Motion sensor circuit

Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
555thingb.png
For the transmitter part. Im going to use +5v , infrared led , and 330ohm.

When the infrared led is released the beam toward the phototransistor. The base of phototransistor will received the energy and now the Q1 is open and the emitter also release voltage to the NPN transistor.
Now the NPN transistor's base have the voltage source + Vcc and now the NPN is open and the voltage will directly flows toward to gnd. At this time, pin 2 is now in "LOW" state and output is "HIGH".

Is my theory correct? But my prof. tell me the output will be "HIGH" when something between it. Or can someone briefly explain it to me.

NOTES:
I'm going to change the speaker to buzzer. So for the transmitter part i'm not going to create fix frequency by using 555 timer.
 

ericgibbs

Joined Jan 29, 2010
18,766
hi Tommy,
When Q1 is receiving the IR beam, it will be conducting current, this current will flow into Q2 Base and turn Q2 ON.
So Q2 and Pin #2 of the 555 will be Low ~0v and the Output pin #3 will be Low.
Only when the IR beam is blocked to Q1 then pin#2 will go high and the 555 Output go High for the timing period.
E
Look at this image of the Trig pulse
 

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AlbertHall

Joined Jun 4, 2014
12,345
When Q1 is receiving the IR beam, it will be conducting current, this current will flow into Q2 Base and turn Q2 ON.
So Q2 and Pin #2 of the 555 will be Low ~0v and the Output pin #3 will be Low.
Only when the IR beam is blocked to Q1 then pin#2 will go high and the 555 Output go High for the timing period.
The '555 trigger is active low so when there is no IR, Q2 will be off and the trigger will be high so the '555 will not be triggered and the output will be low.
If the IR beam is continuous (not pulsed) then the trigger will be held low and the output will be high.
If the IR beam is pulsed then the '555 be triggered repeatedly and the output will be a series of pulses.
 

Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
The '555 trigger is active low so when there is no IR, Q2 will be off and the trigger will be high so the '555 will not be triggered and the output will be low.
If the IR beam is continuous (not pulsed) then the trigger will be held low and the output will be high.
If the IR beam is pulsed then the '555 be triggered repeatedly and the output will be a series of pulses.
So if nothing block between it..the output is high. Then any idea to modify into when theres nothing block between it..the output is low?
 

AlbertHall

Joined Jun 4, 2014
12,345
It depends just what you want to do with the output. For instance, if it is to light an led, then connect the LED and its resistor between the '555 output and the supply (instead of ground) so the LED will light when the output is low.
 

Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
It depends just what you want to do with the output. For instance, if it is to light an led, then connect the LED and its resistor between the '555 output and the supply (instead of ground) so the LED will light when the output is low.
How about putting a not gate ic at output? So when trigger is "high" and output will be "high" as well.Because im going to use buzzer and relay for the output.
 

ericgibbs

Joined Jan 29, 2010
18,766
hi T,
Using a PNP transistor is another option.
LED lit when beam blocked. [ or Relay operated]
Choose the timing R1 and C2, to suit your required ON time.
E
 

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ericgibbs

Joined Jan 29, 2010
18,766
hi T,
Its not a silly question.
It is possible while the circuit devices are switching they will cause 'electrical noise' on the +9v power rails.
This noise can cause other parts of the circuits to malfunction.

Consider that you are using say a 9v battery for the project and that also powers a Relay.
When the relay energises it will draw current from the battery and will cause a 'dip' in the 9v voltage level.
This dip could be sensed by the other devices and cause misoperation.
The caps supply the additional momentary current for the relay as it switches ON and so reduces the drop in the 9v.

E
If you want a more detailed techno explanation, just ask.
 

Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
hi T,
Its not a silly question.
It is possible while the circuit devices are switching they will cause 'electrical noise' on the +9v power rails.
This noise can cause other parts of the circuits to malfunction.

Consider that you are using say a 9v battery for the project and that also powers a Relay.
When the relay energises it will draw current from the battery and will cause a 'dip' in the 9v voltage level.
This dip could be sensed by the other devices and cause misoperation.
The caps supply the additional momentary current for the relay as it switches ON and so reduces the drop in the 9v.

E
If you want a more detailed techno explanation, just ask.
Really appreciated it. I will try to build the circuit with my little knowledge and according to your circuit diagram tomorrow. And if anything happened i would like to ask you again. I might be a annoy person but thanks for spending your precious times teaching me haha.
 

ericgibbs

Joined Jan 29, 2010
18,766
hi,
If you add a relay you MUST connected a diode across the relay coil.
When an inductor ie: the coil is carrying current and the current is quickly turned OFF, it will produce a 'back EMF' voltage pulse, which could be a high enough voltage to damage the driving transistor.
So the diode is connected across the coil with diode Cathode symbol 'bar' pointing towards the transistor Collector [ as its a PNP]
The diode Anode is connected to 0V/Gnd.

There are other methods of suppressing this back EMF, but a diode is simple solution.

OK.?
 

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Last edited:

Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
hi,
If you add a relay you MUST connected a diode across the relay coil.
When an inductor ie: the coil is carrying current and the current is quickly turned OFF, it will produce a 'back EMF' voltage pulse, which could be a high enough voltage to damage the driving transistor.
So the diode is connected across the coil with diode Cathode symbol 'bar' pointing towards the transistor Emitter [ as its a PNP]
The diode Anode is connected to 0V/Gnd.

There are other methods of suppressing this back EMF, but a diode is simple solution.

OK.?
Got it. If what if i want to add a buzzer? is it just connects parallel with with D1?
 

ericgibbs

Joined Jan 29, 2010
18,766
OK,
I assume that it operates from the same DC voltage as your power supply rail and requires a 'lowish' current.?
The speaker you have shown in your first post would not make a continuous sound, as its coupled via that capacitor C3.
It may make a short beep while the cap charges.

You could connect a beeper across the relay coil.
 

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Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
OK,
I assume that it operates from the same DC voltage as your power supply rail and requires a 'lowish' current.?
The speaker you have shown in your first post would not make a continuous sound, as its coupled via that capacitor C3.
It may make a short beep while the cap charges.

You could connect a beeper across the relay coil.
Alright. Thanks!! Much appreciated !!
 

Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
OK,
I assume that it operates from the same DC voltage as your power supply rail and requires a 'lowish' current.?
The speaker you have shown in your first post would not make a continuous sound, as its coupled via that capacitor C3.
It may make a short beep while the cap charges.

You could connect a beeper across the relay coil.
If i accidently connect a 9v battery from collector to the output of ic. Will the ic damage? Seems like the circuit doesnt work well. The led is just keeps light up whether it is blocked or not blocked .
 

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ericgibbs

Joined Jan 29, 2010
18,766
hi T,
A direct connection of 9v to the 555 output would most most likely damage the 555.
You may also have damaged the transistor.

As a test, disconnect everything from pin #3 of the 555, then connect a 470R resistor and a LED in series between pin#3 and 0v.
Power up the IC and check if blocking and unblocking the IR beam makes the LED light or not.

Lets know what you see.
E
 

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Thread Starter

Tommy Lee_1489743474

Joined Mar 17, 2017
39
hi T,
A direct connection of 9v to the 555 output would most most likely damage the 555.
You may also have damaged the transistor.

As a test, disconnect everything from pin #3 of the 555, then connect a 470R resistor and a LED in series between pin#3 and 0v.
Power up the IC and check if blocking and unblocking the IR beam makes the LED light or not.

Lets know what you see.
E
20170320_195609.jpg
The led wont light up even blocked or no blocked
 
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