I'm a hobbyist trying to learn the basics of the 555 timer, and am working off of the 50% duty cycle circuit from the National Semiconductor data sheet for the 555 (PDF pg. 10 Fig. 14) and have successfully prototyped the circuit on a breadboard, but am having difficulty identifying why my soldered protoboard version isn't working properly. Here's what I've looked at:

Schematic from the datasheet:

Datasheet Figure 14

Below is how I've laid out the components and soldered them to the protoboard (as part of a larger design which is why explicit V_cc and GND are omitted):

enter image description here

(Bold lines are soldered connections, arrows are wire based connections. Red for V_cc, Black for ground, blue for intermediate connection and gold for primary output.)

  • Pins 1, 5 are connected to ground (I recognize that 5 should go to a capacitor to ground, but have omitted it here as it worked without it on the breadboard and space was limited)
  • Pins 2, 6 are connected together and their junction is placed between R_A = 10 k Ohm and C = 100 uF.
  • Pin 3 is outbound.
  • Pin 4 and 8 are connected together and to R_A with a 5V regulated source.
  • Pin 7 is connected to R_B = 2.2 k\Ohm which is in turn placed between R_A and C.

I've gone through and taken voltage measurements and from ground

  • Pins 2 and 6 are 0.9V
  • Pin 3 is 0V
  • Pins 4, 8 are 5.01V
  • Pin 7 is 0.01V

My first thought was that I damaged something during the soldering process, so I went through and verified that the resistors and capacitor were functioning. The 555 sits in a socket, and each of the socket connections is going through.

So the big question is, what did I screw up in translating from the datasheet to this design? And, in general, what tests can I conduct myself to identify how to solve "why isn't my protoboard working" kind of problems in the future?

  • \$\begingroup\$ Replace Cform 2,6 to 1 with a large as available NON electrolytic and see if it oscillates. Electrolytic leakage may vause problems. Also check cap polarity (2) You can safely connect pin 6 to gnd or Vcc and IC should respond. If not the problem is elsewhere than the cap chg/discg (or as well as :-) ) \$\endgroup\$
    – Russell McMahon
    Apr 26, 2013 at 3:02
  • \$\begingroup\$ "It worked without the capacitor on the breadboard", ha! Famous last words. Maybe the connection was bad on the breadboard and the capacitor wasn't actually there. Sometimes things work without a capacitor, but don't work if a capacitor is replaced by a short. Unless you still have the breadboard version, forget about what worked on the breadboard. The breadboard version is gone and so you can't confirm whether something really worked, and whether the breadboard circuit was really what you think it was. \$\endgroup\$
    – Kaz
    Apr 26, 2013 at 3:38

2 Answers 2


The "Control Voltage"

There are probably several problems here... but I'd start by not grounding pin 5.

This is a simplified schematic of what's inside your 555 timer:

enter image description here

The flip-flop is RESET (nominal output low) when "Threshold" (pin 6) > "Control Voltage" (pin 5)

The flip-flop is SET (nominal output high) when "Trigger" (pin 2) < "Control Voltage" (pin 5) divided by 2

By shorting pin 5 to ground, you have forced "Control Voltage" to equal 0 volts.

In your case, you have prevented the timer from ever being able to SET its output by guaranteeing that "Trigger" can't ever be lower than "Control Voltage"/2 -- no possible positive voltage being lower than 0 ("ground").

Float (disconnect) pin 5 to fix. When you do not connect anything to pin 5, the "Control Voltage" defaults to 2/3 of Vcc.

Further Reading

There are a bunch of app notes and datasheets available here.

General Electrical Network Debugging

As to your general question about debugging. I tend to advise students and clients to proceed in this general order:

  1. Follow the power (ensure power is being supplied, delivered, and propagated as expected)

  2. Control the inputs (use a range of controlled inputs to explore the boundaries of the problem scenario)

  3. Isolate the elements (disconnect key elements to create smaller mini-circuits in the overall network)

Good luck!

  • \$\begingroup\$ Removing the short from pin 1 to 5 produced the desired result. I'll keep your debugging recommendations in mind as I continue my work. Thanks! \$\endgroup\$
    – GEL
    Apr 26, 2013 at 22:50

this shows what you have built


simulate this circuit – Schematic created using CircuitLab

you should have pin 1 going to pin 5 (ground to control) through a capacitor, but you have a straight short from 1 to 5 junction with the capacitor. and you seem to be missing the second capacitor from the original diagram.


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