i finished my NE555 circuit and created some pcbs. After a few attempts i could get the circuit working, and tested it successfully. I placed my circuit between a 12V 1.5Hz square wave signal and a 12V LED strip. Everything worked fine, the circuit delayed the on-time of my led by a few hundred milliseconds. Then I went for a week on vacation and when i came to test my circuit a second time, and nothing worked... (The circuit was laying on my desk and no one was at home) The output voltage was not delayed any more and the output voltage was synchronous to the input voltage.

Luckily, i had ordered 10 pcbs and had parts to solder a few more... I did so, and tested my circuit with my lab bench power supply and measured every pin/trace with my oscilloscope. Everything was fine and with my R1 potentiometer I could change the delay time of the output signal.

I decided to connect the new circuit between the +12V square wave and my led strip and now nothing worked any more!? But in difference to the first pcb, the output voltage of this one stays at 0 volt for the whole time. Back on my lab bench power supply and my oscilloscope i could reproduce this problem. The circuit did not work any more.

The strange thing now is, that the capacitors voltage changes if i change the resistance of the potentiometer. If i set R1 to 0Ohm, the Capacitor i charged up to 12V. If i set R1 to 5k Ohm, the capacitor only charges up to something like 8V!? The higher I set R1, the lower gets the Capacitor voltage? If is test the same combination of capacitor and potentiometer on my breadboard, everything works just fine and the capacitor gets charged up to 12v every time, just the the charging time varies when i set the resistance higher or lower (How it should be).

Maybe someone can tell me what is going on and what i did wrong? I am really freaking out because even with my oscilloscope i can't find the root of the problem.


I attached my scheme and my pcb design ;)

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  • \$\begingroup\$ You might want to try adding 100uF/16V across the input voltage. What exactly stopped in the first example- is the MOSFET shorted, for example? Did you do any failure analysis at all? \$\endgroup\$ Commented Jul 14, 2018 at 16:40
  • \$\begingroup\$ @SpehroPefhany Thanks for your reply. The mosfets are okay, in both circuits. I checked their behaviour in relation to the gate voltage, which was just as expected. I just checked both circuits and determined that the behaviour of the C1 voltage is really strange. The higher I set the potentiometer resistance, the lower the voltage of the capacitor gets. You might think that it does not have enough time to get fully charged, but it has. The curve goes up really fast and then stays at the same voltage. This is what it looks like with 10k Screenshot \$\endgroup\$
    – chixlol
    Commented Jul 14, 2018 at 17:25
  • 1
    \$\begingroup\$ If the capacitor is not installed backwards, it's possible the 555 got damaged due to some transient on the supply as a result of you not having any bypass capacitor as I suggested above. \$\endgroup\$ Commented Jul 14, 2018 at 18:25
  • 2
    \$\begingroup\$ Are you powering the 555 with the 1.5 Hz signal? That is a very strange thing to do - no telling how the 555 will react. \$\endgroup\$ Commented Jul 14, 2018 at 23:58
  • \$\begingroup\$ the VCC traces seem narrow compared to GND not the cause of this problem, just a comment. \$\endgroup\$ Commented Jul 15, 2018 at 5:20

1 Answer 1


it looks like the problem is the energy stored in C1 is being discharged through the IC when the 12V quits, add an upwards pointing diode in parallel with R2 and a 10K resistor between the +ve end of C1 and the 555 (replacing the pcb trace)

you can probably fit those parts on the underside of the board.

  • \$\begingroup\$ Bingo. Absolute maximum rating for the NE555 input high voltage on THRES and TRIG is VCC. That limit is exceeded every time the input drops. The suggested fix makes things a lot better, but the input will still be a diode drop above VCC at the start of every input low. \$\endgroup\$
    – user128351
    Commented Jul 15, 2018 at 9:57
  • \$\begingroup\$ the 10K resistor will keep the current into pins 2 and 6 below 70uA, so that should be fine. but the timing task might be better done using something like LM358 which is not fussed by voltage (up to 32V) on the inputs even while powered down. \$\endgroup\$ Commented Apr 26, 2022 at 23:25

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