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Light activated circuit

audioguru

Well-Known Member
Most Helpful Member
The polarity of your LED is shown upside down (the triangle is positive and the line is negative) and it is missing a series resistor to limit the current and prevent destruction of the LED and the 555. The LED with its series resistor should connect between pin 3 of the 555 and GND, not to Vcc. R2 has no resistance shown.
If you connected the LED polarity correctly and without the important series resistor then the 555, LED and battery must be replaced.

Your circuit is completely different to the original monostable circuit. When the original circuit is triggered with a short duration pulse of light, it turns on the LED and Piezo beeper tor a timed duration. The LDR must be in the dark for the timer to turn them off.
The LDR is supposed to trigger pin 2 when the LDR has light and it drives pin 2 to near GND. When pin 2 triggers the 555 the pin3 output goes near Vcc and pin 7 is turned off.
Pin 6 should connect to pin 7 and have a capacitor to GND and a resistor to Vcc so pin 7 can discharge the capacitor before the LDR triggers pin 2, then when the resistor charges the capacitor to near Vcc the 555 turns off the LED and Piezo beeper. So the 555 circuit is a triggered timer.
Please read the datasheet of an LM555 or NE555.

You are using the 555 as a simple on-off single transistor.
 

fixit7

Member
As drawn the LED is backwards. When pin 3 is low nothing will happen and when pin 3 is high nothing will happen. Why are you taking the out and routing the out (pin 3) back to the trigger (pin 2)? Note I said "as drawn" the LED is backwards.

Take your LDR and just place it on the bench. Measure the light and dark resistance and note what they are. Learn how to use a transistor on the output of the 555 for increased current loads.

Ron
Ok, I will reverse the led symbol.

Instead of taking the out and routing pin 3 to pin 2, what else should I do?
 

Reloadron

Well-Known Member
Most Helpful Member
The polarity of your LED is shown upside down (the triangle is positive and the line is negative) and it is missing a series resistor to limit the current and prevent destruction of the LED and the 555. The LED with its series resistor should connect between pin 3 of the 555 and GND, not to Vcc. R2 has no resistance shown.
If you connected the LED polarity correctly and without the important series resistor then the 555, LED and battery must be replaced.
The OP pointed out it was a 12 V LED with an internal current limiting resistor. That does not change the fact that as drawn the LED is shown backwards though.

Ron
 

audioguru

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Most Helpful Member
Since the LED has a built-in resistor then you should draw it with the resistor. You should also say battery or power supply on the schematic.
Is the LED bright enough and piezo loud enough when the 9V battery has dropped to 6V?

The original circuit has the 555 as a timer but the timer parts are not used.
What do you want the circuit to do? Simply turn the LED and Piezo on when the LDR is lighted then turn them off when the LDR is dark?
A simple transistor will do it.
 

Reloadron

Well-Known Member
Most Helpful Member
The LED is now correctly drawn. Still no idea why you have pin 2 (Trig) tied to Pin 3 (Output) using that SPDT switch? What is that supposed to accomplish?

Ron
 

fixit7

Member
Since the LED has a built-in resistor then you should draw it with the resistor. You should also say battery or power supply on the schematic.
Is the LED bright enough and piezo loud enough when the 9V battery has dropped to 6V?

The original circuit has the 555 as a timer but the timer parts are not used.
What do you want the circuit to do? Simply turn the LED and Piezo on when the LDR is lighted then turn them off when the LDR is dark?
A simple transistor will do it.
The resistor is insulated and if I strip it back it will destroy the led.

My circuit would not work with my current piezo, so I substituted the led.

The led also gives a visual indication that my circuit is energized.

I simply want my circuit to sound an alarm when the laser beam is interrupted.

I did some reading.

Do I need a field effect transitor to be able to use a buzzer?
 
Last edited:

Reloadron

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Most Helpful Member
I simply want my circuit to sound an alarm when the laser beam is interrupted.
OK, so here is a few things to consider. You mention a LASER. What your drawing is showing is just a basic LDR (Light Dependent Resistor). A common everyday LDR is sensitive mostly to ambient light be it room light or daylight. What you really want is a something photo sensitive to LASER light. What you need is a LASER Emitter (Light Source) and a LASER Detector. The links are merely examples and they can be had inexpensive on Amazon. The linked detector modules output a logic high when beam is present going low when beam is broken. With any LDR or LASER detector you should shield the sensor so ambient light has no effect.

Depending on your load, be it a buzzer, LED or whatever the 555 only can drive so much so we place a transistor at the 555 output and use the transistor as a a switch for the load. Less an external transistor you can only get so much current (source or sink) from the 555 alone.

Finally, I am not saying a basic generic LDR is a bad choice for LASER light, only that there are much better choices. Especially if you want a reliable circuit.

Ron
 

fixit7

Member
For now I would like to get my current circuit working with a buzzer.

I plan on putting the detector inside a tube.

I would like to try using a transistor as a switch.

I already have a laser. I may get the laser detector later.
 

audioguru

Well-Known Member
Most Helpful Member
The resistor in series with the LED should be shown on the schematic, not on a photo.

You say you want the alarm to light and sound when a laser beam is broken.
For how long, what turns off the alarm? Do you need a timer to turn off the alarm after a few minutes if a big truck blocks the beam for weeks? The original circuit's timer does not time-out if the beam is continuously blocked.

Why are you talking about a Fet? A Fet, a transistor, a 555 or an LDR could activate a piezo beeper. Any one of them can be used. If the light beam is dim or if the LDR has poor sensitivity then it must be used to drive a Fet, a transistor or a 555.
 

fixit7

Member
The resistor in series with the LED should be shown on the schematic, not on a photo.

You say you want the alarm to light and sound when a laser beam is broken.
For how long, what turns off the alarm? Do you need a timer to turn off the alarm after a few minutes if a big truck blocks the beam for weeks? The original circuit's timer does not time-out if the beam is continuously blocked.

Why are you talking about a Fet? A Fet, a transistor, a 555 or an LDR could activate a piezo beeper. Any one of them can be used. If the light beam is dim or if the LDR has poor sensitivity then it must be used to drive a Fet, a transistor or a 555.
I am learning and read where a FET can be used as a switch.

I plan on having the beeper inside my apartment with the laser and detector outside.

I may need to put the detector in a tube.

I would like the alarm to turn off after 5 minutes.

I would be nice to use my current piezo beeper.

My cadmium LDR is quite sensitive but it takes a while to stabilize.
 

fixit7

Member
I replaced the led with my buzzer.

It does work, but it beeps with a much higher rate than it does when wired directly?

The max current for the buzzer is 10 ma.
 

audioguru

Well-Known Member
Most Helpful Member
Like a mechanical switch, a Fet or an ordinary cheaper transistor can turn on and off or act like a linear amplifier.
5 minutes is a long time for an alarm beeper to scream.
Your last schematic uses a 555 as an ordinary cheap little transistor and you have pin 2 floating. Connect pin 2 to Vcc to properly disable it.
 

Reloadron

Well-Known Member
Most Helpful Member
Yes, you can use a FET or MOSFET as a switch, you can also use a transistor as a switch. The decision to use anything as a switch depends on how much current you want to switch. Take your buzzer and place it across your current source and measure the current it takes at a given voltage. Your meter, when measuring current, goes in series with the load, your buzzer. This will tell you right away if the buzzer works and how much current it draws. I have an old Radio Shack buzzer laying here. The package tells me 3.0 to 7.0 VDC and at 6 VDC 20 mA. Look at a 555 data sheet. A typical 555 can sink or source current up to about 200 mA which is about ten times what the average little buzzer will draw. There is a good possibility you don't need a transistor or any external switch.

Ron
 

audioguru

Well-Known Member
Most Helpful Member
Pin 2 on a 555 is its trigger pin. When it goes low then the output pin 3 goes high and the discharge pin 7 removes its low on the timing capacitor. If you leave pin 2 floating then interference can cause triggering. Pin 2 is disabled if you connect it to Vcc.

Your alarm runs continually because either you damaged the 555 or connected it wrong.
 

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