Re: Calculating LED resistors- long winded

On Nov 16, 10:46 pm, "GPE" <GPEnos...@xxxxxxx> wrote:
"Taylor-VA" <tkre...@xxxxxxxxxxx> wrote in message

news:4c22885d-f675-4e04-b1ae-03478480d963@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx





On Nov 16, 12:08 pm, Taylor-VA <tkre...@xxxxxxxxxxx> wrote:
On Nov 16, 12:05 pm, kenny_ii <kenny_...@xxxxxxxxx> wrote:

I agree with Ed, just make sure the reverse voltage rating of the LED
is not exceeded. Rememebre, the peak voltage of AC is 1.414 times the
advertised AC rating.

K2

On Nov 16, 9:56 am, beaver <e...@xxxxxxxxxxxx> wrote:

My personal opinion with mods IS to use LEDs.

Just put two LEDs back-to-back with a single resistor. The
conducting
diode will protect the other one. You will need to drop the resistor
value as you will now only have 50% duty cycle.

Edward Cheung CARGPB26

On Nov 16, 10:42 am, PT <zeeca...@xxxxxxxxxxxxx> wrote:

Yes, the GI is AC not DC. It also runs through triacs for dimming.
Not that it won't work, but running LEDs on AC is not the best way,
since there may be some flicker - especially when the GI is dimmed.
LEDs don't have persistence like a real bulb does.

Why not just tag into the GI string(s) and use real 44 or 47 bulbs
as
Chris did? It will be MUCH simpler and most likely look a lot
better
in the end. You'll have a hard time back-lighting something as big
as
the Earth and Mars with the pinpoints of light that come from LEDs.

You can put up to 18 bulbs on a GI string. Start counting the
bulbs
on each string to find ones with less than 18. This will also
allow
you to use blinking bulbs (455) behind the ships as Chris does.

John

I think I mispoke about the 6.3vdc, I think it is 6.3 vac. I do
no
think that should change anything, I think. And yes thank you for
clarifying I will be running them in parralel.

Well I was only completely wrong with two things. :)

I think I get it now, thanks.

-Taylor- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Scratching my head again.- Hide quoted text -

- Show quoted text -

So that would mean that 6.3vac would peak at 8.9082. So do I calculate
my resistor using 9v instead of 6.3?

Just to toss my 2cents into the ring --

Unlike other diode's, LED's do NOT like reverse bias voltages. Most
breakdown at around 5 volts.
What -I- would do is feed the 6.3VAC into a small bridge rectifier, hang a
small electroltyic cap on the output then use this to bias your LED
circuits. The DC voltage would be about 8.9 volts. Since you rectified the
AC and have a cap on it - you won't see any flickering.

For calculating three resistors in series - resistor values:
Resistor = (8.9 - (Vf_LED1 + Vf_LED2 + Vf_LEDn)) / 0.010
Typical VF for green is 2.1 volts, forward current of 10mA

Resistor = (8.9 - (2.1 + 2.1 + 2.1)) / 0.010 = 260 ohms -- use 270

Wattage of this resistor = (8.9 - (2.1 + 2.1 + 2.1)) x 0.010 = 0.026 -- use
1/4 watt.

So, you have one 270 ohm, 1/4 watt resistor for each three LED's connected
in series.

Ed- Hide quoted text -

- Show quoted text -

I agree with Ed, that's why I posted about peak voltage. It doesn't
take much to kill an LED if you exceed the peak inverse rating by even
a little bit on every half cycle. White and blue LED's are even more
delicate that reds and greens. Sometimes a bridge rectifier and
filter cap are more than is needed, and for those applications I put a
1N4937 type diode in series with the LED. Without getting too deep
into the BS, a fast swtiching diode will block the reverse current and
protect the LED. A bit simpler than a full wave, but the same idea.

K2
.

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