Main > Main Forum

LED powering question

<< < (5/6) > >>

RandyT:
I'll send you an email...

mvolke1:
Awesome! Thanks Randy!

PL1:

--- Quote from: RandyT on October 20, 2014, 11:38:30 am ---
--- Quote from: PL1 on October 20, 2014, 12:45:24 am ---:banghead:  Didn't think about this earlier, but you could just wire the two CDRs in series to drop 4.5v on each.

--- End quote ---
This probably won't provide the effect desired, unless the wiring is chained without removing the inline resistors.  These are what make them 5v.

--- End quote ---
Yes, I'm referring to using them as delivered with the inline resistor.

If 4.5v is enough to light a CDR -- and it should be -- placing two of them in series will light them just fine with a 9v power supply.

The fact that each is dropping 4.5v instead of 5v means that they are not running at peak current flow.

For example, a 12v LED/resistor combination that draws 20mA when powered by 12v only draws 5.85mA when powered by 5v.

The higher resistance of the 12v LEDs/current limiting resistors in parallel with the CDRs will result in more current flowing the part of the circuit with the CDRs.

To ensure the CDRs aren't damaged, you may need add a resisor in series with the CDRs so that that part of the parallel circuit draws 100mA or less. (verify with your multimeter)


Scott

mvolke1:
Scott and Randy,

Thanks so much for your help!

I put in an order with Randy, and he's tossing in some extra 12v LED bulbs. I can just switch them out and I should be good to go with my other 12 12v LED pushbuttons.

Thanks again for all your help! When I get them wired up, probably end of week, I'll post some pictures.

Matt

mvolke1:
Randy,

Thanks for the bulbs! They worked great!

Navigation

[0] Message Index

[#] Next page

[*] Previous page

Go to full version