Anonymous321

Hi guys. My first post and thread.

Ok, so I'm making a laser alarm. And I need help. The schematic of what I have is attached. This is what (I believe) happens:

- Photocell is given voltage through laser beam AND 20 1.5 V batteries.
- Voltage flows to the transistor. When there's enough voltage for the base of the transistor (it requires 30V), it opens and voltage flows through the emitter of the transistor. I'm a little sketchy here. So the current from the 9V battery cannot power the siren when the base of the transistor is getting 30V? But how? Like is it that b/c the switch is open, there is not a complete circuit or something for the 9V to power the siren?
- And when the transistor is off (beam is broken), there is a complete circuit (I guess?) and the siren rings.

The only part of that circuit that I haven't done yet is the 20 AA batteries connected to the photocell. They should give 1.5V each (so 30V in total) to the transistor so when I have the 9V clipped in, the alarm shouldn't ring, right? (I'm gonna try after this post to see if this happens or not)
But then there's also the problem that if it does work correctly, I don't think that the beam would be doing anything useful b/c it's the 20 AA batteries providing all the 30V. So the only way to turn the transistor off and get the siren ringing is to remove one of the batteries or something. I couldn't tell how much voltage the laser pointer would be giving to the photocell with my multimeter on its lowest-10V-range (the needle just wouldn't move) so I don't know exactly how many batteries I would need. Am I even connecting it properly? I just attach the red and black leads of the multimeter to the 2 leads coming out of the photocell and shine the laser pointer onto the photocell.....The voltage the laser is giving is probably like 0.3V or something anyway. But even then, how would I get something to provide the rest of the 29.7V?? 19 AA batteries would provide 28.5V and the laser's voltage on that will still not be enough for the transistor to turn on.
So I'm really confused on how to go on from here...I would really appreciate it if any of you helped out. Thank you in advance.

Attached Images

My circuit right now.JPG (19.0 KB, 39 views)

Anonymous321

Also, I may just be reading the transistor's ratings incorrectly.

This is what is says:

Silicon..........NPN
Typical hFE...200

Maximum ratings
VCE.............30V
IC................800mA
Dissipation.....1.8W

Does maximum rating mean it needs that much? Or you should only provide up to that much? I would think the latter but then even 1V should open the transistor's switch...

Anonymous321

May someone please at least verify my schematic?

john1

Maybe like this ....
try to give a little more info.

John

Attached Images

laser-sensor.JPG (14.3 KB, 29 views)

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Anonymous321

Thank you for the reply, John. Ok so I see how you have 4 connections from the photocell: the resistor, the base of the transistor, the emitter of the transistor, and the ground. But how come you have the photocell connected to the emitter and base of the transistor? Shouldn't it just go to the base? Also, in my original circuit, I had only one wire coming from the photocell. It was to the base of the transistor. Does a photocell have a + and - system so that I'd have to have a wire from the other lead going to the ground? Also, how is the transistor powered? By the 9V as well? Also, for the resistor--do you know any specific resistor that I'd need to use? Or would any "dinky" one work?

john1

If your little siren doesn't take much current,
then the circuit i've shown may work.

The resistor has to be a low enough value to turn the
transistor on, with no light on the sensor
and a high enough value so as not to prevent the sensor
turning the transistor off, when there is light on it.

I would expect to see a darlington pair in this configuration,
or two transistors in direct cascade instead,
with the resistor around the 80K area.

See what others say, before making the unit up.

John

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john1

Yes, almost all types of photo-cell are polarity concious.

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john1

The resistor should be chosen by its Ohms.
Usually colour-coded on the body of the resistor.

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john1

Using a single transistor like that,
I would guess at around 10 K,
but there is really nothing to go on.

See what others say.

John

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Boncuk

Hi Anonymous321,

John was right with his suggestion. I altered it to be an NPN-transistor.
Using a photo resistor the circuit is really easy to make. A photo resistor has a high resistance in the dark (300KOhms and more) and a very low resistance at bright sunlight (represented by R4 in the schematic).

This circuit activates alarm whenever the lightsource is interrupted. If you want it to react the opposite way just change the positions of R1 and R4.

The switch is only for simlution of dark and bright. Omit it in your circuit.

Use a self contained piezo buzzer (built in oscillator circuit). They have low power consumption and a noise level up to 105db.

Regards

Hans

Attached Images

Photo-Alarm.png (53.9 KB, 27 views)

crutschow

The maximum ratings (usually declared as "Absolute Maximum Ratings") are just that, the maximum at which you should operate the device. Anything beyond them may zap the device. Typically you want to operate well below (perhaps 50-75%) of the maximum for best reliability.

Anonymous321

Thank you for your reply.
So even 1 amp of current should open the transistor? On this website,
http://www.kpsec.freeuk.com/components/tran.htm
it tells me that Vce is the "Maximum voltage across the collector-emitter junction. You can ignore this rating in low voltage circuits." So I don't understand. How is the transistor supposed to work (what voltage does it require) so that the siren doesn't ring?

john1

I am trying to compose an explanation for you.
But some things are puzzling me.
Are you English ?
I think you are.
Is this a school project ?
... off line at a quarter to nine ? ...
If so, then i think you should not stay up so late.

I will try to compose an explanation for you, but for now
i have to get on with removing an engine from an unhappy
mini.
Hopefully it will get happily fixed.

Later, John

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crutschow

1A of current through the transitor could fry it.

The voltage across the collector-emitter is determined by the supply voltage and whether the transistor is on or off. When off, it will be equal to the supply voltage and when on, it typically is a few tenths of a volt. The collector current, when on, is determined by the value of the supply voltage and collector load resistance (in this case the resistance of the sounder/siren).

The base current controls the transistor. A bipolar transistor is current controlled with the base-emitter drop being about 0.7V when the transistor is on. The required current to turn the transistor on depends upon the transistor gain (hfe). Good design practice is to use a gain of 10 to completely turn on the transistor when it's used as a switch. Thus, for example, if the collector current, when on, was 100mA, you would size the base resistor to provide 10mA of base current.

When the photoresistor pulls the base-emitter voltage below about 0.6V the transistor will be in the off state and the siren is silent.

Torben

What they mean is that in low-voltage circuits, you're not going to have voltages high enough to exceed the Absolute Maximum Vce, which is between 20V and 100V in the chart shown.

The Absolute Maximum ratings are the highest levels at which the device is guaranteed to work. It's not guaranteed to work for long if one or more of the Maximums is met for very long, though. As crutschow noted, make sure you're operating the device with some headroom--you don't want to hit the Absolute Maximums in normal usage. Some people refer to these ratings as the Fry Points--go past them and you risk frying your device.

I would second John1's suggestion of a Darlington pair--I find that I get a much sharper on/off. For a light-trigger alarm, what I would do is put the LDR from base to ground, a 10k* resistor from base to Vcc, hook the emitter to ground, and then hook up the buzzer/LED/whatever from the collector to Vcc.

* - The value of the 10k resistor depends on the value of the LDR. I actually used a pot instead of the 10k resistor to find a good operating range and then measured the resistance of the pot, found it was near 10k, and then replaced the pot with the 10k resistor.

Also, the alarm will only buzz as long as something is breaking the beam and will immediately go dark/quiet again once the beam is restored. You can hook up a 555 triggered via an inverter from the collector of the Darlington to control the length of time the alarm stays on for if the beam is broken (say something breaks the beam for 0.1 second; you can set the 555 to keep buzzing for, say, 5 seconds, so you have time to notice the alarm.

Make any sense?


Torben