# Connecting nine LEDs in series.. why the results? [duplicate]

I asked this before but I'm having problems still. I need to connect nine LEDs in parallel with a 9 volt battery. The wire leading away from the 9V separates into three and on each three are three LEDs in series. A calculator told me that if I have 2.25 volt, 28 mA LEDs on each branch I need an 82 ohm resistor. If I have an 82 ohm resistor and 9/82 is .10976... wouldn't my current be too low? I just don't know how to do it. How do I decide which resistor I need to use with an LED?

• Each string of 3 needs it's own resistor. The voltage drops across each of the 3 LEDs in a string adds up: 3x 2.25V. The current through the string remains 28mA. From here, follow formula in this thread - "Correct formula for LED current-limiting resistor" – Nick Alexeev Jan 9 '14 at 0:49

In order to decide the current of the led you need to know the max current specs and select a value below that.
In order to calculate a current limiting resistor for a led you need to know the forward voltage drop of the led and the intended current.

When you know all these you use the following equation

$$R= \frac {V_{in}-V_f}{I_{led}}$$

Where Vin is the supply voltage, Vf is the led forward voltage drop and Iled is the intended led current. When you connect multiple leds in series then Vf will be the sum of the forward voltage drop of all connected leds.

For your specific example
Vf=2.25V (for three leds in series 2.25v*3)
Iled=28 mA

$$R= \frac {V_{in}-V_f}{I_{led}} = \frac {9V-(2.25V*3)}{0.028A} = 80.36 Ohm$$

If you round the resistor to 82 Ohm

$$I_{led}= \frac {V_{in}-V_f}{R} = \frac {9V-(2.25V*3)}{82} = 0.027A$$

simulate this circuit – Schematic created using CircuitLab