Gave it a go but ultimately I'm too thick. - Page 1
Forum section: electronics / beginners

author: nx084f001
author_rank: Newbie
timestamp: April 24, 2013, 06:30:53 pm
content: Hi all,I'm trying to hack together a rudimentary circuit to drive a high power LED. I have a mere basic understanding of electronics and electrical theory and I am turning to you brighter chaps for support with my calculations as I don't want to switch my circuit on and watch it smoke! These high power LEDs ain't cheep...I waffle a bit to begin but only because I want to demonstrate that I've tried and not just posting on here for a quick answer without giving it a go first!I hacked apart an old 300W computer PSU putting a dummy load between the PS_ON and GND wires. I've done this so that I can use the 5v/12v rails.My LED has a Vf of 3 to 3.7V and forward current of 700 to 750mA. I would like it to take 730mA (to be persnickety!) but because I expect to use a number of these at a later date I have opted to use the 12V rail and figure therefore that I'll need a substantial resistor in the interim.On the back of some paper my circuit is simple: Vcc --- LED --- R --- GND.To my understanding KVL states that the sum of all voltages across components is equal to Vcc. I'll use 3V as my LEDs Vf, then the voltage drop across the resistor is 12 - 3 = 9V. I'd like to operate the LED at 730mA, so the wattage of the resistor is going to be 9 * 0.73 = 6.57W, but doubling up as per convention means I'm going to need a 12 to 15W resistor. AFAIK its value will need to be (12 - 3) / 0.73 = 12.32R, ~13R. So a 13R 12W resistor... a ceramic one comes to mind. And I have one, but at 10R 20W.The problem:LEDs are sensitive as you well know and so I need to determine what effect this resistor value is going to have in terms of both the new voltage across the LED (I am assuming that as the resistor I have is a lower value then what I require based on the previous calculation I'm guessing it will not sink the voltage I require it to and therefore the voltage drop over the LED will increase?) and the current through the LED (note, my DMM says the resistor is 10.14 ohms)Am I correct in calculating the current through the LED with the following: (12- 3) / 10.14 = 0.888A, or ~890mA ?I don't know how to determine the voltage, how would I do this?General questions:Is it more appropriate to place the resistor before or after an LED, e.g. Vcc --- R --- LED --- GND or Vcc --- LED --- R --- GND?Is the shunt resistor described in the tutorial (here:http://www.electronicspoint.com/current-divider-circuits-t222484.html) "Current Divider Circuits" more appropriate for what I'm trying to do?ThanksQF.

author: c4757p
author_rank: Super Contributor
timestamp: April 24, 2013, 06:39:43 pm
content: If you're just going to use a resistor to limit the current (I'd use a buck converter, but that's probably a bit advanced at this point), why not run it off the 5V rail? I promise the 5V rail on that power supply can give an amp.

author: c4757p
author_rank: Super Contributor
timestamp: April 24, 2013, 06:56:54 pm
content: You still have to use a resistor to drop the voltage, but it will wastesignificantlyless power.

author: IanB
author_rank: Super Contributor
timestamp: April 24, 2013, 07:07:51 pm
content: If you assume the Vf of the LED is about 3 V, then with a 10 ohm resistor and a 12 V supply the current will indeed be about (12 - 3) / 10 = ~900 mA.Note that you can do experiments to confirm your numbers. For instance, do you have two 10 ohm resistors? Put them in series to give about 450 mA and then measure the voltage drop across the LED. This will give you a more accurate estimate of Vf.Your resistors will get hot and waste a lot of power, but this need not bother you for bench test purposes.Experimenting always complements calculations. A good rule of experimenting is to start small and safe, and work up, measuring as you go. So begin with a high resistance for lower currents and gradually increase the current in steps.

author: JackOfVA
author_rank: Unknown
timestamp: April 24, 2013, 07:49:54 pm
content: If the data sheet for your LED has a plot of Vf versus If you can solve the equation graphically for a given resistor and voltage. That is, given the applied voltage (12V) and the series resistor, and the plot of Vf versus If, you can determine the LED current.However, that's going to be a static value and the LED Vf versus If will change with temperature to some degree.Without getting into something complicated like a constant current drive, I would design so that in the worst case the LED operates within the safe area and take some degradation in the "average" operating case.I think you have done that to a reasonable degree already, so you are on the right track.If you only have a 10 ohm resistor and need 730 mA, and if the LED has a 3V drop at that current, but you don't have a 3 ohm resistor to add, perhaps you have some standard 1A silicon diodes, such as 1N400x type. Those will have around 0.8V or a bit more drop at 730 mA. If you series two 1N4001 (or any of the 1N4001-4007 series) diodes with the LED (all arranged for forward conduction, of course) you will have a net voltage drop across the diodes of 3 + 0.8 + 0.8 = 4.6V. With a 12V source, the 10 ohm resistor will see 12-4.6 = 7.4V and the total current will thus be 740 mA. Resistor dissipation will be a bit over 5 watts and the 1N400x diodes will each dissipate about 0.6 watts. That's OK for the 1N400x diodes, but they will get hot. ( I say that without pulling the 1N400x series data sheet, but it should be within the maximum current and dissipation rating.)But, I suspect the LED will have a bit more than 3V drop which is not a bad thing as it will drop the current, not increase it.

author: IanB
author_rank: Super Contributor
timestamp: April 24, 2013, 07:51:33 pm
content: Quote from: qwertyface on April 24, 2013, 06:30:53 pmIt doesn't matter whether you put the resistor before or after--the result is the same.Shunt resistors decrease resistance and increase the total current in the circuit. This is not what you want here unless you want to use the LED as a firework

author: sleemanj
author_rank: Super Contributor
timestamp: April 25, 2013, 07:17:26 am
content: Quote from: qwertyface on April 24, 2013, 06:30:53 pmThis is where you're going wrong and confusing yourself. For all intents and purposes you can consider the voltage drop across the diode will stay the same no matter what, it will ALWAYS (try to) drop the "same" voltage (somewhere within that 0.7v range) [ok hand waving away messy details, but that's not important].If you change the resistance of your limiting resistor it will change the current, but the voltage drops across diode and resistor will still remain as they were.Now work your calculations based on the voltage of the diode being fixed at say an average of 3.5v12v - 3.5v = 8.5v drop across your resistordoesn't matter what value resistor it is, it will by necessity drop 8.5v.So knowing that, simply applying ohms law on the resistor you can see the current in the circuit (and by extension the power dissipation especially in the resistor), it's this current which is the important factor to the LED.So with a 10R (convenient for maths) we get 0.85A at a 3.5v drop on the diode.But that diode could "want" to drop anywhere from 3.0 to 3.7, so we better look at the extremes too.At 3.0v we get 0.9A, and at 3.7v we get 0.83AClearly, 10R is going to make your LED very bright, for a brief period of time, and then very dark