Try replacing the chip, and making sure it's not backwards, and that it's the right chip.
Sometimes we miss the obvious stuff.
I spent a whole weekend troubleshooting, then realized I had the wrong chip the whole time. Not fun.

 

Imagine crying over a circuit for two weeks, figuring out why that darn 555 won't work, and then realising it's a PIC!

 

thanks for some insight magnum...i checked the chip...everything is what it should be..i.e..the chip being the 555...etc...only thing to do is get some more chips...ill have to do that tomorrow...well im gonna keep putzin with it regardless...maybe i can get it to work again....not to stray off topic but how hard is it to program a PIC?

 

Imagine crying over a circuit for two weeks, figuring out why that darn 555 won't work, and then realising it's a PIC!

it was only 2 days... and it was a TL071...

 

Hahaha.

To answer your question, it's not very hard to programme a PIC. It's not somethind I do, but that's down to a combination of not needing to and being too lazy to learn it for sake's sake. Give it a try, I think I will be very shortly.

 

i will try it sometime...most of these items are cheap enough that if i mess them up im not losing any $$$....i need some help troubleshooting this alt. flash circuit i made...i built the same circuit on a 556 last night and had it working again...with the alternating leds..however...when i changed some resistors and such to find some different flash rates the device ended up malfunctioning again after the 4th time i changed the values of the components and only one of the leds lit up and did not flash ...I wanna troubleshot it to see why this keeps happening...does it sound like there is a spike in current or something that is happening and blowing something out?...i would need a voltmeter to properly troubleshoot the circuit correct?...im gonna get one today...and where to start when i start troubleshooting...as always thanks for your time...

 

Well, you shouldn't just change parts at random. Decide what frequency and duty cycle you want to see, calculate the values of resistance and capacitance that you'll need, and then put in the correct parts and see if it works.

One item that will destroy the timer is if you accidentally connect pin 7 (discharge) to +V somehow. You should always have at least 100 Ohms per volt of Vcc between pin 7 and +V; if you were using a 12v supply, you'd use at least 1.2k Ohms.
Connecting pin 3 (out) directly to +V or ground can also destroy it.

If pin 5 is accidentally shorted to +V or GND, the timer will most likely get "stuck" with the output high or low; removing the short will allow resuming normal functioning.

 

Make sure you aren't mixing and matching CMOS and conventional 555 timers. It matters a lot.

Another couple of aids for designs I've come up with over the years.

First the posters...

http://forum.allaboutcircuits.com/picture.php?albumid=41&pictureid=308

http://forum.allaboutcircuits.com/picture.php?albumid=41&pictureid=1233

and now a very incomplete cookbook...

My Cookbook

Basic Multivibrator

This circuit has multiple experiments, here and here for example.

C2 is optional, and can be eliminated if the power supply is quiet.

Conventional 555 Characteristics

Good

• Excellent stability over power supply range.
• Can drive LEDs without affecting performance.
• Can be adjusted from 50.1% to 99% duty cycle.

Bad

• R1 is across the power supply when output is low, drawing unnecessary power.
• If R1 is too low IC will be damaged.

Basic Multivibrator with Diode

C2 is optional, and can be eliminated if the power supply is quiet.

Conventional 555 Characteristics

Good

• Can cover a wide range of duty cycles, 1% to 99%.
• Can drive LEDs without affecting performance.

Bad

• The diode makes the math fairly imprecise.
• Duty Cycle and Frequency will drift with power supply voltage variations (also due to the diode).
• If R1 is too low the IC will be damaged.

Use the schematic off the 555 Hysteretic Oscillator to alternate LEDs, it is a very standardized design.

I also have some other ideas in .LEDs, 555s, Flashers, and Light Chasers.

It is the end of my day, I may dithering and repeating myself here. Time for bed.

 

You should always have at least 100 Ohms per volt of Vcc between pin 7 and +V; if you were using a 12v supply, you'd use at least 1.2k Ohms.

thanks Sgt...this sounds easy enough...basically if im using a 9v supply i would want a 1k resistor between pin7 and (+)..right?....and this might sound a bit slow..but..do all IC's generate sound tones to drive the led or whatever it is that may be hooked up?..and what qualifies as a quiet power supply...thanks

 

thanks Sgt...this sounds easy enough...basically if im using a 9v supply i would want a 1k resistor between pin7 and (+)..right?....

Sure, that would work. 9*100 = 900 Ohms. 910 Ohms is the closest standard value. Since 1k Ohms is greater than 900 Ohms, it would be OK.
What you're actually trying to do is keep the current flow from Vcc to pin 7 at 10mA or less. 9v/10mA = 9/0.01 = 900 Ohms.
9v/1k = 9/1000 = 9mA, which is less than 10mA.

Pin 7 can actually sink a good bit more than 10mA, but at higher currents, the IC starts dissipating more power as heat, and also it could prevent the pin from discharging the timing capacitor through R2 low enough to trip the trigger input.

and this might sound a bit slow..but..do all IC's generate sound tones to drive the led or whatever it is that may be hooked up?

I'm not quite certain what you're asking here.
There is a bewildering assortment of ICs available from multitudes of manufacturers.
You really need to be a bit more specific.

..and what qualifies as a quiet power supply...thanks

It's more or less the ripple voltage on the output of the supply.
Switching regulators are generally very efficient, but are inherently noisy. Getting the noise down can be problematic.
Linear regulators are generally very quiet, but rather inefficient. They dissipate a lot of power as heat, as the regulator basically acts like a voltage controlled variable resistor.

 

Try laying it out as per this diagram.

This isn't right. DISCH should be hooked up between the 2 resistors, like in Bill_Marsden's post.

 

It looks okay to me - but I think it's quite confusing.

Try laying it out as per this diagram.

It's far easier - in this case - to have one positive and one negative rail.

Also, some people might wrap LED and resistor legs together, but that's something I really hate. Why bother with the breadboard if you're going to wrap components together? Just use the holes.

I hadn't looked at your breadboard layout before, but Gage's response prompted me to do so.

You have pin 7 connected directly to +V.
As I mentioned in a recent reply, that will destroy the 555 timer.
You also have the resistors connected incorrectly as Gage mentioned.

 

I re-created a viable layout using Pebble. Link: http://www.rev-ed.co.uk/picaxe/pebble/
The layout is nearly identical to what the kidson first posted.

 

thanks again everyone....so i take it my timers are fried...good thing just got a couple more..considering ill be stuck at home all night...Sgt. Wookie..is the layout you have provided a working revision of what i had originally??...and so i understand for future reference...does pin 7 wire into the Vcc pin8??...or do i just wire a resistor from (+) to pin7 then from pin7 to pin6 with another resistor??...i think i confused myself...as alwaysthaks...

 

thanks again everyone....so i take it my timers are fried...

The timers you tested with Sparky49's layout will certainly have pin 7 burned up. They might be salvageable if a different circuit layout is used, but not for this particular circuit.

good thing just got a couple more..considering ill be stuck at home all night...

More is good.

Sgt. Wookie..is the layout you have provided a working revision of what i had originally??...

Yes, it is. What you showed on your breadboard in the photo should have worked unless pin 4 was staying low for some reason; maybe you had accidentally shorted the resistor leads together, shorting pin 7 to pin 8 and thus Vcc.

and so i understand for future reference...does pin 7 wire into the Vcc pin8?

No!! Not unless you want to burn up more timers!

...or do i just wire a resistor from (+) to pin7 then from pin7 to pin6 with another resistor??...i think i confused myself...as alwaysthaks...

+ to pin 7 with a 10k, then pin 7 to pin 6 with a 100k.

Alright, let's just pull EVERYTHING off of your breadboard except from your + and - buses to your power supply (turn it off, BTW), stick in a fresh 555 timer, and start building the circuit one piece at a time, like this:

On the breadboard image I created, it shows:
1) Pin 1 connected to the ground bus via a black jumper.

2) Pin 2 connected to pin 6 via a green jumper, then a 470uF cap from pin 2 to the ground bus. Cap polarity is important; the stripe is the negative lead. If you reverse the polarity on an electrolytic cap, you can destroy it.

3) Pin 3 has a resistor to the anode of the LED over on column 15, then the LED's cathode connects to the ground bus. The resistor is shown as 1k Ohms, but it probably could be around 470 Ohms.

4) Pin 4 is connected to the positive bus via a red jumper.

5) Pin 5 has been left disconnected.

6) Pin 6 to pin 2 has already been mentioned; a 100k resistor connects from pin 6 to pin 7.

7) Pin 7 to pin 6 jumpered by a 100k resistor already documented; a 10k resistor connects from pin 7 to the positive bus.

8) Pin 8 connected to the positive bus by a red jumper.

9) Go back and double-check all of the connections. Make certain that no bare leads are shorting against other leads.

10) After you're double-certain, turn on the power and see how it does.

With the timing cap being 470uF, it will take a long time for it to charge through the 10k and 100k resistors; about a minute. Then it will take about 30 seconds to change states on/off after that.

If you measure across the cap, or from pin 2 to ground, you should see the voltage slowly rising and falling. When you first power it up, pin 2 will start off at 0v, and eventually reach about 2/3 of what your positive bus (Vcc) reads; at that point the output will change states, and the voltage will fall to ~1/3 of Vcc; the output will change states and the cycle will repeat.

It will take longer to charge the cap than to discharge the cap. There is 110k of resistance in the charge path, but only 100k in the discharge path; so it will take ~1.1 times as long to charge as it does to discharge.

If you measure from pin 5 to ground, you should get about 2/3 Vcc.

 

Now, here's a way you MIGHT be able to salvage those burned-up 555 timers, using ALMOST the same layout:

The differences between this layout and the previous layout, are that the 10k resistor from the + bus to pin 7 is gone, and the 100k resistor has been moved down to jumper between pins 2 and 3. I shrunk the size of the cap so that everything would fit so that you can see all of the connections.

This way, the capacitor is charged and discharged from the output, pin 3, instead of charged via 2 resistors in series and then discharged by pin 7.

 

great!..im gonna try to use the old ones first...then im gonna try the new scheme you have provided...i might have blown my capacitors...when i was trying to figure out what the problem was originally i was taking my capacitors out and shoving them in the bread both ways....lol...had no idea...so right now im about to salvage parts from an old VCR since it is littered with capacitors and such...ill let everyone know how things come about...thanks again..

 

thanks Sgt. Wookie...it works!...i didnt try the old ones yet..thats next...a few notes though...when the light would go out it wouldnt flash back on cleanly...it reminded me of how an HPS light fires up...if that makes sense...and another thing..if i may...on a capacitor..the lower the number μF the faster the flash rate the light would perform?...and what exactly does the voltage number on the capacitor indicate??...thanks

 

thanks Sgt. Wookie...it works!...

Great! It's always nice when you're winning...

I didnt try the old ones yet..thats next...

OK, you sure can try them with the 1st breadboard layout that I posted; but if that doesn't work, go to the 2nd breadboard layout.

a few notes though...when the light would go out it wouldn't flash back on cleanly...it reminded me of how an HPS light fires up...if that makes sense...

Like a high-pressure sodium lamp? That's a bit odd... well, a couple of things that are missing from your circuit as it is right now is an 0.1uF cap across pins 1 and 8, and another larger cap (1uF or larger) also across pins 1 and 8. I didn't have you put them on as you were having so much trouble with just the basic circuit. However, if you look in National Semiconductor's datasheet for the LM555, you will see that they recommend both 0.1uF and 1uF or larger caps across the power pins.

and another thing..if i may...

NO! .... Ahhh ... well, ok - go ahead.

on a capacitor..the lower the number μF the faster the flash rate the light would perform?...

Yes. It's a function of the resistance and capacitance.
In the first layout with both of the resistors, the time to charge the capacitor is roughly:
(R1+R2)*C1*.7
and the time to discharge it is roughly:
R2*C1*.7

But, in the 2nd layout with just the 1 resistor between pins 3 and 2/6, it's always roughly:
R1*C1*.7
... but not quite, because pin 3 doesn't go higher than about Vcc-1.3v, but it does go almost all the way to 0v.

and what exactly does the voltage number on the capacitor indicate??... thanks

That's the maximum voltage that you can put across the terminals. If you try to exceed that voltage, the "leakage current" will become excessive; the capacitor acts almost like a Zener diode and "clamps" the voltage at the capacitor's voltage rating. This isn't good for the capacitor; it's not designed to do that. The capacitor will heat up, and if it gets hot enough, it will rupture forcefully, spewing aluminum confetti everywhere and smelling like a stink bomb. It's a lot less fun than it sounds like.

 

i might have blown my capacitors...when i was trying to figure out what the problem was originally i was taking my capacitors out and shoving them in the bread both ways....lol...had no idea...

They may very well be damaged (I tried that when trying to build a plasma speaker, which is essentially the same circuit... desperation is a funny thing), they only have a very thin aluminum oxide coating inside that prevents current flow, all it takes is 1.5 volts the wrong way to destroy that, and had you left them on too long they may very well have exploded, they're essentially a short, so you're dissipating energy inside a sealed metal capsule full of saltwater(that's an oversimplification), which will get hot... then boil

http://www.teravolt.org/capboom.php