| Author |
Message |
Bob (Bobb) (69.19.4.131)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 12:14 am: | |
Ok.. here is one for you electronic guys..
I have 12v LED lights that I wish to run on 24v. I cannot take them apart to change the resistors, so I was thinking of adding a resistor in series with the + lead.
At 12v, the lights draw 120 milliamps. What value of resistor would I need to put in series to have the light operate at 24v?
Thanks! |
Derek (24.85.245.203)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 12:34 am: | |
http://www.theledlight.com/LED101.html
This site should help, if you can find some of the information about your specific LEDs. |
CoryDane RTSII (66.155.188.202)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 4:00 am: | |
Perhaps you can figure out with this calculator.
http://www.quickar.com/noqbestledcalc.htm
Since you know the MA's and voltage for 12v, figure the resistor, then get another of the same size, putting two in series should do it. Figure the circut with the two resistors on 24v. If the number jive, then you are in.
"Imagine"
cd |
Craig Smith (Craigs) (65.202.123.254)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 9:07 am: | |
Hi Bob,
The series resistor load must drop 12V. Using the formula P=VI, 1.44W = 12V*.12I. This is the power requirement of the resistor. The resistor value R is calculated with the formula V=IR. In this case, 12V = .12I * 100R. This size resistor, 100 ohm 1.5 watt may be hard to find. To substitute, you could use 2 200 ohm 1 watt or 4 400 ohm 1/2 watt in parallel. This is where I'd start. Of course, we are not at exactly 24V so these calcs are approximate.
FWIW, Craig S |
Abajaba (12.217.192.62)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 9:32 am: | |
Using Ohms law to find out what resistance equals
120 milliamps at 12 volts one will divide volts
by current. 12V / 0.120mA = 100 Ohms
Now, lets use some more math and come up with
another important number here. How much power is
going to be dissipated in the series resistance.
There are two ways to do this and they both come
up with the same answer.
Voltage multiplied by current will result in
wattage. 12V x 0.120A = 1.44 watts.
The other way if you don't know the voltage
is to use the current value squared, in this case
0.120 x 0.120 = 0.0144 then multiply that by the
value of the resistor it is passing through for
the following equation. The formula for this one
is I^2 x R = P where I stands for current flow in
Amps.
0.0144 x 100 = 1.44 watts.
A 1.5 watt resistor would do the trick at 100 ohms
but I would recommend going to a 2 watt resistor
or maybe even a little bit bigger. The idea is
that the resistor will be someplace where air
can't get to it readily and so it will not have
that to cool it down. 1.44 watts may not seem
like much but in an insulated area it could build
up a substantial amount of heat. The larger
resistor is designed to handle dissipation of more
heat and should be able to do so in a more
confined space while keeping it's surface
temperature down. More surface area equals less
heat per square inch.
Sorry to get into the math so early in the morning
but I had not seen anyone mention power
dissipation by the series resistor yet. I have
been caught by surprise sometimes when I did all
the math on a series dropping resistor and then
did the power dissipation and figure there might
be a problem with my 1/8 watt resistor. They
really glow bright for a short time when one is
dissipating one watt in them. :-o |
Gary Stadler (Boogiethecat) (68.7.217.217)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 1:00 pm: | |
You could also find a 12 volt 2 watt(or larger) zener diode and stick it in series with the lamp. More expensive and harder to find, but it's another way to do it.
Or you could simply put two of your lamps in series....
If you do the 100 ohm thing, just get a 5 watt resistor. They're cheap and plentiful from many sources, and having the extra few watts capacity will be helpful if you try to put the resistor in an enclosed space...
Cheers
Gary
Cheers
Gary |
Sean Welsh (Sean) (64.81.73.194)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 1:07 pm: | |
Bob,
Another option is to double the lights up in series. We were able to find 24v models for many of our lights, but some of the smaller markers were unavailable. We bought 12v units, cut the ground wire away from the mounting screw, then mounted them in pairs wired in series. Works like a charm, and more of our power is going into making light rather than heat.
If you choose to go the resistor route, mount the resistors away from the plastic lamp housing -- they get very hot.
-Sean |
BrianMCI96A3 (198.81.26.45)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 3:05 pm: | |
I was going to do just what Bob is wanting to do and came up with the same numbers as Abajaba.
I ordered a number of 100 ohm 2 watt resistors to wire to each light... BUT in a test of the finished light with it hooked up to 24V... I really was unhappy with just how hot the resistor got after a very short time.
After only a few seconds, you could still hold the resistor in your fingers, but only just!
I believe that I will try to run a 12V power source to that circuit and run the lights on 12 volts.
Brian |
bill moldenhauer (12.74.13.122)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Monday, April 19, 2004 - 9:22 pm: | |
I had this question before and finally just ran all of them in parralell and picked up 12 volts in front junction box and changed the hot side of switch.works great
mci-9 almost complete |
Bob (Bobb) (69.19.4.131)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Wednesday, April 21, 2004 - 12:39 am: | |
I like the idea of using a 12v zener diode. I found a cheap supplier (actually 13v/5w diodes were cheaper, and more in line with what an auto voltage would be).
www.futurelec.com has them for .50 in a qty of 25. They also have 100 ohm/5 watt resistors cheap.
I like the idea of reducing the voltage through a semiconductor, rather than reducing it and producing heat.
Going to give it a try.. thanks for all the suggestions.
Bob |
Abajaba (12.217.192.62)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Wednesday, April 21, 2004 - 6:05 pm: | |
Bob:
The voltage drop through the zener diode will
cause the same heat as if it was going through a
resistor. The only difference will be that the
voltage dropped across the zener will be very
constant.
If the supply voltage goes to 28 volts the light
will see 16 volts. With the resistor the light
might not see as much voltage depending on how the
internals of the light are set up.
Notice in the ratings of the zener that they are
rated in both volts and watts? They dissipate the
energy that they drop as heat. One other
important rating on a zener is the minimum current
they need to reach their zener action. If there
isn't enough current they can actually drop less
voltage than they are rated at.
Hope this helps a little bit.
Abajaba |