Why is my 555 Timer not blinking?

In summary: Okay, I was assuming Vo=Vs when the output is high -- that has been my (limited) experience. I was unaware that this is not necessarily the case for all 555 timer variants.
  • #1
Twinfun2
32
0
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
 
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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
  • #5
negitron said:
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 :cry:*)
 
  • #6
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.
 
  • #7
negitron said:
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.
 
  • #8
negitron said:
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 Vs?
  • 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
 
  • #9
Redbelly98 said:
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.
 
  • #10
negitron said:
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 Vo=Vs 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.
 
  • #11
Redbelly98 said:
Okay, I was assuming Vo=Vs 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.
 
  • #12
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
 
  • #13
chroot said:
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.
 
  • #14
Redbelly98 said:
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 Vs?
  • 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.
 
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  • #15
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
 
  • #16
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?
 
  • #17
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.
 
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  • #18
Twinfun2 said:
1) 330 Ohm
2) 6v
3) Yes
Okay. I was fishing for some obvious error, but everything sounds right here.

Twinfun2 said:
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.
 
  • #19
negitron said:
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
 
  • #20
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.
 
  • #21
I strongly suspect that you're somehow shorting the output to ground or +Vs.

- Warren
 
  • #22
chroot said:
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 :biggrin:.

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.
 
  • #23
Twinfun2 said:
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.
Twinfun2 said:
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.
 
  • #24
Twinfun2 said:
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
 
  • #25
@Chroot

Im using a breadboard. :bugeye:

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!
 
  • #26
Twinfun2 said:
Im using a breadboard. :bugeye:

There are both soldered and solderless breadboards.
 
  • #27
Twinfun2 said:
@Chroot

Im using a breadboard. :bugeye:

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.
 
  • #28
Redbelly98 said:
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...:frown: (lol)

Yah I understand how the solderless breadboard works and how its laid out, good thing to just ask; I understand.
 
  • #29
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. :smile:
 
  • #30
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.
 
  • #31
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
 
  • #32
vk6kro said:
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.
 
  • #33
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.
 
  • #34
vk6kro said:
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. :rolleyes:

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

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! :!)
 
  • #35
Twinfun2 said:
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.
 

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