Hi All
I'm looking at getting some replacement light fittings and note that
they have a stated limit of 60 watts.
Does this only apply when the old incandescent type bulbs are to be used?
The LED bulbs display the equivalent old wattage e. g. 75w, but also
show the required power at 9.5w.
I expect that when the cooler operation LEDs are fitted, the 60 or 75w
value is no longer relevant.
Thanks
Phil
Hi All
I'm looking at getting some replacement light fittings and note that
they have a stated limit of 60 watts.
Does this only apply when the old incandescent type bulbs are to be used?
The LED bulbs display the equivalent old wattage e. g. 75w, but also
show the required power at 9.5w.
I expect that when the cooler operation LEDs are fitted, the 60 or 75w
value is no longer relevant.
Thanks
Phil
I'm looking at getting some replacement light fittings and note that
they have a stated limit of 60 watts.
Does this only apply when the old incandescent type bulbs are to be used?
In article <115a621$2e3s3$1@dont-email.me>,
Generalie <Generalie@nospam.com> wrote:
I'm looking at getting some replacement light fittings and note that
they have a stated limit of 60 watts.
Does this only apply when the old incandescent type bulbs are to be used?
The limit is because of the heat generated, which is the actual
wattage, not the "equivalent". No household LED bulb will approach
the limit.
Although LED bulbs in some fittings have been known to gradually cook >themselves which shortens their life, well below the amount of heat needed
to damage the fitting. It's worth thinking about ventilation of the fitting >- eg an upturned bowl will collect the heat, whereas a cylindrical lampshade >open at the top and bottom will allow fresh air to circulate to cool the >bulb.
In article <aWd*2GINA@news.chiark.greenend.org.uk>,
Theo <theom+news@chiark.greenend.org.uk> wrote:
Although LED bulbs in some fittings have been known to gradually cook
themselves which shortens their life, well below the amount of heat needed >> to damage the fitting. It's worth thinking about ventilation of the fitting >> - eg an upturned bowl will collect the heat, whereas a cylindrical lampshade >> open at the top and bottom will allow fresh air to circulate to cool the
bulb.
I don't have any figures, but I'm doubtful about how significant this
is. I suspect that the internal temperature of the bulb depends more
on the design of the bulb itself than on the surrounding air flow.
Cheap bulbs are likely to lack enough metal to conduct the heat away
to the air.
I have some small ceiling lamps designed for G9 halogens. They point
down, but can be angled slightly as well, are made of glass, and have no ventilation at the socket end. when I changed to G9 leds I was surprised
at how quickly those bulbs died. I assumed they overheated, and tried a different make, but that also failed fairly quickly.
Hi Allessentially yes
I'm looking at getting some replacement light fittings and note that
they have a stated limit of 60 watts.
Does this only apply when the old incandescent type bulbs are to be used?
The LED bulbs display the equivalent old wattage e. g. 75w, but also
show the required power at 9.5w.
I expect that when the cooler operation LEDs are fitted, the 60 or 75w
value is no longer relevant.
Thanks--
Phil
Jeff Layman wrote:
I have some small ceiling lamps designed for G9 halogens. They point
down, but can be angled slightly as well, are made of glass, and have no ventilation at the socket end. when I changed to G9 leds I was surprised at how quickly those bulbs died. I assumed they overheated, and tried a different make, but that also failed fairly quickly.
With incandescent and halogens, they produce 90-95% of their rated power
as infrared rather than visible light, which can escape somewhat.
LEDs tend to generate heat within the LED itself, so even if they are
lower actual power to begin with, 40-60% of that power ends up as heat,
with limited airflow for cooling, can tend to cook ...
Andy Burns <usenet@andyburns.uk> wrote:
Jeff Layman wrote:
I have some small ceiling lamps designed for G9 halogens. They point
down, but can be angled slightly as well, are made of glass, and have no >>> ventilation at the socket end. when I changed to G9 leds I was surprised >>> at how quickly those bulbs died. I assumed they overheated, and tried a
different make, but that also failed fairly quickly.
With incandescent and halogens, they produce 90-95% of their rated power
as infrared rather than visible light, which can escape somewhat.
LEDs tend to generate heat within the LED itself, so even if they are
lower actual power to begin with, 40-60% of that power ends up as heat,
with limited airflow for cooling, can tend to cook ...
Also a tungsten filament is by nature hot up to about 3000K. Semiconductor junctions are good to about 125C,
but higher temperature causes faster
ageing, well below the tungsten temperature.
Persistently elevated temperatures can also damage other components like capacitors that are included as part of the lamp - if your capacitor dropper loses capacitance then it's not dropping so much voltage, you start to overvolt/overcurrent the LEDs, they run hotter, and conditions continue to deteriorate until something finally goes.
Theo--
Andy Burns <usenet@andyburns.uk> wrote:
Jeff Layman wrote:
I have some small ceiling lamps designed for G9 halogens. They point
down, but can be angled slightly as well, are made of glass, and have no >>> ventilation at the socket end. when I changed to G9 leds I was surprised >>> at how quickly those bulbs died. I assumed they overheated, and tried a
different make, but that also failed fairly quickly.
With incandescent and halogens, they produce 90-95% of their rated power
as infrared rather than visible light, which can escape somewhat.
LEDs tend to generate heat within the LED itself, so even if they are
lower actual power to begin with, 40-60% of that power ends up as heat,
with limited airflow for cooling, can tend to cook ...
Also a tungsten filament is by nature hot up to about 3000K. Semiconductor junctions are good to about 125C, but higher temperature causes faster ageing, well below the tungsten temperature.
Persistently elevated temperatures can also damage other components like capacitors that are included as part of the lamp - if your capacitor dropper loses capacitance then it's not dropping so much voltage, you start to overvolt/overcurrent the LEDs, they run hotter, and conditions continue to deteriorate until something finally goes.
Theo
Persistently elevated temperatures can also damage other components like capacitors that are included as part of the lamp - if your capacitor
dropper loses capacitance then it's not dropping so much voltage, you
start to overvolt/overcurrent the LEDs, they run hotter, and conditions continue to deteriorate until something finally goes.
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other components
like capacitors that are included as part of the lamp - if your
capacitor dropper loses capacitance then it's not dropping so much
voltage, you start to overvolt/overcurrent the LEDs, they run
hotter, and conditions continue to deteriorate until something
finally goes.
Capacitance dropping will underrun the leds.
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other components
like capacitors that are included as part of the lamp - if your
capacitor dropper loses capacitance then it's not dropping so much
voltage, you start to overvolt/overcurrent the LEDs, they run
hotter, and conditions continue to deteriorate until something
finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
On 13 Aug 2026 at 10:43:33 BST, "Joe" <joe@jretrading.com> wrote:
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other components
like capacitors that are included as part of the lamp - if your
capacitor dropper loses capacitance then it's not dropping so much
voltage, you start to overvolt/overcurrent the LEDs, they run
hotter, and conditions continue to deteriorate until something
finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
Or maybe more. So - doesn't alter the statement that that reducing the voltage
dropper capacitance will under run the LEDs.
On 13 Aug 2026 at 10:43:33 BST, "Joe" <joe@jretrading.com> wrote:
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other
components like capacitors that are included as part of the lamp
- if your capacitor dropper loses capacitance then it's not
dropping so much voltage, you start to overvolt/overcurrent the
LEDs, they run hotter, and conditions continue to deteriorate
until something finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
Or maybe more. So - doesn't alter the statement that that reducing
the voltage dropper capacitance will under run the LEDs.
On 14 Aug 2026 21:12:31 GMT
Roger Hayter <roger@hayter.org> wrote:
On 13 Aug 2026 at 10:43:33 BST, "Joe" <joe@jretrading.com> wrote:
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other
components like capacitors that are included as part of the lamp
- if your capacitor dropper loses capacitance then it's not
dropping so much voltage, you start to overvolt/overcurrent the
LEDs, they run hotter, and conditions continue to deteriorate
until something finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
Or maybe more. So - doesn't alter the statement that that reducing
the voltage dropper capacitance will under run the LEDs.
And reducing the larger value capacitor in the dropper will *increase*
the LED drive.
It works pretty much like a resistive voltage divider if the current
drawn is small, remembering that the larger the capacitor, the smaller
the reactance.
On 15 Aug 2026 at 11:45:40 BST, "Joe" <joe@jretrading.com> wrote:
On 14 Aug 2026 21:12:31 GMT
Roger Hayter <roger@hayter.org> wrote:
On 13 Aug 2026 at 10:43:33 BST, "Joe" <joe@jretrading.com> wrote:
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other
components like capacitors that are included as part of the lamp
- if your capacitor dropper loses capacitance then it's not
dropping so much voltage, you start to overvolt/overcurrent the
LEDs, they run hotter, and conditions continue to deteriorate
until something finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
Or maybe more. So - doesn't alter the statement that that reducing
the voltage dropper capacitance will under run the LEDs.
And reducing the larger value capacitor in the dropper will *increase*
the LED drive.
It works pretty much like a resistive voltage divider if the current
drawn is small, remembering that the larger the capacitor, the smaller
the reactance.
No it doesn't. The voltage across the LED is largely governed by the properties of the LED, and putting a large capacitor across it might vary the reactive current but doesn't act as a voltage divider at all.
On 15/08/2026 13:45, Roger Hayter wrote:
On 15 Aug 2026 at 11:45:40 BST, "Joe" <joe@jretrading.com> wrote:The problem is that once you introduce a non linear element like an LED
On 14 Aug 2026 21:12:31 GMT
Roger Hayter <roger@hayter.org> wrote:
On 13 Aug 2026 at 10:43:33 BST, "Joe" <joe@jretrading.com> wrote:
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other
components like capacitors that are included as part of the lamp >>>>>>> - if your capacitor dropper loses capacitance then it's not
dropping so much voltage, you start to overvolt/overcurrent the
LEDs, they run hotter, and conditions continue to deteriorate
until something finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
Or maybe more. So - doesn't alter the statement that that reducing
the voltage dropper capacitance will under run the LEDs.
And reducing the larger value capacitor in the dropper will *increase*
the LED drive.
It works pretty much like a resistive voltage divider if the current
drawn is small, remembering that the larger the capacitor, the smaller
the reactance.
No it doesn't. The voltage across the LED is largely governed by the
properties of the LED, and putting a large capacitor across it might vary the
reactive current but doesn't act as a voltage divider at all.
you can essentially throw all of the linear analyses out of the window.
A large part of circuit design consits in arranging things so you dont
have to deal with that.
On 15 Aug 2026 at 14:06:44 BST, "The Natural Philosopher" <tnp@invalid.invalid> wrote:
On 15/08/2026 13:45, Roger Hayter wrote:
On 15 Aug 2026 at 11:45:40 BST, "Joe" <joe@jretrading.com> wrote:The problem is that once you introduce a non linear element like an LED
On 14 Aug 2026 21:12:31 GMT
Roger Hayter <roger@hayter.org> wrote:
On 13 Aug 2026 at 10:43:33 BST, "Joe" <joe@jretrading.com> wrote:
On Thu, 13 Aug 2026 01:57:56 +0100
me9 <me9@privacy.net> wrote:
Theo <theom+news@chiark.greenend.org.uk> wrote:
Persistently elevated temperatures can also damage other
components like capacitors that are included as part of the lamp >>>>>>>> - if your capacitor dropper loses capacitance then it's not
dropping so much voltage, you start to overvolt/overcurrent the >>>>>>>> LEDs, they run hotter, and conditions continue to deteriorate
until something finally goes.
Capacitance dropping will underrun the leds.
There are two capacitors.
Or maybe more. So - doesn't alter the statement that that reducing
the voltage dropper capacitance will under run the LEDs.
And reducing the larger value capacitor in the dropper will *increase* >>>> the LED drive.
It works pretty much like a resistive voltage divider if the current
drawn is small, remembering that the larger the capacitor, the smaller >>>> the reactance.
No it doesn't. The voltage across the LED is largely governed by the
properties of the LED, and putting a large capacitor across it might vary the
reactive current but doesn't act as a voltage divider at all.
you can essentially throw all of the linear analyses out of the window.
A large part of circuit design consits in arranging things so you dont
have to deal with that.
For instance by setting bounds within which linear approximations are valid.
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