I need to hook up 2-1.5 volt 30ma bulbs in parallel and need to know what size resistor to use for a 12v source (the decoder).
1. Is the total current 30ma for the two bulbs, or do I add the current of each bulb to get a total of 60ma (as in a series circuit)?
2. In R=E/I, is E 12v or do you use the voltage to be dropped, E=10.5?
So, should I end up with:
R = 12v / .03a or R = 12 / .03 + .03 or R = 10.5 /.03 or R = 10.5 / .03 + .03
I've been using LEDs for so long and just throwing on 1K resistors, I've forgotten some of my ohm's law.
Jay
C-415 Build: https://imageshack.com/a/tShC/1
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I use this site to figure that stuff out: http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator
The calculator is for LED's but it works the same for bulbs.
TONY
"If we never take the time, how can we ever have the time." - Merovingian (Matrix Reloaded)
Two bulbs in parallel will double the Amp draw but voltage will stay the same; in series the voltage draw doubles.
So, for two 1.5 Volt 30ma bulbs in parallel you need a resistor of 175 Ohms. This is not a standard value of resistor, so use the next higher value, which is 180.
Personally, I'd go with at least a 220 Ohm resistor to prolong bulb life.
In a parallel circuit, you add the current draw, in a series circuit, you add the voltage rating. If you put 12 volts across a bulb rated for 1.5 volts, it will look like a flash bulb.
Other than that, you have the Ohm's Law formula correct, you just need to remember series circuits. If each bulb requires 1.5 volts to light fully, and you want to tie them into a 12 volt power supply, you have a couple options. First, you can make a series circuit. with 8 bulbs, which will give you a proper 1.5 volts across each bulb. Second choice, you can attach a dropping resistor in series with each bulb. By Ohm's Law, it's easy to calculate that each bulb has a resistance of 50 ohms, so you'd need a 350 ohm resistor in series with each bulb to drop the voltage. A half watt resistor should be sufficient.
The figures above will run the bulbs at their rated voltage. Most people will run bulbs at less than their rated voltage, they will not run as hot (important if they're in a plastic building), they don't burn as bright, and they will last longer. Options are to put about 10 bulbs in series, or use a larger resistor. Use some clip leads and experiment with number in series and check for heat and brightness.
As someone said, ohms law is the same for LED's and light bulbs. Below is a link to lighting that may be of some use. Store the link for future use.
Anyone reading this message should also store the link for reference. I see this question a lot in different forums.
The below person has a lot of good info. He is in some of the Yahoo DCC Groups and helps many people. Look at his home page.
http://www.members.optusnet.com.au/nswmn1/Lights_in_DCC.htm
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
TomDiehl In a parallel circuit, you add the current draw, in a series circuit, you add the voltage rating. If you put 12 volts across a bulb rated for 1.5 volts, it will look like a flash bulb.
Thanks guys! I was forgetting to add the current draw. I can't believe my grey matter failed me, as I did earn an AA in elctronics, but then again that was forty years ago and I've been using LEDs for 10 years or so - much simpler.
modelmaker51 TomDiehl: In a parallel circuit, you add the current draw, in a series circuit, you add the voltage rating. If you put 12 volts across a bulb rated for 1.5 volts, it will look like a flash bulb. Thanks guys! I was forgetting to add the current draw. I can't believe my grey matter failed me, as I did earn an AA in elctronics, but then again that was forty years ago and I've been using LEDs for 10 years or so - much simpler.
TomDiehl: In a parallel circuit, you add the current draw, in a series circuit, you add the voltage rating. If you put 12 volts across a bulb rated for 1.5 volts, it will look like a flash bulb.
Been there, done that, have the T shirt.
I have a very simple 3 step procedure for calculating the value of the dropping resistor for lights, LEDs or anything I want to operate from a higher voltage than it's intended for.
1) Subtract the desired voltage from the supply voltage.
2) If necessary, change the current draw of the device from milliamps to amps by dividing by 1,000.
3) Divide the answer from step 1 by the answer from step 2. Use the next highest standard resistor value.
It works for me!
..... Bob
Beam me up, Scotty, there's no intelligent life down here. (Captain Kirk)
I reject your reality and substitute my own. (Adam Savage)
Resistance is not futile--it is voltage divided by current.
Since I always have an active PC, around, I have an on line LED resistance calculator specifically for this issue.
There are many on line calculators available for all kinds of calculations.
Metric conversion is one useful one I have handy.
One for LM317 voltage regulators resistor settings is another.
Many people have no idea they can search the 'Net for on line calculators or formulas.
Looks like you had it right. It comes out to 175 ohms. As noted earlier I'd try a 220 ohm or maybe even a 330 ohm resistor to start. Should be plenty bright enough and add life to the bulb. It will also protect you from voltage that rides a little high or spikes.
Springfield PA
Why not post it for all to use???
MOAVBILLY Since I always have an active PC, around, I have an on line LED resistance calculator specifically for this issue. Why not post it for all to use???
Dude, someone already did post a link to an on line calculator in this thread. I use the same calculator.
Learn to read all the messages and also click on the links in a message. Never know what kind of useful info you might find. Store the links in Favorites or Bookmarks for future use.
It is so easy to do a Google search for an on line calculator you desire.
Here's the caclulator link again
http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator
Remember the series and parallel methods are different so pay attention when using it.
If you have a series and parallel configuration things get a little trickier but not much harder. The calculator supplied will handle most MRR applications.
Also remember the key difference between a light bulb and an LED. With a light bulb, the specified current is how much it will use at the rated voltage when lit. With an LED, the current rating is a MAXIMUM the LED can stand without damage, so you always want to calculate a resistor for an LED using something less than the current LIMIT printed in the specs. Half usually works, with the white LEDs used in locos - a 1K resistor with a typical DCC decoder gets about 9-10ma to the LED, and most white LEDs are rated about 20ma.
The coltage is just the opposite - with an LED, the voltage rating is how much will be dropped by the LED. 2.1 volts for red/green/yellow is typical, and 3.2V is typical for white. That voltage will be 'lost' in the LED, always. For a light bulb, the voltage is essentially the maximum it can handle - sure they can go higher, for a while. Same with the current in an LED. Put 25ma through one that is rated for 20ma isn't going to make it go poof instantly, but like running 14v through a 12v light bulb, the lifespan will be reduced.
More technically, when calculating a resistor for a light bulb, you are adjusting the voltage drop. The current is known. Voltages in series add up, so if the bulb is rated at 3 volts and yout power supply is rated at 12 volts, you know you need to drop 9 volts in the resistor. And since current in series is shared, you know that if the light bulb is rated for 50ma, then 50ma will flow through the resistor. You now know voltage and current, and thus can calculate the resistor required.
For an LED, you know the voltage drop. You have to pick a current level, something under the LEDs rating. Since again we have a series circuit, voltages add, so if the LED drops 3.2 volts and the power supply is 12 volts, we know 8.8 volts must be dropped in the resistor. Current is shared, so whatever we limit the LED to is how much current will flow through the resistor. Once again we have a voltage and current and can calculate resistence from that.
With a light bulb, using a slightly larger resistor will reduce the voltage, since the current is fixed. With an LED< using a larger resistor will reduce the current, since the voltage is fixed.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Dude, where's my car?
or
The Dude abides.
Hey dudes,
don't post questions to the answers some of us already know.
Geez, what a bummer
Already answered I believe but
12V - 1.5V = 10.5 V drop
10.5V = 60 ma * R
10.5V/.060 = R = 175 Ohm
Another way to do it:1.5V = .030 * R(lamp)R(lamp) = 50 ohms
Since they are in parallel, that cuts the resistance in half (25 ohms)
.060 = 2 x 30 mamps
12 = .060 * (R1 + 25)R1 = 175 Ohms
Don - Specializing in layout DC->DCC conversions
Modeling C&O transition era and steel industries There's Nothing Like Big Steam!
rrinker With a light bulb, using a slightly larger resistor will reduce the voltage, since the current is fixed. With an LED< using a larger resistor will reduce the current, since the voltage is fixed.
The current in the bulb circuit is not fixed. If the resistance is increased in any series circuit, the current will decrease as long as the supply voltage remains the same--Ohms Law. That holds true for light bulbs as well as LEDs. But it is true that the voltage across the bulb will decrease, while it will remain essentially the same across the LED. The current will change in both cases.
Recovering former former model railroader.
Ok, dudes and dudettes, a very useful link for light bulb and LED lighting.
Obvious Man says, store the link in Favorites or Bookmarks depending on which browser you use. It should be obvious.