Normally home appliances are controlled by means of switches, sensors, etc. However, physical contact with switches may be dangerous if there is any shorting. The circuit described here requires no physical contact for operating the appliance. You just need to move your hand between the infrared LED (D2) and the phototransistor (Q1). The infrared rays transmitted by D2 is detected by the phototransistor to activate the hidden lock, flush system, hand dryer or else. This circuit is very stable and sensitive compared to other AC appliance control circuits. It is simple, compact and cheap. Current consumption is low in milliamperes. The circuit is built around an IC CA3140, D2, phototransistor and other discrete components.
Circuit Diagram:
Large ViewParts:
R1 = 470R
R2 = 100K
R3 = 3.3K
R4 = 10K
D1 = 1N4007
D2 = IR LED
Q1 = L14F1
RL = 5Vdc Relay
IC = CA3140
Q2 = BC548
Circuit Operation:
When regulated 5V is connected to the circuit, D2 emits infrared rays, which are received by phototransistor Q1 if it is properly aligned. The collector of Q1 is connected to non-inverting pin 3 of IC1. Inverting pin 2 of IC1 is connected to voltage-divider preset R4. Using preset R4 you can vary the reference voltage at pin 2, which also affects sensitivity of the phototransistor. Op-amp IC1 amplifies the signal received from the phototransistor. Resistor R3 controls the base current of transistor BC548 (Q2). The high output of IC1 at pin 6 drives transistor Q2 to energies relay RL1 and switch on the appliance, say, hand dryer, through the relay contacts. The working of the circuit is simple. In order to switch on the appliance, you simply interrupt the infrared rays falling on the phototransistor through your hand. During the interruption, the appliance remains on through the relay. When you remove your hand from the infrared beam, the appliance turns off through the relay. Assemble the circuit on any general-purpose PCB. Identify the resistors through colour coding or using the multimeter. Check the polarity and pin configuration of the IC and mount it using base. After soldering the circuit, connect +5V supply to the circuit.
R1 = 470R
R2 = 100K
R3 = 3.3K
R4 = 10K
D1 = 1N4007
D2 = IR LED
Q1 = L14F1
RL = 5Vdc Relay
IC = CA3140
Q2 = BC548
Circuit Operation:
When regulated 5V is connected to the circuit, D2 emits infrared rays, which are received by phototransistor Q1 if it is properly aligned. The collector of Q1 is connected to non-inverting pin 3 of IC1. Inverting pin 2 of IC1 is connected to voltage-divider preset R4. Using preset R4 you can vary the reference voltage at pin 2, which also affects sensitivity of the phototransistor. Op-amp IC1 amplifies the signal received from the phototransistor. Resistor R3 controls the base current of transistor BC548 (Q2). The high output of IC1 at pin 6 drives transistor Q2 to energies relay RL1 and switch on the appliance, say, hand dryer, through the relay contacts. The working of the circuit is simple. In order to switch on the appliance, you simply interrupt the infrared rays falling on the phototransistor through your hand. During the interruption, the appliance remains on through the relay. When you remove your hand from the infrared beam, the appliance turns off through the relay. Assemble the circuit on any general-purpose PCB. Identify the resistors through colour coding or using the multimeter. Check the polarity and pin configuration of the IC and mount it using base. After soldering the circuit, connect +5V supply to the circuit.
Source : www.electronicsforu.com
12 comments:
plz clearly mansion from where we can connect the device which we want to protect.
Means there should be some output & it should be clearly shown so that the wires of the device could be connect.
plz help
regards,
rajesh
yes plz answer the raj's question... n i have to ask... is there any alternative of CA3140... and L14F1... 2ndly can i use 6 v relay intead 5V...waiting for reply
Muhammad Faizan
Recommends: INTERSIL brand CA3240 (CA3240A, CA3240EZ). This is a dual Version of CA3140.
@Raj, Dj_Fix, Yongsik
Hey guys !
I am really sorry that i am unable to ans your questions, because currently i am working on my other site. ExtremeTricks.Net
Keep visiting and have fun because this site is nothing without all of you. I will add more and more interesting projects in future and will try to help you out, whenever i am free.
King Regards
Izhar Fareed
@Dj_Fix
Yes! you can use a 6 volt relay instead of 5 volt.
i need help i want to make a static ram for my final project any one can guide me???
sj_98diablo@!hotmail.com
can u tell me where the IC is.............
@Soumyajit
IC1 (IC) is shown in diagram, and it is CA3140.
how to connect relay switch in diagram. cant understand circuit connection of relay. in relay 8 pins are there how to connect and send pin connection description. thanks.
hey can u tel me any substitute ic insted of ic CA3140 plz rply soon.........
i want 6v wireless on/off switch diagram for my project pls help me... (for led)
email id : ulagamesivamayam@gmail.com
Can I know the practical applications of this circuit?
One i can mention is dat it can be used in hotels,while one moves inside the room,he has to put the card in given slot which further gives the supply to the room.This will thus avoid wastage of electricity when no one is there in room.It also lessens the possibility of misplacing the keys....pls reply soon.
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