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First of all, I will go through my concept which make the doubt.

When we connect positive side of LED to a positive terminal of a battery and negative of LED to floor (not to -ve of battery), current will not flow, as -ve of battery is not at a potential equal to that of floor (0V). But if I hold by standing on floor of phase line of commercial electric circuit, I get shock because neutral line is at same potential as ground. (Correct me if I'm wrong.)
Now look to this circuit:

image http://www.circuitsgallery.com/wp-content/uploads/2013/04/555-touch-sensor-circuit.png
Figure above shows a touch sensor using 555 timer. It work only if a low pulse applied to pin 2. But how when we touch the plate it get low pulse? Although we consider -ve terminal to be 0v, in practice, I think it is not. As like my first example, I expect the current never flow and pin 2 stays at high voltage. But it is not so.

Then what is my mistake? Pls help me I am a beginner. Sorry for my English.

EDIT

Some answers tell me "how the 555 works" or "how sensitivity of circuit increase". But my question is different from that. May be this is due to unclear data from my question. My question is how the pin 2 trigger (or the voltage at pin 2 become less than \$\frac{1}{3}V_{cc}\$ ) when I touch the plate?

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    \$\begingroup\$ What you are seeing is a capacitive effect. At rest there is a charge sitting on the touch pad like a capacitor. The human body also acts as a capacitor. When you touch the pad you effectively connect two capacitors in parallel, a charged one (pad) and a discharged one (body). When this happens the charge is distributed evenly between them causing the voltage to drop since the pad is such a small capacitance compared to the human body. This gives you the low voltage on pin 2. \$\endgroup\$ – vini_i Dec 31 '15 at 15:13
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    \$\begingroup\$ If you build this circuit, likely Q1 will fail when it turns off -- the back emf from the inductance of the relay will generate a very high voltage and overstress Q1. A diode needs to be added -- anode = collector; cathode = +12V to protect it. \$\endgroup\$ – jp314 Dec 31 '15 at 15:48
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Three things to note:

  1. Your body can be modeled as a resistor.
  2. The touchpad is likely made up of 2 pads; one connected to pin 2 of the 555 IC and the other connected to ground.
  3. The 555 is triggered by taking pin 2 lower than 1/3 of the supply voltage.

Pin 2 is being pulled up by R3 which is 1M ohms. This means that it is extremely weak. When your finger makes contact with both pads of the touchpad, the circuit sees essentially a resistor between pin 2 and ground.

Using the voltage divider equation: $$ 1/3 = R_{finger}/(R_{finger}+R_3) $$ $$ \therefore R_{finger} = 500K $$

The resistance provided by your finger needs to be less than 500K ohms, which is quite achievable if the 2 pads are close together.

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I remember when I was in high school, I built a kit that works exactly like this. I remember being disappointed that it didn't work for my intended purpose: I wanted a touch sensor hidden below a car's door handle for security purposes. I digress...

The idea is fairly simply. Pin 2 has a very high input impedance. The 1M-ohm pull-up resistor thus pulls it up almost all the way to the supply rail, thus preventing the 555 from triggering. The circuit relies on the ground of the circuit ACTUALLY being connected to real ground. Now, when you touch the touch plate, your body (and the earth) creates a resistance that's much lower than 1M-ohm. It pulls pin 2 down to below a third of the supply, which is what the 555 needs to trigger. The 555 then pulls pin 3 high for the period as determined by R1 and C1. I forgot how you do that and I'm sure you know how that works.

For interest sake, a human body has a resistance of around 10k-20k. I'm not sure about earth, might depend on how recently it rained :-)

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You don't actually need a negative pulse but take the lvl of pulse less then 5/3 Volt due to comparator action. The specific O/p of plate depends on type of h/W used. If I'm correct in the diagram you are exploiting the monostable operation of the 555 timer.Thus on negative pulse on 2 qbar of ff becomes 0, by effect on comparators and sr flip flop capacitor C1 charges to level of 2/3 Vcc and then quickly discharges. This allows an internal transistor(pnp type) to be on "high" till charging occours and qbar to be on 0 thus allowing a 1 on O/P port 3 which powers on LED.enter image description here

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  • \$\begingroup\$ OP is not asking how the 555 timer works. He's asking specifically about how the touchpad will trigger the 555 timer. \$\endgroup\$ – efox29 Dec 31 '15 at 16:01
  • \$\begingroup\$ It seems you are saying about the working of 555. But my question is not that. \$\endgroup\$ – Muhsin Ibn Al Azeez Dec 31 '15 at 16:27
  • \$\begingroup\$ what kind of touch pad r u using is it just a mere press break switch or a sensor I.C. plz. tell \$\endgroup\$ – manav.tix Dec 31 '15 at 16:48
  • \$\begingroup\$ About the first part of ur question i'm still not able to get what u mean. \$\endgroup\$ – manav.tix Dec 31 '15 at 16:59
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Personally, i would add a resistor shown as Rx. Its value will be chosen in conjunction with R3 such that the resting (untouched ) voltage will sit near the trigger level point, see pic...enter image description here

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  • \$\begingroup\$ Perhaps try the low power version of the 555, the ICM7555. Then increase R3 to 10M or even 20M. This will make it much more sensitive and therefore help trigger! \$\endgroup\$ – pcmedic Dec 31 '15 at 16:17
  • \$\begingroup\$ but my question is how it trigger when we touch it. \$\endgroup\$ – Muhsin Ibn Al Azeez Dec 31 '15 at 16:28
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    \$\begingroup\$ Because we act like antennas and pick up stray voltages. Now, when we touch the plate (which has a high input resistance - NOT damping the signal voltage too much) the voltage from out bodies enters pin 2 bringing its level above and below the trigger point. \$\endgroup\$ – pcmedic Dec 31 '15 at 16:33

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