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Old 16th January 2007, 09:21 AM   #1
Default Beam break (dark alarm) question

Hi all,

I've got a beam-break detector working. The circuit below works, but I don't think it's a good circuit. I'm happy with the 555 (not saying it couldn't be improved); it's the transistor trigger section I'm not sure about.

The purpose is to cause pin 2 on the 555 to go LOW when the beam striking the phototransistor is broken. The 555 should pulse out for about 3 seconds when that happens. So far, this works fine. I'm just not sure how goofy the circuit is.

Here's the circuit:



It's all +5VDC. The far right + output is the trigger out, it actually goes to a resistor then an inverter into the counter.

What I need to know is this: if I'm reading my books right, an NPN should have the collector positive relative to the emitter...but in the above it seems it's the other way 'round. However, it demostrably works. What I can't predict with what I know is how reliable it is.

Here's my noob explanation of what's going on:

When PT1 is lit, it robs the base of T1 of its current by diverting it to ground, which causes T1 not to conduct, pulling the base of T2 to ground via R1. When the beam is broken, PT doesn't conduct and T1's base is free to go high via R2. Therefore T1 conducts, also pulling T2's base high, which brings the 555's pin 2 to ground, triggering the monostable.

T2 is just there as an inverter/buffer because it was for a piezo trigger as originally designed, but piezos turned out to be fiddly to position and fragile. Still good for simple drum triggers though. It also seems upside down.

Anyway, would I be OK leaving it like this? If not, would simply swapping in 3906s make it safer? I'm still researching, but at times in circles. Learning as a hobbyist has it plusses but no prof/mentor makes it...interesting.


Thanks,

Torben

Last edited by Torben; 16th January 2007 at 09:47 AM.
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Old 16th January 2007, 11:56 AM   #2
Default

hi torben,
can you recheck your circuit diagram. Add your resistor and capacitor output circuit.

where are you taking your output from ?
is Q connected to +v?

Regards
EricG

Last edited by ericgibbs; 16th January 2007 at 12:00 PM.
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Old 16th January 2007, 03:37 PM   #3
Default

hi torben
I have re drawn the transistor section of your circuit.
Can you now see why I think you should recheck your drawing.

If you need more guidance, ask.

regards
EricG

excuse the rough sketch, its just to indicate whats wrong.

Last edited by ericgibbs; 7th July 2008 at 12:18 PM.
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Old 16th January 2007, 09:18 PM   #4
Default

Quote:
Originally Posted by ericgibbs
hi torben,
can you recheck your circuit diagram. Add your resistor and capacitor output circuit.
OK, I'll check it again. I've checked it and rechecked it and gone back and forth to the 3904 datasheet and it looks like what I have, but maybe I've got that weird blindness one gets from staring too long at one thing.

There isn't a cap on the output (should there be?) The output right from the 555's pin 3 out is:

Code:
 
                                   VCC
                                    +
                                    |
                                    |
                                   .-.
                  VCC              | |
                   +               | |
                   |               '-'
                   |                |
                   |                o-----o
                   |                |
                   |                |     74LS90 pin 14 (CKA)
                   |    ___       |/
               o---o---|___|------|
        555 pin 3                 |>
                                    |
                                    |
                                   ===
                                   GND

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
So in my original circuit, the two leftmost circles-with-plus-signs are +5V and the rightmost one leads to the inverter above.

Quote:
where are you taking your output from ?
From the 555? It goes from pin 3 out the far right circle-with-a-plus-sign.

Quote:
is Q connected to +v?
Q on the 555? Yeah I didn't notice until later that those weren't labeled--sorry). From the +5V source second from the left.

Quote:
Regards
EricG

Thanks,

Torben
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Old 16th January 2007, 09:27 PM   #5
Default

Quote:
Originally Posted by ericgibbs
hi torben
I have re drawn the transistor section of your circuit.
Can you now see why I think you should recheck your drawing.

If you need more guidance, ask.
Great, thanks! That looks a lot more like what I originally planned to build, based on my Art of Electronics. Then I wired it up while way overtired, went to bed, woke up, and found the transistor circuit I originally posted. So I figured what the heck and tested it--and it worked, despite the fact that it breaks at least rule #1 of transistors (collector should be above emitter, potential-wise).

I guess I didn't word my original question correctly: I thought all along that this is wrong, but want to know why the dang thing works.

So...I'm going to redesign, but can anybody explain why the above works? Can the transistors work backward like this, and if so, why? Or does my datasheet just not match the pinout on my 3904 (which I hope is not true)?

Quote:
regards
EricG

excuse the rough sketch, its just to indicate whats wrong.
Something comes to mind about equine dentristry. The fact that you are helping is more than enough, I don't need great artwork too!


Thanks!

Torben

Last edited by Torben; 16th January 2007 at 10:14 PM.
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Old 16th January 2007, 11:44 PM   #6
Default

A transistor with a current gain of 230 when connected correctly will have a current gain of about 5 when its collector and emitter are connected backwards, so some circuits will work. The backward circuit's max voltage before avalanche breakdown of the reverse-biased emitter-base junction is about 6V.
__________________
Uncle $crooge
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Old 17th January 2007, 08:04 AM   #7
Default

hi torben,
as audioguru points out, transistors will work connected the 'wrong' way
around, but at a much lower gain.

If my redraw of your original circuit is correct, then T2 is not connected
to the rest of the circuit. ie T2 collector is to 0v and its base is to 0v
via a resistor, no other connection.

Q is the 555 output and should not go directly to +v, I would suggest a
4k7 pullup resistor from Q to +V.

You do not need a capacitor in the output of your transistor to the counter.

I would recommend a decoupling capacitor of about 100mF 6v between the +5v supply and 0v gnd and a 0.1mF in parallel with 100mF

Keep trying, we will get there

Regards
EricG
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Old 17th January 2007, 10:24 AM   #8
Default

Quote:
Originally Posted by audioguru
A transistor with a current gain of 230 when connected correctly will have a current gain of about 5 when its collector and emitter are connected backwards, so some circuits will work. The backward circuit's max voltage before avalanche breakdown of the reverse-biased emitter-base junction is about 6V.
OK, that makes sense. Using that I found some info on avalanche breakdown which should be very helpful. Thanks!


Torben
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Old 17th January 2007, 10:36 AM   #9
Default

Quote:
Originally Posted by ericgibbs
hi torben,
as audioguru points out, transistors will work connected the 'wrong' way
around, but at a much lower gain.

If my redraw of your original circuit is correct, then T2 is not connected
to the rest of the circuit. ie T2 collector is to 0v and its base is to 0v
via a resistor, no other connection.

Q is the 555 output and should not go directly to +v, I would suggest a
4k7 pullup resistor from Q to +V.
:/ There were some errors in my ascii schematic, I'm afraid. Pin 3 does
not go directly to +5V, it goes to the base of another 3904 which triggers the score lamp, and via a resistor to yet another 3904 which inverts the pulse for the 74LS90. Would I still need a pullup on that?

Here is what I have so far (checked). This is just on the breadboard, but it's working for me when clipped into the main circuit:

Code:
                                   +5V
                                    +
                                    |
              +5V                  .-.
               +                   | | 100R
               |                   | |
             |/                    '-'
           ->|  PT                  |
          /  |>                  ---o------> Out
               \                |
               |      ___     |/  T1
               ------|___|----|   3904
                      1K      |>
                                |
                                |
                                |
                               ===
                               GND

(PT = phototransistor)

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
Part of my requirement for this section is that it be as low on parts as possible. The weird thing at the far left is supposed to be the phototransistor.

Is this design fatally flawed, and if so, how? It works for me and seems like it should work.

Quote:
You do not need a capacitor in the output of your transistor to the counter.

I would recommend a decoupling capacitor of about 100mF 6v between the +5v supply and 0v gnd and a 0.1mF in parallel with 100mF

Cool, thanks.

Keep trying, we will get there

Regards
EricG

Thanks for the help!

Torben

Last edited by Torben; 17th January 2007 at 11:37 AM.
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