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I've got a project I'm working on, which is essentially a set of MCU-controlled relays. I'd like to set up a test harness for the system, using something visual on the outputs. I was thinking some small, low-power light bulbs.

I've got 28V AC output, and I need a bulb that would draw ~0.2A. How do I go about locating and choosing a suitable bulb?

Mark
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2 Answers2

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Farnell, RS, etc stock plenty of suitable bulbs. You need 28V * 0.2A = 5.6W so a bulb rated for 28V at around this wattage.

Here is one rated for 28V/6.44W, so 0.23A. Here is a page with similar options. If you decide you need one of a different rating (although I selected a few different wattages between ~4-10W) simply change the search filters.

Noticed you are in the US, so here is one from Newark too.

Oli Glaser
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An LED solution is almost certainly going to suit you better than using filament lamps. LEDs take somewhat more effort to use than lamps in this application - but you will always be happy with the result once you forget what it took to implement.

BUT - lamps first

  • Digikey sell 28V and other lamps

  • Including lamps from Chicago miniature lighting.
    This spec sheet from CML shows 7 x 28 V lamps in T-3 3/4 wedge base design with currents of 70 MA to 200 mA. That's only one Digikey inspired example. Many more.

  • You could use 24 V automotive bulbs with a small series resistor (R~= 4/I ohms)

  • You could use 2 x 12V automotive bulbs in series plus a small R (perhaps more available).

  • You could use 1 x 12V bulb with larger R. (R~= 16/I. Power rating or resistor = 2 x P bulb.)

BUT

LEDS:

As you have MCU controlled relays, operating the LEDs from the MCU supply or similar low voltage supply and driving them from the MCU drive to the relays makes most sense.

Power at 28V and 0.2A = V x I = 28 * .2 = 5.6 Watts.
A 5 Watt lamp is about car tail light bright.
If you used 5 Watts for a LED you could utterly dazzle your viewers.
Far less power will be adequate.

I'd suggest that even 1 Watt would be very adequate for an LED (lights up a small room) and even much much less would probably suffice. If you wished you could use several LEDs in series, which has some advantages.

If you did need to operate the LEDs from 28 VAC, starting with a resistively driven LED, 50 mA would give you "lots of light". Something like a 270 ohm resistor and series diode and probably a capacitor at the LED would drive this. R value needs to be checked due to half wave AC effects on average current, For an LED an NSPWR70CSS-K1 from Nichia will produce approaching 20 lumen at this drive level.

Ideally you'd use a buck converter to reduce the voltage from 28 VAC. LED driver boards that will drive up to eg 350 mA LEDs can be had on ebay and elsewhere for under (or well under) $5. Deal Extreme sells a number of potentially suitable parts.

Example only - ebay 3 Watt LED driver, up to 24V in - use a series R and diode and cap.

$1.94 each.

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Operation from eg 5 Volts requires an eg ULN2803 (costs a few dollars US) to drive up to 8 LED drivers plus a resistor per LED. ULN2803 input can be driven by microcontroller directly OR from 28 VAC using 1 diode, 1 resistor and 1 small capacitor.

Russell McMahon
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  • @ Russell - I was thinking of LEDs also, but I assumed he wants to draw 0.2A for a reason. – Oli Glaser Aug 26 '11 at 13:57
  • @Oli Glasser - you could well be right. I assumed it was a perceives adequate bulb brightness level. May have to add a resistor :-). – Russell McMahon Aug 26 '11 at 14:50
  • I'm attempting to simulate solenoids (I use the word "simulate" loosely here...) on the outputs of my setup. I'll only ever need this test rig once (probably...) and I have 16 outputs to test, so $2 per output for LED drivers is, well, wasteful. The goal here is not to get LEDs lit up by the MCU, that's easy. The goal here is to test the entire system, MCU, shift registers, SSRs, connectors, everything, without having to purchase 16 solenoids or disconnect/disassemble 16 sprinkler valves. – Mark Aug 26 '11 at 14:55
  • You can use a resistor as a load in place of solenoids. You can drive LEDs with (as noted) a series diode and resistor and a capacitor. At 20 mA you can get a VERY bright LED. If you are driving solenoids then it would be a REALLY good idea to have an inductive load in their place. Inductors cause interesting noise problems. Best to find out about them now. You may be able to find some surplus inductors (transformers, solenoids, ...) for about free. // Are solenoids switched at zero crossings of waveforms? If not, are they "snubbed"? (Sounds like recent questions :-) ). – Russell McMahon Aug 26 '11 at 15:16
  • So I could use, f.e. a simple 12V automotive bulb, and calculate the series resistor as ((Vin - Vbulb) / Ibulb ohms), and ((Vbulb * Ibulb) * 2) watts? – Mark Aug 26 '11 at 15:28
  • My SSRs are non-zero crossing type. I'm looking for light bulbs in the test rig because they make it really easy to visually identify whether any given SSR is on or off. I've tested individuals with an extra solenoid I have laying around, and certainly ideal would be to have 16 solenoids all wired up, but unfortunately I don't have 16 laying around... – Mark Aug 26 '11 at 15:52
  • @Mark - yes on bulb resistor R. Actual Watts in R is (Vin-Vbulb) x Ibulb I added a safety factor od about 2x for a 12V bulb - a good idea so resistor runs not too hot. You can buy 24V bulbs for truck systems. BUT an LED is better power and heat wise. Half wave R will be ABOUT half DC or full wave so R~= 1/2 (Vin-Vled) / Iled. 28V, 20 mA, white LED = 3V, R = 1/2 x 25 /0.020 = 625 ohm say 680 ohm to 1 k ohm. Use series diode and say 470 uF cap (6.3V or better) across LED. – Russell McMahon Aug 26 '11 at 17:33
  • @ Mark - just for interests sake, here is a LED based 28V AC/DC bulb: http://uk.farnell.com/cml-innovative-technologies/18602452/led-ba9s-28vac-dc-yellow/dp/1105170 bit expensive though. Personally I'd go with the incandescent option unless the extra power loss is a big deal, in which case I'd follow Russell's good advice. – Oli Glaser Aug 26 '11 at 19:49