The circuit which I made shows the following behavior, when power is given to the circuit:

  • Sometimes it lights up, sometimes it doesn't.
  • When it does light up, it flashes properly for a few seconds but then becomes dimmer and dimmer.

What is causing this unpredictable behavior? Is it the components? Or is it the breadboard?

This is my first project using an IC

Yellow wire shorts 4 and 8. Orange wire shorts 2 and 6

UPDATE: The circuit is now working. All I had to do was change the battery!

Circuit diagram from a YouTube video where I got the circuit:

Circuit diagram from a YouTube video where I got the circuit

555 pins 1, 2, 3 and 4:

555 pins 1, 2, 3 and 4

555 pins 5, 6, 7 and 8:

555 pins 5, 6, 7 and 8

  • 1
    \$\begingroup\$ What's the quality of the battery? \$\endgroup\$
    – Huisman
    Dec 14, 2019 at 17:33
  • 3
    \$\begingroup\$ Lovely heart jumper +1. can you measure Voltage on each pin starting with power when Ok \$\endgroup\$ Dec 14, 2019 at 17:35
  • 1
    \$\begingroup\$ @TonyStewartSunnyskyguyEE75 What is 220? What is Pd? I dont think there is a temperature issue here \$\endgroup\$
    – user29463
    Dec 14, 2019 at 17:50
  • 3
    \$\begingroup\$ You said "Yellow wire shorts 1 and 8', presumably you meant 4 and 8 \$\endgroup\$
    – HandyHowie
    Dec 14, 2019 at 19:00
  • 1
    \$\begingroup\$ Battery voltage. You ought to take measurements in future and calculate power dissipation and sense heat with finger. Vbat = = battery voltage, batteries are like precharged supercaps with mAh capacity Ic * dt = C * dV that operate down to -10% or more until series Rs rises sharply and charge is depleted \$\endgroup\$ Dec 15, 2019 at 14:52

3 Answers 3


Looks okay to me. Try a fresh battery (almost 100% sure that’s your immediate issue), and increase the series resistor to the LED to draw less current (try 2K or so), at least temporarily. I also can’t see the color code clearly on the existing part.

You seem to have two 1uF caps, put the remaining one across the battery, observing polarity, of course.

The ancient bipolar 555s draw a large pulse of current when switching, so a bypass cap is desirable.

  • \$\begingroup\$ Im using only one 1uF capacitor \$\endgroup\$
    – user29463
    Dec 14, 2019 at 18:02
  • \$\begingroup\$ There is an orange one in the top photo and a blue one in the below photo, so did it change color between photos? \$\endgroup\$ Dec 14, 2019 at 18:03
  • \$\begingroup\$ Oh, the top photo is from a YouTube video from where I got the circuit. The blue one is the one which Im using \$\endgroup\$
    – user29463
    Dec 14, 2019 at 18:28
  • 1
    \$\begingroup\$ Thanks a lot sir! I took a new battery and it worked! \$\endgroup\$
    – user29463
    Dec 15, 2019 at 8:43

Examine your specs

Battery 5~12V and LED = Red is 20mA rated @ 2.1V,

Vbat Pd (Pd=V(^/270R)

This is OK for 5V but not 12V

Voltage across R ~ 10V for Vbat=12V Pd=10^2/270= 370 mW for a 1/4 W resistor with 125'C rise at 1/4W. You ought to limit temp rise to 60'C and thus only use 50% of power rating.

Also with 10V/270R=I= 37 mA, you have not only exceeded 20mA rating but absolute maximum 30mA rating and are heating up diode molecules and melting epoxy (slightly) .Can you smell it? Epoxy fumes are toxic.

Change 270R to 1k and live with less bright LED or get an LED that has more efficient Luminous Intensity @ 10mA e.g. 10,000 mcd @ 20mA


Measure voltage between pins 1 & 8 (right at the pins) to make sure it stays constant at battery voltage level. If it droops then bad connection between battery and I.C.

Measure voltage at pin 3. See if 'pulse high' level stays at constant amplitude or reduces over several pulses. If it stays constant or rises then bad connection between pin 3 & ground (via LED).


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