Beginners Questions about LED's and DPST switches

[rewonderman]

Hello everyone, this is my first post and I am an electronics newbie.

I have a midi controller hooked up to 16 switches. The switches are double pole single throw. I am interested in using the second pole to wire an LED to every switch indicating when it is on.

My first question is: Are DP switches built in such a way that there is no chance (through wear and slippage or just cheap design) of the two poles interacting with each other?

The reason I ask is because the pole of the switch currently in use requires very little power and is probably rather sensitive. I assume that additional power accidentally being sent from the new led circuit might damage the midi controller portion of the device. I hope that is clear enough to understand.

Second question:

Is wiring the LED to the switch as simple as closing the circuit in line with an appropriate battery? There will be 16-18 leds each attached to a switch. Between 1 and 3 will ever be active at one time.

Please forgive my laymans terminology in this. Any help would be greatly appreciated.

Thanks,
Alex

 

[Antlion]

Most multipole switches that I've seen are failsafe. The worst that will happen is they won't switch at all, it is nearly physically impossible for cross-connections to occur. Frequently the entire purpose of a multipole switch is to switch a high and low voltage at the same time; they can be considered safe for this unless your fault tolerance is really zero, e.g. if you're switching the 110V to the power supply of a device on at the same time as you're switching on a 5V logic signal it is not a good idea to use a single switch, but if you're talking about switching two low-voltage circuits then your tolerances are probably wider. Depends on the circumstances, but in yours I doubt that the current from the LED could damage the MIDI controller - the reverse is actually more likely.

LEDs are very current-sensitive, in fact, and how you connect them is critical. The worst thing that can happen is you'll blow up an LED, which isn't a catastrophic event, so you have some leeway to experiment. I don't know your experience level so if you don't know anything about Ohm's Law or resistance you may want to ask somebody for a crash course. Generally speaking "indicator" LEDs, the low-power red, green and yellow ones you can look at directly, require about 10 milliamperes of current. If they go above that by a couple mA they explode and if they go below that they don't operate. Since you will have varying numbers of LEDs lit at any given time you will need to attach a resistor to each LED individually. The resistors value is based entirely on what voltage you're running your LED on.

The best source for power for your LEDs is a good DC supply. What MIDI controller are you using? Is it one of those do-it-yourself MIDI controller kit boards that has a row of contacts that you just plug switches into? If so, I'm sure it has its own power supply, probably 12, 9 or 6 volts. The best approach to this would be to splice into that power supply and use it to power your LEDs.

As I said, the voltage influences what value resistors you use substantially. If you know your supply voltage you can determine what resistor value you need with a simple formula. It's known as Ohm's Law, but if you're not quite up on your theory yet I'll simplify it further. Your LEDs probably want about 1.5 volts at about 10mA. Take your supply voltage, subtract 1.5 volts, divide by 0.01. The result is the resistance you need, measured in Ohms, the symbol for which is Ω. I calculated the values you'd need for some common voltages:

1050Ω 12V
750Ω 9V
450Ω 6V
350Ω 5V

So if you want to just skip to the chase, here's your circuit. For each LED, connect the positive voltage from your power supply to the center of one side of the appropriate switch, connect a resistor to the correct contact (normally closed or NC, I expect), connect the other lead of the resistor to the positive leg of your LED (the long one), then connect the negative lead (the short one) to the other lead from the power supply. If your LED blows up, try a resistor with a higher value, if it's too dim try recalculating the resistor values for 20mA (same formula but divide by 0.02 at the end), and if it doesn't work at all then you have the LED backwards or I need more information.

Good luck!

 

[rewonderman]

Hi,
And thank you for the detailed reply. I am actually attaching switches to replace the buttons on this device: ww.sweetwater.com/store/detail/nanoKONTROL

It is strictly powered by usb and I was hoping I could power the LED's via an external battery (a 9v or something small like that) instead of figuring out how to lead power from the board of the device.

What I read of your post seems rather complex for me. Maybe theres a pre-calculated combination of the said components? I also saw LED's with built in resistors.

Basically I am looking for a cookbook formula that I can easily apply to 18 blue LED's.

I deeply appreciate your response. I will soon take a picture of the PCB as I have some other questions about what I am seeing on there.

Best,
Alex

 

[sirkituk]

Blue led needs about 3 to 3.5 volts. Assume 10mA. On 9V supply from battery you need to drop just under 6V at 10mA. Ohms law says R = V/I. So 6/0.01 = 600 ohm. The nearest value you can get easily is 560 ohm (470 ohm is suitable too, just a bit brighter)