DIY laser schematic

A friend of mine dug up this schematic he wants to build with his son.

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If you scroll down a bit, you'll see an annotated schematic. It shows the +9V battery and on the right side it has Positive side of B1 and on the left side it has Negative.

I'm not much of an electronics whiz, but that only makes sense to me if the - side of the battery is connected to chassis ground. If you connect the - side of the battery to the voltage divider consisting of the pot and the photocell Q would never trigger.

Am I totally out in left field?

Reply to
yan
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No, I think you're in the ballpark. Unless I have my head up my donkey*, as wired the circuit can deliver no more current than will flow through the 5K pot and the photocell.

I _suspect_ that the 9V battery is supposed to go from ground to the transistor collector, but if you do that then your siren will go off until the photocell starts conducting (presumably it conducts when it gets lit up).

  • I never did understand that turn of phrase :-).
--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

You are correct.

The schematic is drawn in a very nonstandard way. The third diagram, the red-lined one, assumes the "B" circle is a 2-terminal battery. It isn't. The big round "B" circle is actually the positive battery terminal, a single circuit node. The battery negative terminal should be grounded.

The buzzer will sound when the photocell is in the dark, if everything works right. But 9 volts through an emitter follower, with a weak base pullup, is pretty wimpy to drive a 12 volt buzzer.

Ugly.

John

Reply to
John Larkin

Do you have a better one? This project was devised by a dad so his enterprising son wouldn't take apart the family dvd player, build a handheld laser, and attack who knows what. ;-)

Seriously, this is a fun project for a father and son who know little of electronics. I got pulled in because I know a few basic principles, but certainly not enough to design a circuit. The requirements are that it has to be simple, be triggered by a "laser" - LED in a wand - and make noise. :-)

Reply to
yan

Well, then it's backwards. If the photocell is a photoresisor as drawn, it beeps in the dark, not when hit by light.

Why not buy a cheap laser pointer? An LED will practically have to touch the photocell, in a dark room, to be detected.

John

Reply to
John Larkin

DUH! That makes sense. I kept trying to figure out how that worked... The circuit is cribbed from a 'burglar alarm' - break the beam and it goes off. What we want is "make the beam and it goes off"

- i.e. reverse the pot and photocell.

I think that's the intent. In my sloppy thinking, that is. :-)

Reply to
yan

That should work; at least it's the right thing to try first.

Play with it a bit to make sure that it'll work reliably -- you may find it goes off in the sunlight, but that doesn't mean it won't be fun inside.

Replacing the transistor with a Darlington, and using a higher-valued pot, may help. Once again -- play with it a bit.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

--
This will work: (view in Courier)


.    +-------+-------------+
.    |       |             |
.    |     [LDR]        [SIREN] 
.    |+      |             |
.  [BAT]     |             C
.    |     [10K]
Reply to
John Fields

The circle labeled "+9V" isn't the battery, it's only the positive terminal of the battery - it's "assumed" that the negative terminal goes to ground.

It's still a pretty cruddy circuit - John Fields's is better.

Cheers! Rich

Reply to
Rich Grise

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