Pilot Light Resistors

[big john]

We have a ton of 20+ year old switchgear by a variety of manufacturers: Westinghouse, ITE, Fuji, etc. On all the indicator lights there are series resistors. So even though everything is fed with 125V the lamps we have to use are a variety of voltages: 6V, 18V, 22V, etc. Some of these lamps are extremely difficult to find.

Why does everything have series resistors?

Is this a safety practice to help limit fault current...? Is there a hazard to simply replacing all these lampholders with a type that accepts 125V lamps? Thanks.

-John

 

[Cold Fusion]

We have a ton of 20+ year old switchgear by a variety of manufacturers: Westinghouse, ITE, Fuji, etc. On all the indicator lights there are series resistors. So even though everything is fed with 125V the lamps we have to use are a variety of voltages: 6V, 18V, 22V, etc. Some of these lamps are extremely difficult to find.

Why does everything have series resistors?

Is this a safety practice to help limit fault current...? Is there a hazard to simply replacing all these lampholders with a type that accepts 125V lamps? Thanks.

-John

Don't know.

Not that I know of.

Not to me, but I'm partial to transformered LEDs - unless they are DC.

cf

 

[gar]

100225-2246 EST

big john:

A high voltage (120 v) incandescent pilot lamp has a very fragile filament. I would not recommend using one. Short life and vibration sensitive.

A common practice is to use about a 6.3 V pilot lamp in a pilot lamp assembly with a built in transformer.

If you have a 120 V source and want to use a series dropping resistor to a 6.3 V lamp @ 150 MA, then the resistor is about 760 ohms. 760 ohms at 114 V is 17 W. No way do you want to use a series dropping resistor for this purpose.

If you have a lamp with a voltage rating approximately equal to the supply voltage, then you might add a series resistance to reduce inrush current to obtain some improvement in lamp life.

Neon bulbs must have a series resistor.

.

 

[big john]

Thanks for the help! I'm replacing everything with LED lamps, so vibration and lamp life is no longer an issue, just wanted to make sure I could run them at full voltage.

-John

 

[gar]

100226-2025 EST

big john:

A single LED has a typical on voltage drop of 1.5 to 2 V. To run this from a 120 V source a series resistor is required. A high brightness, non-diffused, LED can possibly be useful with a 250,000 ohm series resistor at 120 V AC or DC. This is not much power dissipation, less than 0.1 W. Might be feasible to operate at 0.25 W in a small package and this would be a resistor of about 56,000 ohms.

.

 

[dbuckley]

Edit: I've just reread your post, and I think you are suggesting running a LED on a much lower current than the standard 20mA.

But for a standard LED on the standard current:

The parameters that are important are the voltage across the LED (2V is ok) and the current through the LED (say 20mA), so calculating the dropper resistor, assuming 120V supply:

R = E (118) / I (.02A) = 5900R.

W = I (.02) x E (118) = 2.36W.

So a 3W 5900R resistor will do it, just - a 5W or 10W would be better from a long term reliability perspective. This resistor will get warm or maybe even hot, so dont put it adjacent to anything meltable. Also most folks agree that there should be a diode in series with the resistor and the LED as well.

You could replace the resistor with a 0.5uF capacitor which at 60Hz has an impedence of about 5300R and this has the advantage of no heat dissapation. I've not tried this myself, but using a capacitor to replace a dropper resistor for low currents is a common technique..

 

[big john]

My mistake, I wasn't clear:

The replacement lamps I'm using are commercial LED lamps designed to run on 125V. They have in integral resistor in the lamps.

The lights I was asking about are for incandescent lamps, but it is the sockets themselves that have resistors, and that's what I wasn't sure about.

-John

 

[gar]

100226-2311 EST

dbuckley:

My math is correct. There is no need for 20 MA.

I use a Stanley EFR 5366X LED with 24.9 K from 12 V and directly on axis it is overly bright. Might be a different story in bright sunlight. This is not a diffused LED, but works very well as an indicator. Roughly I am exciting this LED with about 0.5 MA.


big john:

Determine the current to your 120 V LED pilot lamps. Calculate the resistance. Measure the resistance in your existing pilot light assemblies. If this resistance is not too large relative to the resistance of the LED, then you probably won't notice the effect.

An ohmmeter probably won't give you a good measure of the LED resistance because its source voltage is too low.

.

 

[SG-1]

The resistors are for safety as you suspected. All of the old bulbs were incandesent or neon encased in glass. If one were broken during replacement and the technician shorted the socket out while digging out the remains of the bulb the resistor limited the short circuit current to milli-amps.

No flash & no taking out the control circuit.

I have seen LED telephone type replacement bulbs for the Westinghouse EZC indicator lights.

I agree you can operate the LEDs at much less than rated current an still have them appear much brighter than incandesents. They look best when the bulb & lens are the same color. Blue is my favorite.

I do NOT recommend them for loss of phase indication or ground detection on delta PTs due to the fact that one can bearly discriminate between full voltage brightness and half voltage brightness. The incandesent is extinguished at the same voltage the LED begins to dim. Line to Ground PTs are fine.

 

[Besoeker]

FWIW
We changed to LED cluster lamps for panel indicators a bit over 10 years ago.
The ones we use are 22mm dia fixing which is pretty common in the EU zone and come in a variety of colours and voltages. They are self contained and directly driven (no transformers).
The reason we started to use them was longevity. To date, I know of no reported failures. I don't know how many we have used, but it is certainly in the thousands.

There are a couple of other benefits for us. They are less expensive than the units they replaced and quicker, thus cheaper, to fit being a single item. The old Telemec units, as anyone has used them would know, came as a kit of parts that had to be ordered separately and assembled.

 

[winnie]

Commecial LED indicators are available in the US; rated for 120V AC or DC, 22mm mounting hole; probably exactly the same beasties as Besoeker is describing.

When considering the circuits for these devices, keep in mind that are not required to stick with a single LED junction. Rather than running a single LED at 20mA and 2V, using a dropping resistor that has to dissipate more than 2W, you can run a series string of 2 LEDs at 10mA and 4V, dissipating about 1W in your dropping resistor and getting more total light output. Extend this; use a cluster of 10 LEDs, at a total voltage of 20-30V, operating at say 5mA...you quickly see where this is going.

-Jon

 

[gar]

100227-1616 EST

20 MA is not a reasonable figure to use. With the high brightness LEDs available today lower current values are realistic to consider. I have already indicated the current level I use is about 0.5 MA.

A quick search of Chicago Miniature produced this part number --- 18620450. This is rated 28 V AC/DC with currents of 5 MA AC, 10 MA DC. Guesstimate 2400 ohms for a drop of 28 to 6.4 V . 21.6/0.010 = 2160 ohms. Power dissipated in the resistor about 0.22 W. Because of the big AC/DC difference it appears that a series diode is used for rectification.

If a series diode is used for rectification, instead of a shunt or bridge, then depending upon the PIV rating of the rectifier diode you may or may not be able to apply an external resistor for use at higher voltages.

.

 

[SG-1]

Another item to consider is if the control circuits are powered from batterys.
Say the control circuit is 125 VDC, the actual voltage will be much higher. The float voltage will be in the range of 138 VDC and the equalize charge will be over 140 VDC.