I'm on my 3rd Hubbell 1203PL 3way pilot light toggle.

(light on - load on)

Honest to God the pilot light only seems to last 8 months. That's terrible. I'm going back to put in a Leviton switch this time and see what happens. I put the meter on there and checked everything in there and it's wired correctly. Nothing funky going on.

Anyone have a successful history with these?

Are you using it for the right load?

Motor loads are hard on switches and they need to be rated for the inrush.

I see from the product datasheet: http://www.hubbellcatalog.com/wiring/section-c-datasheet.asp?FAM=Switches&PN=HBL1203PL
this switch is rated for a 1/2 HP motor at 120 volts.

I see on the schematic, a supplemental ground connection on the pilot.

I have a couple of those ( of the single pole leviton variety) on a furnace and power vent water heater circuit and have noticed the indicators have become really dim, but still visible. Mind you they have been on almost 24/7 for the past 8-10 years.
Perhaps the hubbell have had a defective run of switches or the little neon indicators?

A.D

From the schematics, the indicator bulbs appear to be neon bulbs. I was under the impression that the common neon bulb was good for 20,000 hours of life.

Geez this is connected to a light. A plain old incandescent lightbulb. This switch is a real disappointment.

I might change it to a standard 3way and add a 120v LED with a flange right over the switch. LEDs have a crazy long lifespan.

Glad you posted.At $65 each,I was looking bad.I have same problem,short lifespan.Next is led pilot light

Bob O.84,Pa.15330

Use a LED, run it at 3 mA and problem is sorted for life.
I use a 66 k.Ohms 2 watt resistor for 230 Volts ( 2 x 33 K ohms 1 watt in series )
Then a 1 N914 diode antiparrallel with the LED.
For anyone knowing basic soldering and tidy working it is a easy to fit in any switch with perfect results. also no heating in the resistors.

For 110 Volts design the circuit to run at about 3 mA which is ample for a high efficiency led. a single 33 or 39 k Ohm 2 watt resistor will suffice.

I have found neons with the incorrect series resistors which means excessive current and a short lifespan.

Above are the schematics how to build the circuit and one application where we use them in the substations for the streetlight and hotwater pilots.

Where:
D1 is 1N4007 diode
D2 is 1N914 or equivalent
D3 is LED,3000, or 5000 Milicandela
R 1 and R 2 series voltage dropping resistors.


Hope they post OK.

Good luck with it

{ Images moved to ECN server }


[This message has been edited by pauluk (edited 10-28-2006).]

Isn't series diode D1 somewhat superfluous, unless the LED being used has an exceptionally low reverse-voltage rating?

PaulUK, all LEDs have an exceptionally low reverse voltage. It's is usually specified around a few volts in the data sheets. D2 clamps the reverse voltage over the LED to D2's forward voltage (0.5 - 0.8 V). As usual, the data sheet specs are minimum ratings so you may get away with not respecting them perfectly but it won't help reliability, especially if you subject them to more than 300 volts, this might terminate them within a few milliseconds. ;-)

Question: How about safety? Usually, the LEDs' innards are molded in a plastic (epoxy) case with the leads sometimes just barely enclosed, especially with the small 3mm LEDs. OK, you got a series resistor but using them in what looks like metal holders (see Rodalcos nice box) could open a way for the electrons to leak out. Even worse, if they can't roam freely because the metal holder is mounted on a plastic board, they might collect and get nasty...

Also watch out for the maximum voltage across the resistors, and their dissipation rating. Metal film resistors don't like either of these ratings exceeded. Again, it will work for some time but it might not fulfill your expectations. The old carbon resistors are much more tolerant here.

For the LED, use a low-current (2-5 mA) type. The 2 x 39k will produce a peak current of 4 mA (Formula: [230V times square root of 2 for peak voltage, 325V] divided by [2 * 39k]) while the average current over time is just 230V / 78k halved (because only one half of the sine is used), or 1.5 mA.

Also note that the LED is only lit part-time (at AC's 50 or 60 Hz), this might cause noticeable/objectionable flickering as the LED does not (or should not) have the thermal inertia of an incandescent bulb (neither does the neon in Haligan's original post).

We use the LED indicators in substation panels and CT kWh meters for potential indication.

As already mentioned above is that the LED's shown, show the actual parts. These need to be heat shrinked or put in tubing of appropiate voltage rating for the application.

In the substation panels these are put in the actual plastic neon holder casing which are fitted in steel cabinets.
In a meter they are fitted and wiring is often airborne with some minor support.

The flickering of the LED is hardly noticeable and not really an issue for a pilot light anyway.

The 1 N 4007 diode could be omitted but i put it in to make it fail safe and forwhat ever reason a spike may pop the LED.

I have fitted around 1500 kWh meters with 3 LED's each and about 200 of these in our substations as replacements for fading Neons.

No faillures over a period of 15 years.

Also pilot wires and meters are subject to nasty spikes sometimes during lighting, The two resistors will probably absorb most of the energy anyway hence two resistors are used to allowe for surges well above 230 Volts (325 peak).

A thing i forgot to ask Haligan is , were the Neons black after they stopped working ?

Startrek, very good comments and good maths.
Thanks.

To get back to the neons, usually they run well below 1 mA about 600µA at around 70 Volts. (depending upon specs).

So for 110 Volts, 40 Volts at needsd to be dissipated in the series resistor at 0.6 mA
usually a 100 kOhm resistor is used.

For 230 Volts a 220 or 270 k.Ohms ¼ Watt resistor is used.
I have seen many neons with only a 100 kOhm resistor fitted for 230 Volts with very early faillure of the neons.

Sure, LEDs have a rating which is much less than the PIV of a regular diode, but I don't recall ever seeing one with such a low rating that it can't withstand 0.6 or 0.7V reverse voltage, which is the most you could get with D2 acting as a clamp to limit the reverse bias.

Belt and braces, I guess.



[This message has been edited by pauluk (edited 10-30-2006).]

Pauluk, you are correct. most LED's will handle a reversed voltage of a few volts. generally less than 5 Volts as per datasheet.
D2 will clamp any reversed voltage and keeps it below around 0.6 volts.

The reason i put D1 in is to reduce heat dissipation in the series resistors.

D2 and D3 acting as an AC type load.
By adding D1 the sinewave will effectively be reduced to 0.707 and power dissipating in R1 and R2 is reduced accordingly.
A diode running at mA's doesn't produce any heat which is better than dissipating it in the resistors.
Also the LED needs the DC component off the AC mains and is not affected in light output via D1.

By under running the LED , well below it's specs, the lifespan will be basically infinite.

I did trials with 47 nF 1000 Volt capacitors and a 1 k.ohm Resistor in series, which will work fine but sometimes a surge may take out the LED or clamping resistor D2.

I did this with a capacitor feeding a LED in an SSR, I used the capacitor for 2 reasons. One, to reduce the load on the dropping resistor and two, I was across the open circuit of a pump switch (interlocking 2 pumps) and I wanted to reduce load through the first pump. I also used both diodes in the original picture you posted. It is still working fine years later.

Just done a few tests at work on our Fluke 5520A calibrator with the led's with and without the series 1N4007 diode.
Ammeter used was a Fluke 75 DMM.

Circuit as per diagramme listed in this thread.

230 Volts 50 Hz 78 kilo ohms and 1N4007 1.27 mA 0.132 Watts.
230 Volts 50 Hz 78 kilo ohms 2.8mA 0.612 Watts.

117 Volts 60 Hz 39 kilo ohms and 1N4007 1.28 mA 0.064 Watts.
117 Volts 60 hz 39 kilo ohms 2.8mA 0.31 Watts.

formula P=I²R

It proves the fact that the 1N4007 series diode reduces current and power dissipation in the series resistors by a quarter with no loss of light output in the LED.

Frequency 50 or 60 hz does not affect the current drawn in the circuit because it is a resistive circuit.

Regards Raymond