I’m trying to figure out how to attach an LED indicator light to my thermostat, so I’ll know when the furnace is on*.
The thermostat is an old 1940’s job with a big open space where a clock used to go, so I can install a battery to power the LED, if that’s the best way.
My idea is to put a 1.5V battery and LED in the space, attach one lead to one side of the switch, and the other lead to the other side, so when the switch closes, the LED lights up. Of course, there’s already 20V across those two leads, so I would have to somehow block that voltage from crossing the LED, while allowing the battery to still light it up. Would a diode accomplish this?
So I would have a wire from the +20v side of the switch, through the battery, to the diode, to the LED, and back to the +0V side of the switch. Does this work, or blow out my furnace controller?
Once the cover is off, there is plenty of room to attach clips to the various parts of the thermostat, without affecting the overall operation. For those looking for background, my furnace is way older than the thermostat, and takes about 30 minutes to get the steam up and heat the radiators. When the house gets a bit of a chill someone decides to goose up the heat* which seems to accomplish nothing but make the boiler run a lot longer and make the house uncomfortably hot, at which point someone turns the heat way down, creating an interesting cycle of hot/cold. My goal is to let that certain someone see that the boiler is on or off before changing the temperature.
**This particular trait could be in the “things you can’t fathom people understanding” thread, but I digress.
Assuming 20v is a good number to run your LED on, wouldn’t attaching both leads to the same side of the switch work? Then no power is running through the circuit unless the switch is on?
I think it might work if I went from the back side of the switch to ground. When the switch closes, the back side would go to 20V, and the LED goes on. But I don’t have a connection to ground available., at least I don’t think I do.
I recently did something similar, added a red LED to my soldering iron station, because it had no other indication that it is turned on.
Since you already have a 20V source there’s no need for a battery. All you need is a resistor in series with the LED to limit the current. Here is a resistor calculator: http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator
Supply voltage is 20, a typical value for voltage drop is 2 Volts and a typical value for LED current is 20 milliamps. (The exact values will be given by the specific LED datasheet.)This gives a 1KΩ resistor with 0.5 Watt rating.
You need to know if your 20 Volts supply is AC or DC. If it is AC then the LED (which is essentially a diode) will block the negative cycle of AC and not light up. The apparent light output will be half in comparison to DC voltage. Therefore you should use a slightly lower value of resistor to allow a bigger current. If you have one of those superbright LEDs it won’t matter either way, stick with the 1kohm resistor.
Also if it is AC then there will be some flickering, but this is acceptable for the specific application.
Pulled out the old multimeter and confirmed two things.
It is an AC power supply at about 26V, though it’s labeled 20V
When I close the thermostat, the furnace controller draws 0.67A.
Would I then wire a resistor in series with the current circuit, that will allow this current to go through with a small (1.5v or so) voltage drop, and then wire the LED parallel to the resistor to use the voltage drop to light up?
If I have my formula right, 1.5V drop with .67A is 2.2Ohm, which would draw 1 Watt.
No.
You wire the LED+dropping resistor combination in Parallel with the furnace controller inputs. You also want to put a blocking diode in series with the LED, to prevent it from being damaged by reverse current.
It’s generally considered bad practice to run an LED with reverse bias, so that it permits breakdown current to flow. The dropping resistor will limit the maximum current, and it probably won’t hurt the LED, but it’s just not a great idea to run an LED on AC.
I think I can work with this. I was assuming that it would be difficult to wire in parallel because I only have access to the furnace hot leg, I don’t have the neutral leg to finish the circuit. However, recall the “big open space where the clock used to go”? Turns out the clock has its own 26V circuit, complete with a neutral leg that I can tap into for the return.
So, I connect one end of my Resistor-LED-blocking diode circuit to the controller side of the switch, and the other end to the neutral for the clock. When the switch closes, I should have 26V across the circuit, and when the switch is open it should be 0V. A big plus is that this wouldn’t require any additional wiring or actually breaking into the current wiring, which is messy enough as it is.