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Need help with touchless on/off switch - detailed working principle

Ayya

Mar 22, 2021
2
Joined
Mar 22, 2021
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2
First of all i would like to say hello to all of you as it's my first post here.

I came here because obviously i need help. To be more specific i'm working with a circuit that will act something like wireless/touchless switch.

I have those 2 schematic. Schematic 1 is the one i'm basing on my project and 2nd is what modifications i've done.
upload_2021-3-22_20-20-6.png
D2 is a IR diode
IC1 is a CA3140 op amp ( in lt spice i used a universal op amp)
Q2 is replaced by the suggestion of teacher on MOSFET transistor ( still dont know which one exactly i need to use)
Q1 is a phototransistor ( in lt spice schematic it'a current source because it's easier to implement it there)
R4 is a potentiometer used to set a sensivity (in lt spice i did a voltage divider because of same reason as Q1)
I replaced D1 and RL1 by inserting a diode
upload_2021-3-22_20-20-18.png
Here i added a D flip flop.
The working principle i'm planning to achieve is when voltage source of 5 V is connected, a IR diode emits it's rays on phototransistor (right?) Q1 and when u pass a hand over IR Diode it should light up the LED behind mosfet and when you pass a hand 2nd time over IR diode it will switch off. Dflop is here to provide such an action.
Next for what i've already searched on and i think i know that when hand is passed over IR Diode pin 3 non inverting it's turning into high state right?. Pin 2 - inverting is connected to potentiometer and it works only for sensivity (right?) So now a result of turning pin 3 ino high state the signal goes on a pin6 output and it is remembered by d flop and led is lighted up and so turn it off second hand move is needed?
I don't know if i explained it well but thanks for any help.
 

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bertus

Moderator
Nov 8, 2019
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Hello,

The D2 , Q1 combination can be a beam break or reflection sensor.
In a beambreak circuit, an object between D2 and Q1 will activate the circuit.
In a reflection circuit, an object close to the reflection sensor will deactivate the circuit.

Bertus
 

Ayya

Mar 22, 2021
2
Joined
Mar 22, 2021
Messages
2
Hello,

The D2 , Q1 combination can be a beam break or reflection sensor.
In a beambreak circuit, an object between D2 and Q1 will activate the circuit.
In a reflection circuit, an object close to the reflection sensor will deactivate the circuit.

Bertus
Thank you for your answer.
So we are clear that this combination is a beam break since IR diode is only emitting a IR and not detecting when it's being reflected but i'm not certainly sure how Q1+R2 works... i know that if its like in the schematic (R2 is above Q1) and Q1 is illuminated it will be in "low state" and when its not illuminated it will be in "high state".. but why? how's that? If u dont want to explain it could you send me some paper or something where i could find an answer? I would be pleased.
 

bertus

Moderator
Nov 8, 2019
2,797
Joined
Nov 8, 2019
Messages
2,797
Hello,

I have attached two papers.
One for the reflective sensor and one with some circuits.

Bertus
 

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  • HOA0708 reflective sensor.pdf
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