DIY laser schematic

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

http://www.diyforums.net/laser-alarm-schematic-noob-questions-459027.html

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?
 
yan@seiner.com wrote:
A friend of mine dug up this schematic he wants to build with his son.

http://www.diyforums.net/laser-alarm-schematic-noob-questions-459027.html

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?
Click to expand...
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
 
On Tue, 14 Apr 2009 10:32:16 -0700 (PDT), yan@seiner.com wrote:

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

http://www.diyforums.net/laser-alarm-schematic-noob-questions-459027.html

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?
Click to expand...
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
 
On Tue, 14 Apr 2009 12:47:22 -0700 (PDT), yan@seiner.com wrote:

On Apr 14, 12:32 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

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.

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. :)
Click to expand...
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
 
yan@seiner.com wrote:
On Apr 14, 1:16 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 14 Apr 2009 12:47:22 -0700 (PDT), y...@seiner.com wrote:
On Apr 14, 12:32 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
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.
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. :)
Well, then it's backwards. If the photocell is a photoresisor as
drawn, it beeps in the dark, not when hit by light.

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.
Click to expand...
That should work; at least it's the right thing to try first.
Why not buy a cheap laser pointer? An LED will practically have to
touch the photocell, in a dark room, to be detected.

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

Play with it a bit to make sure that it'll work reliably -- you may find
Click to expand...
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
 
On Apr 14, 12:32 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

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.
Click to expand...
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. :)
 
On Apr 14, 1:16 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 14 Apr 2009 12:47:22 -0700 (PDT), y...@seiner.com wrote:
On Apr 14, 12:32 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

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.

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.  :)

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

Why not buy a cheap laser pointer? An LED will practically have to
touch the photocell, in a dark room, to be detected.
Click to expand...
I think that's the intent. In my sloppy thinking, that is. :)
 
On Tue, 14 Apr 2009 12:47:22 -0700 (PDT), yan@seiner.com wrote:

On Apr 14, 12:32 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

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.

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. :)
Click to expand...
---

This will work: (view in Courier)


.. +-------+-------------+
.. | | |
.. | [LDR] [SIREN]
.. |+ | |
.. [BAT] | C
.. | [10K]<--[1k]--B 2N3904
.. | | E
.. | | |
.. +-------+-------------+

In ambient light, adjust the 10k pot in the direction which just causes
the siren to go off, and then a little more. Then, when the laser plays
on the LDR, the siren should sound.

JF
 
On Tue, 14 Apr 2009 11:25:56 -0700, Tim Wescott wrote:
yan@seiner.com wrote:
A friend of mine dug up this schematic he wants to build with his son.

http://www.diyforums.net/laser-alarm-schematic-noob-questions-459027.html

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?

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).

The circle labeled "+9V" isn't the battery, it's only the positive
Click to expand...
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
 
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