Could it work

[Grim-Reaper]

I'm new doing electronic circuits and I don't know if it's right or I just ? the thing. 

[Unimportant]

@Grim-Reaper Lighting 2 LED's, each with their own series resistor, from a 9V supply ? Yes, that layout is fine.

 

However, you need to take into account what kind of LED's you're going to be using. Is the forward voltage really 3V ? If it's lower then 3V then there'll be more then 20mA with those resistor values. For your average 5mm LED, 20mA is pushing the maximum, typically one aims more toward 10mA, which will create almost the same amount of light but making the LED last much longer.

[mariushm]

You have the basic formula :  V = I x R   voltage = current x resistance

 

 

So (Voltage battery - forward voltage of led )  = Current x Resistance

 

You have 9v  and the forward voltage of the green led let's say it's 2v and the current is 0.02A (20mA) ... then 9v - 2v = 0.02 x R  therefore your R is 7 / 0.02 = 350 ohm , and you can round that down or up to a more standard value like 330 ohm or 370 ohm

 

Your circuit has 300 ohm, because probably assumes that the led has a forward voltage of 3v (which is more common for white or blue leds). The lower the resistance the more current will flow through the resistor and through the led, so if your green led has lower forward voltage then more current will flow through it. LEDs aren't that sensitive, but you also shouldn't push 20mA through the leds because most likely the amount of light produced at 20mA will not be much higher than let's say 15mA - modern leds are quite efficient these days.

 

A LED is a current driven device, it has a voltage from where it (in a very simplified way) turns on and produces light as long as there's some current flowing through it. If the voltage goes too high, it burns out. If the voltage goes too low, it stops emitting light. If too much current flows through it, the led overheats and dies.

So you want to limit the current going through the led, while keeping the voltage at least a bit above the forward voltage of the led.

If the voltage is higher than the forward voltage, you need to put a component before the led to "drop" some voltage on it, so that the led will only see some voltage a bit above the forward voltage. A basic component that would do this and also limit the current is a resistor.

 

 

 

 

[Grim-Reaper]

So. I have to use between 20mA and 10mA to make the led last longer

[mariushm]

No, use as much as you want and as much as it's needed and as much as the led supports according to its datasheet.

 

For example, the intensity of a led will depend on the lens on the led (how much it spreads the light around, the view angle).. there are some leds that have very wide angle, like let's say 120 degrees, which are good for some things like panel indicators, because the light spreads nicely around. There's also leds which have a very narrow view angle, they have a lens or something that concentrates all the light like a laser pointer, and such leds could be used for example with light pipes (to direct light from led through a fiber optic cable or some plastic tube to some place on a weird shape)

Because the light is so concentrated in a narrow angle, you may need much less current to produce the same amount of visible light.

 

Then, the efficiency will vary with the color you choose and what you actually need to do with the led. For example, red leds are the most efficient so if you need a led just as a sort of power on indicator for a product , even 1mA of current could be enough to get the led bright enough to be visible when the device is outside in the sun or something like that.

You may find some High Brightness Red leds that light up enough even at 0.2 mA

 

That value of 20mA is more or less a value that companies came up with, which basically says "if you put 20mA through the led in this particular series we sell, all the leds in this box are binned so that they will produce this much light give or take 5-10% and the main color emitted will be this many nm give or take 5 nm or something like that"

 

For the 3.5mm and 5mm through hole leds, 20 mA is a safe value, the amount of heat that's generated by the led die inside the led is small enough that it can radiate through the case of the led and the leads of the led and the circuit board, so the led will basically never burn out by itself. This is what you'll often find in datasheets, listed as test current.

You CAN use more current than 20mA and often datasheets specify this with phrases like "100mA for 5 ms, then off for at least 20ms"  - this simply says that if needed the led can be driven with way more current to produce more light but in those 5ms the led die inside get so hot and the leads of the led get hot and the case of the led gets hot so unless you turn it off for at least 20ms to allow for everything to cool down (the leads of the led behave like a heatsink, like a radiator), the led may actually burn up due to too much heat.

 

It's not necessarily the best value to use. The led is not the brightest or the most efficient at 20mA or whatever value they say in datasheet. If you really measure, you may find that if the led is 100% bright at 20mA, it may be 95% bright at 15mA and maybe 80% bright at 10mA ... so you may prefer to use just 10mA to extend battery life, because just 80% of the light would still be enough for you.

 

 

One more thing you should be aware of. Your circuit, as you designed it, it's very inefficient. Unless you need those two leds to be turned on and off independently, you can get a much higher efficiency out of the circuit by placing the two leds in series and use a single resistor to limit the current going through both leds. This is possible because the battery voltage is much higher than the voltage drop of each led.

 

In your example, you have a 3v led and 6v dropped across the resistor, and the current is 20mA (0.02A) ... so you have 3v x 0.02A = 0.06w consumed by the led, and you have P = I²xR = 0.02 x 0.02 x 300 = 0.12w consumed by the resistor (as heat in the resistor) ... so basically each pair of your resistor + led consumes  0.18w  but only 0.06w are used to produce light (33%)

 

If you put your two leds in series and use only one resistor to limit the current, then you have

 

9v --- [ resistor ] ---- + led -  + led -  ---- 9v ground

V = I x R

(V battery - 2 x forward voltage led ) =  0.02A x R   =>  R =  (9v - 2 x 3v ) / 0.02A  = 3/0.02 = 150 ohm 

So now you're going to have 2 x 0.06w = 0.12w in the leds, as light and heat, and your resistor will only dissipate P = IxIxR = 0.02*0.02*150 = 0.06w

 

Your circuit is not twice as efficient, just by placing the two leds in series and using a different resistor value.