# LED's with switch on a 12v supply

I hope there is someone who will assist me with this. I have 5 LEDs, with a switch in front of them. I use a 12v supply. Switches can be randomly turned on or off. LED's are 2v, 10 mA.

I can not get my head around how it should be assembled. I am afraid to send too much current through the LED's. Should I put a resistor in front of the LED's?

You can probably read that I am not very good at electronics. But when has it stopped anyone :)

I am very sorry I have written something wrong. There is a switch in front of each LED. So each of them can be switched on and off. Again, I'm sorry, and thanks for the replys.

• "Should I put a resistor in front of the LED's?" Yes, always do that, if you connect 12V directly on a 2 V LED the LED will be destroyed. For each LED use a resistor which needs to drop: 12 V - 2 V = 10 V, 10 V / 10 m A = 1 k ohm. So you need 5 resistors of 1 k ohm (1000 ohms), one in series with each LED. – Bimpelrekkie Mar 29 '16 at 13:11

Series circuit

simulate this circuit – Schematic created using CircuitLab

Figure 1. Five LEDs in series.

LEDs aren't linear (current is not proportional to voltage as in a resistor) and small increases in voltage can cause a large increase in current to the point of destruction. A series resistor will limit the current to a safe value provided voltage doesn't deviate too much.

Since you require 2 V per LED at 10 mA that means we need to drop another 2 V in the series resistor. From Ohm's Law, $V = IR$, we can calculate $R = \frac {V}{I} = \frac {2}{0.01} = 200~\Omega$. 220 Ω is the next highest standard value.

Parallel connection

simulate this circuit

Figure 2. Parallel circuit.

This time each LED requires its own resistor which needs to drop 10 V across it. Using Ohm's Law again we get $R = \frac {V}{I} = \frac {10}{0.01} = 1000~\Omega = 1~k\Omega$.

Use a 220 ohm resistor, assuming you connect the LEDs in series. If they are parallel, use a 1.2Kohm for each. There are several good websites you can use to calculate this, here is the first Yahoo! link: ledcalc.com