one? Remember,
Go to your favorite distributor and plug "neon" into their search** Folk normally use neon lamps for that job.
That would be fine. Can you indicate where I can get one? Remember,
it has to be 240 VAC, not 120 VAC.
The difference is that for 240V, a higher value resistor is needed in** Folk normally use neon lamps for that job.
That would be fine. Can you indicate where I can get one? Remember,
it has to be 240 VAC, not 120 VAC.
Thanks,
John Bobinyec
For UK 240V, (with usually assumed voltage of 90V for the neon start voltage)Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
The resistor will get hot if you run that much current. With the LED, ifhal-usenet@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote in
news:a4Gdna_IA_hoYufUnZ2dnUVZ_rrinZ2d@megapath.net:
Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
For UK 240V, (with usually assumed voltage of 90V for the neon start
voltage) a 270K resistor in series with it (Though I've also seen 470K and
as low as 150K). But you could just as easily use an LED, if you calculate
on RMS voltage minus Vf of LED to get the resistance needed for a given
current (usually 20 to 40 mA for a signalling LED), and don't mind it
lighting up only on alternate half cycles. So long as your LED mounting is
properly specified to operate on a mains supply, the worst you could do is
burn out a cheap LED.
True. Ok, forget my advice on LED's. The neon stuff is right though, I justLostgallifreyan wrote:
hal-usenet@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote in
news:a4Gdna_IA_hoYufUnZ2dnUVZ_rrinZ2d@megapath.net:
Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
For UK 240V, (with usually assumed voltage of 90V for the neon start
voltage) a 270K resistor in series with it (Though I've also seen 470K and
as low as 150K). But you could just as easily use an LED, if you calculate
on RMS voltage minus Vf of LED to get the resistance needed for a given
current (usually 20 to 40 mA for a signalling LED), and don't mind it
lighting up only on alternate half cycles. So long as your LED mounting is
properly specified to operate on a mains supply, the worst you could do is
burn out a cheap LED.
The resistor will get hot if you run that much current. With the LED, if
the LED drops say 3 Volts, then the other two or three hundred odd volts
(time dependent) will be dropped across the resistor. This is therefore
very wasteful and the resistor power dissipation may be awkward. Very new
LEDs that run ok with less than a milliamp may be more convenient in this
case. In any case with the LED you should connect another diode (or even
another LED) in parallel with the LED but with the opposite polarity so
that the LED does not break down in the reverse direction which could
damage the junction over the long term. You could also connect a diode in
series with the resistor so that there is no current through the resistor
when the LED is not conducting. I would not leave off the diode across the
LED in this case in case the series diode is a bit leaky.
Chris
Chris Jones <lugnut808@yahoo.com> wrote in
news:ALLel.106580$2w.67826@newsfe28.ams2:
Lostgallifreyan wrote:
hal-usenet@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote in
news:a4Gdna_IA_hoYufUnZ2dnUVZ_rrinZ2d@megapath.net:
Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
For UK 240V, (with usually assumed voltage of 90V for the neon start
voltage) a 270K resistor in series with it (Though I've also seen 470K and
as low as 150K). But you could just as easily use an LED, if you calculate
on RMS voltage minus Vf of LED to get the resistance needed for a given
current (usually 20 to 40 mA for a signalling LED), and don't mind it
lighting up only on alternate half cycles. So long as your LED mounting is
properly specified to operate on a mains supply, the worst you could do is
burn out a cheap LED.
The resistor will get hot if you run that much current. With the LED, if
the LED drops say 3 Volts, then the other two or three hundred odd volts
(time dependent) will be dropped across the resistor. This is therefore
very wasteful and the resistor power dissipation may be awkward. Very new
LEDs that run ok with less than a milliamp may be more convenient in this
case. In any case with the LED you should connect another diode (or even
another LED) in parallel with the LED but with the opposite polarity so
that the LED does not break down in the reverse direction which could
damage the junction over the long term. You could also connect a diode in
series with the resistor so that there is no current through the resistor
when the LED is not conducting. I would not leave off the diode across the
LED in this case in case the series diode is a bit leaky.
Chris
True. Ok, forget my advice on LED's. The neon stuff is right though, I just
didn't think through the 'translation' properly.
Could be sometimes, but I've run them constantly for years with the standardIn article <Xns9B9DDDB5A8EB0zoodlewurdle@216.196.109.145>,
Lostgallifreyan <no-one@nowhere.net> wrote:
Chris Jones <lugnut808@yahoo.com> wrote in
news:ALLel.106580$2w.67826@newsfe28.ams2:
Lostgallifreyan wrote:
hal-usenet@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote in
news:a4Gdna_IA_hoYufUnZ2dnUVZ_rrinZ2d@megapath.net:
Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
For UK 240V, (with usually assumed voltage of 90V for the neon start
voltage) a 270K resistor in series with it (Though I've also seen
470K and as low as 150K). But you could just as easily use an LED, if
you calculate on RMS voltage minus Vf of LED to get the resistance
needed for a given current (usually 20 to 40 mA for a signalling
LED), and don't mind it lighting up only on alternate half cycles. So
long as your LED mounting is properly specified to operate on a mains
supply, the worst you could do is burn out a cheap LED.
The resistor will get hot if you run that much current. With the LED,
if the LED drops say 3 Volts, then the other two or three hundred odd
volts (time dependent) will be dropped across the resistor. This is
therefore very wasteful and the resistor power dissipation may be
awkward. Very new LEDs that run ok with less than a milliamp may be
more convenient in this case. In any case with the LED you should
connect another diode (or even another LED) in parallel with the LED
but with the opposite polarity so that the LED does not break down in
the reverse direction which could damage the junction over the long
term. You could also connect a diode in series with the resistor so
that there is no current through the resistor when the LED is not
conducting. I would not leave off the diode across the LED in this
case in case the series diode is a bit leaky.
Chris
True. Ok, forget my advice on LED's. The neon stuff is right though, I
just didn't think through the 'translation' properly.
Almost every neon I have seen, goes to NO SEE EM at some point.
MAybe they are driving them too hard.
Could work, but still needs same average current. Or just possibly not, if itHow about some capacitor in series with the Led system ??
I never tried that yet.
Hi ChrisLostgallifreyan wrote:
hal-use...@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote in
news:a4Gdna_IA_hoYufUnZ2dnUVZ_rrinZ2d@megapath.net:
Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
For UK 240V, (with usually assumed voltage of 90V for the neon start
voltage) a 270K resistor in series with it (Though I've also seen 470K and
as low as 150K). But you could just as easily use an LED, if you calculate
on RMS voltage minus Vf of LED to get the resistance needed for a given
current (usually 20 to 40 mA for a signalling LED), and don't mind it
lighting up only on alternate half cycles. So long as your LED mounting is
properly specified to operate on a mains supply, the worst you could do is
burn out a cheap LED.
The resistor will get hot if you run that much current. With the LED, if
the LED drops say 3 Volts, then the other two or three hundred odd volts
(time dependent) will be dropped across the resistor. This is therefore
very wasteful and the resistor power dissipation may be awkward. Very new
LEDs that run ok with less than a milliamp may be more convenient in this
case. In any case with the LED you should connect another diode (or even
another LED) in parallel with the LED but with the opposite polarity so
that the LED does not break down in the reverse direction which could
damage the junction over the long term. You could also connect a diode in
series with the resistor so that there is no current through the resistor
when the LED is not conducting. I would not leave off the diode across the
LED in this case in case the series diode is a bit leaky.
Chris
I would use a neon.On Jan 24, 1:30 pm, Chris Jones <lugnut...@yahoo.com> wrote:
Lostgallifreyan wrote:
hal-use...@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote in
news:a4Gdna_IA_hoYufUnZ2dnUVZ_rrinZ2d@megapath.net:
Neon lamps are (roughly) constant voltage, typically about 60 volts
for small ones.
For UK 240V, (with usually assumed voltage of 90V for the neon start
voltage) a 270K resistor in series with it (Though I've also seen 470K
and as low as 150K). But you could just as easily use an LED, if you
calculate on RMS voltage minus Vf of LED to get the resistance needed
for a given current (usually 20 to 40 mA for a signalling LED), and
don't mind it lighting up only on alternate half cycles. So long as
your LED mounting is properly specified to operate on a mains supply,
the worst you could do is burn out a cheap LED.
The resistor will get hot if you run that much current. With the LED, if
the LED drops say 3 Volts, then the other two or three hundred odd volts
(time dependent) will be dropped across the resistor. This is therefore
very wasteful and the resistor power dissipation may be awkward. Very
new LEDs that run ok with less than a milliamp may be more convenient in
this
case. In any case with the LED you should connect another diode (or even
another LED) in parallel with the LED but with the opposite polarity so
that the LED does not break down in the reverse direction which could
damage the junction over the long term. You could also connect a diode
in series with the resistor so that there is no current through the
resistor
when the LED is not conducting. I would not leave off the diode across
the LED in this case in case the series diode is a bit leaky.
Chris
Hi Chris
I can tell you that the life of an LED without the reverse bypassing
diode is quite
short. I saw this done and the fellow asked me to take a look at the
strange behavior.
The LED glowed orange and was quite hot to the touch. It failed
completely in about
10 minutes ( this was the third LED he'd burned out ).
One can use a capacitor to drop the voltage but this has some
problems. If just
using a single capacitor, it will pass higher frequency spikes on the
power line
as well.
As mentioned, just using a resistor for a 20ma LED would be about 5
Watts.
It would be more practical to use a series capacitor to drop the
voltage of
about 0.22 uF ( 400V, big ). Then, add a resistor to limit the current
of high frequency of about 510 ohms. Now place another capacitor of
about
0.02 uF in parallel with the LED, that has a diode in parallel with it
to handle
the back voltage and keep the capacitors working as basically AC
devices.
This should give the average light of about 5-10 ma for the LED. This
was
just designed off the top of my head so a spice run would give better
values.
One could improve things by using a full wave bridge between the last
capacitor and the LED, removing the need for the back voltage diode.
Another thought might be to create a small switcher to drop the
voltage.
Dwight