I have made a circuit with two (3v) LEDs in series. Connected to a 5v supply from an adapter with usb. (1.5Amp) I belive in series circuit each led gets around 2.5v which was enough to light them. However they both burned out after a short while. With Ohms low it would mean i dont need a resistor in this circuit! But now i think I should need one. Confused as to which resistance to choose and how to place this in this series circuit!?!!!? Please help I know its might be a very basic question!
Red,yellow leds usually need only 1.8-2V to glow depending on the led. so you need to use a resistor in series based on your brightness requirements. this forward voltage needed depends on the led.
for red, I suggest around 20mA current which will give you a resistance of 50ohms.
They are almost certainly not '3V LEDs'. LEDs can operate over a variety of voltages so long as the current through them is kept in check.
A resistor is needed to do that.
Your LEDs most likely have a forward voltage of 2.1V and want around 20mA to be at their brightest. Exceeding 30mA would probably cause them to burn out.
So let's stick them in series with your 5V supply and see what resistor we need:
Ohms law: V = I x R can be rewritten as R = V / I
Our voltage is 5 - 2.1 - 2.1 = 0.8V and if we want maximum brightness our current is 20mA.
So R = 0.8V / 0.02A = 40 Ohms.
Thus you should build (increase R1 to reduce brightness):
An LED should, in most cases, have a device to limit the current.
An LED does not have a single specific forward voltage (Vf). There is min, typ, and max. The there is a curve based on the amount of current flowing through the LED.
This is the I-V curve for a 3.2 Vf, 20 mA LED Cree 5-mm Round LED C512A-WNS/WNN
This I-V curve is made by measuring the Vf as the current is increased. Vf is a function of current. You cannot reverse this to where current is a function of voltage. You cannot apply a voltage and have a predictable current.
So when you connect directly to a voltage source, you will have no idea how much current is flowing. You are forcing the LED to a fixed voltage rather than allowing it to operate at the natural I-V point putting stress on the LED causing it to fail or decrease its lifespan.
When two LEDs are in series the current through each LED must be the same. Driving a string of two 3V LEDs with a voltage source, the source voltage will be divided between the two LEDs. It's not likely the characteristics of both LEDs will ever allow the LEDs reach a point of equilibrium where both can operate harmoniously at the same current.
No two LEDs have the exact same I-V characteristics. LEDs have a dynamic resistance which changes according to the amount of current flowing through it.
When the current starts flowing through the string each LED's dynamic resistance will attempt to adjust to the current. Because each LED has will have a different dynamic resistance for the instantaneous current, the source voltage cannot be split evenly.
As one LED readjusts its dynamic resistance to the instantaneous current the change in resistance cause the current to change. This causes the other LED to have to its dynamic resistance change with the change in current. Again. And again.
The likelihood of two LEDs having I-V characteristics such that their dynamic resistances will evenly divide the source voltage is not statistically a reality.
If the characteristics of the two LEDs are nearly matched the current fluctuations are reduced and may withstand the stress. Their lifespan will be reduced.
The other problem you have is the 5V is not enough voltage to drive two 3V LEDs in series. The supply voltage must be greater than the Vf.
You should drive them separately, each with their own 82Ω resistor.
When the supply voltage is very close to the Vf, you can use an LDO constant current regulator (CCR). It acts as an adjustable resistance. It will dynamically change it's resistance to regulate the current.
This is a 25mA CCR On Semi NSI45025AZ