220v LED wiring on kettle

[wuttheheeck]

I would like to have an LED on my gangbox on outside of HLT and BK.

the led on control panel will remain on all the time, however the led on outside of kettle will flash when SSR turns it on.

Do i need a neutral wire to run this led? I am only bringing in 2 hots and a ground to my box on 10/3 wire.

cheers

 

[IslandLizard]

If you have to ask these questions, you shouldn't tinker with it, me thinks.

 

[ajdelange]

If you want a positive indication that the element is drawing current then slip a current transformer over one of the element leads and wire the secondary to the led through a current limiting resistor. No need to run any additional wires from the control point in that case. If all you want is an indication that the SSR has been sent a trigger signal then send that signal to the led through a pair of low voltage wires. Again be sure to use the current limiting resistor so that the external led does not load down the SSR gate input too much.

 

[jeffmeh]

You could also get a 240v led and wire it in parallel with the element. A 240v led should only require the 2 hots (not neutral), and to wire in parallel you could use a 2 position terminal block, and stack three ring terminals (incoming, out to the element, out to the led) for each hot line, assuming it fits in the gang box.

 

[craigmw]

How about this:

http://www.electroschematics.com/3623/

220V (e.g. +120 on one hot and -120 on the other) will light the LEDs. It has a couple of electrolytic capacitors, which likely prevent the LEDs from flickering. A full wave rectifier converts the AC to DC, and this is done after a couple of 470k resistors to drop the current to this rectifier. Note that the resistors are 1W so you need rather beefy resistors at this point since they connect directly to the two hots. Any reason to do this with LEDs instead of a neon indicator bulb?

 

[jeffmeh]

How about this:

http://www.electroschematics.com/3623/

220V (e.g. +120 on one hot and -120 on the other) will light the LEDs. It has a couple of electrolytic capacitors, which likely prevent the LEDs from flickering. A full wave rectifier converts the AC to DC, and this is done after a couple of 470k resistors to drop the current to this rectifier. Note that the resistors are 1W so you need rather beefy resistors at this point since they connect directly to the two hots. Any reason to do this with LEDs instead of a neon indicator bulb?

Sure, but you can buy a 22mm LED indicator already configured to run at 240v for under $4. http://www.auberins.com/index.php?main_page=product_info&cPath=7_33&products_id=248

 

[ajdelange]

http://www.electroschematics.com/3623/

220V (e.g. +120 on one hot and -120 on the other) will light the LEDs. It has a couple of electrolytic capacitors, which likely prevent the LEDs from flickering.

I know this forum isn't Circuit Theory 101 but the capacitor at the left of the diagram is actually the impedance across which the voltage is dropped. That will certainly work but I don't see any advantage to doing it this way. A resistor will suffice. The second capacitor does reduce ripple but is un- necessary as the led's will flicker (pulse) at 120 pps - much higher than your eye can perceive.

A full wave rectifier converts the AC to DC, and this is done after a couple of 470k resistors to drop the current to this rectifier.

As noted above the voltage drop or, if you prefer, current limiting is by the reactance of the capacitor. The upper resistor is indeed 470K which, if it were the current limiting element, would limit current to about half an ma. It's function is to bleed residual charge remaining on the capacitor after the circuit is de-energized. Not a bad idea from the safety point of view but un-necessary if the current limiting is done with a resistor rather than a cap. The other resistor is 470Ω and there to limit current inrush when the thing is turned on. Again, unnecessary. Is he worried the panel breaker will trip?

Note that the resistors are 1W so you need rather beefy resistors at this point since they connect directly to the two hots.

The bleeder resistor across the cap will see essentially the full 220V and as such will dissipate 220*220/470000 = 0.1 Watt. A quarter watt resistor will do. The inrush limiting resistor would, if the led current was 20 ma, dissipate 0.2 watt. A half watt resistor will do. The fact of being connected to the hots is not relevant here. The current flow through the resistors and their resistances is what determines power dissipation. Twenty ma through 470Ω dissipates just as much power at 1 V as it does at 220V.

Any reason to do this with LEDs instead of a neon indicator bulb?

Neon bulbs are pretty dim. The traditional resistor size for them is 470K so the current is half a mil as noted above. You can easily push 10 or more times that through leds (or a chain of leds as shown in the diagram). Or you can put a pair of leds back to back (in paralell) and, provided they have the necessary PIV, eliminate the bridge.

This post from yesterday discussed how to calculate the size of resistors in a circuit that used leds with high voltage.

If the LED consists of a single diode (as opposed to a back to back pair) and you connect another diode with the wrong polarity then you will prevent any current flow. If the LED is a single diode it will only conduct on one half cycle. If you want it to conduct on both halves install a diode bridge in the AC line. This converts the AC to a sequence of single polarity DC pulses. You don't need to do this if the 'lamp' consists of a pair of LEDs or has its own built in bridge. Check with your ohm meter in 'diode test' mode.

A transformer is not necessary unless you want the voltage in the LED circuit to be low. Otherwise determine (from the specs) the current draw of the led. If it is i then you want a total resistance of R = E/i kΩ in the circuit. So, for example, if i = 20 ma and E = 120V R = 120/20 = 6 kΩ. If the LED is a 12V 20 ma LED then it already has a resistor of about 12/20 = 600Ω in the package and you only need 5.4 kΩ in the external cirucuit. Clearly, you can probably ignore this internal resistor and just use 6 kΩ. The final thing you need to do is check the power dissipation in the resistor. It is i*i*R. In this example, 20 ma is 0.02 amps and that squared is 0.0004 amps^2. Multiplying by 6 kΩ gives 6 E3*0.4 E-3 = 2.4 watts. Be sure to get a fat resistor.

If you use a transformer (a doorbell transformer from Radio Shack is the most likely choice) just use 28 V in the calculations instead of 120. A smaller resistor (both ohms and dissipation rating) will be required.

 

[wuttheheeck]

How about this:
Any reason to do this with LEDs instead of a neon indicator bulb?

I have already purchased a set of 220v 20ma led from ebrewsupply. In Kals diagram he wires his In parallel with the heating element without any resistors or capacitors. I'm already cloning his set up Ill probably do the same with these.

-thanks kal

 

[kal]

I would like to have an LED on my gangbox on outside of HLT and BK.

The only thing I'd caution about doing this is that the typical LEDs people use (myself included) are somewhat fragile and kettles are heavy. The LEDs usually look like this:

Part of the reason I use the 2-gang boxes I do on my kettles to protect the elements is that they're robust and can take a beating, but a plastic LED sticking out of the 2-gang box will not. There's enough momentum (weight) behind a kettle that if you knock into the LED it'll probably shatter.

I'd look at adding two little panel switch guards - one on either side. Something like this:

In use (pretend it's an LED light instead of a switch):

The other option is to spend a few bucks and get a name brand LED light that is really tough, like something from Allen Bradley as they have models that are considerably more heavy duty. For example:

Some of them also have guards that are available.

Good luck!

Kal

 

[craigmw]

Sure, but you can buy a 22mm LED indicator already configured to run at 240v for under $4. http://www.auberins.com/index.php?main_page=product_info&cPath=7_33&products_id=248

Yep, that looks like the simplest (and cheapest) way to go.

 

[craigmw]

I know this forum isn't Circuit Theory 101 but the capacitor at the left of the diagram is actually the impedance across which the voltage is dropped. That will certainly work but I don't see any advantage to doing it this way. A resistor will suffice. The second capacitor does reduce ripple but is un- necessary as the led's will flicker (pulse) at 120 pps - much higher than your eye can perceive.



As noted above the voltage drop or, if you prefer, current limiting is by the reactance of the capacitor. The upper resistor is indeed 470K which, if it were the current limiting element, would limit current to about half an ma. It's function is to bleed residual charge remaining on the capacitor after the circuit is de-energized. Not a bad idea from the safety point of view but un-necessary if the current limiting is done with a resistor rather than a cap. The other resistor is 470Ω and there to limit current inrush when the thing is turned on. Again, unnecessary. Is he worried the panel breaker will trip?


The bleeder resistor across the cap will see essentially the full 220V and as such will dissipate 220*220/470000 = 0.1 Watt. A quarter watt resistor will do. The inrush limiting resistor would, if the led current was 20 ma, dissipate 0.2 watt. A half watt resistor will do. The fact of being connected to the hots is not relevant here. The current flow through the resistors and their resistances is what determines power dissipation. Twenty ma through 470Ω dissipates just as much power at 1 V as it does at 220V.



Neon bulbs are pretty dim. The traditional resistor size for them is 470K so the current is half a mil as noted above. You can easily push 10 or more times that through leds (or a chain of leds as shown in the diagram). Or you can put a pair of leds back to back (in paralell) and, provided they have the necessary PIV, eliminate the bridge.

This post from yesterday discussed how to calculate the size of resistors in a circuit that used leds with high voltage.

I stand corrected! Thanks for the thorough description of that circuit.