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Basic LED circuit, 9v battery, what resistors to use? 
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#1
May1112, 03:40 AM

P: 3

Hi all, im a new guy to the forum, glad to meet you all. iv been doing a little rc project and now need to wire up some lights. iv never done circuit work before but im willing to gain knowledge on how to.
the main problem is that no matter how much i read or calculate, i cannot work out what resistors i will need or if my circuit is even possible. i want to run 3x 2 led circuits off one 9v power supply. is this possible? here is a quick sketch of what i was thinkin..any info is much appreciated. thanks 


#2
May1112, 04:44 AM

P: 3

Also, would i be better off using 2 circuits of 3 leds? i would need a red, white and UV on one circuit which would add up to 9v exactly, would i need to run a resistor or not, would it even work? iv drawn that many diagrams im starting to confuse myself so sorry for the 'novice' questions



#3
May1112, 05:14 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

hi alh1987! welcome to pf!
(btw, never reply to your own first post, it takes you off the "No replies" search ) will that damage them?if you're trying to keep the voltage drop across each the same as the rated voltage, you need to put each one in series with a resistor, or to put them all in series with each other and a resistor (I = 9/∑R, V_{1} = IR_{1} etc) 


#4
May1112, 05:45 AM

P: 3

Basic LED circuit, 9v battery, what resistors to use?
Thanks Tim, thats was something i wasnt sure of, thought it was best to check with you guys first. is there any chance u could draw me up a little diagram and list of components please, i just dont get which way to go about it.
should i run a resistor to each LED? 


#5
May1112, 09:04 AM

P: 834

Calculating LED resistors is very straight forward if you help yourself out and use the known variables of the LED. This is assuming you have voltage supply > resistor > LED(s) in series with no current branches between the LED and the series resistor (you can have currents branch from the voltage supply into other pairs like in your circuit, this won't change the calculation).
The three voltages you must use: Supply Voltage  known  you pick this, but realize it must be high enough to drop across a resistor and LED. LED Voltage  known  You kind of pick this too, its the values you have in your drawing, and its given in the datasheet. This is the desired operating voltage for the color LED you chose. It may be different in the end from what you designed it for because you might choose a lower current, and because of the IV relationship of its diode nature, but you don't worry about that. Resistor Voltage  unknown  This is the voltage you don't know initially, and so you must calculate it given the other 2 knowns. Using KVL: Resistor Voltage = Supply Voltage  LED Voltage (if you have more than 1 LED in series, sum their voltages together) Now, you know all 3 voltages, and the only thing left to do is calculate your resistance value. This is determined by choosing how much current you want to put through the LED. You have the maximum current ratings for all the LEDs already found in the datasheet, so you can use those values, but they'll be very bright and on the edge of recommended operating current. You might want to lower the brightness or power consumption, and so choosing 1015mA should be more than bright enough in most uses (you can always reduce your resistor values later to get more current and make it brighter). Your LED datasheet probably has an I vs V curve which is common to other diodes too, and so your minimum current to get the desired LED voltage drop is found there, but you don't need worry about this detail if it doesn't make sense to you; if you pick anything more than 5 mA, you're probably safe. So, let's say we want 10mA through the LED. Applying KCL says that the same current in the LED(s) must pass through the resistor. Now we just use Ohm's law to calculate the resistor value. R = V/I = Resistor Voltage/10mA So that's how you can calculate the resistance value. The only thing left is to make sure that your physical resistor package is appropriate. If you choose a bigger voltage supply value, then you are going to have a larger voltage drop across your resistor for a given LED drop, and so the resistor's power dissipation will increase for a given LED current. Calculate the power the resistor dissipates using R*I^2 or equivalently, Resistor Voltage * I, and make sure the physical resistor package is rated for that much power. 


#6
May1112, 09:18 AM

P: 834

Also, your original circuit can work, you just have the wrong resistor values. For example, you're putting about 100mA through your red LEDs, and that will destroy them and possibly burn up the resistors if they're not rated for enough power dissipation.
Using what I told you in my first post, I'll do the calculation for your red LEDs, and then you can use this example to calculate the resistor values for the other 2 pairs. Voltage Supply = 9V approximate LED drop = 2.5V x 2 = 5V Resistor Voltage = 9V  5V = 4V Let's say we want 10mA in the LEDs, then we can calculate the resistor value: R = 4V/10mA = 400 ohms Now, the last thing to check is if we can find a resistor capable of dissipating the power it drops: P = I^2*R = 0.01 * 0.01 * 400 = 0.04 = 40mW. You can buy resistors rated for this much power, so your circuit can work at 10mA. 


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