How to Make a Simple Touch Sensitive ON OFF Relay Switch

A very accurate touch sensitive switch can be built using a single IC 4093 and a few other passive components. The shown circuit is extremely accurate and fail-proof. and will turn ON and OFF a load with finger touch operations.

The circuit is basically a flip-flop that may be triggered through manual finger touches.

Using Schmitt Trigger IC 4093

The IC 4093 is a Quad 2-input NAND Gate with Schmidt trigger. Here we employ all the four gates from the IC for the proposed purpose.

Simple Touch Sensitive ON OF Relay Switch

How the Circuit works

Looking at the figure the circuit may be understood with the following points:

All the gates from the IC are basically configured as inverters and any input logic is transformed into an opposite signal logic at the respective outputs.

The first two gates N1 and N2 are arranged in the form of a latch, the resistor R1 looping from the output of N2 to the input of N1 becomes responsible for the desired latching action.

Transistor T1 is Darlington high gain transistor which has been incorporated for amplifying the minute signals from the finger touches.

Initially when power is switched ON due to the capacitor C1 at the input of N1, the logic at the input of N1 is pulled to ground potential making N1 and N2 feedback system latch with this input producing a negative logic at the output of N2.

The output relay driver stage is thus rendered inactive during initial power switch ON. Now suppose a finger touch is made at the base of T1, the transistor instantly conducts, driving a high logic at the input of N1 via C2, D2. C2 charges instantly and blocks any further faulty triggers from the touch, making sure the de-bouncing effect does not disturb the operation.

The above logic high instantly flips the condition of N1/N2 which now latches to produce a positive at the output, triggering the relay drive stage and the corresponding load.

So far the operation looks pretty straightforward, however now the next finger touch should make the circuit collapse and return to its original position and for implementing this feature, N4 is employed and its role becomes truly interesting.

After the above triggering is done, C3 gradually gets charged (within seconds), bringing a logic low at the corresponding input of N3, also the other input of N3 is already held at logic low through the resistor R2, which is clamped to ground. N3 now becomes stationed in a perfect stand by position “waiting” for the next touch trigger at the input.

Now suppose the next subsequent finger touch is made at the input of T1, another positive trigger is released at the input of N1 via C2, however it does not produce any influence over N1 and N2 as they are already latched in response with the earlier input positive trigger. Now, the second input of N3 which is also connected to receive the input trigger via C2 instantly gets a positive pulse at the connected input.

At this instant both the inputs of N3 goes high. This generates a logic low level at the output of N3. This logic low immediately pulls the input of N1 to ground via the diode D2, breaking the latch position of N1 and N2. This causes the output of N2 to become low, switching OFF the relay driver and the corresponding load. We are back into the original condition and circuit now waits for the next subsequent touch trigger in order to repeat the cycle.

IC 4093 Internal Gate Connection Diagram

Parts List

Parts required for making a simple touch sensitive switch circuit.

R1, R2 = 100K,

R6 = 1K

R3, R5 = 2M2,

R4 = 10K,

C1 = 100uF/25V

C2, C3 = 0.22uF

D1, D2, D3 = 1N4148,

N1---N4 = IC 4093,

T1 = 8050,

T2 = BC547

Relay = 12 volts, SPDT

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Prince Ghotra said…
Where to connect the load or leds for which,,this switch is prepared???
Prince Ghotra said…
Plz describe the connection of led on relay poles and where to connect pin#11 of ic 4093 i.e. N4 output????
Swagatam said…
to the contacts shown beside the small blue box at the left which is a relay.
Swagatam said…
pls read this post:
Swagatam said…
you may also see the example below, it shows how a load should be connected with a relay given contacts:
Prince Ghotra said…
So the pole of relay is connected to one end of the load and the other end of the load,and the N/O ofthe relay BOTH are connected to the DC supply

Is it right or not???
Prince Ghotra said…
Sir. Do you mean that led should be conected in series with relay coil or with pole and N/C of relay..
Becoz , In the example, you gave, led is connected in series with relay coil but
In my first question you told me to connect led at the contacts beside the relay (pole and others)...
Plzzz helpppp I'm confusedddd I have to prepare this for science fair project which is on 25 december
Swagatam said…
the relay contacts should close the circuit for the load when it gets activated and vice versa that's what we need...just as we do manually with an ON/OFF switch
Swagatam said…
What do you want to use as the load? Usually a high wattage AC load is connected with the relay, nobody will want to connect an LED with contacts because for an LED a relay won't be required, it can be directly connected with the transistor.

the led shown in the link circuit is irrelevant to your question, so don't bother about it.
Prince Ghotra said…
Sir all is done but tell me where to connect output of n4???
2:-question is
Relay is too much clicking sound . What to do ? I also changed the diode to 1n4007 but nothing works and touch is working sometimes..
Hlp plzz
Swagatam said…
Connect a 10uF/25V capacitor parallel to the relay coil, positive of the capacitor should connect with the positive end of the relay coil, his will stop relay clicking.
Make R6 = 1k and then check the response, it should be quicker.
Prince Ghotra said…
Where to connect output of n4?
Swagatam said…
only the input of N4 should be connected as shown, the output should be left open, it's unused.

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