# lighting circuit help

Discussion in 'Electronic Basics' started by DH, Feb 21, 2005.

1. ### DHGuest

Hello,
I was wondering if any here could help with a bit of info. I've
been asked to help wire up a kinda chandelier "fallen on the floor"
lighting display. Basically it's going to be a mess of bulbs, fixtures
and structure (copper tube, glass, etc...) on the floor. The "look"
has not yet been completed by the designer and I'm just trying to be
ready for whatever is thrown at me. It will mostly be made of standard
25 - 40w bulbs that I will wire up to a variac (silent running is key)
to control brightness. The designer also has her eye on some lower
voltage bulbs she would like to mix in the display. Is it ok to add in
parallel to the 120v lights a series of (lets say) ten 12v bulbs?
Would the 12v bulbs on the neutral side be dimmer than those on the
hot side? As this display will be frequently turned off and on (a few
times a day for a few months) I would like to add an inrush current
limiter to extend bulb life but am unsure of the placement. Right at
the variac on the hot or neutral side? Or multiple locations? The
display will use upward of 1000w and possibly more. Go big or go home!
LOL!

Thanks!!!
DH

2. ### Robert MonsenGuest

You means series, I think.

I'm not sure what would happen (I've never tried it). I know that bulbs
are very low resistance until they heat up. This causes a surge of
current which quickly heats the bulb. Once it's at temperature, the
resistance is such that it limits current to whatever is required for
the power rating. If your 12V bulb is in series with the big bulb, and
subjected to 120V, it's possible the 12V bulb would be destroyed by the
initial surge of current. Once the bulbs are at temperature, though,
what'll happen is that the voltage across the 12V will be affected by
the relative wattages of the bulbs. Consider two 100W bulbs, one 12V,
one 120V. The resistances will be

120^2/100 = 144 for the 120V bulb, and

12^2/100 = 1.44 for the 12V bulb.

Thus, the voltage across the 12V bulb would be

120 * 1.44/(1.44 + 144) = 1.18V

For a 100W 120V bulb and a 10W 12V bulb, we have

120 * 14.4/(144+14.4) = 11V, which is closer.
No. Elements in a resistive circuit are generally independent of where
they are in the circuit. It's possible the non-linearities of the bulb
heating will muck with this, but it'll only be a factor at startup.
Thus, it may affect whether the 12V bulb will survive.

As this display will be frequently turned off and on (a few
Again, it shouldn't matter. These devices can be seen as a resistance
which gradually diminshes with time. Initially, the resistance is much
greater than the resistance of the bulbs, and gradually subsides. Thus
the voltage divider which is created causes most of the voltage to be
across the limiter. However, again, non-linear effects may be an issue.
This device may also enable your 12V bulb thing.

However, if you are using a variac, why not just use it to slow start
the whole system? That would protect your 12V bulbs.

The

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

3. ### Rich GriseGuest

DO NOT put bulbs in series unless they came that way on an Xmas tree
string.

Get a low-voltage (i.e. 12V) lawn-light transformer, and drive the
12V bulbs with it.

Give each bulb its own two-wire feed - you can use surprisingly small
and inconspicuous wire to power ONE BULB, whether it's a 120V 150W
or a 12V garden-path light. Put a terminal block in the base, and
connect each bulb to its own terminals, much like a breaker box.

But if you want to use 12V bulbs, get a proper transformer.

Good Luck!
Rich