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'[EE:] Lights and power supplies'
2004\07\31@013922 by James Tu

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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?

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

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


Thanks.

James

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2004\07\31@043920 by Russell McMahon

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> 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.)

> 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?)

> We are doing this because we need to interface the button to my PIC.

STOP!
Please explain simply and clearly what you actually want to *achieve*.
While this sounds like a simple enough task it still has good and bad and
cheaper and dearer ways of doing it. It seems likely that you are heading
for a dearer and worse solution :-(.

IF the problem is -

- Control N x 15W, 12 VAC bulbs from 12 VAC using manual switches.

- Detect when bulbs ae turned on using a PIC

Then you are doing it harder and dearer and wronger than necessary.

The following assumes the above.
Please explain any bad assumptions i have made.
__________

Operate the bulbs from an 12 VAC AC source.
Wattage is N x 15W. 30 bulbs = 450 W.

The changeover (SPDT) switch you describe can be used to also signal the PIC
if you wish (OR you can detect the AC on the lamp - easier and cheaper than
what you propose.

Switch action detect.
- Common to ground
- NO to bulb
- NC to PIC pin
- Other side of bulb to 12 VAC.
- PIC pin set to input with internal pullup.

Swicth open - PIC sees ground. Bulb sees O/C
Switch closed - PIC sees high. Bulb sees 12 VAC.

Debounce etc on PIC pin as per usual. Switch arcing or breakdown can destroy
PIC. Extra protection easily provided if desired.
Beccomes part of usual PIC switch interfacing discussion.

OR take connection from "bottom" of bulb via diode, 100k + 33K res in series
to ground. 0.1 uF cap 33k/100k point to gnd. Same point to PIC pibn. Some
resistor changes may be needed depending on final parameters.

____________

Bulbs on DC can be fun: The 12 V rating of 12 VAC is the RMS value. It is
the equivalent to 12V DC in a PURE LINEAR RESISTIVE load. eg if you have a
24r resistor then 12V DC or 12 VAC RMS sine wave will cause the same heating
in each (power = V^2/R = 12^2/24 = 6 watts

****** BUT ****** a 12 VAC bulb is NOT a pure linear resistor. It increases
in resistance as it heats. As the voltage rises the bulb draws more power,
heats more and the resistance rises somewhat. The thermal time constant of
mains is such that the bulb retains most of its temperature across the low
voltage portion of thehalf cycle - but there is some variation. Run a bulb
on DC and the heating is constant at all times BUT the bulb will see no peak
voltages or low voltages. There will be an equivalent voltage (eg ABOUT 12
VDC for a a2 VAC bulb ut it wont be exactly 12V.

NBNBNBNB        An automotive bulb is rated for DC voltage - actually about
13.8 VDC as this is the notional float point fora "12V" lead acid battery. A
mains AC bulb and a low voltage bulb intended for AC use will of course be
AC rated.

For practical purposes you can assume that 1 12VAC bulb will run OK on 12
VDC and vice versa. If you apply 12 x 1.414 = 17 VDC to a 12VAC bulb it will
run very bright and very white - possiblty followed by very black and never
going again :-)

If you had to run your bulbs on rectified AC for some reason then the
'stiffness" of your DC would affect results. If you just rectify 12 VAC and
use very large amounts of capacitance so your voltage is very smooth then
the bulbs will die. If the DC is poorly filtered and ripples you may be OK.


- What is yor application?
- What are your REAL needs?


       Russell McMahon



















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'[EE:] Lights and power supplies'
2004\08\02@021859 by James Tu
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Hi Russel:

Thanks for the suggestions and helpful info!

Please see below for response.


IF the problem is -

{Quote hidden}

Here's the complete picture of what we're trying to do:

-We have 30 buttons which have bulbs.  We want the buttons to be lit in
their normal state.  When someone presses a SPDT button, we want the light
to go off.  When the button is released we want the light to go back on.
-There will be 3 user stations with 10 buttons each.  When a user presses
ANY of the 10 buttons, I want to PIC (which is using pullups) to detect a
button push.  Because of the application, we don't care which button is
pushed, as long as one or more of them are being pushed.  So, essentially
we just need 3 PIC pins...one for each station.  So the 10
button/light/switch combination will be connected in parallel for each
station.  When any button(s) are pressed, I want the PIC to see GND.


{Quote hidden}

OK.  I've always kept my AC and DC parts of circuits separate when I have
to work with both.  I would use a relay to keep the AC circuitry
separate.  Ex:  using a PIC to turn on a toaster.

Here, you are suggesting that I connect my PIC circuit DC GND to the 12 VAC
GND?  So the COMMON of my switch is connected to the GND of my DC ground
_and_ AC ground?  Could this cause damage to my pic circuit in anyway?  So
you're saying GND is GND, no matter if it's AC or DC?



>OR take connection from "bottom" of bulb via diode, 100k + 33K res in series
>to ground. 0.1 uF cap 33k/100k point to gnd. Same point to PIC pibn. Some
>resistor changes may be needed depending on final parameters.

This is a cool method, but since we are using switches I will use your
first suggestion.




James Tu
spam_OUTjamesTakeThisOuTspam2-bit-toys.com

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2004\08\02@031613 by Russell McMahon

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> Here's the complete picture of what we're trying to do:
>
> -We have 30 buttons which have bulbs.  We want the buttons to be lit in
> their normal state.  When someone presses a SPDT button, we want the light
> to go off.  When the button is released we want the light to go back on.
> -There will be 3 user stations with 10 buttons each.  When a user presses
> ANY of the 10 buttons, I want to PIC (which is using pullups) to detect a
> button push.  Because of the application, we don't care which button is
> pushed, as long as one or more of them are being pushed.

Answering your question about grounds 1st.
REAL ground is REAL ground wherever you find it.
BUT, as you suggest, ground around power equipment can be less real than low
power electronics ground. If you can keep low and high power separated,
within reason,  your circuitry is liable to be happier.

Think about the following solution and then rethink interconnections.

Per bulb etc in a group of 10 bulbs:

- Bulb top to 12VAC
- Bulb bottom to NC
- Switch common to ground
- NO to signalling common from all 10 switches.

When any switch in the group is closed the signalling common (SC) will be
grounded. Otherwise it will be floating, although it will have a degree of
AC coupled into it via switch capacitance etc.

The SC *COULD* be taken to a PIC pin via an eg 100k resistor with a 1M
pullup at the pin and an eg 0.01 uF to 0.1 uF, pin to ground. (Capacitor
will vary depending on button press times). This does produce a degree of
low to high power coupling but is probably "safe enough".
It's what I would most probably do myself, probably with the addition of a 1
transistor buffer.

For complete isolation try this.
Use above circuit for bulbs and switches.

Make a nominal 12VDC supply (you may have one already).
But eg use a bridge rectifier from your 12 VAC supply or use a single diode
(1N400x etc).
Smooth this with eg 100 uF to 1000 uF.
This will power 3 optoisolators / optocouplers / optos.

+12 VDC nominal (probably about 16v) to 2k2.
Other side of 2K2 to opto input diode anode.
Cathode of opto input diode to SC (signalling common) of 10 switches.
Whenever SC has a ground on it the opto input diode will conduct ABOUT
(16-1.5)/2200 =~~ 7 mA
Adust resistor to suit desired opto current.

The output of the opto can be used to drive the PIC as desired.
Opto input current depends on CTR (current gain) of chosen opto.
4N25 is an industry standard starting point.

This is overkill for this application but it is cheap and workable overkill.

The arrangement you describe is moderately annoying to test for 1 or more
unlit bulbs if you do not have the changeover switches that you originally
described. It can be done easily enough but needs one device (or more) of
some sort per bulb. eg you could power an opto from each bulb and have all
10 opto outputs in series so if one turns off the circuit is broken. And so
on.
Note that you can buy quad optos in a 16 pin DIP package - these are cheap
and compact.

You could drive a photoresistors from each bulb.
etc

Whatever you do you are effectively implementing a 10 input AND gate.

A non AND solution (more or less) is to measure total bulb current and trip
a detetcor when this reduces.
Could be cheapish with care. Also sensitive to bulb parameter variation,
supply variation etc.

For interest only - can you tell us what the application is ?




       Russell McMahon

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