newbie Lamp-dropper

[enthalpystudios]

so is this the newbie forum? Because I have a very green question about voltage dividers, might as well not waste a thread elsewhere.....


Basically, my first attempt at applying ohms law to a real situation has failed

I'm trying to use a series resistor to drop the voltage going to a "power on" lamp. It's a 6V lamp, and I've got 15vdc and 15vac to feed it.

I hooked it up without a voltage divider and it was very dim on the 15vdc, but looked about right with 15vac. this was just for a split second to see if a) the lamp burst in my face, and b) to figure out if it likes ac or dc better.

Looks like ac. Maybe i'm wrong and it just needs the correct voltage.

Anyway, I tried a series resistor that I calculated with ohms law, 9v drop, and I (obviously incorrectly) assumed it would pull 10mA.

So, what I'm gathering (watch the slow new guy learn, eh?) is that I need to hook up my common to the bulb, and connect the V+ to the lamp with my dmm set to mA, which would connect the dmm in series. Then I use that measured current (since, as I learned in CJ's electronics 101, current across a series circuit is constant throughout) to calculate the resistor value for a 9v drop. And then use watts law to determine what type of power i'll need for the resistor.

Am I close?

billy

 

[Larrchild]

Heck yeah, yer close. Let's say the bulb draws 100 ma. R= E divided by I.
So if E is 9 and I is .1 or 100 ma, then R = 90.

Find the bulb's current and do that.

When the bulb looks normal but the resistor starts smoking, come back and we'll discuss watts.

 

[enthalpystudios]

;]

Yeah.

This is on a 4 ch. green pre. I hooked up my common to the bulb, turned on the unit, and connected probes of the DMM from 15vac to the lamp. Looked like way less current than 10ma. Which is good, as I happen to have high wattage resistors in only like 5 and 10 ohms and about 1k and 1k2 ohms. Before I get into the fire bomb mic preamp, quick question:

If I parallel 4 100k 1/4 watt resistors, do I have a 25k 1 watt?

anyway, the light didn't light. Maybe it fried.. i have 3 more to try again.

Except that I burned out 4 resistors on 2/4 channels, but still, I suppose I can disconnect power to the mic pre's and get the bulb going (and now, maybe troubleshoot the psu for any issues.)

At least I know what a fried resistor looks/smells like now.

back to the chalk

thanks for the quick reply larrchild

billy

 

[hejsan]

To figure out stuff like that, would it work to put a pot in series with the bulb full on and then turn it down until you have enough light, and then measure the resistance of the pot at that instance?

 

[mwkeene]

[quote author="enthalpystudios"]
If I parallel 4 100k 1/4 watt resistors, do I have a 25k 1 watt?
[/quote]

Yes...you will have 4 times the current (4 resistors in parallel), but you will also be dividing the total power by 4 for each individual resistor

 

[Scodiddly]

[quote author="hejsan"]To figure out stuff like that, would it work to put a pot in series with the bulb full on and then turn it down until you have enough light, and then measure the resistance of the pot at that instance?[/quote]

Yup, though of course you'd want to do it quickly before the pot has a chance to overheat (assuming enough current for that). I have one of those resistor-selector boxes, basically a bunch of 2-watt resistors in a box with knobs & switches to select which one the two output terminals are connected to. Very handy for this sort of thing, a quick trial & error to find the right resistance for the desired light output.

 

[enthalpystudios]

whoa.... i was wondering where this thread went.

i like the pot idea. but i need to figure this one out on paper for my own sake.

besides, i have to retest the psu first. it's actually the first thing i ever soldered, and I did so with big thick 60/40 and a crappy pencil iron. And then I resoldered all the regs, the bridge rectifier, a couple caps, etc etc.

its a nightmare. I should just make a new one. i was going to but never did. I think that would be smart.

billy

 

[The Kid]

If I parallel 4 100k 1/4 watt resistors, do I have a 25k 1 watt?



Yes...you will have 4 times the current (4 resistors in parallel), but you will also be dividing the total power by 4 for each individual resistor

I think you meant to say 1/4 times the current. The current stays the same, it just divides among the 4 resistors, like the power. :wink:

At least I know what a fried resistor looks/smells like now.

Now move up to a cap or diode, for some real fun! (Don't forget your safety glasses!) :twisted: :sam:

 

[alk509]

[quote author="The Kid"]

Yes...you will have 4 times the current

I think you meant to say 1/4 times the current.[/quote]

No, he meant 4 times the current... If one 100K resistor pulls X current, two of them in parallel will pull 2X, three pull 3X and 4 in parallel will pull 4 times X.

Peace,
Al.

 

[The Kid]

If one 100K resistor pulls X current, two of them in parallel will pull 2X, three pull 3X and 4 in parallel will pull 4 times X.

I think we're talking about the same thing. But, in case we're not, current going into a resistor equals current going out of resistor. If you replace that resistor with 2 of the same value, the total current going in will divide, not double. So, when someone says 4 times the current, that makes me think the current has quadrupled. :sam:

 

[enthalpystudios]

damn, i am the short-meister. i think i may be destined not to have a lamp ;]

seriously....i am trying like all get up to measure current over here. it is a sorry site to watch me try and measure current. i have a fluke 8060a. it just says 1, no matter what scale i set A to. Anyway, all I've accomplished is a couple fried resistors and another fried lm317 48v reg. somehow i keep tripping up that damn 48v.

i'll take a pic tomorrow or something, maybe someone could tell me where to put my probes to measure current, because I'm just missing something. besides it's after 1 in the morning, and i'm sure there are at least 2 people pissed at me right now.

;] all in good fun, i say.

besides, i think this dual 1176 will "lap" my 4 channel green pre, so i'm not all that disappointed about replacing my 6th 48volt regulator (seriously). I work on it when the green pre just isn't going well. And thats obviously been quite a lot ;]

thanks for all the help you guys.

billy

 

[Brian Roth]

[quote author="enthalpystudios"]
I'm trying to use a series resistor to drop the voltage going to a "power on" lamp. It's a 6V lamp, and I've got 15vdc and 15vac to feed it.

I hooked it up without a voltage divider and it was very dim on the 15vdc, but looked about right with 15vac. this was just for a split second to see if a) the lamp burst in my face, and b) to figure out if it likes ac or dc better.

Looks like ac. Maybe i'm wrong and it just needs the correct voltage.

billy[/quote]

SOMETHING is amiss here!! A 6V lamp fed from 15V (AC or DC) should make the lamp VERY bright, if not burning it out quickly.

What am I missing here??

Also, what markings are on the bulb's metal base? Typically it will be a two to four digit number or else a voltage plus either milliamps or Watts.


Bri

 

[enthalpystudios]

checking the base right now..... sadly i'm still at the computer

anyway, i'm actually trying to find an appropriate series resistor.

if there is a current/wattage stated on the bulb, I'll be happy. very.

billy

 

[Brian Roth]

Well, as I mentioned, a 6V lamp (if that is indeed correct!) should be VERY bright when powered from 15V.

I'm now guessing that the lamp *might* be actually rated for 24V or higher....

Bri

 

[enthalpystudios]

i cant be sure, but i believe it is stamped ".15A"

so something close to 60 ohms for a 9v drop?

BUT... i may have bee duped! It appears to be stamped 3V. Then again, this metal base is really hard to read. Maybe I can just get a new lamp or something. I just picked these up because they were a bargain at apex jr.


billy

 

[pstamler]

[quote author="The Kid"]

If one 100K resistor pulls X current, two of them in parallel will pull 2X, three pull 3X and 4 in parallel will pull 4 times X.

I think we're talking about the same thing. But, in case we're not, current going into a resistor equals current going out of resistor. If you replace that resistor with 2 of the same value, the total current going in will divide, not double. So, when someone says 4 times the current, that makes me think the current has quadrupled. :sam:[/quote]

Only if you hang the resistors on a constant-current source. Connect them to a voltage source, and...well, let's say it's 10V. Hang 100k on it and you get 10/100000 = 0.1mA. Hang another in parallel and it will draw another 0.1mA for a total of 0.2mA. Add two more and you're drawing 0.4mA through your total of 25K.

Peace,
Paul

 

[pstamler]

[quote author="enthalpystudios"]seriously....i am trying like all get up to measure current over here. it is a sorry site to watch me try and measure current. i have a fluke 8060a. it just says 1, no matter what scale i set A to. Anyway, all I've accomplished is a couple fried resistors and another fried lm317 48v reg. somehow i keep tripping up that damn 48v.[/quote]

The smoking gun here, if you'll pardon the expression, is when you said "LM317 48V reg." The LM317 isn't really designed to handle 48V; you can trick it by floating the adjust terminal, but it still fries when you ask it to drop that much voltage between its input and its output. So while the LM317 has current limiting in it and therefore ought to be immune to short circuits, it's not immune when you're floating it for 48V.

Since you've already fried it, try substituting a TL783 high-voltage regulator, once again available from TI in a TO220 package. It's designed for this sort of thing.

Meanwhile, get a good magnifying glass and figure out what voltage that bulb wants to see, then measure its resistance with your handy Fluke. Ohm's Law tells you Vbulb/Rbulb = Ibulb; figure out how much voltage you need to drop from your power supply (Vdrop = Vsupply - Vbulb), and calculate Rdrop = Vdrop/Ibulb. Ibulb(squared) x Rdrop = power dissipated in the resistor; make sure the resistor's power rating is at least double that.

Peace,
Paul

 

[Larrchild]

Meanwhile, get a good magnifying glass and figure out what voltage that bulb wants to see, then measure its resistance with your handy Fluke.

The resistance of the filament is lower when cold, however.

 

[Svart]

before you make the mistake of using bulbs, which you will have to change fairly often (as i have learned the hard way..), I bought a bag full of warm white leds. these are so close to incandescent color temp it's uncanny, especially when used through a filter like a jewel. I use these exclusively now. they are well worth the money and much easier to work with than bulbs.

 

[kruz]

Hey Svart, may i ask u more details about these warm white leds?
a link?
Thank you!