Why is my 555 Timer not blinking?

[Twinfun2]

Hey Everybody!

I have been using this page as a guideline to learning how to use a 555 timer:

http://www.kpsec.freeuk.com/555timer.htm

I have been trying to make an LED flasher by using a 555 Timer for the first time, but after 2 days, I have been unable to see my problem. I am using the astable mode schematic seen in the astable mode section of the page I have been using. I am using C = 10uF Capacitor,
R1 = 10k, and R2 = 100k. I understand that this should yield about 40 blinks by the light according to the chart I am using.

My problem is that after I build the circuit onto my breadboard, the light constantly stays on, and does not blink. I have also noticed that the 555 gets terribly hot, but I have no idea why. Also, I saw posts with people who were able to get voltage ratings on each pin. How would I do that, and is an analog multimeter accurate enough?

Thanks in advance! Don't hesitate to ask me any necessary questions.
Brandon

 

[negitron]

The 555 should not get hot at all and the circuit diagram you linked to should work as drawn. Re-check each and every connection (assume nothing and double-check the pinout on the data sheet for your specific device); if you're absolutely certain they're correct then it's possible the 555 is bad. If you still can't get it working, a good-quality photo of your breadboard setup might be helpful to us.

 

[Twinfun2]

I have no access to a digital camera at this time, but when I do, I'll most definitely post those images. I used the timer in another circuit layout that drives a PNP transistor to make a speaker buzz like a siren, and that worked without any issues. Also, in my last post i meant "about 40 per minute", which I'm sure was pretty obvious anyway.

Any recommendations will be most appreciated, as I can't be 100% sure if anything is correct if it isn't working as planned, or even if its wired correctly.

Again, I will try and post images as soon as possible.

Thanks

 

[negitron]

Silly question: Are you wiring the LED between pin #3 (Output) and ground (0V) or between pin #3 and +Vs, and are you including an appropriately-sized current-limiting resistor?

 

[Twinfun2]

Silly question: Are you wiring the LED between pin #3 (Output) and ground (0V) or between pin #3 and +Vs, and are you including an appropriately-sized current-limiting resistor?

Between Pin #3 and +Vs, with a resistor just before the LED.

(*Awaits emotional slap in the face *)

 

[negitron]

There's your problem, I suspect. Connect the LED instead between pin #3 and ground with the cathode to ground. Assuming a +5 V output level and a 2.0 V drop across the LED (typical for standard red and green LEDs,) use a (5-2) / .010 = 300-ohm resistor (standard value 270-ohms is close enough) in series on either side.

 

[Twinfun2]

There's your problem, I suspect. Connect the LED instead between pin #3 and ground with the cathode to ground. Assuming a +5 V output level and a 2.0 V drop across the LED (typical for standard red and green LEDs,) use a (5-2) / .010 = 300-ohm resistor (standard value 270-ohms is close enough) in series on either side.

LED still doesn't blink sadly; solid red light.

I am using a 6v (4aa) power supply, if it helps or explains anything.

 

[Redbelly98]

Silly question: Are you wiring the LED between pin #3 (Output) and ground (0V) or between pin #3 and +Vs,

Does it matter? The duty cycle is nearly 50%, so either way should work.

... and are you including an appropriately-sized current-limiting resistor?

A few more questions to help with the troubleshooting:

• What is the value of the current-limiting resistor (the one in series with the LED)?
• What is the value of the source voltage V_s?
• Is there a 0.01 uF cap between pin 5 and ground, as shown in the circuit here:
  http://www.kpsec.freeuk.com/555timer.htm#astable

 

[negitron]

Does it matter? The duty cycle is nearly 50%, so either way should work.

Depends on what +Vs is and what the logic level output of pin #3 is. If +Vs - Vout is more than the Vf of the LED, it will always be biased on. Best practice is to connect through to ground.

An additional question: which flavor of 555 are you using? There are, for example, CMOS versions and TTL versions and they are not necessarily directly compatible.

 

[Redbelly98]

Depends on what +Vs is and what the logic level output of pin #3 is. If +Vs - Vout is more than the Vf of the LED, it will always be biased on. Best practice is to connect through to ground.

Okay, I was assuming V_o=V_s when the output is high -- that has been my (limited) experience. I was unaware that this is not in general true.

An additional question: which flavor of 555 are you using? There are, for example, CMOS versions and TTL versions and they are not necessarily directly compatible.

Or better yet, provide the manufacturer and exact part number.

 

[negitron]

Okay, I was assuming V_o=V_s when the output is high -- that has been my (limited) experience. I was unaware that this is not in general true.

Looks like for most flavors of 555, Vout(high) = Vs - ~1.7 V.

 

[chroot]

I don't have much time at the moment to thoroughly consider the deductive paths the others here are taking you down, but I'm only concerned right now about your 555 timer getting very hot. I won't go into details, but often the only way for this to be possible is to run the part way above its rated power supply voltage, activating its ESD structures and drawing as much current as your (presumably) small wall-wart transformer can provide.

Look up the datasheet for your exact 555 part on the manufacturer's website, and find its intended power supply range. Also, please explain to us how you're powering the part.

(It's also worth mentioning that activating the ESD structures on a chip generally derates it -- damages it measurably, making it less reliable -- and continuously powering it above its intended supply range could destroy it pretty quickly.)

- Warren

 

[negitron]

Look up the datasheet for your exact 555 part on the manufacturer's website, and find its intended power supply range. Also, please explain to us how you're powering the part.

He did say 4 AA cells for a total of 6 V. This seems to be well within the tolerances of all the common types of 555.

 

[Twinfun2]

Does it matter? The duty cycle is nearly 50%, so either way should work.
A few more questions to help with the troubleshooting:

• What is the value of the current-limiting resistor (the one in series with the LED)?
• What is the value of the source voltage V_s?
• Is there a 0.01 uF cap between pin 5 and ground

1) 330 Ohm
2) 6v
3) Yes

EDIT: I am using an NE555N; also tried NE555P, but yields no difference. The data sheets for each model seem to allow 6v, where max volt is around 15v-18v for each. I am unsure of finding the exact manufacturers of each timer except by searching for the entire model number shown on the timer itself. The NE555P has nothing but NE555P on it, while the NE555N has "H2L937", which yields no results on Google.

 

[chroot]

Okay, from the look of it, the Fairchild NE555N does not have a current-limited output driver, and does not provide a plot of output voltage versus current (shame on them). I can only assume that if you short the output to ground or +Vs, you will draw enough current to damage the device. Check for this.

- Warren

 

[Twinfun2]

I bring up again, how would I check for loose connections on my breadboard, and how to check ratings on each pin out properly with my multimeter.

In other topics on this forum, many people have posted that information.

Do I simply need to identify the voltage on each pin?

 

[negitron]

Well, you can easily check the voltages on each pin, but you can't so easily check the currents. Best thing to do is to go to each wire and gently pull on it; if it comes out of the breadboard very easily, it's probably too loose and you should use a different connection point. You can also use your multimeter on ohms (or better still in audible continuity modem, if you have it) and ohm out between circuit terminals which should be connected with the circuit unpowered.

However, since you have an overheating problem, it's probably not an open connection which is the issue, but a short. Check for continuity between points which should NOT be connected and correct as necessary.

 

[Redbelly98]

1) 330 Ohm
2) 6v
3) Yes

Okay. I was fishing for some obvious error, but everything sounds right here.

I bring up again, how would I check for loose connections on my breadboard, and how to check ratings on each pin out properly with my multimeter.

In other topics on this forum, many people have posted that information.

Do I simply need to identify the voltage on each pin?

To check with a multimeter, I would first remove the diode, and also try a larger cap or resistors (about twice the value you have) so that the device should be on for at least 1.5 seconds, and off for at least 1.5 seconds. See if the output does what it should without the LED connected, and make sure that Vs is what it should be too.

 

[Twinfun2]

Well, you can easily check the voltages on each pin, but you can't so easily check the currents. Best thing to do is to go to each wire and gently pull on it; if it comes out of the breadboard very easily, it's probably too loose and you should use a different connection point. You can also use your multimeter on ohms (or better still in audible continuity modem, if you have it) and ohm out between circuit terminals which should be connected.

However, since you have an overheating problem, it's probably not an open connection which is the issue, but a short. Check for continuity between points which should NOT be connected and correct as necessary.

Just now made sure nothing is connected to anything else that it is not meant to be connected to; I'm assuming that is called continuity. I'm also quite sure that the timer isn't very loose since I have move the the timer 3 times from its original position since yesterday, and am now holding the IC in place with the end of my pencil's eraser just in case :shy:.

@Redbelly: Doing that now, brb

 

[negitron]

Also check your 10 uF cap. Remove the device from the circuit and fully discharge it. If it's an electrolytic, place the (+) probe of your meter in ohms mode to the (+) side of the cap and the (-) on the (-). If the cap is not shorted or pen, you should see an initially low reading which rises rapidly to infinity. If it's short, you'll see it read persistently near zero ohms; if open, you'll see an immediate reading of near infinity.

 

[chroot]

I strongly suspect that you're somehow shorting the output to ground or +Vs.

- Warren

 

[Twinfun2]

I strongly suspect that you're somehow shorting the output to ground or +Vs.

- Warren

Wouldn't connecting the output to a resistor, then to an LED, then to ground be the correct way to make the LED blink, and would this be a correct example of shorting Pin 3 to ground?

Trying to make sure I have the lingo correct, can't take any chances .

I sadly don't have much other resistor options, and only 1 100uF capacitor in which I tried in response of Redbellies request, but no difference. I'll have to make a stop at Radioshack when I get the chance.

EDIT: Also, it has been rather difficult for me to get voltage reading on each pin out, because the darned 555 is getting WAY WAY too hot for comfort. Is it safe for me to let it build up so much heat? After I leave the circuit on for about 10 seconds, I am unable to leave my finger on it for even a second.

 

[negitron]

Wouldn't connecting the output to a resistor, then to an LED, then to ground be the correct way to make the LED blink, and would this be a correct example of shorting Pin 3 to ground?

That's the correct way to do it. Unless the LED and the resistor both are shorted, you're fine.

Is it safe for me to let it build up so much heat? After I leave the circuit on for about 10 seconds, I am unable to leave my finger on it for even a second.

No, that's way too hot. Allowing the device to run that hot for too long WILL destroy it in short order. Sometimes these devices have some integral overcurrent protection, but you don't ever want to count on such.

 

[chroot]

Wouldn't connecting the output to a resistor, then to an LED, then to ground be the correct way to make the LED blink, and would this be a correct example of shorting Pin 3 to ground?

Oh, heh, I use lingo without even realizing it. "Shorting to ground" means (accidentally) putting some kind of direction connection (short) to ground. If you really have things correctly wired, and correctly soldered -- with no globs of solder accidentally connecting things -- the series combo of the resistor and diode, one end tied to the output and the other tied to ground, is correct.

Is it safe for me to let it build up so much heat? After I leave the circuit on for about 10 seconds, I am unable to leave my finger on it for even a second.

No, it's not safe, and it has probably already destroyed the part. Even if you fix your circuit, the part may continue to do this. You probably should get a new one, after you've figured out what's wrong with your circuit.

Do you by any chance have a glob of solder accidentally connecting, say, Discharge and VCC?

- Warren

 

[Twinfun2]

@Chroot

Im using a breadboard.

I'll probably pick up a new timer tomorrow if this all means the timer is simply defective or destroyed.

I might just post again tomorrow, but please, feed me more information, for I am starving for a blinky red light!

 

[negitron]

Im using a breadboard.

There are both soldered and solderless breadboards.

 

[Redbelly98]

@Chroot

Im using a breadboard.

Hey, another thought here. Do you understand how the bread board holes are automatically connected together underneath (where you can't see it)?

Sorry if the question seems obvious to you, I'm trying to see if there's something obvious that has been overlooked.

 

[Twinfun2]

Hey, another thought here. Do you understand how the bread board holes are automatically connected together underneath (where you can't see it)?

Sorry if the question seems obvious to you, I'm trying to see if there's something obvious that has been overlooked.

You insult me... (lol)

Yah I understand how the solderless breadboard works and how its laid out, good thing to just ask; I understand.

 

[Twinfun2]

Sorry that I didn't mention earlier, but I have great gratitude for you guys who make learning about this stuff so much easier for those who wish to learn it. I would have no idea what to do next if people like you were missing.

Great gratitude unto you guys.

 

[Redbelly98]

You're welcome!

I'm logging off for the night (it's just past 10 pm in my time zone), good luck and I should be around tomorrow.

 

[vk6kro]

Check that the power source + only goes to pins 4 and 8 of the 555 and does not go to pin 3. Make sure the polarity is right.

Note that pins 4 and 8 are diagonally opposite each other on the chip.
Pins number anticlockwise around the chip starting at the top left if you look at the top of the chip with the notch at the top.

You can do this without applying power. The chip may have blown up already, so don't put a new one in until you find the fault.

See this link from a few weeks ago. There are pictures of a similar setup and a list of voltage measurements.
https://www.physicsforums.com/showthread.php?t=311178

 

[Twinfun2]

Check that the power source + only goes to pins 4 and 8 of the 555 and does not go to pin 3. Make sure the polarity is right.

Note that pins 4 and 8 are diagonally opposite each other on the chip.
Pins number anticlockwise around the chip starting at the top left if you look at the top of the chip with the notch at the top.

You can do this without applying power. The chip may have blown up already, so don't put a new one in until you find the fault.

See this link from a few weeks ago. There are pictures of a similar setup and a list of voltage measurements.
https://www.physicsforums.com/showthread.php?t=311178

Yes, I checked so many times if my connections were correct that I'm beginning to stutter when I say "555".

I have checked that thread before I posted here but the poster's solution is inapplicable to my setup; I have the correct resistors.

POSSIBLY USEFUL INFORMATION: When I use a 9v battery supply instead of my 4xAA battery 6v supply, the circuit doesn't work at all! The timer doesn't warm up or anything! From what I understand, the timer should work exactly the same way as it would the 6v supply; correct me if I'm wrong.

Driving to philly, I'll be back in 2 hours perhaps.

 

[vk6kro]

Are you going to plug in a new 555 to see if it works?

How about unplugging the present 555 and measuring the voltages on all the pins relative to the negative terminal on the battery? A multimeter, especially a digital one, is fine for this. Just poke a piece of wire into the socket holes to measure the voltages.

Make a list of 8 voltages from pin 1 to pin 8.
Put the list here if you can't see already what is wrong.

That should tell you if it is safe to put the new chip into the circuit.

You should get a list like this with a 6 v supply.
1 0V
2 6V
3 0V
4 6V
5 0V
6 6V
7 6V
8 6V
Pins 6 and 2 could be a bit lower depending on your multimeter's internal resistance. Maybe 5.4 V if the meter is a digital with 1 M input resitance because of the 100 K in series.

Your 9 V battery probably can't deliver much current into a short circuit so it doesn't cause heating.

 

[Twinfun2]

Are you going to plug in a new 555 to see if it works?

How about unplugging the present 555 and measuring the voltages on all the pins relative to the negative terminal on the battery? A multimeter, especially a digital one, is fine for this. Just poke a piece of wire into the socket holes to measure the voltages.

Make a list of 8 voltages from pin 1 to pin 8.
Put the list here if you can't see already what is wrong.

That should tell you if it is safe to put the new chip into the circuit.

You should get a list like this with a 6 v supply.
1 0V
2 6V
3 0V
4 6V
5 0V
6 6V
7 6V
8 6V
Pins 6 and 2 could be a bit lower depending on your multimeter's internal resistance. Maybe 5.4 V if the meter is a digital with 1 M input resitance because of the 100 K in series.

Your 9 V battery probably can't deliver much current into a short circuit so it doesn't cause heating.

When I get home, I will do this for sure, but I am not quite sure if I understand how to do what your saying 100%.

"How about unplugging the present 555 and measuring the voltages on all the pins relative to the negative terminal on the battery? A multimeter, especially a digital one, is fine for this. Just poke a piece of wire into the socket holes to measure the voltages."

Can you be a little more specific? I don't understand what you mean by socket holes, or what you mean by measuring the voltages on all the pins relative to the negative terminal on the battery. Do I just remove the 555 and test an incomplete circuit with the multimeter? Because from what I understand, no voltage can travel through an incomplete circuit.

Sorry about being so newbie, but I simply don't understand what you mean.

But yes, I have a new 555 timer, but I will first do what you recommend (once I figure out exactly how) before inserting the new timer.

Thank you so much for your patience! :!)

 

[negitron]

Can you be a little more specific? I don't understand what you mean by socket holes, or what you mean by measuring the voltages on all the pins relative to the negative terminal on the battery. Do I just remove the 555 and test an incomplete circuit with the multimeter? Because from what I understand, no voltage can travel through an incomplete circuit. :uhh

Yes, just leave the chip out (but all the ancillary components in place) and check the voltages at the points where the pins of the chip would be. No, the circuit won't function under these conditions, but you will still have some voltages, either directly from +Vs or through a resistor.

 

[Twinfun2]

Yes, just leave the chip out (but all the ancillary components in place) and check the voltages at the points where the pins of the chip would be. No, the circuit won't function under these conditions, but you will still have some voltages, either directly from +Vs or through a resistor.

Good deal; makes sense now. I'll post right after I get the ratings.

 

[ShadowPho1]

Rule of thumb - if the part gets too hot for comfort than it either needs a heatsink or you are doing something wrong (like reverse the polarity). For many chips if your reverse the polarity it will act as a short and will heat up very quickly.

Does your voltmeter have a current measuring ability? If so, try with that (making sure that you connect it in SERIES). Now, also disconnect everything and connect only power and ground and see if it gets even warm. If so then somethign is bad there.

 

[vk6kro]

The 555 is especially vulnerable to external voltage on pin 3, its output.
So, it is important not to plug a new chip in if there is voltage on this socket pinhole.


I found a great website on 555 applications. Sometimes someone just does a superb job with making a website and this is one of them.
http://www.uoguelph.ca/~antoon/gadgets/555/555.html

Anyone here could probably fix that circuit in a few minutes but it can take hours on this Forum to ask the right questions without seeming to be insulting. So, please excuse any questions that seem to be too simple.

I have a couple of breadboards , but I prefer to solder up a circuit on a bit of printed circuit board.
Maybe you would like to try it. You turn the chip so its legs are facing upwards. (this construction method is called "dead bug construction" for this reason).
Any legs that are to be grounded are bent downwards and soldered to the PC board. Also any capacitors etc that need to be grounded are soldered in.
Power leads etc are attached to tag strips and not directly to the IC. (I use header pins cut to shape as tag strips).
This way, you can lay out a circuit as it is in a diagram and quickly see if something is wrong. It is very safe and the IC is in no danger of being pulled apart. And you can easily change a component value if you want to.

 

[Twinfun2]

Rule of thumb - if the part gets too hot for comfort than it either needs a heatsink or you are doing something wrong (like reverse the polarity). For many chips if your reverse the polarity it will act as a short and will heat up very quickly.

Does your voltmeter have a current measuring ability? If so, try with that (making sure that you connect it in SERIES). Now, also disconnect everything and connect only power and ground and see if it gets even warm. If so then something is bad there.

Well, I think it's possible that this would explain my problem. After I only connected the V+ and 0v, it still got very hot (people who can verify this will be greatly appreciated)! I guess the bottom line would be that the timer is down right fried/defective. Before I plug in my new timer, I'm going to take a few pictures of my bread boarded circuit before I go ahead and connect the 6v supply.

Again, thank you all for being of so much help! :!)

 

[Twinfun2]

I have a couple of breadboards , but I prefer to solder up a circuit on a bit of printed circuit board.
Maybe you would like to try it. You turn the chip so its legs are facing upwards. (this construction method is called "dead bug construction" for this reason).
Any legs that are to be grounded are bent downwards and soldered to the PC board. Also any capacitors etc that need to be grounded are soldered in.
Power leads etc are attached to tag strips and not directly to the IC. (I use header pins cut to shape as tag strips).
This way, you can lay out a circuit as it is in a diagram and quickly see if something is wrong. It is very safe and the IC is in no danger of being pulled apart. And you can easily change a component value if you want to.

Does current travel differently in a soldered circuit than in a bread boarded circuit? And also a question, I mind as well ask, in relation is: Can the magnetic field of electricity traveling in a bread boarded circuit be severe enough to effect activity going on in adjacent bus stripes in the bread board?

 

[negitron]

And also a question, I mind as well ask, in relation is: Can the magnetic field of electricity traveling in a bread boarded circuit be severe enough to effect activity going on in adjacent bus stripes in the bread board?

Nah, the breadboard will melt long before there's enough current to create large magnetic fluxes.

 

[vk6kro]

Does current travel differently in a soldered circuit than in a bread boarded circuit?

No, but the connections are more secure, you can see the whole circuit and wire lengths are shorter meaning you get less inductance in the circuit.

Inductance of a few inches of wire doesn't matter for a 555 light flashing circuit but for any serious circuits at high frequencies, it can mean the difference between a circuit working or not working.
For example, I just measured the reactance of a 4 inch piece of wire at 15 MHz. It had a reactance of 17 ohms. That would be enough to upset most circuits operating at that frequency.

Can the magnetic field of electricity traveling in a bread boarded circuit be severe enough to effect activity going on in adjacent bus stripes in the bread board?

Two such wires anywhere near each other at high frequencies would behave like a transformer and pass power from one to the other and possibly cause an amplifier to oscillate.

At VHF (frequencies above 30 MHz) construction techniques that involve short wire lengths are very important and I build all equipment as if it was for VHF. Good ground planes and short wires on components. It is a good habit to get into.

 

[chroot]

Two such wires anywhere near each other at high frequencies would behave like a transformer and pass power from one to the other and possibly cause an amplifier to oscillate.

I think we can agree that this is way too small an effect to be of concern here to Twinfun2.

Twinfun2, what you have is some kind of very large current flowing through your device, either from the supply terminals, or through the output pin (which appears to be the only pin which is not internally current-limited in the NE555N).

- Warren

 

[vk6kro]

The poster did ask about the effect of lead length.
Can the magnetic field of electricity traveling in a bread boarded circuit be severe enough to effect activity going on in adjacent bus stripes in the bread board?

Even if you are making an audio amplifier, most transistors these days can operate at 300 MHz or more and they may well oscillate at those frequencies if the layout is poor.

I used a 2N3563 in a battery charger for NiMH batteries recently. This transistor has a ft of 900 MHz, so I cut the leads very short to try to avoid parasitic oscillation. Good layout never goes astray.

I am hoping the fault will be a wiring error or a faulty IC, but VHF oscillation is a real problem in breadboard circuitry. The 555 may not operate there but it has transistors in it that probably can. It can produce some pretty fast square wave edges.

 

[Redbelly98]

At the frequencies Twinfun2 is using, a 555 should work just fine on a solderless breadboard.

 

[Twinfun2]

Well, I think it's possible that this would explain my problem. After I only connected the V+ and 0v, it still got very hot (people who can verify this will be greatly appreciated)!

Still wanting to know if this means that the timer is officially bad.

EDIT: I'm still trying to get a hold of a camera . I'll wind up drawing my bread board circuit layout on paint or something.

 

[Twinfun2]

Here are my pin ratings according to instruction given by vk6kro, and negitron.

For Reference:
__________

R1:10k
R2:100k
C:10uF
__________

1: 0v
2: 2v
3: 0v
4: 5v
5: 0v
6: 2v
7: 4.5v
8: 5v

(I'm, sadly, using an analog multimeter, and those are the best ratings I have been able to grab.)

 

[chroot]

Still wanting to know if this means that the timer is officially bad.

Yes, the part is bad.

- Warren

 

[Twinfun2]

Yes, the part is bad.

- Warren

Muchos gracias.

Now just awaiting if those volt readings are correct, or near correct, before inserting my new timer.

 

[chroot]

3: 0v

This worries me. Pin #3 is the output pin, which supplies current to the load. If nothing is connected to it, it should be floating, and will probably not go all the way down to 0V. If you've connected pin #3 to your LED and resistor, it should have a voltage of around 4-5V on it.

If pin #3 is really all the way down at ground (0V), then a huge amount of current is likely flowing out of it. Something's screwed up with that pin. Disconnect everything from it, and make absolutely sure there's no continuity between it and the ground rail.

- Warren