Hooking up a power LED

Community General Discussions
1

Hi Guys,

I installed a powerswitch on my bot that runs of an 11.1v lipo. Now I would like to hook up a LED that switches on with the power. Can I just connect a LED with a resistor to the switch (if so what value should I use)? Or do I need to use a voltage regulator to reduce the power?

It's roaming by now, I'll post some pictures etc. later this week. Did some serious liposuction on the thing and removed near to 7 kilo's. It's now small enough to run arround the house without moving the furniture or cause some serious damage to the kids :)

2

"Can I just connect a LED

"Can I just connect a LED with a resistor to the switch (if so what value should I use)?"

you can, a 600R resistor should do

3

Voltage does not kill LEDs

Current does.

Use the usual calculations to reduce that current to acceptable levels. Search LMR for “led blow glow”.

However, you are right to be suspicious. The usual series resistor needs to resist a 9.9 V tension to a 30 mA current (let’s say those are your numbers). This means the 330 Ohm resistor is dissipating a hot 300 mW. That is one expensive little power indicator!

Alternative might be to use much higher resistance. Feed that current to the LED through a little transistor tht is powered from a separate source. For example two AA batteries or a voltage regulator.

4

Wait a sec…

You are not running any kinda robot brains on 11 volts are you? Just run it off of the volt regulator.

-or-

Shove 11.2 volts into your pic and at that point the on/off light won’t matter.

5

It’s twice the votage

It’s twice the votage recommended and in some cases 3 times so your proc should run that much faster!! AMIRITE?!?!?   :smiley:

j/k, don’t run a proc with that voltage.

 

6

You can use a simple

You can use a simple resistor, calculation goes here: http://led.linear1.org/1led.wiz

Or you calculate it by hand:

Operating voltage UO = 11.1 V

For 3mm and 5mm LEDs, the diode forward current ID is usually 20 mA or close to it

The diode forward voltage UD depends on the color, let’s say, it is red, then we have 2.2 V

UR = UO - UD =11.1 V - 2.2 V = 8.9 V

So we have to eliminate 8.9 V

By Ohm’s law the resistor for the LED RLED is then:

RLED = UR / ID = 8.9 V / 0.02 A = 445 Ω

So we use a 470 Ω resistor

The current is then I = 8.9 V / 470 Ω/ ≈ 0.01894 A = 18.94 mA

The power dissipation P is simply UR · I, thus 8.9 V · 18.94 mA = 168.566 mW

 

 

7

Thanks!

Thanks guys, it works by now! Great post on the calculation!

I’m using the robot control board from parallax, you can hook up any power source up to 24 volts (well depending on the motor you use, they have to match the power source). I had the bot roaming for few hours on a fully charged lipo (11.1v 2500mah) and it uses very little power so far.