Re: FOAK: LEDs and connecting

>> I am after some high intensity LEDs to produce a specialised
>> interior/exterior light unit. The spec is
>> Lens Type: Water clear
>> Reverse Voltage: 5.0 V
>> DC Forward Voltage: Typical: 3.2 V Max: 3.6V
>> Luminous Intensity MCD: Type: 11,000 mcd Max: 13,000 mcd
>> DC Forward Current: 20mA
>
>> I also have with them the same qty of resistors (510 ohm) for DC 12V
>> operation
>
>> All I need to know is if I wire them in parallel where do I put the
>> resistors?
>> Are they on each + leg or where?
>
>> I'm mounting them in a section of white wall trunking (10mm) with a
>> connnector to the van 12v supply.
>
> YTC!

Maybe on this stuff, but what do you know about Lorry Cranes?

>
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> 510 ohm Vf=3.2 V
> Thus: nominally (voltage across resistor) = 12 V - 3.2 V = 8.8 V
> So: current through resistor = 8.8 / 510 A = 17.25 mA == current through
> LED. (In reality, (13.6 -3.2)/510 = 20.4 mA is closer.)
>
> Now, if you consider the resistor/LED pair as a single lamp,
> how do you think you would wire up N of them? The same as N filament
> lamps:
>
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> +12V ----/\/\/\/\/\/\----|>|--------o GND
> ...
> +12V ----/\/\/\/\/\/\----|>|--------o GND
>
> It's not advisable to wire multiple LEDs to one (smaller) resistor
> as the manufacturing tolerances would mean that the LED with the highest
> Vf would take all the current, causing its demise, followed by the next
> highest, etc.
>
> It _is_ possible to wire more than one LED in series though, to
> reduce the power load -- note that above 3/4 of the power is used just to
> heat
> the resistor and is thus wasted. An alternative would be:
>
> +12V ----/\/\/\/\/\/\----|>|-|>|-|>|------o GND
> R [Vf = 3x3.2 = 9.6 V]
>
> For the nominal 12 V: voltage across R = 2.4 V; since desired current is
> 20 mA, R = E/I = 2.4/.02 = 120 ohm.
> (For a "12 V" lead-acid battery: R = 4/.02 = 200 ohm -- multiple LEDs'
> current
> will vary more strongly with source voltage than asingleton's since Vf is
> approximately constant because it's the residual voltage drop across R
> alone
> that determines operating current.)
>
> This reduces the power for each triplet of LEDs from 720 mW to 240 mW.
> This may not be important to you.


Now *that's* what I needed.
Perfect.
Looks like I'll be getting out the soldering iron and some very thin wire to
make an LED array of around a metre to provide a no heat/low current draw
light source


--


R o b b o

Trophy 1200 1998
BotaFOF #19. E.O.S.M 2001/2002/2003/2004.
B.O.S.M 2003, 2004, 2005
FURSWB#1 KotL..YTC449
PM#7
..


.

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