Possibly dumb question from a novice. Please help!
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Possibly dumb question from a novice. Please help!
I'm attempting my first kit with lighting. Very basic, just 2 LEDs in series with one resistor. Having just soldered some wire to the LEDs, I tested them on the breadboard. The odd thing is, when I test 1 LED with a 200 ohm resistor it is noticeably brighter than when I test both LEDs in series with a 100 ohm resistor.
I was under the impression that both of these circuits would have the same brightness. Am I wrong? What's going on here?
I was under the impression that both of these circuits would have the same brightness. Am I wrong? What's going on here?
I'll assume you are using 3 or 5mm LEDs, probably white and as such likely require 3 to 3.4 volts each. 2 such LEDs in series would require 6 volts to get them to light nicely. ( 3+3=6 )
I use 330 ohm resistors for my projects, though I usually see others using 470 for protection. 200 won't be having a negative effect though and since both light up, it sounds like wiring is fine. That leaves voltage.
200 versus 100 won't be an issue as neither provide much resistance.
I use 330 ohm resistors for my projects, though I usually see others using 470 for protection. 200 won't be having a negative effect though and since both light up, it sounds like wiring is fine. That leaves voltage.
200 versus 100 won't be an issue as neither provide much resistance.
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Let's say you're using a 12 volt power supply and white LEDs with a 3-volt forward voltage.
The amperage going through the circuit is (battery voltage- LED voltage) / ohms.
With one led and a 200 ohm resistor, the amperage you'd get would be (12-3)/200 = 0.045 amp (i.e, 45 milliamps)
With 2 LEDs and a 100-ohm resistor, the amperage would be (12-6)/100 = 0.060 (60 milliamps)
In the particular case above, the second setup would be brighter (more amps).
But let's try again with a lower voltage battery. Say we were using 7.5 volts (5 AA batteries in series).
One LED with 200 ohms: (7.5-3)/200 = 22.5 milliamps
Two LEDs with 100 ohms: (7.5-6)/100 = 15 milliamps (so, dimmer than one LED with 200 ohms)
The amperage going through the circuit is (battery voltage- LED voltage) / ohms.
With one led and a 200 ohm resistor, the amperage you'd get would be (12-3)/200 = 0.045 amp (i.e, 45 milliamps)
With 2 LEDs and a 100-ohm resistor, the amperage would be (12-6)/100 = 0.060 (60 milliamps)
In the particular case above, the second setup would be brighter (more amps).
But let's try again with a lower voltage battery. Say we were using 7.5 volts (5 AA batteries in series).
One LED with 200 ohms: (7.5-3)/200 = 22.5 milliamps
Two LEDs with 100 ohms: (7.5-6)/100 = 15 milliamps (so, dimmer than one LED with 200 ohms)
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Thanks for responding guys. I think Rocketeer has the answer. I'm using 5mm White LEDs. Not sure about the forward voltage as I don't have a volt meter. I'm using 4, 1.5v AA batteries, so 6V in total. Because the LEDs light up, I'm assuming that means they have a forward voltage of less than 3V.
If I assume a forward LED voltage of 2.2v I get 19 milliamps for the 1LED + 200 ohm resistor and 16 milliamps for the 2LED + 100 resistor.
If I assume the LED voltage is 2.9 those values change to 15mA and 2mA respectively.
I really want that extra brightness, whether it's 19mA, 15mA or some other value. This brings up another question. In trying to work out the minimum safe resistor value, I used the series equation R = (battery V - combined LED voltage) / amps
Is this how you determine the minimum safe resistor value? Because it seems like these two equations are circular, given that the resistance and amperage values change based on each other.
Also, is it true that wiring 2, 100ohm resistors in parallel will result in 50ohms resistance?
If I assume a forward LED voltage of 2.2v I get 19 milliamps for the 1LED + 200 ohm resistor and 16 milliamps for the 2LED + 100 resistor.
If I assume the LED voltage is 2.9 those values change to 15mA and 2mA respectively.
I really want that extra brightness, whether it's 19mA, 15mA or some other value. This brings up another question. In trying to work out the minimum safe resistor value, I used the series equation R = (battery V - combined LED voltage) / amps
Is this how you determine the minimum safe resistor value? Because it seems like these two equations are circular, given that the resistance and amperage values change based on each other.
Also, is it true that wiring 2, 100ohm resistors in parallel will result in 50ohms resistance?
Yes, that's the correct equation to determine resistor value. You decide how much amperage you want flowing through the LED, and determine what resistor to use. I usually decide on about 20 milliamps, which gives a nice bright glow; but you may want to look up your diodes to determine whether they have a maximum allowable current that's less than that.Rogue Leader wrote:... This brings up another question. In trying to work out the minimum safe resistor value, I used the series equation R = (battery V - combined LED voltage) / amps
Is this how you determine the minimum safe resistor value? Because it seems like these two equations are circular, given that the resistance and amperage values change based on each other.
Also, is it true that wiring 2, 100ohm resistors in parallel will result in 50ohms resistance?
Resistors in parallel are governed by the following equation:
1/total resistance = 1/resistance1 + 1/resistance2 + 1/resistance3...
So, using 100 ohm resistors:
1/total resistance = 1/100 + 1/100
1/total resistance = 2/100
Inverting both sides:
total resistance = 100/2 = 50 ohms
In fact, if I were you, I'd wire the LEDs in parallel, each with their own resistor. This page shows what i think is a fairly typical white 5mm LED: https://www.superbrightleds.com/moreinf ... 0-mcd/261/
The forward voltage is about 3.7 volts; since two times 3.7 is actually greater than your available battery voltage of 6V, I think it's only by pure luck that you're managing to get two of them in series to light.
Wiring the LEDs in parallel would draw twice as much current, halving your battery life, but your available battery voltage would be comfortably above the LED forward voltage.
If you wanted 20 miliamps to flow through each LED, you'd wire each of them in series with a (6-3.7)/0.020 = 115 ohm resistor.
The forward voltage is about 3.7 volts; since two times 3.7 is actually greater than your available battery voltage of 6V, I think it's only by pure luck that you're managing to get two of them in series to light.
Wiring the LEDs in parallel would draw twice as much current, halving your battery life, but your available battery voltage would be comfortably above the LED forward voltage.
If you wanted 20 miliamps to flow through each LED, you'd wire each of them in series with a (6-3.7)/0.020 = 115 ohm resistor.
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Thanks for the help and suggestion, Rocketeer. I don't know what the specifications are for my white LEDs or how I'd find out the max current. They came with a starter set I got from an eBay seller in China. For my current model I'm at the point where it's probably easier to stick with a series circuit. But I'll almost certainly be using a parallel circuit for my next lighting project.
- Mr. Engineer
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Rule of thumb (for me) for any white LED is:
Forward voltage: 3.2v to 3.5v (I am using 3.2v)
Forward current: 20mA
So, using those information, the closest resistor value is 150 Ohms, 1/4watt and in a parallel wiring setup.
Different LED lens also play into the space required and its effective brightness but that would be another story for another day
I love to use this calculator because.... I;m too lazy nowadays
http://led.linear1.org/led.wiz
Forward voltage: 3.2v to 3.5v (I am using 3.2v)
Forward current: 20mA
So, using those information, the closest resistor value is 150 Ohms, 1/4watt and in a parallel wiring setup.
Different LED lens also play into the space required and its effective brightness but that would be another story for another day
I love to use this calculator because.... I;m too lazy nowadays
http://led.linear1.org/led.wiz
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- Posts: 1009
- Joined: Wed Oct 21, 2009 4:14 am
- Mr. Engineer
- Posts: 440
- Joined: Mon Dec 10, 2007 6:01 am
- Location: Malaysia
- Contact:
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- Posts: 1009
- Joined: Wed Oct 21, 2009 4:14 am
Just in case you guys don't frequent the Star Wars sub-forum, here is a link to my finished build:
http://www.starshipmodeler.net/talk/vie ... 58#1687958
Lighting my first model (as rudimentary as it is) was such a rewarding experience and a big part of that was due to the help I received here. Now I'm looking forward to incorporating fiber optics and SMDs in future builds. Cheers!
http://www.starshipmodeler.net/talk/vie ... 58#1687958
Lighting my first model (as rudimentary as it is) was such a rewarding experience and a big part of that was due to the help I received here. Now I'm looking forward to incorporating fiber optics and SMDs in future builds. Cheers!