I want an indicator lamp to show when my garage door is open. I have a rail mounted magnetic switch rated at 30v max. I want to use my 16vac output door bell transformer for the power supply. I have a few questions:

1. Some LED indicator lamps are rated for VDC and others for AC/DC. I presume I need an AC/DC rated lamp?
2. I can find 24v AC/DC lamps and 12v AC/DC lamps. Will the 24v lamps likely be too dim? Would it be better to use a 12v lamp with a resistor? If better to use the resistor, what size?
3. Given that this is being wired into an AC circuit, does the polarity of the installation of the indicator lamp matter?

Thank you!

 

Welcome to AAC!
The "16V" AC transformer will give that voltage when passing its rated current (bell current). If your lamp draws less current then the voltage will almost certainly be somewhat higher, say 18V (rms).
Given that an indicator LED drops ~ 1.8V - 3V (depending on its colour), that leaves at least about 15V rms to be dropped across a current-limiting resistor in series with the LED. Commercial LED-based indicators sold for use with a specific supply voltage, e.g. 12V, incorporate the resistor and, if rated for AC also incorporate another diode in inverse-parallel with the main LED to prevent damaging reverse voltage appearing across the main LED. This other diode may well be another LED.
You can use a 12V rated indicator with an additional resistor to drop from 18V to 12V, the resistance being chosen according to the desired indicator current. Can you link to a 12V indicator of your choice? An alternative is a DIY indicator consisting of a resistor in series with two LEDs in anti-parallel.

 

If they are high brightness types, then the 24Vac rated would likely be bright enough.
You don't need much brightness for an indicator unless it's in bright light or sunlight.

 

Welcome to AAC!

Thank you!

Can you link to a 12V indicator of your choice?

http://amzn.to/2uiUdtA is a 12v version which doesn't specify AC or DC. So do I presume DC only?
http://amzn.to/2sFH8IM is a 12v ac version, with a bigger diameter
http://amzn.to/2szr8gz is a 24v AC/DC version.

FWIW, I'm looking to power 2 LEDs at the same time. I have two garage doors and I want to have a red lamp light when a door is open and a green lamp light when a door is closed, so at any given moment there will be two lamps lit. The garage door track switches have both a normally open and a normally closed loop for me to connect to.

The "16V" AC transformer will give that voltage when passing its rated current (bell current). If your lamp draws less current then the voltage will almost certainly be somewhat higher, say 18V (rms).
Given that an indicator LED drops ~ 1.8V - 3V (depending on its colour), that leaves at least about 15V rms to be dropped across a current-limiting resistor in series with the LED. Commercial LED-based indicators sold for use with a specific supply voltage, e.g. 12V, incorporate the resistor and, if rated for AC also incorporate another diode in inverse-parallel with the main LED to prevent damaging reverse voltage appearing across the main LED. This other diode may well be another LED.
You can use a 12V rated indicator with an additional resistor to drop from 18V to 12V, the resistance being chosen according to the desired indicator current.

Ok, I am definitely showing my ignorance of electricity here, which I hope you'll pardon...I thought that voltage was a constant, as is the resistance of a load, with the current draw being the dependent variable in the relationship V=IR. So, how does a lamp draw "less current" when wired in a circuit with a given voltage?

I'm thinking that my problem is that I have very little understanding of how an LED works! Since most LEDs are DC, then perhaps what is going on with the 24v AC/DC LED is that it has rectifying circuitry inside that converts to DC?

 

Since most LEDs are DC, then perhaps what is going on with the 24v AC/DC LED is that it has rectifying circuitry inside that converts to DC?

Yes.
It doesn't have to be smooth DC, just current that goes in only one direction, whether smooth, pulsed, or a half sinewave.

This can be provided by a adding diode in series with the LED.

Alternately you can connect two LEDs in inverse parallel.
For that connection, one LED conducts for one polarity (half-cycle) of the AC and the other conducts for the other half-cycle.

 

LED 1 is red, 12V DC, and would need a series diode plus a resistor (~330Ω).
LED 2 is 12V AC/DC and either green or blue depending on which bit of the ad you read! Use with resistor (~330Ω).
LED 3 is 24V AC/DC and either green or blue depending on which bit of the ad you read! Usable without added resistor or diode.

 

LED 1 is red, 12V DC, and would need a series diode plus a resistor (~330Ω).
LED 2 is 12V AC/DC and either green or blue depending on which bit of the ad you read! Use with resistor (~330Ω).
LED 3 is 24V AC/DC and either green or blue depending on which bit of the ad you read! Usable without added resistor or diode.

Thanks much. Is the 330Ω if two lamps will always be lit at the same time? And, can a single resistor be wired on one side of the circuit from the transformer to the switches or should there be separate 330Ω resistors at each of the 4 lamps?

To be more specific, I was intending to wire one side of the transformer to one side of each of two magnetic SPDT switches (one NC one NO so two circuit loops–one for each garage door–will always be closed), with the other sides of each of the two switches running to one side of each of 4 LEDs, with the other side of the LEDs tied together going back to the second side of the transformer. The question is, where should the resistor(s) go in this arrangement?

 

Post a schematic of how you plan to wire all the lamps, switches, and transformer together, so that there's no misinterpretation of your scheme, and we'll see if lamps can share resistors (though at $0.01 each it's hardly worthwhile ).

 

OK, here's my proposed schematic. 1st time using an online schematic drawing app...cool!

I wasn't trying to augment my retirement savings by using only one resistor...just trying to save a little time in the install!

I was thinking of putting the resistor between one side of the 16vac transformer secondary and the SPDT switches. Will that work?

 

I was thinking of putting the resistor between one side of the 16vac transformer secondary and the SPDT switches. Will that work?

That won't work because the two color LEDs have distinctly different ON voltages so the one with the lower voltage will hog most of the current.

You could use two resistors, one in series with each wiper of the switches.

You also need a diode in series with the transformer to avoid reverse biasing the LEDs (they can only tolerate a few volts reverse bias).
But that may cause a noticeable flicker at 50/60Hz since the diode only conducts for each half cycle.
Better would be a bridge rectifier, which allows conduction for both half cycles and doubles the flicker frequency.
The bridge input is connected the the transformer and the bridge output goes to your circuit.

So I recommend a resistors in series with each switch wiper and a bridge rectifier at the transformer output.

 

I agree with having a respective resistor in series with each switch common (wiper), to drop the volts down to 12V, providing the red and green indicators draw the same current and providing each indicator is rated for 12V. If the indicators are AC-rated then the bridge rectifier would be unnecessary unless the indicators only use internal half-wave rectification to make them 'AC'-compatible.

 

That won't work because the two color LEDs have distinctly different ON voltages so the one with the lower voltage will hog most of the current.

You could use two resistors, one in series with each wiper of the switches.

I presume you mean a circuit like this:

I don't see how this circuit would take care of the problem of different colored LEDs drawing different currents. Since either garage door could be open or closed, there are four lamp-on states: RR,RG,GR,GG. If the red and green LEDs draw different currents, then wouldn't the issue with "current hogging" occur in either the RG or GR situations? What am I missing here?

Assuming, for the sake of an example, that red LEDs have lesser voltage than green ones, wouldn't I need to have a circuit that looks like this:

Then R1 and R2 serve the role of reducing the 16vac voltage to 12vac, and R3 and R4 serve to equalize the current draw of the red LEDs to the green ones. (Though I don't see why R1 and R2 have to be separate resistors, since they both go to the wiper terminal of the switches...why wouldn't one resistor here be enough?)

Short of calling the manufacturer (if that's even possible), how do check to see if the red and green indicator lamps have internal circuitry to equalize the voltages? Can I just connect them in parallel to the transformer secondary and measure the voltage across each LED to see if it's the same?

You also need a diode in series with the transformer to avoid reverse biasing the LEDs (they can only tolerate a few volts reverse bias).
But that may cause a noticeable flicker at 50/60Hz since the diode only conducts for each half cycle.
Better would be a bridge rectifier, which allows conduction for both half cycles and doubles the flicker frequency.
The bridge input is connected the the transformer and the bridge output goes to your circuit.

If I understand the purpose of the rectifier, this is only needed if the LEDs are NOT rated AC/DC. Correct?

 

I agree with having a respective resistor in series with each switch common (wiper), to drop the volts down to 12V, providing the red and green indicators draw the same current and providing each indicator is rated for 12V. If the indicators are AC-rated then the bridge rectifier would be unnecessary unless the indicators only use internal half-wave rectification to make them 'AC'-compatible.

How do I calculate the resistance necessary to reduce the voltage from 16vac to 12vac?

Is there any way to figure out, with a preassembled indicator lamp assembly, whether the LED. is using half- or full-wave rectification?

 

That won't work because the two color LEDs have distinctly different ON voltages so the one with the lower voltage will hog most of the current.

As an aside, does this explain the phenomenon I've observed with strings of LED Xmas lights, that different colored lights have different brightness levels?

 

How do I calculate the resistance necessary to reduce the voltage from 16vac to 12vac?

You need to know the LED current drawn at 12V and then R= 4v/Iled.

Is there any way to figure out, with a preassembled indicator lamp assembly, whether the LED. is using half- or full-wave rectification?

You could apply 12Vdc to them and see if they light with either polarity applied.

As an aside, does this explain the phenomenon I've observed with strings of LED Xmas lights, that different colored lights have different brightness levels?

Not likely.
Those bulbs are generally in series so the difference in brightness would be due to the difference in efficiency of the different color LEDs at the same current.

 

I presume you mean a circuit like this:
View attachment 130420
I don't see how this circuit would take care of the problem of different colored LEDs drawing different currents. Since either garage door could be open or closed, there are four lamp-on states: RR,RG,GR,GG. If the red and green LEDs draw different currents, then wouldn't the issue with "current hogging" occur in either the RG or GR situations? What am I missing here?

The actual currents taken by the red and green LEDs is not really relevant - you probably wouldn't notice the difference between, say, 10mA and 20mA if they are just indicators. The problem is that red and green LEDs drop a different voltage. So, let's keep it simple and say that when you apply power to a red LED, it drops 2V; when you apply power to a green LED it drops 4V. Now, if one of your doors is open, its red LED will be on, so the voltage appearing between the switch contacts and ground will be 2V. If the other door is open, you are now applying just 2V to its green LED - which is not enough to illuminate it at all, so the red LED will 'hog' all the current! By using a separate resistor in each door circuit, you can drop 2V on one and 4V on the other.

Assuming, for the sake of an example, that red LEDs have lesser voltage than green ones, wouldn't I need to have a circuit that looks like this:
View attachment 130421
Then R1 and R2 serve the role of reducing the 16vac voltage to 12vac, and R3 and R4 serve to equalize the current draw of the red LEDs to the green ones. (Though I don't see why R1 and R2 have to be separate resistors, since they both go to the wiper terminal of the switches...why wouldn't one resistor here be enough?)

This would be technically correct. However, as I said above, the difference in current through the two LEDs (14/330 = 42mA for the red, 12/330 = 36mA for the green) is so small it's not worth worrying about.

Short of calling the manufacturer (if that's even possible), how do check to see if the red and green indicator lamps have internal circuitry to equalize the voltages? Can I just connect them in parallel to the transformer secondary and measure the voltage across each LED to see if it's the same?

Don't bother! It won't make any noticeable difference.

If I understand the purpose of the rectifier, this is only needed if the LEDs are NOT rated AC/DC. Correct?

Yup.

 

I presume you mean a circuit like this:

I don't see how this circuit would take care of the problem of different colored LEDs drawing different currents. Since either garage door could be open or closed, there are four lamp-on states: RR,RG,GR,GG. If the red and green LEDs draw different currents, then wouldn't the issue with "current hogging" occur in either the RG or GR situations? What am I missing here?

The circuit won't prevent one LED drawing slightly more current then the other when on, but that only causes a small change in brightness, which is usually not a problem.
"Hogging" only occurs when two LEDs of different voltages are in parallel with a single resistor in series, and that does not occur in that circuit.

 

Hello,

When using AC higher than 5 Volts on the leds may destroy them, as most leds can not handle more than 5 Volts in reverse.

Bertus

 

"Hogging" only occurs when two LEDs of different voltages are in parallel with a single resistor in series, and that does not occur in that circuit.

Thanks for clearing this up. It was this fact that I didn't understand.

What I know of electrical circuitry goes back to my semester of E&M in Physics class in college (I graduated in 1973!) but I do remember a lot of it. Though this isn't relevant to building my project, if it isn't too much to ask, I'd love to understand, more for curiosity than anything else, why one of these two circuits results in "hogging" and the other does not (assuming that the two LEDs drop different voltages).

Thanks for all of your help. I really appreciate the time you've taken.

 

The actual currents taken by the red and green LEDs is not really relevant - you probably wouldn't notice the difference between, say, 10mA and 20mA if they are just indicators. The problem is that red and green LEDs drop a different voltage. So, let's keep it simple and say that when you apply power to a red LED, it drops 2V; when you apply power to a green LED it drops 4V. Now, if one of your doors is open, its red LED will be on, so the voltage appearing between the switch contacts and ground will be 2V. If the other door is open, you are now applying just 2V to its green LED - which is not enough to illuminate it at all, so the red LED will 'hog' all the current! By using a separate resistor in each door circuit, you can drop 2V on one and 4V on the other.



This would be technically correct. However, as I said above, the difference in current through the two LEDs (14/330 = 42mA for the red, 12/330 = 36mA for the green) is so small it's not worth worrying about.



Don't bother! It won't make any noticeable difference.



Yup.

So, if I understand you correctly, the only thing I have to worry about is dropping the voltage from 16vac to 12vac (assuming the LEDs are ac/dc), for which I need two resistors in series with the switch wipers, and there is no need to equalize the voltage of the individual LEDs because the effect is negligible. Yes?

Now, I'm curious why different colored LEDs drop different voltages...is it that to generate a give color the resistance of the LED circuity must be different?

Thank you VERY much for giving me a detailed explanation of what's going on.