-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of James Tu
Sent: 31 July 2004 09:40
To: PICLIST@MITVMA.MIT.EDU
Subject: [EE:] Lights and power supplies

I'm trying to size the appropriate power supply for a bunch of 12 VAC,
15
Watt bulbs, which are connected to buttons.  There is a momentary single
pole double throw push button (it has a normally closed lead, a normally
open lead and a common lead.)
The bulbs go into their own housing and has two leads for power.

A rectifier bridge to hooked up to a 12 VAC power supply to convert it
to
DC power.  This DC voltage from the rectifier is used power the 12 VAC
lights (will this pose a problem?)  I guess the 12 VAC is actually the
RMS
value so the peak value is 12/.707 = 16.9 V.  (Does this mean the 12 VAC
bulb can handle a 16.9 VDC?)

We are doing this because we need to interface the button to my PIC.
The
+12 VDC of the rectifier is connected to the bulb and the other end of
the
bulb is connected to the NC lead of the button.  The negative lead of
the
12 VDC from the rectifier is connected to COM of the button.  The NO
lead
of the button is connected to a pin on my PIC.  My PIC's GND is also
connected to the COM of the button.  Every time someone presses this
button, the light goes out and my PIC will see a "0." (there's a pullup
on
the PIC pin.)

I've never had to think about lights in as much detail as on this
project
so here are a few questions regarding bulbs and electronics.

Since it is a 12VAC/15W bulb, the resistance is P=V*V/R so R is 12*12/15
=
9.6 ohms.
I = V/R = 12 / 9.6 = 1.25 Amps...I guess bulbs need to draw that many
amps
to get hot enough to emit light.  :)

We have about 30 bulbs that we are powering with the power supply.

Here are the questions:
(1) The bulbs are 12 VAC so does this mean that it can handle 16.9 VDC?

I assume you have no smoothing capacitor on the DC (rectified AC). Hence
the ave voltage applied to the lamps is Vpeak (16.9) X SQrt 2 (0.707) =
12 V so the lamp is working on 12 v nominal with a peak voltage of 16.9v
its actually slightly lower that the 12 vac you started with as you have
to remove the 0.8v x 2 for the rectifier forward voltage conduction. The
16.9 volts only becomes apparent when you start to add smoothing
capacitors and is then a function of how much you can store versus the
load consumption

(2) Can I drive these bulbs with a DC power supply?

Yes !

(3) If these bulbs can be driven with DC should I just use a 15W*30 =
450 W

assuming the voltage is 12v it's a function of ohms law P=VI but allow
some overhead say 10% and ensure that the power supply can support the
cold starting current. A tungsten lamp has a very low cold resistance
and only increases to its rated value when hot. It can be as low as 10%
of the hot resistance for some Tungsten halogen lamps for instance a 12v
50w dichroic lamp has a cold resistance of 0.7R when cold but when hot
draws a current of 4.1A this indicates a resistance of 3.0R. Some power
supplies cant take this type of starting current.

Steve...

DC power supply for the bulbs?


Thanks.

James

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