Really EASY electronics help needed
- Thread starter armisiano
- Start date
Your LEDs should come with a data sheet or you should be able to find one online at the manufacturer's site. Last time I put a circuit for an LED together I had to put some resistance in the circuit between the battery and LED in order to keep from destroying the component. Also, they operated at 5 VDC so we probably would have put a zener diode in the circuit and connected multiple LEDs in parallel. Too many technical terms I know, so try this: google 'led circuit' and you can find some circuit building tutorials.
Dick Campbell
Veteran
What are you trying to do? Do you just want two LEDs to be lit? If so, connect the (+) of one LED to the (-) of the other, then connect the other leads across a 9V battery. If they don't light up, reverse the battery connection. J Michael is right, LEDs are 5V. By connecting them in series, you will drop 4.5V across each LED.
Doc Bernard
Indie Arms Member
This is the formula you need to calc LED resistor requirements
This is the formula you need to calc LED resistor requirements
R = (VS - VL) / I
VS = supply voltage
VL = LED voltage (usually 2V, but 4V for blue and white LEDs)
I = LED current (e.g. 20mA), this must be less than the maximum permitted
This is the variation on the formula for multiple LED's:
A red, a yellow and a green LED in series need a supply voltage of at least 3 × 2V + 2V = 8V, so a 9V battery would be ideal.
VL = 2V + 2V + 2V = 6V (the three LED voltages added up).
If the supply voltage VS is 9V and the current I must be 15mA = 0.015A,
Resistor R = (VS - VL) / I = (9 - 6) / 0.015 = 3 / 0.015 = 200,
so choose R = 220 (the nearest standard value which is greater).
This is the formula you need to calc LED resistor requirements
R = (VS - VL) / I
VS = supply voltage
VL = LED voltage (usually 2V, but 4V for blue and white LEDs)
I = LED current (e.g. 20mA), this must be less than the maximum permitted
This is the variation on the formula for multiple LED's:
A red, a yellow and a green LED in series need a supply voltage of at least 3 × 2V + 2V = 8V, so a 9V battery would be ideal.
VL = 2V + 2V + 2V = 6V (the three LED voltages added up).
If the supply voltage VS is 9V and the current I must be 15mA = 0.015A,
Resistor R = (VS - VL) / I = (9 - 6) / 0.015 = 3 / 0.015 = 200,
so choose R = 220 (the nearest standard value which is greater).
BobDiaz
The Professor
Sorry, to say, that you will blow both LEDs if you do that.
Unlike light bulbs, LEDs do require a current limiting resistor. In general, most LEDs run at 10mA (0.010 A) to 20mA (0.020 A) maximum current.
Let's assume that you want both LEDs to light at near maximum, we'll use the 10mA limit.
Red, Blue, Green, Yellow, and Orange LEDs have about a 2 V drop across each LED, so: 2 Volts x 2 LEDs = 4 V
9V - 4V = 5V (This is the voltage that will be across the current limiting resistor.)
5V / 0.010A = 500 ohms minimum (You can always use a slightly higher resistor and the LEDs will still light, but will not be as bright ...)
IF the LEDs are Blue or White, the voltage drop across each LED will be 4V. Thus, 4V x 2 = 8V
9V - 8V = 1V across the current limiting resistor.
1V / 0.010A = 100 ohms minimum
Bob Diaz
Dick Campbell
Veteran
I've connected single LEDs across 3V battery packs and never blown one. However, I agree your method a current limiter is more robust.
Doc Bernard
Indie Arms Member
3v will reduce LED life (given that it is a color other than blue or white), but will not immediately blow them. Take that same LED and place it across a 9v battery, and you will get an immediate flash then nothing. The formula I gave and Bob re-iterated is the way to do this. Electronics is an exact science, if you want it to work right.
BobDiaz
The Professor
Depending on the LED you can connect it to a voltage higher than 2 volts and it will work without blowing right away, BUT there is some risk. Not all LEDs have the same current - voltage drop curve. Here are 4 charts:
In the case of the Red and Yellow LEDs, 3V works (@80+ mA), BUT it stresses the LED, so it won't come close to running 50,000 to 100,000 hours. It might even die after an hour or so...
In the case of the Green LED, you can go just a little beyond 3.2 volts and still stay within the 20mA limit.
In the case of the Blue LED, you can go up to about 3.3V and still be within the 20mA limit.
These charts are for 5mm LEDs (OVLFx3C7 Series) made by OPTEK Technology Inc. Other LEDs made by other companies will have a different curve, but the general idea will still hold.
Bob Diaz
Jim Brennan
VHS Member Old Skool
Charts, graphs and Big Brains....
I love this site.
I love this site.
armisiano
Veteran
I've finished the prop/costume piece I needed this info to build. You can read about it and see some cool pics here: http://www.dvxuser.com/V6/showthread.php?t=148081
BobDiaz
The Professor
Looking at the photo:
http://www.dvxuser.com/V6/showpost.php?p=1427408&postcount=6
I see your resistor is 150 ohms, +/- 5% and it feeds 3 white LEDs in parallel.
9 volts / 150 ohms = 0.03333... = 33.3mA through the resistor.
The current must be split through the 3 LEDs, so each LED gets about 1/3 or 33.3mA = 11.1mA
The current will not be exactly 1/3 of 33.3mA, because each LED is a little bit different. However, for what your are doing, I don't think the difference matters.
Bob Diaz
armisiano
Veteran
In the future Bob, I may actually end up just begging you to wire this stuff for me, lol, seriously. You're electronically brilliant, and I honestly bow before your knowledge.
