Resistance effects on LED when connected to battery

Question

I have this doubt whether to go with theoretical method or practical.

Lets say I'm using a 3.7V battery 350mAh to light up a LED which has forward voltage of 3.3V (which means LED lights up if 3.5V is provided) and forward current of 30mA.

So it lasts 350mAh/30mA = 5.83 hours.

The remaining 0.2 volts is reduced using a resistor. So resistor required for LED would be,

(Battery voltage - LED voltage)/(Battery current) = (3.5 - 3.3)/(30mA) = 6.66 ohms

My question is, even though after connecting the resistor to the LED, the intensity of the LED is more than expected. So a higher resistance is preferred here to reduce the intensity of the LED. Suppose say at 50ohm, I have matched my required intensity of light.

But if I put a 50 ohm resistance over there, voltage at the LED reduces to a value where it is less than the forward voltage (3.5v, that is required). So there is a chance LED doesnt light up.

Is this the case or is there any solution for this?

A brief explaination would be very useful, since these minute details are much important everywhere

Answer 1

Why do you think it needs 3.5V? The forward voltage is 3.3. If you use a 50ohm resistor, the current through the LED is (with full battery): (3.7-3.3)/50 = 8mA

That is all you need. The LED should light up fine with 8mA. That is the thing you need to be looking at, the current through the LED. LEDs don't follow Ohms Law so don't worry yourself with voltage drops etc, as a beginner (which it sounds like you are) just worry about forward voltage and current for now.

Answer 2

We approximate that the voltage across the LED is always 3.3V no matter what the current is. If the LED is on, its voltage is 3.3V. End of story. *

The resistor voltage must be 0.2V because the voltage has to add up (Kirchhoff's Voltage Law). But the current will change depending on the resistance.

You also forgot that the battery voltage changes depending on how charged it is. When the battery voltage isn't much higher than the LED voltage, this causes a big change in the current. When the battery voltage is much higher than the LED voltage, the change is relatively small. This is because the resistor has to compensate for the voltage change - if the battery goes from 3.7V to 3.9V, the resistor has twice as much voltage across it, which means the current is twice as much.

* In reality, the voltage across the LED does change depending on the current, but it changes a lot less than the voltage across the resistor does. We pretend it's always the same, because it makes the math much easier, but we must remember this is approximate!

Answer 3

There is no one correct resistor that will work for all supply voltages. This is especially true when your supply is a cell where the voltage drops as the cell discharges. A fully charged LiIon cell gives about 4.2V. By the time the cell is flat, it's as low as 3.0V.

To a rough approximation, LEDs have a constant voltage drop over a wide range of currents. An LED will light at currents considerably below their maximum. So an LED rated for 30mA will still glow at 3mA, and will even glow dimly at 0.3mA.

If we calculate the correct resistance for a 4.2V cell to deliver 30mA to a 3.3V LED, we get (4.2 - 3.3) / 0.03 = 30 ohms. But if the battery drops to 3.7V, the current now becomes (3.7 - 3.3) / 30 = 3.6mA. The LED will still glow, but will be a lot less bright.

If you take apart an LED torch, you will usually find a little circuit board inside, including a microchip and a number of other components. The only way you are going to get a constant LED brightness across a wide range of cell voltages is to use a boost/buck converter feeding a constant current power supply. That's a lot more than just a resistor. The design is a bit easier if your battery voltage is always lower than the LED voltage, or always higher. Most people will use the reference design in the microchip's data sheet.