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very high voltage across 555 output!!!!!!

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vinodstanur

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Hi all, ones while i was making an LDR circuit using 555 (for automatic night lamp), i used the pin number 4(reset) to connect with LDR. Output of ic become high(7v) when the LDR is covered with hand while the output become low (0v) while LDR is exposed to light.
But the interesting thing i noticed is that,when i slowly remove my hand above the LDR then at a particular position (at the stage of threshold point where the o/p is tends to become off, but when i hold my hand at that particular position, THEN I OBSERVED THE VOLTAGE SHOWN IN MY ANALOG MULTIMETER Is ABOVE 1200V AC. (actually very very larger than 1200 volt since the needle strikes the other end very fast)..
I never used any inductor in the circuit!!!
1>without inductance is there any chance of building such a high voltage at 555 output?
2>Is the multimeter coil inductance making this ?
3>or any other reason?
Actually am confused...
 

your body can generate a electrostatic discharge of thousands of volts.
How are you holding the meter ?
 

I don't think you can be generating any high voltages. I think your meter is misinterpreting the high frequency square wave.

Keith
 
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actually the high voltage in NO load voltage. When I touch the o/p with my hand and also touch the ground, then I think the meter shows 0 volts...
Some vibrations will be there at o/p at the threshold point . But still I cant believe it! If I connect a small inductor in parallel wit hthe o/p then there is a chance of showing such high voltage due to self induction...is it...
But in this case, the o/p very high voltage is some thing like a high voltage source with very high internal resistance. So it could not be used .
6_1286180879.png
 
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As I said earlier, I think your meter is misinterpreting the high frequency square wave.

Keith.
 

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As I said earlier, I think your meter is misinterpreting the high frequency square wave.

Keith.
so u r telling that my multimeter at 1200v AC range will show high voltage if i give low voltage high frequency square wave to the multimeter input?
Then could u explain it?
 

It seems to be a ground bounce.
In your circuit, there is some inductance in the very small lead wires between the chip itself and the lead carrier of the package and the PCB trace or wires. This inductance is very small, but it is significant. Consider what happens the moment your internal Q turns on/off. A spike of current flows from the output through Q to ground. This current flows through the inductance in
the lead and PCB trace or wires. The voltage across this inductance is directly related to the change in current:
V=L*di/dt

try to put a caps on the output.
 

I don't believe you will generate 1200V from a bit of stray inductance in the setup described! You might get a volt or two of ringing. Without some better test equipment it is impossible to be certain to know why you meter is reading that value, but I know the 555 isn't generating a high voltage AC.

Keith
 

It seems to be a ground bounce.
In your circuit, there is some inductance in the very small lead wires between the chip itself and the lead carrier of the package and the PCB trace or wires. This inductance is very small, but it is significant. Consider what happens the moment your internal Q turns on/off. A spike of current flows from the output through Q to ground. This current flows through the inductance in
the lead and PCB trace or wires. The voltage across this inductance is directly related to the change in current:
V=L*di/dt

try to put a caps on the output.
Yes, i already done that, and the voltage vanishes. I also think the reason is same as that told by U.
 

Sorry, I just don't believe that you meter is measuring correctly. The attached is a ground bounce scenario (rather extreme to show the effect). As I said, a volt or two but not 1200V. I doubt your meter accurately measures small spikes.

Keith.
 

Attachments

  • Ground bounce.pdf
    4.1 KB · Views: 94

have you tested with some other meter?
 

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