[CorradoKidd]

In your schematic like thing for the MFA light you used a 2000 ohm resistor in parllel with the LED. What was that for? Did you use it instead of a smaller resistor in sieries?
Good mod, I'm in the middle of doing it now. Gotta give you mad credit! http://****************.com/smile/emthup.gif

 

[Iceman666]

Re: Blue LED Q's (CorradoKidd)

Someone said thats how to do the mfa so thats how i drew it....i havent tried it yet but check the original post...I dont know what resistance the LEDs have i just checked a 12V led and got a bigger resistor(in size) with the same resistance. So i cant figure out what resistor to use Im pretty sure you can use the same resistor in the MFA as on the string though....
also do you know what resistance that resistor is....i totaly forgot

 

[CorradoKidd]

OK, I used a 680 ohm resistor on the string w/3 5v 30mA LED's. The LED's I got for the MFA are 3.7v 20mA, so I'll just use one of the 680 ohm resistors on them. I'll let you know how it works. I got different LED's for the MFA and the odometer, they are more intense of a blue than the others.

 

[Iceman666]

Re: Blue LED Q's (CorradoKidd)

you should only need one LED for the MFA and how are they more intense?? you mean brighter?
And i would recomend usng more then 3 LED's.....they might be brighter then the bulb but they have narrower feild of light


[Modified by Iceman666, 8:45 PM 11-27-2001]

 

[CorradoKidd]

Oh, I have MFA and odometer in LCD, its a '91 and up thing, sorry. The more intense blue are like a deeper darker blue, I think they are the same ones you used in your string. The ones I used in my string are a lighter shade of blue, and a 36 degree field of light as opposed to a 30 degree. Only reason I took this route is cause its all Radioshack had in stock

 

[Iceman666]

Re: Blue LED Q's (CorradoKidd)

Ah i see...ok are you doing the LCD's with one resistor or two?
let me know how it works

 

[Iceman666]

Ok from what i have been hearing doing it with the 2 resistors will make them brighter and last longer...so im thinking you may want to do it that way
like this

[Modified by Iceman666, 9:39 PM 11-27-2001]

 

[Iceman666]

Re: Blue LED Q's (redrocketrado)

quote:[HR][/HR]what a bunch of techie dorks!

[HR][/HR]​

Your point

 

[beenana]

Sunrise Semester # 2 - Ohm's Law and Semiconductors
diodes(semiconductors) don't really have a "resistance" like a resistor.. they kinda do, but it's not really something that you need for calculations.. diodes are like one way valves that let current go one way and not the other.. they do this at a specific voltage.. a typical 4000 series diode like a 1N4007 has a turn on voltage of 0.7V... and that is fixed.. you might get slightly more than .7 but it won't be very noticeable... if you try hooking up a diode like an led or regular diode to more than it's rated voltage, with no resistor in series, you'll blow it up.. the resistor acts like a sponge to soak up the current and burn it off(they actually give off heat, how do you think your stove elements work?).. soooo... what does all this mean? well, by looking at Iceman666's diagram, that 750ohm resistor is actually taking current away(shunting) from the LEDs!! If you were to measure the the voltage across the LEDs with that 750ohm resistor in there, it would be less than the rated voltage of the LEDs.. why? because the 2Kohm resistor is fixed, and the 12V power supply is "fixed"(i know it goes up to 14V, but everything else goes up proportionately).. since these are fixed values, you can use Ohm's Law to verify that the LED voltage goes down.. say for example you have 5V LEDs.. the current through the 750ohm resistor would be 5V/750ohms = 6.67mA.. hold on a second, that doesn't seem right does it? of course it isn't right because if you were to multiply 6.67mA x 2Kohms(the other resistor), you would have 13.34V?? that is not possible because the 12V power supply is 'fixed'.. and Ohm's Law doesn't lie, so something has to give.. the LEDs .. If you pull out that 750ohm resistor your LED's will be brighter, and they will be operating at their normal forward voltage.. That's what you want.. by the way, the 2Kohm resistor seems a bit high.. I would use a 1K or less.. ie, for one LED @ 20mA @ 5V, from a 12V supply, you would need a 350ohm resistor in series(12-5=7, 7/0.02=350).. for 5 LEDs @ 20mA @ 5V, you will need a 70ohm resistor(5 x 20ma = 100ma, 7V/100mA = 70ohms)...V = IR my friends.. Ohm's Law is the Law..
beenana

 

[Iceman666]

Re: Blue LED Q's (beenana)

except we are dealing with 12V not 7
the 2k ohm resistor is used for current regualtion and the 750 ohm resistor is used to get the voltage across the LED down to 3.27
I just spent 30 min arguing with redDevil about this
10:06 PM 11-27-2001) RedDevil: Uaually when you get Blue LEDs they are tated at 3.4V at 20mA
V(across 750R) = (750/ (750+2000)) * 12Volts
V(across 750R) - 3.27V
So if they are paraller with 750ohm resistor they are getting 3.27Volts


(10:07 PM 11-27-2001) RedDevil: Current is limited by 2000ohm reistor
I = E/ R
current = 12/ 2000
current = 6mA perfect
I ignored 750ohm resistor because it is paralleled with LED and LEDs have very small resistance and current always flows through least resistan path.

 

[beenana]

it's good that you're using the voltager divider rules properly, but with the LEDs in there, voltage divider doesn't apply.. semiconductors have a "dynamic" resistance.. it changes with voltage... if you look at a current versus voltage plot of a diode, there's a point where it "turns on" as you increase the voltage...
i know we are dealing with 12V... but you don't have 12V/2000ohms... if the LEDs are 3.27V, you're only getting 8.73V across the 2000 ohms.. Kirchhoffs current law says that the current going in must equal the current going out.. if you say you have 3.27V across the LEDs and 750ohm resistor, divide 3.27V by 750ohms to find out the current through the 750ohm resistor.. what do you get? 4.36mA.... current going into the circuit must be equal to the current going out.... if you multiply 4.36mA by 2000ohms, you get 8.72V... 8.72V + 3.27V = 11.99V, round it to 12V... so your point is proved, right? well no, because Kirchhoff says that the current going in has to equal the current going out.. if you have all your current going through the 750ohm resistor, you have none going through the LEDs.. that can't be... because some of it has to go through the LEDs if you can see that they are on.. you put that 750ohm resistor in there and it sucks the current away from the LEDs.. i would go measure the voltage at the middle wire with respect to the top wire and write it down.. then measure the voltage across the LEDs and 750ohm resistor and write it down.. then do your calculations.. the Laws are there for a reason.. if you want to have 20mA of current running through each LED @ 3.4V(typical operating forward voltage and current), 12-3.4=8.6... 8.6V/100mA= 86ohms.. the LEDs "regulate" their own voltage(that's what diodes do) once they have enough current to turn on.. just make sure you don't make too much current go through them because they do have a maximum current spec, and they will burn out if you go beyond it..
beenana