G-100: DIY LED Interior Part X: Cupholder Illumination!
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Healingduck (11-02-2012)
07-29-2010, 02:43 AM
#43
Always Working In
Okay death this is how it goes. Input your 47 ohms into that formula I gave you and solve for Ir (Current rating of the led). If the led is not rated for that current guess what? You get a burnt out led because of current overload. If you work with the fixed values of the led(forward voltage and current rating) then you will be fine.
Long story short your leds were burning out because of current overload. You up the resistot to a 1k and they last longer. Doesn't mean they are operating at designed current rating either just resists more overload current so they last longer.
Here's what you do. Ok you have the forward voltage about right. Most leds fall in the 2-4v range. Ok now you need to know their current rating. It should be right there on the package listed with forward voltage rating and soetimes the mcd value(light output). With those values and the current value of the circuit you're installing them into you can't go wrong.
Long story short your leds were burning out because of current overload. You up the resistot to a 1k and they last longer. Doesn't mean they are operating at designed current rating either just resists more overload current so they last longer.
Here's what you do. Ok you have the forward voltage about right. Most leds fall in the 2-4v range. Ok now you need to know their current rating. It should be right there on the package listed with forward voltage rating and soetimes the mcd value(light output). With those values and the current value of the circuit you're installing them into you can't go wrong.
07-31-2010, 07:30 PM
#45
Looking into the cup holder the LEDs are hidden from sight by the rubber flaps.
When the flaps are pressed all the way in by my finger or an extra large cup they are just short enough that they don't completely cover the LEDs. Like I said before, my sister was very thoughtful in coming up with this location and I could 't be happier with it.
10-06-2012, 04:56 PM
#46
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I just did this one today and it looks amazing, a little on the bright side but a resistor should do the job. i actually wired mine in parallel as they were pre-wired with resistors, so it did the trick. i plan on posting some pics of the wiring set-up and how it turned out (i drilled the top holes as to avoid glare 
). but thanks so much for the info!! it really got me going on it and without this page i couldn't have done it! Thanks so much for your help!!
). but thanks so much for the info!! it really got me going on it and without this page i couldn't have done it! Thanks so much for your help!!
10-30-2012, 07:39 PM
#47
I just did this one today and it looks amazing, a little on the bright side but a resistor should do the job. i actually wired mine in parallel as they were pre-wired with resistors, so it did the trick. i plan on posting some pics of the wiring set-up and how it turned out (i drilled the top holes as to avoid glare ). but thanks so much for the info!! it really got me going on it and without this page i couldn't have done it! Thanks so much for your help!!
). but thanks so much for the info!! it really got me going on it and without this page i couldn't have done it! Thanks so much for your help!!
10-31-2012, 03:38 PM
#50
Moderator
iTrader: (7)
2 pages of comments and no one has given a proper value of resistors?
standard forward voltage of 5mm leds is 3.2V
standard current for LEDs=20mA=0.02A
battery- 14.4v
V=IR
subtract the battery voltage by the forward voltage and do the math. DON'T forget to take account into any LEDs in series wiring, parallel connections can be neglected.
do this right and your LEDs will NOT burn out!
standard forward voltage of 5mm leds is 3.2V
standard current for LEDs=20mA=0.02A
battery- 14.4v
V=IR
subtract the battery voltage by the forward voltage and do the math. DON'T forget to take account into any LEDs in series wiring, parallel connections can be neglected.
do this right and your LEDs will NOT burn out!
Last edited by paperboy42190; 10-31-2012 at 03:41 PM.
11-01-2012, 06:01 AM
#51
2 pages of comments and no one has given a proper value of resistors?
standard forward voltage of 5mm leds is 3.2V
standard current for LEDs=20mA=0.02A
battery- 14.4v
V=IR
subtract the battery voltage by the forward voltage and do the math. DON'T forget to take account into any LEDs in series wiring, parallel connections can be neglected.
do this right and your LEDs will NOT burn out!
standard forward voltage of 5mm leds is 3.2V
standard current for LEDs=20mA=0.02A
battery- 14.4v
V=IR
subtract the battery voltage by the forward voltage and do the math. DON'T forget to take account into any LEDs in series wiring, parallel connections can be neglected.
do this right and your LEDs will NOT burn out!
11-01-2012, 06:51 AM
#52
**The Catfish**
iTrader: (11)
14.4 volts running. that should be the output from the alternator when the car is on. When the car is off, the car is running directly off the battery which should be ~12v. You have to take into account the 14.4v for the max operating voltage to wire up led's
11-01-2012, 08:53 AM
#53
Gotcha. Even though my 12V 0.028A LEDs are 12V, I'll still need a resistor for maximum life (for parallel LEDs).
V = IR
(14.4V - 12.0V) = 0.026A (R) ==> R = 2.4V/0.028A = 85.7 Ohms
So I'd need a 1/4W resistor value of gray, blue, and black (in that order). Hmmm, not sure if they make 'standard' resistors with these values.
Edit: So it looks like the highest standard value I might have in my inventory is a 100 Ohm resistor. I'll see what effect those have on the reliability and brightness.
V = IR
(14.4V - 12.0V) = 0.026A (R) ==> R = 2.4V/0.028A = 85.7 Ohms
So I'd need a 1/4W resistor value of gray, blue, and black (in that order). Hmmm, not sure if they make 'standard' resistors with these values.
Edit: So it looks like the highest standard value I might have in my inventory is a 100 Ohm resistor. I'll see what effect those have on the reliability and brightness.
Last edited by gatrhumpy; 11-01-2012 at 09:04 AM.
11-01-2012, 08:35 PM
#55
Fearless DIY Guy
Thread Starter
iTrader: (2)
Ruh-roh...looks like you're making a bit of a mistake here, my man.
Vfa = 2.4V. This is the forward voltage of the diode.
I = 20mA. This is the current of the diode, and drives how bright it will be.
R = Unknown
(14.4V - (2)(2.4V) = 9.6V/0.0020mA = 480ohms for 2 in series
(14.4V - 2.4V) = 12.0V/0.0020mA = 600ohms for 1 in alone
SO,
for 2 in series, use 560-ohms; for 1, use 680-ohms. I recommend using 1/2W.
11-01-2012, 10:34 PM
#56
Engineer in Training
I see a lot of formulas being presented without a reason. I hope I can help.
Let's use attainable parts to begin with to keep things as realistic as possible. Let's say...RadioShack. If you purchase a NTE30036 Blue LED then the common specs will be listed on the packaging, or you can look up a datasheet for the exact product you're buying, provided here: http://www.datasheetarchive.com/NTE30036-datasheet.html
NTE30036
Forward voltage: 3.5V
Continuous forward current: 30mA
Peak forward current: 100mA
Power dissipation: 120mW
Now, let's say we have a brand new battery. Typical difference potential unloaded (nothing connected to the battery) is at a generally agreed upon value of 12V, with a new or healthy alternator in our car's bringing it up to a maximum of about 14.4V. With this given, we can assume the typical operating voltage is 13V. Just so our numbers are not at one extreme (too low) or another (too high).
A good rule of thumb is to start with your power dissipation to figure out how big of a resistor to use thermal-wise. I like to go for a bit of overkill so I figure out what my power dissipated by the circuit (not the resistor) will be and use a resistor power rating of at least double that. ex: If my circuit will put out 10mW of power, I want resistors that are rated at least at 20mW of power dissipation.
ok, so a brightness from the LEDs seen in the original post is - and this is a very rough estimate since this is easy to adjust or play with to get just to how bright or dim you want it - about 20mA of current. It is current that kills a resistor, not necessarily voltage.
So, now we have at least 2 pieces of the puzzle and can use it to solve 2 questions
First:
I = 20mA
V = 13V
Using Ohm's law I = V/R
This gives us: 20mA = 13V/R
Which when rearranged gives us R = 13V/20mA
Therefore: R = 650Ω
Well, we're forgetting something here. An LED is a Light Emitting DIODE. This means that the spec listed as the Forward Voltage is the Voltage Drop. This means that the voltage we will use to figure out the current in the circuit is not 13V, but rather 13V - Forward Voltage.
Therefore: V = 13V - 3.5V = V = 9.5V
So with the new numbers we get:
20mA = 9.5V/R
Which when rearranged gives us R = 9.5V/20mA
Therefore: R = 475Ω
Since 475Ω is not a standard Resistor value we can pick either 470Ω(less resistance, more current, brighter light) or 560Ω(more resistance, less current, dimmer light). I personally like to start with 1kΩ. I've just grown used to using it as a stable starting point which develops a point of reference.
Second:
We can also figure out power generated/dissipated from the circuit overall.
P(power in Watts) = Vs(Voltage Source) x I(Current)
*I like to use source voltage here instead of the operating voltage because it will give me a larger value than necessary for better safety rating*
Using our given values we have: P = 13V x 20mA
Which is what our circuit will produce/dissipate: P = 0.2603W or 260mW
That's about a quarter of a Watt. Power safety from before tells me that I want to then use a 1/2Watt resistor.
Summed up:
If you purchase the LEDs at RadioShack that I linked up a above, you can follow my numbers and use different size resistors to play with the brightness.
Personally, I like use a Potentiometer (also cheap at RadioShack) to dim the lights to the brightness level I want, then read the resistance between terminal connected to my Voltage Source(13V) and the wiper. This gives me the exact resistance needed for that level of brightness/dimness. From here on I can decide to build a combination of resistors to get as close as I can to my measured value, us the potentiometer itself, or use a resistor above/below my measured amount.
If you want me to go deeper, and explain what to do with Series, Parallel, or Series-Parallel circuits just let me know
P.S. There are LEDs that are meant to be used as "Fast-Switching" LEDs but most aren't, so it's best to use capacitors in parallel with the Voltage Source of the circuit to keep the LEDs from flickering upon engine start-up. This helps prolong their life.
Let's use attainable parts to begin with to keep things as realistic as possible. Let's say...RadioShack. If you purchase a NTE30036 Blue LED then the common specs will be listed on the packaging, or you can look up a datasheet for the exact product you're buying, provided here: http://www.datasheetarchive.com/NTE30036-datasheet.html
NTE30036
Forward voltage: 3.5V
Continuous forward current: 30mA
Peak forward current: 100mA
Power dissipation: 120mW
Now, let's say we have a brand new battery. Typical difference potential unloaded (nothing connected to the battery) is at a generally agreed upon value of 12V, with a new or healthy alternator in our car's bringing it up to a maximum of about 14.4V. With this given, we can assume the typical operating voltage is 13V. Just so our numbers are not at one extreme (too low) or another (too high).
A good rule of thumb is to start with your power dissipation to figure out how big of a resistor to use thermal-wise. I like to go for a bit of overkill so I figure out what my power dissipated by the circuit (not the resistor) will be and use a resistor power rating of at least double that. ex: If my circuit will put out 10mW of power, I want resistors that are rated at least at 20mW of power dissipation.
ok, so a brightness from the LEDs seen in the original post is - and this is a very rough estimate since this is easy to adjust or play with to get just to how bright or dim you want it - about 20mA of current. It is current that kills a resistor, not necessarily voltage.
So, now we have at least 2 pieces of the puzzle and can use it to solve 2 questions
First:
I = 20mA
V = 13V
Using Ohm's law I = V/R
This gives us: 20mA = 13V/R
Which when rearranged gives us R = 13V/20mA
Therefore: R = 650Ω
Well, we're forgetting something here. An LED is a Light Emitting DIODE. This means that the spec listed as the Forward Voltage is the Voltage Drop. This means that the voltage we will use to figure out the current in the circuit is not 13V, but rather 13V - Forward Voltage.
Therefore: V = 13V - 3.5V = V = 9.5V
So with the new numbers we get:
20mA = 9.5V/R
Which when rearranged gives us R = 9.5V/20mA
Therefore: R = 475Ω
Since 475Ω is not a standard Resistor value we can pick either 470Ω(less resistance, more current, brighter light) or 560Ω(more resistance, less current, dimmer light). I personally like to start with 1kΩ. I've just grown used to using it as a stable starting point which develops a point of reference.
Second:
We can also figure out power generated/dissipated from the circuit overall.
P(power in Watts) = Vs(Voltage Source) x I(Current)
*I like to use source voltage here instead of the operating voltage because it will give me a larger value than necessary for better safety rating*
Using our given values we have: P = 13V x 20mA
Which is what our circuit will produce/dissipate: P = 0.2603W or 260mW
That's about a quarter of a Watt. Power safety from before tells me that I want to then use a 1/2Watt resistor.
Summed up:
If you purchase the LEDs at RadioShack that I linked up a above, you can follow my numbers and use different size resistors to play with the brightness.
Personally, I like use a Potentiometer (also cheap at RadioShack) to dim the lights to the brightness level I want, then read the resistance between terminal connected to my Voltage Source(13V) and the wiper. This gives me the exact resistance needed for that level of brightness/dimness. From here on I can decide to build a combination of resistors to get as close as I can to my measured value, us the potentiometer itself, or use a resistor above/below my measured amount.
If you want me to go deeper, and explain what to do with Series, Parallel, or Series-Parallel circuits just let me know
P.S. There are LEDs that are meant to be used as "Fast-Switching" LEDs but most aren't, so it's best to use capacitors in parallel with the Voltage Source of the circuit to keep the LEDs from flickering upon engine start-up. This helps prolong their life.
Last edited by LockDots; 11-01-2012 at 10:38 PM.
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paperboy42190 (11-03-2012)
11-02-2012, 01:47 PM
#58
PREDICTION = They instantly burn out.
Ruh-roh...looks like you're making a bit of a mistake here, my man.
Vfa = 2.4V. This is the forward voltage of the diode.
I = 20mA. This is the current of the diode, and drives how bright it will be.
R = Unknown
(14.4V - (2)(2.4V) = 9.6V/0.0020mA = 480ohms for 2 in series
(14.4V - 2.4V) = 12.0V/0.0020mA = 600ohms for 1 in alone
SO,
for 2 in series, use 560-ohms; for 1, use 680-ohms. I recommend using 1/2W.
Ruh-roh...looks like you're making a bit of a mistake here, my man.
Vfa = 2.4V. This is the forward voltage of the diode.
I = 20mA. This is the current of the diode, and drives how bright it will be.
R = Unknown
(14.4V - (2)(2.4V) = 9.6V/0.0020mA = 480ohms for 2 in series
(14.4V - 2.4V) = 12.0V/0.0020mA = 600ohms for 1 in alone
SO,
for 2 in series, use 560-ohms; for 1, use 680-ohms. I recommend using 1/2W.
Yep, they have an integrated 680 resistor in them, so they did not burn out. Also, I had 12V, as in twelve volts. Not 2.4V. Twelve.
Last edited by gatrhumpy; 11-02-2012 at 01:50 PM.
11-02-2012, 02:12 PM
#59
Engineer in Training
I believe the reason he subtracted from 12 instead of using 12 alone is because of the voltage drop accross the LED leaving residual voltage used to calculate current and resistance
11-02-2012, 03:27 PM
#60
I guess I would need another 86 Ohm resistor if I want to play it safe, but they work fine now.
These are the LEDs I got from Radioshack: http://www.radioshack.com/product/in...ductId=2062568
Last edited by gatrhumpy; 11-02-2012 at 03:35 PM.
11-02-2012, 03:35 PM
#61
Engineer in Training
Why are you subtracting 12V from 14.4V?
What you should be doing (as outlined in my bible-length post) is subtracting the LED forward voltage from the supplied voltage (14.4V). The remaining voltage is what you divide by the current you want driving the diode (20mA) and that will give you the resistance you want to aim for.
What you should be doing (as outlined in my bible-length post) is subtracting the LED forward voltage from the supplied voltage (14.4V). The remaining voltage is what you divide by the current you want driving the diode (20mA) and that will give you the resistance you want to aim for.
11-02-2012, 03:50 PM
#62
Why are you subtracting 12V from 14.4V?
What you should be doing (as outlined in my bible-length post) is subtracting the LED forward voltage from the supplied voltage (14.4V). The remaining voltage is what you divide by the current you want driving the diode (20mA) and that will give you the resistance you want to aim for.
What you should be doing (as outlined in my bible-length post) is subtracting the LED forward voltage from the supplied voltage (14.4V). The remaining voltage is what you divide by the current you want driving the diode (20mA) and that will give you the resistance you want to aim for.
Once again, the forward voltage of the LED itself, is 12.0V. The LED was designed for a car application (~12V). It says it can go all the way to 16V with no external resistor needed.
11-02-2012, 03:55 PM
#63
Engineer in Training
Holy God. Maybe I'm not explaining it well. Sorry. The voltage of the LED is 12.0V not 2.4V.
Once again, the forward voltage of the LED itself, is 12.0V. The LED was designed for a car application (~12V). It says it can go all the way to 16V with no external resistor needed.
Once again, the forward voltage of the LED itself, is 12.0V. The LED was designed for a car application (~12V). It says it can go all the way to 16V with no external resistor needed.
11-02-2012, 04:03 PM
#65
Engineer in Training
Well hell, so am I......how do we explain this now? lmao I'm the same way though "see what you've got to do is.....a screw it let me just do it myself!"
11-02-2012, 05:35 PM
#69
Fearless DIY Guy
Thread Starter
iTrader: (2)
It's got a 680-ohm resistor built in...don't you find it the least bit coincidental that my math calling for a 680-ohm resistor is PRECISELY what I recommended? You're adding impedance to the circuit to compensate for the remaining flow (the balance 12V)...simple as that. You will INSTANTLY kill it.
Hell, I'll even play devils advocate here. If your LED is Vf = 12.0, how about this? Cut the resistor out and slap your originally calculated 85.7ohm resistor in there, fire up your car, and see if it still lights.
I don't mean to beat a dead horse here, man...but you're NOT using an LED with Vf = 12.0. You're using an LED with a 12V label on the bag because it's easier for the market to understand. The 12V title on the bag is for the APPLICATION...NOT the Vf,.
PS = Third engineer throwing his hat in the ring. I make wind turbines...the big ones.
11-02-2012, 05:59 PM
#71
Engineer in Training
He's actually adding resistance, not impedance. Impedance refers to AC circuits, whereas this is a DC circuit.
Truth
Truth
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DeathMetal (11-02-2012)
11-02-2012, 07:17 PM
#73
You do not have a 12V LED, PERIOD. What you have is an LED running at a FORWARD VOLTAGE of 2.4V, which is a nominal value for this style LED.
It's got a 680-ohm resistor built in...don't you find it the least bit coincidental that my math calling for a 680-ohm resistor is PRECISELY what I recommended? You're adding impedance to the circuit to compensate for the remaining flow (the balance 12V)...simple as that. You will INSTANTLY kill it.
Hell, I'll even play devils advocate here. If your LED is Vf = 12.0, how about this? Cut the resistor out and slap your originally calculated 85.7ohm resistor in there, fire up your car, and see if it still lights.
I don't mean to beat a dead horse here, man...but you're NOT using an LED with Vf = 12.0. You're using an LED with a 12V label on the bag because it's easier for the market to understand. The 12V title on the bag is for the APPLICATION...NOT the Vf,.
PS = Third engineer throwing his hat in the ring. I make wind turbines...the big ones.
It's got a 680-ohm resistor built in...don't you find it the least bit coincidental that my math calling for a 680-ohm resistor is PRECISELY what I recommended? You're adding impedance to the circuit to compensate for the remaining flow (the balance 12V)...simple as that. You will INSTANTLY kill it.
Hell, I'll even play devils advocate here. If your LED is Vf = 12.0, how about this? Cut the resistor out and slap your originally calculated 85.7ohm resistor in there, fire up your car, and see if it still lights.
I don't mean to beat a dead horse here, man...but you're NOT using an LED with Vf = 12.0. You're using an LED with a 12V label on the bag because it's easier for the market to understand. The 12V title on the bag is for the APPLICATION...NOT the Vf,.
PS = Third engineer throwing his hat in the ring. I make wind turbines...the big ones.
:deadhorse: Let's move on.
Bottom line, the LEDs work and they're spectacular!
I even wired in two other wires to light up three small LEDs under the front seats.
11-03-2012, 01:52 PM
#76
Engineer in Training
01-16-2013, 12:51 PM
#79
Well thats what I'm trying to figure out. I would like to wire up lights in other areas but is that going to consume the power flowing through the line.
The LED's are wired in a series and so its 12 volts that are flowing through. Do the LED's eat all 12 volts of the power or whats the deal there? For that matter if you're wiring up driving lights then why is a line pulled from the battery rather than powering it from a line running within the car? Thats where my lack of understanding stems from...
The LED's are wired in a series and so its 12 volts that are flowing through. Do the LED's eat all 12 volts of the power or whats the deal there? For that matter if you're wiring up driving lights then why is a line pulled from the battery rather than powering it from a line running within the car? Thats where my lack of understanding stems from...
01-16-2013, 01:16 PM
#80
**The Catfish**
iTrader: (11)
Well thats what I'm trying to figure out. I would like to wire up lights in other areas but is that going to consume the power flowing through the line.
The LED's are wired in a series and so its 12 volts that are flowing through. Do the LED's eat all 12 volts of the power or whats the deal there? For that matter if you're wiring up driving lights then why is a line pulled from the battery rather than powering it from a line running within the car? Thats where my lack of understanding stems from...
The LED's are wired in a series and so its 12 volts that are flowing through. Do the LED's eat all 12 volts of the power or whats the deal there? For that matter if you're wiring up driving lights then why is a line pulled from the battery rather than powering it from a line running within the car? Thats where my lack of understanding stems from...
leds run off 12v-14.4v but this does not mean that they consume all that power. With regard to your example of running driving lights, you need to grab a 12v power wire from the battery to run the relay. if every accessory consumed 12 volts, the first accessory off the battery would be the only accessory you could run.