R
Roger
Guest
If I connect a 270 nf capacitor in series with an LED and the 230v
mains what happens?
mains what happens?
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B
Bill Bowden
Guest
On Apr 1, 7:16 pm, Roger <dewhurst.ro...@gmail.com> wrote:
With no resistor or rectifier diodes, the LED will smoke. Try this and
use a 270nF cap rated at 400 volts or more, and a 1 watt 1000 ohm
resistor.
http://www.bowdenshobbycircuits.info/lineled.gif
-Bill
If I connect a 270 nf capacitor in series with an LED and the 230v
mains what happens?
With no resistor or rectifier diodes, the LED will smoke. Try this and
use a 270nF cap rated at 400 volts or more, and a 1 watt 1000 ohm
resistor.
http://www.bowdenshobbycircuits.info/lineled.gif
-Bill
J
Jamie
Guest
Roger wrote:
R = 230/0.023 = 10000
C = 1/(6.28 x 60 10000) = 265nf or .265uf
But, there are other problems, you need two LEDS, connected
back to back. There is two reasons for this, 1.) you need to
have current flowing in both directions so the CAP can work and
2.) at that voltage, if you don't have another LED in the reverse
orientation, the reverse break down on the single diode will get
exceeded and be destroy, in all likely hood.
Also there is this slight problem with noise spikes on the AC line
which happens all the time, this will cause the capacitor to react at
a much lower R and thus, drive your poor little LEDS to hell!
that should give you something to think about.
Jamie
On Apr 2, 3:23 pm, Bill Bowden <bper...@bowdenshobbycircuits.info
wrote:
On Apr 1, 7:16 pm, Roger <dewhurst.ro...@gmail.com> wrote:
If I connect a 270 nf capacitor in series with an LED and the 230v
mains what happens?
With no resistor or rectifier diodes, the LED will smoke. Try this and
use a 270nF cap rated at 400 volts or more, and a 1 watt 1000 ohm
resistor.
http://www.bowdenshobbycircuits.info/lineled.gif
-Bill
Thank you. I used the equation C=1/(2 pi.f.X) where X=V/i.
Given V = 230v and i = 23mA X = 1000.
Thus C = 1/(6.28 x 60 x 1000)
=2.7 microfarads
I got the decimal point in the wrong place to give me 270 nanofarads.
Where have I gone wrong? Surely the LED will pass 23 milliamperes?
Roger
I think your math is off a bit..
R = 230/0.023 = 10000
C = 1/(6.28 x 60 10000) = 265nf or .265uf
But, there are other problems, you need two LEDS, connected
back to back. There is two reasons for this, 1.) you need to
have current flowing in both directions so the CAP can work and
2.) at that voltage, if you don't have another LED in the reverse
orientation, the reverse break down on the single diode will get
exceeded and be destroy, in all likely hood.
Also there is this slight problem with noise spikes on the AC line
which happens all the time, this will cause the capacitor to react at
a much lower R and thus, drive your poor little LEDS to hell!
that should give you something to think about.
Jamie
R
Roger
Guest
On Apr 2, 3:23 pm, Bill Bowden <bper...@bowdenshobbycircuits.info>
wrote:
Given V = 230v and i = 23mA X = 1000.
Thus C = 1/(6.28 x 60 x 1000)
=2.7 microfarads
I got the decimal point in the wrong place to give me 270 nanofarads.
Where have I gone wrong? Surely the LED will pass 23 milliamperes?
Roger
wrote:
Thank you. I used the equation C=1/(2 pi.f.X) where X=V/i.On Apr 1, 7:16 pm, Roger <dewhurst.ro...@gmail.com> wrote:
If I connect a 270 nf capacitor in series with an LED and the 230v
mains what happens?
With no resistor or rectifier diodes, the LED will smoke. Try this and
use a 270nF cap rated at 400 volts or more, and a 1 watt 1000 ohm
resistor.
http://www.bowdenshobbycircuits.info/lineled.gif
-Bill
Given V = 230v and i = 23mA X = 1000.
Thus C = 1/(6.28 x 60 x 1000)
=2.7 microfarads
I got the decimal point in the wrong place to give me 270 nanofarads.
Where have I gone wrong? Surely the LED will pass 23 milliamperes?
Roger
R
Roger
Guest
On Apr 3, 9:43 am, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
It does indeed.Roger wrote:
On Apr 2, 3:23 pm, Bill Bowden <bper...@bowdenshobbycircuits.info
wrote:
On Apr 1, 7:16 pm, Roger <dewhurst.ro...@gmail.com> wrote:
If I connect a 270 nf capacitor in series with an LED and the 230v
mains what happens?
With no resistor or rectifier diodes, the LED will smoke. Try this and
use a 270nF cap rated at 400 volts or more, and a 1 watt 1000 ohm
resistor.
http://www.bowdenshobbycircuits.info/lineled.gif
-Bill
Thank you. I used the equation C=1/(2 pi.f.X) where X=V/i.
Given V = 230v and i = 23mA X = 1000.
Thus C = 1/(6.28 x 60 x 1000)
=2.7 microfarads
I got the decimal point in the wrong place to give me 270 nanofarads.
Where have I gone wrong? Surely the LED will pass 23 milliamperes?
Roger
I think your math is off a bit..
R = 230/0.023 = 10000
C = 1/(6.28 x 60 10000) = 265nf or .265uf
But, there are other problems, you need two LEDS, connected
back to back. There is two reasons for this, 1.) you need to
have current flowing in both directions so the CAP can work and
2.) at that voltage, if you don't have another LED in the reverse
orientation, the reverse break down on the single diode will get
exceeded and be destroy, in all likely hood.
Also there is this slight problem with noise spikes on the AC line
which happens all the time, this will cause the capacitor to react at
a much lower R and thus, drive your poor little LEDS to hell!
that should give you something to think about.
Jamie