Resistor Value Calculation Needed

No and errrr, I'm not sure you've got the circuit right.
If it's a 240v relay it's designed to work on 240v you don't need a resistor.
But the relay needs to be in series with the Lamp, in which case it won't have 240v across it.
Say the lamp and relay coil are in series across 240v you want say 220v across the lamp and the remaining 20v across the coil, so that would be say a 24v coil...BUT I dunno if it would like the switch on surge. (The lamp fillament would be effectively zero ohms while it's cold, so the relay coil would get 240V across it)
There are probably standard circuits for this sort of thing, hopefully someone who is used to this sort of electrical stuff (I'm electronics rather than electrical)

Actually it's all so confused the only sensible thing is to post a circuit diagram.
My furry brain is hurting. (I'm prob missinterpretting you)
Del

Putting a 160 ohm coil in series with a 1 ohm lamp will pull in the relay but the lamp will not come on. Unless I'm missing something.

I assume that the lamp is tungsten type. My suggestion is to connect a 14ohm 30W resistance in series with lamp. The resistance would drop approx 14V. Now connect a small bridge rectifier parallel of that résistance and convert that 14VAC into 12.5VDC. Now enable a 12VDC relay and achieve your goal.

Neat, that would work but a touch involved. Yes, it's a tungsten IR lamp used to keep newly hatched chicks warm. It fails, they die, hence the need for a backup. Is it not possible though to get a suitable voltage to drive a 240 VAC relay (that I have) using an appropriate resistor? Lamp current 1 Amp

Put your bulb in front of an optocoupler, feed an NPN transistor with optocoupler and drive a 12volts relay with this transistor, do whatever you want to do with relay.

Rakesh Semwal. Thank you for the thought but the concept was to make something simple that "failed safe" Your idea "Put your bulb in front of an optocoupler, feed an NPN transistor with optocoupler and drive a 12volts relay with this transistor, do whatever you want to do with relay" would require a separate DC power supply and a whole load of components. Not terribly practical for a device that only has to keep a few chicks warm.

So why not do temperature control on the chicks instead?

No no no, you guys have it all wrong. You place my cat next to the lamp and it will stay there as long as it's warm. When the lamp burns out, my cat will get up and step on a switch to activate a latching relay and that'll light up the spare lamp. After getting warm again, my cat can go play with the chickens .

Yes, go to Digi-Key and put in "current transformer" and you'll get to a selection of parameters. Continue to select 5A and 1:1000 and you'll have your the transformer to begin interfacing between the output of the current transformer and the relay. You are not going to do get from here to there with just a few parts! Myself, I would search for a solid state relay with high sensitivity ... then you'll get the parts count down if that's an objective.

There's a lot of ways to do skin a cat (sorry Rosie). Rosie is still volunteering as your heat detector, she didn't like my "skin the cat" comment.

This wont work as for the coil could not supply the necessary load to light the bulb correctly. A better option would be to use a simple photo circuit in which IF Bulb A is lit, Bulb B is not. A photocell, transistor, potentiometer(calibration of sensitivity) and a 12 volt voltage to drive photo detection/trigger and relay coil power. Then have each bulb fed through a single DPDT relay. There are tons of photo/relay circuits out on the web that can be made with radio shack parts inexpensively.

You can also use a oil furnace control unit with thermostadt. The flame detection circuit is happy with photocell pointed at a bulb. This would give you the ability to control the temp in their living space. Connect 2 bulbs parallel, so that if one fails the other still functions. The latch is intended to run 10kv transformer and oil pump motor. I'd use a heavy duty relay off of this to isolate load to the heavy duty relay vs 500 watts through the furnace relay module. You could also set up a thermostadt with a alarm if the bulbs both fail to set off an alarm. I made a temp alarm for a server room once with a $10 mechanical thermostadt for oil furnace and $10 radio shack parts for 120dB siren and 9VDC battery snaps + battery. Added a 120dB alarm in series with the latch and 9VDC battery so that if the temp went above 80 degrees in server room, it let out a very loud alarm and worked no matter if there was a power failure or not as well.

Place a small resistor in series to provide a small but acceptable voltage drop (possibly around 5V). Obtain a normally closed relay with a 240 VAC set of contacts and a low coil voltage. Probably end up with a 5 VDC coil voltage so rework initial sentence to add bridge and filter cap. Place coil across resistor, add surge protection and observe clearances for everything since 5 VDC with respect to relay coil is not 5 VDC with respect to ground (or your fingers).

Now, main bulb good = main circuit has current flow. With main circuit having current flow the normally closed relay will be open. If main bulb burns out or fails for any reason then main current should drop to zero and normally closed relay should close. Thus, safety bulb turns on.

Your count of chicks should remain constant even after they are hatched.

P.S. EDIT: Oops, I am basically saying what #3 already posted. Sorry for the duplication. I'm not quite following why #3 and my posting would be too involved?

Seems with what you have it doesn't work. I have read the posts and some are doable. If you want to work with your existing relay, you also can get and old filament transformer for 6,3 volts - and put this low voltage output in series with lamp 1. Or a transformer 240 to a few volts. Low voltage coil ca. 4 amps. The High voltage output (secondary winding) can deliver the voltage to your relay. You can play with the relay - if the voltage output of your transformer is not high enough, you can always put a DC rectifier before your relay. You can also try with a output transformer from a dump tube radio or amplifier.

Yep, using an old audio output transformer as a current transformer might work if it operates sufficiently at 60Hz; good idea. A "real" current transformer is what I would select, but my design solutions tend to be too exotic; worked too many years designing defense electronics . And I tend never to have cost as a design restriction. As an engineer rips nichrome wire out of a toaster, I'll be ordering a spool of the stuff <g>.

Friends,

I question the effectiveness of a photocontrol. When it turns on the backup lamp, the control will then see normal "daytime" light and shut the lamp off. Thus, depending on the time delay involved, you will have a blinker. I agree with the earlier post suggesting the use of a current sensing relay. There are many manufacturers for them. Since they (almost) all use a current transformer to sense the flow of current to the lamp, they do not need any series or parallel direct connection to the lamp circuit. Their relay output can then directly or indirectly (through an interposing relay if the lamp inrush current is too high) turn on the backup lamp.

Simple, off-the-shelf, and well-proven.

--JMM

Only if you don't cause feedback to the photocontrol device. Otherwise, the light will blink.

It's just a matter of time before someone suggests using a PLC. Leave it to a bunch of engineers to complicate a simple circuit!

Depending on how robust your relay coil is, it may do just fine with starting surge, and you certainly do not need to use a bunch of electronics to switch bulbs.

When you say your relay is 240VAC, do you mean the contacts are rated for 240V or the coil operates at 240V? Because 160Ω sounds more like a 24V coil.

If your lamp uses 1A at 240V, then presumably it has a resistance of 240Ω, and putting a 160Ω coil in series with it might cause a problem. However, depending on the required coil voltage you could put resistance in parallel, as you suggested, to reduce the total circuit resistance while maintaining the minimum drop-out voltage on the coil without taxing the lamp too much. If the lamp will operate at 220V, then you just need to make sure the coil and parallel resistance doesn't drop more than 20V at steady state, so you need a total resistance of no more than 20Ω, which of course could be achieved with a 20Ω resistor in parallel with the coil. I'm just making up tolerances here, so check your data sheets.

Fortunately with incandescent lamps there's quite a bit of margin for error. Nevertheless, you should probably put the same total resistance in series with the back-up lamp so that it doesn't end up running hotter than the primary.

If you already have a relay that you intend to use, post the manufacturer and part number. Otherwise, I'd say go with a smaller relay that is robust enough to handle the starting surge.

Mitsurati; Thank you. The relay I have is a "IMO" HY41PN230AC. Very good point about the back-up lamp, the extra output could have caused a problem.

Ok, here an idea. Take a N.C. relay with a coil voltage that the same as the bulb voltage. Insert the coil in series on one leg of the light. When light is working the relay is energized keeping the relay contacts open. When the lamp fails opening the circuit voltage/current through the relay coil allowing the contacts to close and energized the second lamp.

The limiting factor as you see is getting enough current through the relay coil to allow the lamp to work correctly. You could use a shunt versus which runs a lot cooler than a high wattage resistor. The shunt would only need to be a few ohm's then you would need to use a relay with a low voltage coil to keep the N.C. contacts(rated for lamp voltage/current rated) open so long the first lamp is good.

Simple ohm law to figure shunt/resistance in parallel with a low voltage coil and its resistance.

The last thing is If you have a power failure then no lamps will work unless you have backup power available. Just something to think about

You brought up some good and valid points, I feel that I am going to bring up another!!

When an incandescent lamp burns out, quite often they make a short circuit and take out the fuse as well, so the two lamps need to be seperately fused as well, not a difficult thing to do....

I also agree with someone's comments about the resistance of the relay, it could be most likely 12 - 24 volts DC......but not 230 AC!

As you mentioned, you need to have a method of knowing if the mains goes out completely as well!!!

Thanks everybody for the thoughts. Much appreciated. Looking around my workshop this morning I suddenly realised that I was guilty of over complicating the problem myself! I have just clamped a Honeywell L641A1039 hot water cylinder thermostat to the primary IR 250 watt lamp holder and set the temperature to 70 degrees centigrade. Power up and both lamps come on, 30 seconds later the back-up lamp goes off. Ha, ha. Feel quite stupid for posting the question now, but at least all your answers made me think. PS: metalSmiths; I can now use my relay to operate a DC powered alarm in the event of a power failure to the chicken rearing area. Cheers all.

There you go, a classic "Keep it simple, stupid" or KISS solution! Amazing what you can find in the garage. I remove and save all the thermostat gas valves from leaky water heaters ... but a gas valve wouldn't help this problem. But anyone could probably drive around and find an old electric water heater for salvaging parts. Rosie the Cat gives you a double-paws up!

Billypil,

I disagree with "Difference in light output at a 10% drop in applied ac voltage was not noticable". Consider standard incandescent lamps sold in the USA. Those rated for 60w at 130v put out 665 lumens at 120v (and are given a 5000 hr life rating) while those rated for 60w at 120v put out 870 lumens (and have a 1000 hr life rating). You are making a significant shift towards the infra-red in your filament's black body radiation spectrum because of the filament's lower operating temperature. These ratings show a very large increase in lamp life (as you correctly state), but at a 20%+ loss in light output.

I have taken published tables of light output as a function of voltage and found that the relationship is close to (V_a/V_b)^3.2. I would agree with a statement that "Our eyes are very adaptable and will often not notice a difference, particularly if the two light sources are in separate areas."

--JMM

Hello JMM, Sorry I have to say, you are wrong regarding this, as despite your tables quoted, several people were involved and the overall consensus was that the difference in output in the trial was, that no one could tell the difference, of course it was only watched by humans and no instrumentation was used, except for a DMM to actually read what voltage was able to be used, without any "noticeable" difference to the effective light output.

I made no claim that there was no difference, only that there was no noticeable and I repeat "no noticeable" difference when viewed by the people present. It would be absolutely stupid to expect that there was no difference, or also that there would be no difference in the heating effect and no tables would be necessary to support that statement, otherwise we could easily be conserving a lot of powere usage.

You could probably usefully find some tables to inform the person asking for a solution as to the improved life of the heating lamp, as well as the change in the rate of heating output and at the same time advise them that the use of a thermostat attached to the lamp directly could be rather dangerous unless precautions are taken to prevent the metal tube of the thermostat making contact with a live conductor in the event of the glass envelope shattering , unless of course, a RCD device is in use, or the whole system was well and truely earthed. Of course as a practical electrician I would ensure that both measures were adopted, because the normal 30 ma sensitivity of the sytems in use here in Australia would ensure that minimal damage would occur and the earth bond would also protect life in the event of a failure, where a MEN system is in use.

Regards

Bill Pilgrim

You have had a lot of suggestions offered both complex and doubtful.

A simple method is to fit a bridge rectifier to supply your lamp DC,wind several turns of wire [No of turns depend on your lamp wattage] in series with one of the wires supplying the lamp round a reed relay [ these can be obtained with no,nc or both contacts ]. This can energise a circuit to keep a relay closed or to energise it depending on your preference, when power fails the relay energises or vise versa and switches in your back up lamp.

You can mount all these parts in a small control box with a plug in and a socket out to whatever lamp you choose to operate.

Hello, Look at an automotive tail light circuit. I know that when my tail light burnt out in my car the dash light came on to alert me. hope this helps.

If you put the relay in series with the lamp the coil resistance won't allow the lamp to light. I'll have to do a test but I think if you put the relay in parallel with the lamp and add resistance in series with the coil you'l find a point where the relay will not operate. When the lamp burns out the voltage will rise and operate the relay. This is kind of a crude way to do it though. There are many ways to detect a burnt out bulb. A very low resistance in series with the coil would drop a volt or two and this voltage could be used to trigger an SCR which in turn could turn on the emergency bulb and maybe give you a warning signal as well. You could also sense the light with a foto-cell and it could activate an SCR or solid state relay on light loss.

I do not believe what you described will work. It seems like you need something like the air-craft warning lights control, or in north America called the Obstruction lights. It consists of two sets of light, when the first light failed, the second one would kick in and at the same time the built-in relay circuit would provide dry contact to initate an alarm indicating the light fails. I had used the Siemens production before, I suggest you log on the Siemens web site and look for more details.

did you mean that the holding power = 1.39VA ?

An AC coil resistance is the resistance value of the winding but the Impedence is the Z ohms including the inductance.

The inrush power will normally be ~ 10 x 1.39 = 13.9VA (?)

In any case: Having this coil in parallel with the lamp and adding with it a resistance in series will not work since the failure of the lamp will not make any significant increase in the current flow to engage the coil.

Yes, thermostat is a very good simple economical idea, provided if you can find a good working one in the junk box. For a simple solution you can use a 6.8 ohms 10watt wire wound resistance in series with neutral lead connected to bulb. connect a bridge rectifier with a currnt limitting resistance of 10 ohms in parallel to this resistance. Connect the bridge rectifier output to a 6volt 100mA DC relay with a protective zener (if you feel required) of 7.5volt 3 watt. use the NC contact of the relay to drive the slave bulb.

I would approach this differently, instead of sticking anything in series with your load, basically reducing the voltage and available power to it, I would use a current transformer (CT) to tap the energy necessary to drive a transfer relay. This way you are not modifying the primary circuit at all. A CT is basically a secondary coil of a transformer wrapped around the primary load conductor, a very neat solution. The supplier of the CT can help you size the correct CT.

Use a relay with a output normall closed contact.

The lamp circuit must have a switching contact to power the lamp.

Connect the relay in parallel with the lamp between the same connection points for the lamp i.e. output point of the power switch to the lamp and the other 240VAC feed.

Using the normally closed pair of contact from the relay connect it series with the backup lamp across the power circuit for the backup lamp.

Have fun.