Help with powering LED

Discussion in 'The Projects Forum' started by Chachie22, Apr 29, 2015.

  1. Chachie22

    Thread Starter New Member

    Apr 29, 2015
    4
    0
    Hello everyone,

    I'm building a simple circuit which basically consists of three 9v batteries wired in series, a switch, a push-button and a load.

    The idea is I use the switch to turn the unit on, which lights up this push-button: http://www.amazon.ca/gp/product/B008MU3OGW?psc=1&redirect=true&ref_=oh_aui_detailpage_o01_s00

    The push-button is also used to activate the load.

    The issue I'm having is powering the LED light in the push-button. Can I simply draw power from the 3 batteries or do I need a separate power source? I don't know a thing about amps, volts, etc... so I'm not sure if 1 or 9v battery(ies) is enough or too much for the LED light.

    Thanks for the help
     
  2. Reloadron

    Active Member

    Jan 15, 2015
    963
    232
    The switch you linked to is a SPDT (Single Pole Double Throw) momentary push button switch. That means the switch has one Normally Open and one Normally Closed contact with a Common. Just so you are aware it is not like a On / Off type switch and is momentary. While I can see the Common on the side and the N/O and N/C contacts on the end I can find absolutely nothing about the LED in the switch? Nothing about LED forward current or voltage? I can not see any connections for the LED either.Do you have a picture of the LED pinout or LED data as to voltage and current?

    Ron
     
  3. Chachie22

    Thread Starter New Member

    Apr 29, 2015
    4
    0
    Hi Ron, thanks for the reply.

    There are actually 5 pins. 3 are on a microswitch which look like this: https://encrypted-tbn1.gstatic.com/...cKzmG_Yr0oxUTRmzijsWbWGHH_9-A3VRay1hJqU3GAGkT

    The button is a plunger arcade button which is spring loaded and closes the circuit momentarily when pressed.

    There are two more pins, one you can see on one side in the photo and the other is hidden but is on the other side. These two pins feed the LED. If you give the base of the button at the microswitch a quarter turn you can remove the inner workings and see the LED.

    I guess my question is can I use a 9v battery to power 1 LED?

    Thanks again
     
  4. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,802
    832
    The answer to you simple question is simply yes, with qualifications.

    The first qualification is that an LED will require some form of current limiting, which can be a resistor. Some switches have this built in, but the description that you link to does not have this information.

    There are many other questions, and is why Reloadron asked questions regarding the LED current and forward voltage. Also, connecting the LED is sensitive to polarity and is why he asked about the LED pinout.

    Additionally, what switch or type thereof is used to turn the light on. Is it also used to turn off the load so the pushbutton does nothing?

    Click on this thumbnail for an example of how it could be wired. It isn't a practical example, because I can't determine the resistor values and not sure of the external switch wiring.
    Capture.PNG


    So the answer is yes, but the question then becomes how...
     
  5. Reloadron

    Active Member

    Jan 15, 2015
    963
    232
    OK, then pretty much as djstfantasi covers above. Just need to know the LED forward current and voltage. With 2 leads for the LED exiting the switch my guess is when powered the LED will be on all the time. While these things pour in from China they make for great inexpensive components for the home enthusiast or hobbyist to use in projects but... they seem to lack in data sheets. :)
    A real rough guess for the LED being a red LED would be forward voltage 1.2 volts and current around 15 to 20 mA maximum. You may try this. Hopefully the LED terminals are marked + and -. Take a 9 volt battery and we get 9 Volts - 1.2 Volts / 15 mA = 520 Ohms so I would try a 560 Ohm resistor in series with the LED using a 9 V battery and see what you get. That is a pure speculative guess with a few maybe tossed in. :)

    Ron
     
  6. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,777
    1,103
    ... and if the LED doesn't light you have the polarity wrong. If it does light, but is very dim, then there is probably already a current-limiting resistor in series with the LED, in which case the 560 Ohm resistor isn't needed.
     
  7. Chachie22

    Thread Starter New Member

    Apr 29, 2015
    4
    0
    Thanks everyone for you help. I dug around a little online and these LED aracde buttons feed off the 5v+ of the circuit board and are 20mA. Many have build in resistors but some are 12v.

    I guess the safest bet is to try with a 560 Ohm resistor first. If it's dim, it may already have a resistor and try without? If it doesn't light at all with or with the 560, it may be a 12v LED? In which case do I try two 9v batteries with a resistor?

    Thanks again
     
  8. Reloadron

    Active Member

    Jan 15, 2015
    963
    232
    If you found they come in 5V and 12V flavors you can try the resistor with the battery and see what you get. If it is very, very dim try it omitting the resistor with just a single battery. That would lend me to believe it is a 12 volt flavor.

    Ron
     
  9. Chachie22

    Thread Starter New Member

    Apr 29, 2015
    4
    0
    Hey everyone,

    I was able to contact the seller and he indicated the following regarding the LED:

    The forward current is 15Amp, 125Volt
    The LED is for 12V

    So that being said, will a single 9v even light it up or would it be too dim?

    Just to clarify the circuit I'm putting together. There's an on/off switch for the unit. When I switch it on I want the LED in the push-button to turn on. The push-button itself closes another circuit which is powered by three 9v batteries.

    I'm wondering if I can use the 9v batteries to power both circuits, the led on/off indicator and the push-button?

    Thanks again
     
  10. Reloadron

    Active Member

    Jan 15, 2015
    963
    232
    No, that is likely the switch contact rating.

    Let's assume the LED has an internal current limiting resistor and is designed for 12 Volt operation. Try placing just one 9 Volt battery across the LED leads observing polarity. If the LED is in fact designed for 12 Volt operation with an internal limiting resistor then 9 Volts should give a dim glow. Try it and see what you get. I wish the actual true LED forward current were known. If you have a DMM (meter) and can measure current place the DMM set to measure current in series with the LED and see how much current it draws at 9 Volts.

    What exactly will your circuit be? Another consideration is those little 9 Volt batteries will only last so long depending on what you are doing. Can you provide a circuit of where this is all going?

    Ron
     
  11. Retiredguy

    Member

    Feb 24, 2007
    28
    5
    Since your LED is rated for 12V, a 9V battery should make it light, somewhat dimmer that with 12 volts. And in answering your second question, Yes you can power it with the 27V (3X9V ) , however, you must place a resistor in series with the LED to limit the current going to it. You do not want to release the "magic smoke" that makes the LED light. Assuming that your switch LED already has a current limiting resistor in series with it within the switch itself, since it is rated at 12V, you need to determine what value the new resistor needs to be. Three 9v batteries will produce 27v if in series. subtracting 12v ( what the LED is rated) from 27v results in 15 volts. Now using Ohm's law, we must limit the current of this extra 15V to about 20ma or .02 amps. V=IR ( ohms law) or V/I=R So 15V/.02= 750 ohms. So a resistor of 750 to about 800 ohm should work.
     
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