I am working on an audio circuit project and want to change a separate toggle switch and LED indicator to single anti-vandal illuminated push button switch. I measured 15vdc at the PCB pads going to the LED. The LED voltages on the switches come in Base Voltage (2.8V), 6V, 12V, and 24V.
I understand these come with a resistor already installed in the LED. I have one of the 2.8V versions on hand and it seems to work fine, but I do not know for how long.
My question is, since it is not available in a 16V version, can I use the 12V version without adding a resistor?
Here is a link to the switch/indicator I am looking at.
https://www.mouser.com/ProductDetail/E-Switch/PV4FWG0SS-344/?qs=uwxL4vQweFM0wezPc85xog==

 

If it has the resistor already included it should be a bit dimmer but otherwise fine at less voltage.

 

The datasheet does not specify what the current is through the LED circuit. To prevent overdriving the LED, I would do the following.
1) Put 12V on the LED as per specifications and measure the current thru the LED circuit
2) With the measured current (in mA) and voltage (12V), you can now calculate the internal resistance of the LED circuit. (V/I = R)
3) You can now re-calculate the required external resistor to make the current the same with 16V on the LED leads. Thus, the current will be the same as when powered by 12V, and you do not risk overdriving the LED. Odds are the resistor value will be fairly small, around 200 ohms +/- a bit (depends on color of LED and its voltage drop)

 

The datasheet does not specify what the current is through the LED circuit. To prevent overdriving the LED, I would do the following.
1) Put 12V on the LED as per specifications and measure the current thru the LED circuit
2) With the measured current (in mA) and voltage (12V), you can now calculate the internal resistance of the LED circuit. (V/I = R)
3) You can now re-calculate the required external resistor to make the current the same with 16V on the LED leads. Thus, the current will be the same as when powered by 12V, and you do not risk overdriving the LED. Odds are the resistor value will be fairly small, around 200 ohms +/- a bit (depends on color of LED and its voltage drop)

Ok, so using the base voltage version (2.8v, 20mA), I calculated I would need a 610 ohm resistor. I have one of the base voltage switches and 680 ohm resistors on hand. If I place the resistor on the positive lead going to the LED and measure the voltage with a meter, shouldn't I see a drop in voltage? I still measure 15V.
I went ahead and soldered in the resistor between the positive lead and the Anode of the LED.

 

I would expect to see ~0.7V drop from the LED.

 

With the resistor, the drop across the LED should be about 2.8V.

But that is not the point, it is the current you should be measuring, keeping it under 20mA.

Bob

 

If the voltage across the 2.8 volt version is 2.8 volts then the current is correct, or close enough. And most LEDs have more thatn a 0.7 volt forward voltage. That is a silicon diode drop, not an LED drop. An adequate drop from 6 volts would be if you used a 560 ohm resistor. To use the 2.8 volt indicator with the 15 volt supply you need a resistor to drop 12.2 volts at 20mA, which ohm's law solved for resistance give R=V/I =12.2/ 0.02=610 ohms. with 10% resistors then your choice is probably 680 ohms, (blue-grey-brown). The indicator will not quite be full brilliance but it should last many years.

 

The Vf of a LED (your 2.8V) varies in big amounts by the color of the LED. Then there are process variations. Then there is the eye's sensitivity to a particular color. So, if you were using a lot of LEDS where the variation in intensity would be noticable, it would be wise to BIN the LED's based on Vf.

You may not need full brightness. So, you could use a 12V led on a 24V system by adding a resistor. 10-20 mA is generally a good rule of thumb for a design current. You can pulse the LED and get a higher brightness. There are some high efficiency LEDs that only need 1mA or so.

So, if you have 5V available, one 2.8V LED and a resistor and your good. For 24V, 10 LEDs in series would be problematic. You might go with two groups of 5, with each group getting it's own resistor.

There are cute driver IC's available. See the BCR431U Low Voltage Drop LED Driver IC.

 

The TS is just wanting to replace a switch with a separate indicator with a button with a built in indicator. What is confusing is the term "base voltage." that might mean the LED voltage without an internal series resistor. I wonder if that word was supposed to be BARE LED, meaning no internal resistor in the button assembly.
In that case an external resistor is mandatory. The external resistor would be in series with the LED, dropping 12.2 volts from the available 15 volt source. In that case a 610 ohm resistor is a good choice, if it is available.

 

base voltage: Say 2.8 V (Bare LED) was nominal and the range was 2V to 3.2. Every 10% change gets assigned a BIN. Those leds are used at the same current.

Even if they are 12V LEDS, they would have a Vf at a specific current.

This is for further understanding of LEDs not specific to the TS's question.

if there were more than one on a panel, the brightnesses in theory could be different.