# Figuring out a circuit for autotech

Discussion in 'Homework Help' started by JB4, Mar 21, 2009.

1. ### JB4 Thread Starter New Member

Mar 21, 2009
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Alright, I'm currently in autotech in high school and my electrical background is very limited. What the project is, I have 10 leds on a strand in series with each other. Each one requires from 2-3volts to be powered. Now our power supply is 12v-15v and has .5-.6 amps. So what I was told is to do 5 leds in series, and then the 2 sets of 5 leds in parallel. So I soldered my circuit and it only lights the first 5. So I can't figure out where I messed up.

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2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Ground is on both "ends" of the right 5, so there is no power.

To put two sets of 5 in parallel, connect the second string of 5 to another resistor, then connect both of the + side resistors together to the + of the battery, and both of the - sides of the string to the - of the battery.

Try this link, it will be very helpful for you, pick "Wiring Diagram" for output, and enter the values you have above: http://led.linear1.org/led.wiz

3. ### JB4 Thread Starter New Member

Mar 21, 2009
4
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Alright thanks. One question though, When I used ohms law I figured it out to 25 or 30 ohms of resistance, but this is saying 56ohms. What is the difference from when I did it, and this calculator? I did:
12v/.5amps (thats the amps of my power source) which totals to 30ohms.

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Most LED's can only handle 20 ma maximum current. For dependable like, using 10 - 15 ma is best. So figure your resistors based on 10 ma and use the next smaller available resistor in the decade.

That is the sum of all the voltage drops of your 5 LED's subtracted from the source voltage to get the effective voltage, and the R = E/I to get the resistance.

5. ### JB4 Thread Starter New Member

Mar 21, 2009
4
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I'm totally confused at what this means. Sorry we didn't really learn anything except ohms law, and got assigned a project

so 12-11 is 1/.5 ? that makes more sense.

6. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
If your LED's are red, their forward drop is more like 1.5 volts. % drops is 7.5 volts. 12 - 7.5 is 4.5 volts. 4.5 / .01 = 450 ohms. Use a 430 ohm resistor, 1/2 or 1/4 watt, whichever is easiest to get. You can also use 470 ohms - it's more common in value.

7. ### JB4 Thread Starter New Member

Mar 21, 2009
4
0
according to the manufactures website, they use 2.2 volts

8. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
To calculate your current limiting resistor:
Rlimit = (Vsupply - (Vf x NumberOfLEDs)) / DesiredCurrent
where Vf is the forward voltage at the LED's rated current. Use the "typical" rating. Let's say for the moment that it's 20mA, which is common.

You need to know what your voltage supply is going to be. 12v to 15v is a pretty wide margin. Let's go with 15v for the moment.
Rlimit = (15v - (2.2v x 5) / 20mA = (15 - 11) / 0.02 = 4 / 0.02 = 200 Ohms.
200 Ohms is a standard value. If it were not, you could go to the next higher value.
Standard values chart: http://www.logwell.com/tech/components/resistor_values.html
E24 are commonly available (the values in the green columns).

But, what happens to the current when only 12v is available? We already have discovered that 11v is dropped across five LEDs in series.
I = E/R, or Current = Voltage / Resistance.
So, I = 12v-11v/200 Ohms = 1/200 = 0.005 = 5mA - the LEDs would be quite dim.

One way around that problem is to use fewer LEDs per string; the other way is to use a current or voltage regulator.

Let's try just using three LEDs at 15v:
Rlimit = (15v - (2.2v x 3) / 20mA = (15 - 6.6) / 0.02 = 8.4 / 0.02 = 420 Ohms. 430 Ohms is the closest standard value.
Iled = 8.4/430 = 19.5mA.

Now what happens when the voltage drops to 12?
I = 12v-6.6v / 430 Ohms = 5.4/430 = 0.0125 = 12.5mA - not so much of a drop.

You also need to calculate the wattage requirement of the resistor.
P = EI, or Power(Watts) = Voltage x Current
At 15v, 6.6v will be dropped across the resistor at 19.5mA
So, 6.6v x 19.5mA = 6.6 x 0.0195 = 0.1287 Watts
For reliability's sake, you always double the wattage requirement, so 2 x 0.1287 = 0.2574 Watts. This is so close, you could actually go with a resistor rated for 1/4 Watts (0.25), but the safe choice would be one rated for 1/2 Watts.