Basic circuit help please

Look at the schematic

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That is the classic 555 astable multivibrator circuit, with the addition of another external trigger. That is NOT what should be there. He would want it as a retriggerable monostable multivibrator. Stays on while light is blocked then times out when light is resumed.

That circuit will not work properly. The better idea is to use the transistor to short the timing cap (threshold pin) to ground and ignore the pin 2 trigger input entirely.

The author of the article, Pankaj Khatri, has no clue as to what he's doing. If he did manage to get it to work (as planned) it isn't because it was designed to work as planned.

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t-diagram-with-alarm

hanging

, that

e.

enough.

being

The 555 can be setup two ways, for pin 3 to either go high for a set time a nd then go back low and wait for the next pin 2 trigger event. Or it will g o high and stay high until it is manually reset. So it sounds like you only want it to beep for say 1 second after it gets triggered. Is that correct? Mikek

Reply to
amdx62

Sorry, I could not see your last post (my problem), had to go to google groups to read and post. This website has the correct 555 schematic to get a single pulse out when pin 2 is triggered. I have not compared it to yours. I could, but you should read the page to understand the 555, the 3rd diagram is a schematic of a one shot circuit.

It's just going to be a change of the resistors, and how they connect to the 555. Let me know where you are at on this. Mikek

Reply to
amdx

A quick once over, and I think? that removing R3 and putting a short there and then removing the connection from pin2 to the 10 uf cap will make it put out a single pulse. But again I think?, the time of the single pulse will be about 11 seconds with a 1Megohm and 10uf. with a

1uf it would be 1 second. Mikek
Reply to
amdx

Yes that is correct, well maybe 2 seconds or for the lenght of time the ldr is shaded or detecting.

Reply to
RobH

Sorry, you can't do, "for the length of time the ldr is shaded or detecting" once pin 2 triggers, the 555 take control and does what it is setup to do, pin 2 can't have control again until the 555 had its cycle. Sofor know we'll just do 2 seconds. Possibly address the other part later. OK, we will work on the 555 timer first and get that to put out a 2 second pulse, when pin 2 is triggered? Remember for NOW, you are triggering pin 2 with you 100 ohm resistor to ground, since that worked. I would put a 200k resistor in place of R5. and put a short where R3 is and remove the connection from pin 2 to the 10uf capacitor.

You want your circuit to be wired like the 3rd image on this page,

labelled Monostable 555 timer. Once you get that, you should be able to insert, then remove your

100 ohm resistor (the pin 2 trigger) and hear a 2 second run of your buzzer.

I watch a dozen or more videos looking for a good monostable 555 timer tutorial, so far this is the best I can find, no completely happy but it does describe a monostable 555 operation. I could not find one with a native English speaker. This is 12 minutes but worth your time to learn.

Mikek

Reply to
amdx

Very true, I was not braining.

NT

Reply to
tabbypurr

Just leave the 555 out and put the buzzer across the LED........"KISS" principle

Reply to
RheillyPhoull

That's a surprise... there should be a resistor below R1.

should be close enough.

If they do they're using damaged parts and it's just dumb luck.

I'd be suprised if anyone could, the current though R2 exceeds the capability of R3 to discharge the capacitor.

remove that wire and replace it with a diode pointing leftwards.

--
  Jasen.
Reply to
Jasen Betts

With 10uF and 1M the timing the 555, it is running at less than 1Hz, so a buzzer is the right choice there, not a loudspeaker.

--
  Jasen.
Reply to
Jasen Betts

that circuit won't work on 5V it needs 9v there's too much voltage drop in the transistor and the 741 output can't go low enough.

this mostly due to the choice of transistor. there should be a NPN part there not PNP - then it will probably work on 5V (eg BC337. BC547) emitter at the bottom this time.

then you might need to change stuff on the left around to get the alarm to sound in the dark. (swap the LDR and the variable resistor positions)

then with a diode rightwards from the collector to pin4 you should ge the alarm when it's dark. else you can have the diode leftwards and get the alarm when the LED is unlit.

--
  Jasen.
Reply to
Jasen Betts

Yes, removing R3 from pins 6 and 7 and then shorting the said pins caused a continuos beep. Also I removed the 10uf capacitor from pin2 at the same time.

Reply to
RobH

Is there a link for the 12 minute video. Thanks

Reply to
RobH

Something is not correct in the wiring. It should only beep for the amount of time it takes for the cap to charge to 2/3 Vc. Mikek

Reply to
amdx

Leave the op-amp, and transistor out and trigger the 555 as a monostable directly from the photo detector. The input impedance to the internal comparator is high, the data sheet gives the current as point five microamps. The sensor would have to connect to VCC and the pot be in the ground leg to sense darkness.

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Mikek

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amdx

Mikek

Reply to
amdx

Hi ROBH, I see others commenting on the first part of the circuit while I has concentrated on the 555 part. I don't care which part you work on first, but decide and then go from there. Don't work on both at the same time. Truly, if you get an understanding of how the 555 works it should be easy. Mikek

Reply to
amdx

I've watched a few videos on 555 Monostable on youtube, and have made up a couple of simple circuits to see how they work. I've also looked at a couple of 555 Astable videos.

The circuit I did from the said website works for me just now and I'll live with it until I know I can improve it. The link you gave me for the hobby-circuits site should help with that.

Thanks for your help.

Reply to
RobH

If I understand, you can now pull pin 2 low and the output make your buzzer buzz for a second of two. Is that what happens? Note: I'm not sure if pin 2 is required to have a momentary pull down to zero or if you can just hold it at zero and the circuit will still do it's timing event. Must situations just give pin 2 a short pull down. It might be you just need to connect the 100 ohm resistor for just a very short period and then remove it. Mikek

Reply to
amdx

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