Resistor Value?

[greenhouse]

I have a switch with a 12V AC indicator lamp.

I need to drop 125V to 12V to power the lamp.

What resistor value do I need? (forgot how to compute it.)

 

[Fitseries3]

you want to drop 120v AC to 12v DC?

 

[W1zzard]

no offense, but if you dont know how to calculate such a resistor your should REALLY stay away from mains power circuits

 

[zAAm]

It all depends on the current/wattage rating on the indicator lamp. Let's assume it's rated for 1Watt (you'll have to change the calculations based on this value).

That means that for 12V drop through the lamp, it will have I=P/V=1W/12V=0.083A=83mA going through it. Obviously, this means that the lamp's resistance will be R=V/I=12V/0.083A=144Ohm.

Through elementary voltage division principles, Vj=(Rj/Req)*Vsource. With Vj the output voltage, Vsource the input voltage, Rj the load resistance and Req the total resistance of the circuit. In our case, Vj=12V, Rj=144Ohm, Req=R+144Ohm (with R the unknown resistance and since voltage division works for SERIES circuits with the lamp and resistor in series) and Vsource=125V. If we calculate for R, we get R=1356Ohm=1.356kOhm.

The closest value to this would probably be 1.47kOhm since we want to round upwards to not overload the lamp.

The power going through the resistor will be P=V^2/R=(125-12)^2/1470=8.68Watt.

Of course, this value for R will only work for the current lamp and would have to be recalculated for every other lamp value.

Edit: I agree with W1zzard, you should know what you're doing before tackling heavy current circuits...

 

[Sasqui]

W1z speaks words of wisdom.

Radio shack, or your local electronics store AC-DC power supply 120v to 12v. You need to know the amperage of the lamp to get the right one.

If you have a spare PC PSU, it does both 12v and 5v

 

[greenhouse]

1.47K. Thanks.

Do not worry zamm and w!zzard, my dad was an electrician and he taught me when I was a kid. Since that Time I have wired about 15 houses.

 

[zAAm]

you want to drop 120v AC to 12v DC?

W1z speaks words of wisdom.

Radio shack, or your local electronics store AC-DC power supply 120v to 12v. You need to know the amperage of the lamp to get the right one.

If you have a spare PC PSU, it does both 12v and 5v

The OP said 125V to 12V AC. Thus the output should be alternating current. PSU's give 12V and 5V DC. There are alot of switches with built-in lamps inside that lights up when it's on and these are known to be AC powered (since you don't want to add diodes and crap to a small switch just for the light).

I think it's hardly worthwhile to use a transformer if the lamp is low enough wattage (less than 2W I would imagine would be fine, although using a 2W lamp would require a much larger resistor). 1Watt and less would be best using this method. Otherwise a small and simple AC-AC transformer would work.

 

[greenhouse]

Thanks zaam. It's a small lamp, 1/4 watt res. should do and the lamp is ac or dc compatable.

 

[zAAm]

1.47K. Thanks.

Do not worry zamm and w!zzard, my dad was an electrician and he taught me when I was a kid. Since that Time I have wired about 15 houses.

ONLY if you have a 1Watt lamp!!!
For another value the value of the resistor will be different! Give me that value and I'll help you with that.

Edit: Actually, if the lamp is less than 1Watt, its resistance will be higher and thus it'll drop more than 12V. So if the resistor value is incorrect it'll just blow. If it's greater than 1Watt it'll be dim. Just check the lamp with a multimeter and measure its resistance. Multiply that with 10.42 and subtract the resistance again from that value to get the final value for R.. So if it's 150Ohm, 150x10.42-150=1413ohm.

 

[greenhouse]

I do not know the AMPs of the lamp. It is not on the spec sheet I have. It's a little tiny thing. Can't be very much.

 

[zAAm]

It has to say SOMETHING on that spec sheet. Whether it's amps or watts or resistance/ohms.
And "can't be very much" isn't accurate enough for calculations unfortunately. For "can't be very much" values of 1mA and 10mA the value of R will differ significantly (113K and 11K respectively).

 

[greenhouse]

All I see is 12V AC/DC incandesant.

Here is the link to the spec sheet:

http://www.philmore-datak.com/mc/Page 165.pdf

 

[Sasqui]

The OP said 125V to 12V AC. Thus the output should be alternating current. PSU's give 12V and 5V DC. There are alot of switches with built-in lamps inside that lights up when it's on and these are known to be AC powered (since you don't want to add diodes and crap to a small switch just for the light).

Oops, missed the AC part... and the only way to caluclate what resistor will work is knowing the resistance of the bulb is and operating wattage... as you need to calc both the bulb and the resistor @ 125v to figure the correct amperage. You cannot control voltage with a resistor in a series circuit, only the apparent voltage drop across any given load, and that would depend on all loads in the series.

That also assumes the power source has no internal resistance.

Edit: BTW, I looked at that spec sheet. I'm quite certain you could find a similar switch that doesn't require an external, separate 12v load for the light - that's kind of cheesy.

 

[pajama]

1.47K. Thanks.

Do not worry zamm and w!zzard, my dad was an electrician and he taught me when I was a kid. Since that Time I have wired about 15 houses.

Are any of the houses still standing?

 

[zAAm]

All I see is 12V AC/DC incandesant.

Here is the link to the spec sheet:

http://www.philmore-datak.com/mc/Page 165.pdf

Better get the multimeter out and measure the resistance between those lamp pins. Only way you can know for certain. Otherwise I'd get a very large value (around 500K) and work down from there until the bulb glows brightly. I'd just get a multimeter. Saves a lot of work and you know it'll work.

Oops, missed the AC part... and the only way to caluclate what resistor will work is knowing the resistance of the bulb is and operating wattage... as you need to calc both the bulb and the resistor @ 125v to figure the correct amperage. You cannot control voltage with a resistor in a series circuit, only the apparent voltage drop across any given load, and that would depend on all loads in the series.

That also assumes the power source has no internal resistance.

Edit: BTW, I looked at that spec sheet. I'm quite certain you could find a similar switch that doesn't require an external, separate 12v load for the light - that's kind of cheesy.

Since he specified the resistor value, it was pretty obvious that he wanted to control the voltage drop across the load and the resistor. The loads in series would only be the lamp and resistor like I said so you can easily work it out using plain ol' voltage division. And I did say that he needed the lamp characteristics. Also, power sources' internal resistance is mostly negligible on this scale of resistances and since he's using direct power from the mains, internal resistance is the last thing you have to worry about.

 

[lemonadesoda]

Just to point out (the OBVIOUS)...

While you can use a resistor to create a 12V drop across the lamp (bulb) from a 120V source, you will need a big fat resistor, and whatever WATTS the lamp uses, the resistor will be "using" 9x that amount as heat.

You will need a big FAT resistor and a heatsink... and you waste 90% of the electricity.

 

[greenhouse]

Edit: I do not think my meter is working correctly. I'll check it with my neighbor's tomorrow.
My multimeter is kindof old, but it looks like the lamp is possibly 49 ohms.

 

[greenhouse]

Are any of the houses still standing?

I can not help but smile. Yes, They are doing quite well.

 

[zAAm]

Just to point out (the OBVIOUS)...

While you can use a resistor to create a 12V drop across the lamp (bulb) from a 120V source, you will need a big fat resistor, and whatever WATTS the lamp uses, the resistor will be "using" 9x that amount as heat.

You will need a big FAT resistor and a heatsink... and you waste 90% of the electricity.

Haha, now that I've slept on it I see your point and my calculation mistake in my first post. Guess I shouldn't be posting at 2am... The power through the resistor is not what I said it would be, it is P=V^2/R=113^2/R=8.68Watt for a 1Watt lamp load.

Edit: I do not think my meter is working correctly. I'll check it with my neighbor's tomorrow.
My multimeter is kindof old, but it looks like the lamp is possibly 49 ohms.

Well, with 49ohms, R=461ohms. Which is WAY WAY WAY too low! There's going to go 28Watts through that resistor. Anyway, that means the lamp draws I=V/R=12/49=245mA. Which is certainly too much for voltage division principles on this scale. You'll have to get a transformer rated for 3VA at 12V. Now, if that lamp was an LED it'd be easy as pie to calculate a suitable resistor.