I'm using Arduino and relay shield to control five 24 VDC Solenoid Valves.
Now I want to place an indicator light (LED) for each valve, so whenever the valve is open, the light comes on.
Do I need 24VDC LED lights in order to do so?
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2 Answers
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There is no such thing as a "24 V LED". However, you can use a resistor in series with a LED so that the combination can be safely lit from 24 V. Let's say you want 5 mA thru the LED, and that the LED drops 2.1 V (typical of normal green LEDs). From Ohm's law, (24V - 2.1V) / 5 mA = 4.4 kΩ, which is roughly a suitable resistance.
Keep the power dissipation in mind. The resistor will drop about 22 V, which times 5 mA is 110 mW. That's getting close to the limit of a "1/8 W" resistor, so either make sure a 1/8 W resistor is properly mounted at the right temperature or use a 1/2 W resistor. Note that the power dissipation in the resitor is proportional to the LED current, which is one consideration for keeping the current low.
answered May 30, 2013 at 13:53
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3\$\begingroup\$ There are pre-built snap in panel indicators that use LEDs and also have built in current limiting. The limiter in many cases is a resistor and results in an LED indicator that is "rated" as optimal for a particular voltage such as 12V or 9V or maybe even 24V. Here is one such example of a "24V LED". chml.com/products/pdf/4-36.pdf \$\endgroup\$ Commented May 30, 2013 at 15:01
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\$\begingroup\$ @Michael: Yes it's possible to make a light assembly that drives a LED such that it lights properly when 24 V is applied (which is the main point of my answer), but that's no longer just a LED. LEDs inherently run on much lower voltages, which are dictated by the semiconductor physics. This is in contrast to incandescent bulbs, which can be made for a wide range of voltages. You can have a 24 V or 2 V incandescent bulb, but not a 24 V LED. \$\endgroup\$ Commented May 31, 2013 at 12:13
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1\$\begingroup\$ Olin, you are being pedantic here. Yes indeed the actual LED behaves as you describe but you need to realize that the emphatic statement you made "There is no such thing as a 24 V LED " is not correct in many contexts. I pointed out just one of many contexts where a "24V LED" indicator is a perfectly reasonable description for a component. Loosen up a bit. \$\endgroup\$ Commented May 31, 2013 at 12:22
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\$\begingroup\$ @Michael: No, it's not correct. There is a difference between a "LED" and "LED lamp assembly". These things matter, and throwing around terms loosely can lead to confusion. For example, someone used to incandescent bulbs might to be aware that LEDs can't be made to operate at arbitrary voltages. That's a important thing to know. You can add a dropping resistor, but that becomes quite inefficient. By being sloppy with the terms, others could get confused. Even the product you linked to isn't called just a "LED". \$\endgroup\$ Commented May 31, 2013 at 12:37
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1\$\begingroup\$ @yogece: I had no problem with the perspective of your answer. The problem was, as I clearly stated, that it had a outright mistake in it. Even then I didn't downvote your answer. I only did that after you refused to fix it, tried to blame someone else, and told me to go report the problem to a third part! Even then, I said I'd undo the downvote if you fixed the error. In this case Michael and I disagree on terms. I don't agree with his downvote, but accept it because it seems to be sincere, and applaud that he explained it so everyone can judge for themselves. \$\endgroup\$ Commented May 31, 2013 at 14:19
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You can place the LED in parallel with the solenoid, or in parallel with the relay input:
simulate this circuit – Schematic created using CircuitLab
Either could work. In either case, you need to select a series resistor (R6 or R7) to limit the current in the LED, but a significant difference is that the voltage across R6 much be must higher than R7, because it is connected to a higher voltage. Say that the forward voltage of the LED is \$2V\$, and you want the current to be \$10mA\$. The power in R6 is then:
$$ P_{R6} = 10mA \cdot (24V-2V) = 0.22W $$
You would want to use a \$0.5W\$ resistor for R6 to handle the heat without danger of of damage. You could also decrease the LED current, which will also reduce the brightness, but maybe that's fine.
In the other case of R7 and D4 being powered by the \$5V\$ input and \$10mA\$:
$$ P_{R7} = 10mA \cdot (5V-2V) = 0.03W $$
An ordinary \$0.25W\$ resistor is in no danger of overheating at this power.
answered Jul 29, 2013 at 19:23