0
\$\begingroup\$

Hello I have built the 555 delay power off circuit by using these 2 guides:

home.cogeco.ca/~rpaisley4/LM555.html#30

http://www.electroschematics.com/5963/adjustable-timer-1-10-minute/

I using a 10kohm potmeter + I was experimenting with 100 microfahrad - 1000 microfahrad caps. With the 10k + 100micro it switches off after 2 minutes something but not always in the same time and some time it stays on forever.

I have tried another new 555P ics in case it would fry or something but no luck. Can somebody see anything wrong with my circuit on the picture?

The red LED coming from the 2nd example (I think that circuit is bogous), even if I disconnect that part it does not make a difference.

enter image description here

\$\endgroup\$
1
  • 2
    \$\begingroup\$ If you want help you must include a schematic. The photo is completely useless to see what is going on. Why not just include the schematic from the site you link to (assuming your schematic is the same). \$\endgroup\$ Commented Mar 10, 2016 at 9:15

1 Answer 1

1
\$\begingroup\$

From the first glance, I see that protoboard power supply could be a problem. So along the long side of your protoboard you have "+" and "-" power rails. You'd say that these are wires going all the way from one end to the other. But it's not - this wire pair is discontinued at half the way. So from pin 0 to 30 you have one "+,-" pair and from pin 31 to 60 you have another. Something like this:

enter image description here

I see that your circuit is supplied partially from one side ("upper half") and partially form another ("lower half"). For instance your 555 pin 2, 100n capacitor or 33k resistor are floating- they're not connected neither to Vcc nor GND.

What should help is that you just bridge these two together with two jumper wires, like here:

enter image description here

Also, you seem to have 10k resistor from 555 pin 4 to Vcc, while your schematics suggests to connect pin 4 directly to Vcc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.