I apologize in advance because I am mostly electronic illiterate.

I am trying to build a set of inline on/off switches for remote control of 3 microphones using 3 illuminated push button latching DPDT 2NO2NC switches the LED in the switch is rated up to 24v 15mA. I'll be powering the switch LED with an AC 100-240v to DC 12v 3A power supply adapter what I need to know is what size resistor needs to be run in series from the adapter to each of the switches LED from my calculations I think I need 700ohm 36watt. Is this right if so I dont have room for a 7 in long wire wound resistor much less 3 of them.

Also there will not be power going to the microphone from the power supply. Any help would be appreciated and if this is absurd and would never work then harsh criticism is welcomed thanks in advance

 

A diagram would help.
As you will be running the LEDs on 12V, and it sounds like they may have a resistor inbuilt for 24V, no extra resistor is needed.
There is no way a 700 ohm 36 W resistor is needed on 12V.
Just out of interest, please show us how you calculated that and we may be able to assist you.

The LEDs are polarity sensitive so must be connected the correct way to avoid popping them.

 

Does the LED in the switch already have a resistor, some do?
For a standard LED from 12V at 15mA you would use (12V - 2V)/15mA ≅ 680Ω. The power in the resistor would be RI^2 = 680 * 15mA^2 = 150mW.

 

As @AlbertHall indicates, I think adding a 680R resistor in series is a good idea as you may not know if the LEDs do include the resistors or not. The extra resistor will not hurt, but will save the LEDs if they do not have internal ones. And you can measure the current and work out what internal resistor is there. That could be a good excercise for you to try your Ohms law out on.

 

there are many options

• the LED has no limiting resistor and is AC/DC (double reverse parallel)
• the LED has a limiting resistor and is AC/DC (double reverse parallel)
• the LED has no limiting resistor and is DC (single unipolar)
• the LED has a limiting resistor and is DC (single unipolar)

if your LED is actually specified 3 to 24 V DC (it has the internal resistor , is likely unipolar , and gives medium brightness at 12V)

otherwise (depending on color , power) the led drops usually 2 to 5 V so \(R_{LIM}=\frac{12V-V_{DIODE\ DROP}}{I_{LED_{FW}}}\)

for the 3.2V , 15mA LED it will be \(\frac{12V-3.3V}{15·10^{-3}A}=\frac{8.7}{15}·10^3=\frac{29}5·10^{3-1}≈600\)

... the power dissipating on the resistor is \(P=I·U=I^2·R=\frac{U^2}R=\frac{8.7^2}{600}≈126mW\) the 250mW resistor likely will be enough

 

Welcome to AAC!

from my calculations I think I need 700ohm 36watt.

Show us your calculations so we can see if you're close. The switches I use don't have a resistor in series with the LED.

You've been given the equations. Presented here in a more readable form:
\( R = \frac{V}{I} = \frac{V_{supply}-V_{LED}}{I_{LED}}=\frac{12V-2V}{15mA}=667\Omega \)
The closest 5% value is 680. You'd normally round up in this case.

Power calculation:
\( P = I^2R = 15mA * 15mA * 680\Omega = 153mW \)
Use a 1/4 W resistor.

If the LED is being used as an indicator, you may not need 15mA. It depends on the environment it'll be used in.

 

I used an online ohms law calculator
I first calculated the dc adapter at 12v 3a and got a result 4ohms and 36watts. Then I calculated the switch at 12v 15mA and came up with 800 ohms .18 watt i subtracted the 4 ohms from 800 and I assumed that I needed a 700 ohm acounting for voltage drop and 36watts acounting for the wattage coming from the power supply. And like I said I'm electronic illiterate

But judging by all of the responses I can use the 12v adapter without the need of a resistor.

This is the information given for the switch on amazon.

The UR197 switch is rated up to 5A/30VDC while the LED is rated for 15mA / 24VDC. As long as you don't exceed the rating amps and volts, it can run anything. If you want to use this with a higher voltage, simply add a resistor in series with the LED connection to keep the LED current at around 15mA.

3V~24V LED: There's a built in resistor so that the LED can be connected to 3V~24V. The higher the voltage, the brighter the LED.

They (Amazon) dont give a diagram of the switch other than wiring it to a battery and controlling an accessory.

My diagram is scribbled on paper and it may be useless to you but I'll explain it as best I can.

+ from power supply to switch 1st pole, C terminal

- from power supply to - switch terminal

Jumper wire from + switch terminal to 1st pole NO terminal on switch

Microphone&amp xlr cable pin 3 to switch 2nd pole, NC terminal

And.

Microphone&amp xlr cable pin 2 to 2nd pole C terminal


Desired effect
switch off, mic is off(pin 2 and pin 3 shorted) and switch light is off

Switch on, mic on and light on

Pic of my poorly drawn diagram is attached

 

3V~24V LED: There's a built in resistor so that the LED can be connected to 3V~24V. The higher the voltage, the brighter the LED.

From this you won't need an external resistor and your diagram does what you want.

 

Ok thanks everyone for the help I appreciate it.

 

I first calculated the dc adapter at 12v 3a and got a result 4ohms and 36watts.

This resistance calculation is meaningless. You have an adapter that can provide 12V at 3A (notice the capitalization).

Then I calculated the switch at 12v 15mA and came up with 800 ohms .18 watt i subtracted the 4 ohms from 800 and I assumed that I needed a 700 ohm acounting for voltage drop and 36watts acounting for the wattage coming from the power supply. And like I said I'm electronic illiterate

These calculations aren't applicable.

But judging by all of the responses I can use the 12v adapter without the need of a resistor.

The UR197 switch is rated up to 5A/30VDC while the LED is rated for 15mA / 24VDC. As long as you don't exceed the rating amps and volts, it can run anything. If you want to use this with a higher voltage, simply add a resistor in series with the LED connection to keep the LED current at around 15mA.

The description implies that the LED has a resistor in series that's suitable for up to 24V. At 12V, you're going to get about half the specified current. If the LED isn't bright enough, you can increase the current if you have access to both terminals of the resistor.

 

Your understanding of how a power supply interacts with its loads is incorrect.

For a simple situation such as yours - fixed-output power supply and a steady load - Ohm's Law will get you there with a little understanding. The power supply output voltage is a constant. The supply will try to impress 12 V across whatever is connect to its output. The 3 A rating is the max steady-state current the supply can deliver ***if needed***. The supply does not try to push 3 A through anything that is connected. Load voltage is impressed by the supply, but load current is drawn by the load. This is why house wiring can power 25 W and 150 W light bulbs without any problems or adjustments; each bulb draws the current it needs from a constant-voltage source.

The 4 ohm resistor value you calculated is the lowest resistance, highest power resistor the supply can drive safely. Your circuit needs only 15 mA, less than 1% of that energy level.

ak

 

Thanks dl324 and Analogkid for the clarification on how power supplies work. I always thought a power supply needed to match the appliance spec for spec or it would fry the appliance. As far as ohms law this is the first time I have used it(very poorly). This is also the first electronic device (if it can be called that) of any kind that I have ever made from scratch. I probably wouldn't be building it at all but I couldnt find this device in a red oak case and brushed aluminum face plate. This is going to be used in a court room so the judge can turn on/off individual mics from the bench. I may also install volume controls.

 

Sounds like an interesting project. Keep us posted.

ak

 

I always thought a power supply needed to match the appliance spec for spec or it would fry the appliance.

That's a common misconception from newbies. The voltage needs to be correct, but the current rating can be higher. The device will take what it was designed to take. That's the part newbies don't seem to comprehend.

An analogy is to consider your household wiring. In my area, regular circuits are rated for 120VAC at 15 amps. If you plug in a lamp with a 100W bulb, it'll draw just under an amp; even though the circuit can provide 15 amps before tripping. If you plug a 1000W hair dryer in the same outlet, it'll draw up to 8.3A. Both appliances just use what they need.