555 Timer...not timing? - Page 1
Forum section: electronics / beginners

author: Mad Mart
author_rank: Contributor
timestamp: July 13, 2014, 04:41:46 pm
content: Hi allAlthough I'm getting on a bit in age I've just started to get interested in Electronics, so have spent the last week or so reading and watching videos on YouTube.Anyway, I bought a starter kit and decided to make a circuit using the 555 chip. The circuit has two LEDs. No.1 LED is on until you press the momentary switch at which time no.1 LED goes out and no.2 LED switches on. The problem is no.2 LED should stay on for a certain length of time before the timing cycle ends and then goes out, turning on LED no.1. What my circuit does is just to toggle between no.1 & no.2 LEDs when I press the switch. I'm using a 100k pot and a 1000uF 25V Cap.I've attached a photo to hopefully clarify my circuit...please be gentle with me.

author: rob77
author_rank: Super Contributor
timestamp: July 13, 2014, 05:19:12 pm
content: posting the schematic would be betterin fact most of the guys here will not even have a look without a schematic diagram posted (and i fully understand them)anyways... is that resistor in the top left corner connected to the + rail ? - if not then that's your problem. and do i see it correctly as 220R ? - should be bigger (few Kohm) and connected to the + rail.

author: Mad Mart
author_rank: Contributor
timestamp: July 13, 2014, 05:43:14 pm
content: Yes, the resister is connected to the +ve rail. It was a 220R but I've now swapped it for a 2K2, it made no diference though.I agree I should have posted a schematic...I have now. Borrowed from a Rob Winter tutorial video.

author: polishdude20
author_rank: Contributor
timestamp: July 13, 2014, 07:25:21 pm
content: It looks to me like in the picture, your ic isnt connected the right way. You should have pins 6 and 7 tied together but it looks likeyou tird together 2 and 3 . Im just looking at that dot there and it looks like its inthe wrong place!Edit: nvm stupid me. The dot confused me but since your chip has a cutout id base it off that. So your picture is actually correct. My bad.

author: polishdude20
author_rank: Contributor
timestamp: July 13, 2014, 07:35:55 pm
content: Try grounding pin 5 with a 0.01uf cap. Also try changing the resistors of the leds around

author: TerminalJack505
author_rank: Super Contributor
timestamp: July 13, 2014, 07:52:35 pm
content: I'm guessing your pot isn't wired-up quite right. To test that theory you can try replacing it with a 4.7K resistor. That should give a ~5 second delay.

author: c4757p
author_rank: Super Contributor
timestamp: July 13, 2014, 08:02:07 pm
content: If the pot's disconnected, the output should stick high. Toggling will be observed if either the discharge pin (pin 7) or the capacitor are disconnected. Check those.

author: Mad Mart
author_rank: Contributor
timestamp: July 13, 2014, 08:07:15 pm
content: I've tried all your suggestions so far. Still no difference. I measured the voltage for pin 6 and when I press the switch it goes immediately to 0V. If, however, I keep the switch depressed then the voltage will gradually build up to ~2.3V over a period of about 16 seconds.I've just noticed that if I take the pot out, it has no effect on the circuit.

author: Bryan
author_rank: Frequent Contributor
timestamp: July 13, 2014, 08:32:26 pm
content: Are you sure you have the polarity of the power supply correct.?

author: rob77
author_rank: Super Contributor
timestamp: July 13, 2014, 08:38:53 pm
content: Quote from: Bryan on July 13, 2014, 08:32:26 pmaccording to the picture he has the polarity right - he's even strictly using RED wires for + rail and black wires for - rail , and other colors for the "signal" wires. furthermore on the picture you see that a wire with a white strip (+) from the wall-wart is connected to the red rail on the breadboard.

author: adi101
author_rank: Contributor
timestamp: July 13, 2014, 08:57:16 pm
content: It looks like the pot is marked 104, right? This means the value is 100K, not 3.3K.My suggestion is to use a 10uF capacitor instead of 1000uF.

author: rob77
author_rank: Super Contributor
timestamp: July 13, 2014, 09:04:35 pm
content: probably the 555 is dead, try a different one.

author: Bryan
author_rank: Frequent Contributor
timestamp: July 13, 2014, 09:18:52 pm
content: Quote from: rob77 on July 13, 2014, 08:38:53 pmOne of my walwart's has a stripe on the negative (-) wires. Worth a double check.

author: eetech00
author_rank: Regular Contributor
timestamp: July 13, 2014, 09:38:35 pm
content: hiHere's something to check.Sima shows R2 as a 3.3k resistor.Note that the output eventually returns to 0v.Now look at simbSimb shows R2 as a 100k resistor.Note that the output never returns to 0v.I'm guessing something wrong with potentiometer.eT

author: Electronics-Repairman
author_rank: Contributor
timestamp: July 13, 2014, 10:07:05 pm
content: Having used the 555 for a long time 1000 uf seems a lot to me ,pin 6/7 are wired together someone suggest they weren't ,pin 5 should have a 0.01 uf ceramic to ground, if you scope pin 3 you should get a timing pulse, as suggested you may have toasted it,pin 3 is the output side , so you should get a signal off that, the pot seems to bother me two, i've always used a 3 pin pot, with the center pin wire to one of the others !!!

author: rolycat
author_rank: Super Contributor
timestamp: July 13, 2014, 11:53:19 pm
content: Your description of the behaviour of the circuit sounds as though it is connected correctly except the timing elements VR1 and C1. If all else has failed, it can be worth reassembling everything from scratch - at least you are likely to encounter a different type of faultSince I have far too much time on my hands at the moment and haven't played with a 555 for a while, I recreated the circuit you gave using a similar layout but shorter wires, which will hopefully show the connections more clearly.If C1 is disconnected it exhibits the same behaviour you describe for your circuit, otherwise it works as expected.VR1 is a 100k trimpot and currently C1 is a 100uF electrolytic, although I tested a 1000uF and that works too, with a long maximum delay. Other resistor values are as in Rob Winter's schematic.With these values, the green LED D1 lights for around 15 seconds after SW1 is pressed with VR1 at its maximum.Hope this helps.

author: edy
author_rank: Super Contributor
timestamp: July 14, 2014, 02:16:22 am
content: Have a look here... Lots of schematics using the 555 that may help:http://www.talkingelectronics.com/projects/50 - 555 Circuits/50 - 555 Circuits.htmlLots of free eBooks on that site, good to learn from!

author: Mad Mart
author_rank: Contributor
timestamp: July 14, 2014, 06:22:16 pm
content: I have just rebuilt the circuit similar to rolycat's, and guess what...it's just the same as before. I've swapped the 555 out about 4-5 times now and that has made no difference either. When pin 2 goes low it should make pin 3 go low also, which it does. But it should also start the timing on pin 6, which it doesn't. That's obviously where the problem lies. It can't be the pot as I've swapped that out several times for a straight resistor. The only other component is the cap and I've swapped that around I don't know how many times.

author: rolycat
author_rank: Super Contributor
timestamp: July 14, 2014, 07:23:01 pm
content: Quote from: Mad Mart on July 14, 2014, 06:22:16 pmIt definitely sounds as though your capacitor isn't charging properly via the pot. When you say you've swapped it around, do you mean you've tried alternative values?Try breaking it down. Attach your multimeter across the capacitor pins, and disconnect the link to the 555 pins 6 and 7. Briefly short the capacitor pins and then watch the voltage. Does it climb rapidly to almost 9V with the pot at its minimum limit, and via the expected charge voltage curve at intermediate settings?Leave the pot at an intermediate value and reconnect pins 6/7. What happens to the voltage? It should start at around 0V, climb steadily to about 6V after SW1 is pressed and while D1 is lit, and then jump back to 0V.Another possible source of problems is the breadboard. Cheap ones especially can introduce mysterious problems due to intermittent or even failed contacts. One way to check this is to move connections systematically to different strips.

author: Mad Mart
author_rank: Contributor
timestamp: July 14, 2014, 09:10:26 pm
content: QuoteDone that and the voltage climbs as you described. Not sure about the curve?QuoteIt doesn't do that. What happens is, if you keep the switch held down the voltage will climb to the limit set by the pot. When you let go the voltage drops to 0V.

author: rolycat
author_rank: Super Contributor
timestamp: July 14, 2014, 10:01:22 pm
content: Quote from: Mad Mart on July 14, 2014, 09:10:26 pmWhen a capacitor is charging the voltage rises quickly at first, then more slowly as it gets closer to the supply voltage. Google "Capacitor charging" for the details.QuoteThe pot isn't setting a voltage limit, it's controlling how much current flows into the capacitor, and thus how quickly it charges up. When the voltage at IC pin 6 reaches 2/3 of the supply voltage it ends the timing period, so the trigger voltage will always be 6V with a 9V supply.You shouldn't need to press SW1 for more than an instant.You will get the behaviour you describe if the pot is at its minimum setting, since the capacitor will be connected almost directly across the supply rails and will charge up immediately. Are you sure you are setting it to a halfway position? If you turn off the supply and disconnect the capacitor, what resistance do you read between IC pin 6 and pin 8?

author: Mad Mart
author_rank: Contributor
timestamp: July 14, 2014, 10:11:47 pm
content: I'm getting 1.1k across pins 6&8.

author: rolycat
author_rank: Super Contributor
timestamp: July 14, 2014, 10:36:45 pm
content: Sounds reasonable. What value capacitor are you using with that?If you try my earlier suggestion again and time how long the capacitor takes to charge to 6V through that resistance when disconnected from the 555, it should be about the same as the monostable timeout when in circuit.It may be worth checking the resistance between the capacitor negative lead and, say, IC pin 1 and the positive lead and pin 6, just in case the breadboard connections are faulty.

author: Mad Mart
author_rank: Contributor
timestamp: July 14, 2014, 11:09:35 pm
content: I'm not sure if I understood correctly but I took the capacitor (470uF) out of the circuit and just connected a 1k resistor between it and the +ve. It didn't really take anytime at all to reach 6V. cap -ve to IC pin 1 is 0 ohms.

author: rolycat
author_rank: Super Contributor
timestamp: July 14, 2014, 11:37:05 pm
content: Quote from: Mad Mart on July 14, 2014, 11:09:35 pmA 1k resistor will take less than a second to charge a 470uF capacitor to 6V, which may explain your problem. Set your 100k pot at about halfway, check the resistance is around 50k, and reassemble your circuit. Press the button briefly and D1 should light, and then go out about 30 seconds later. If not, double check that the capacitor leads are making good contact.

author: Richard Head
author_rank: Frequent Contributor
timestamp: July 15, 2014, 06:03:46 am
content: There are two glaring issues with this circuit. The first is that the 1000uF capacitor is massive for this circuit. The problem is that electrolytic capacitors have a relatively large leakage current that needs to be taken into account. This limits how large you can go in timing resistor value.The second problem is that the internal NPN discharge transistor has a limited current carrying capacity. To limit the current (and protect the transistor) you should have a low value resistor in series with the discharge pin.

author: rolycat
author_rank: Super Contributor
timestamp: July 15, 2014, 09:54:22 am
content: Quote from: Richard Head on July 15, 2014, 06:03:46 amThis is largely nonsense, and not very helpful to the OP.As I imagine you are aware, the output pulse width for a 555 in monostable mode is given by:t = RC ln(3)where t is in seconds, R is in ohms and C is in farads. ln(3) is roughly 1.1, so for a 1000uF capacitor and a 100k resistor, the pulse duration will be:1.1 x 0.001 x 100000, which is 110 seconds.This is perfectly suitable as a maximum value for a variable delay timer of the sort being tried. The 3.3k resistor given in the schematic would give a maximum of around 3.6 seconds, which is maybe a little brief.As for the capacitor leakage current, I charged a cheap Chinese 1000uF capacitor to 9V and after 60 seconds the voltage had dropped by 0.02 V. I really don't think that's going to be an issue in this circuit.Lastly, 555 timers have internal limits on the collector current for the discharge pin. You aren't going to damage it by not using a discharge resistor.

author: Mad Mart
author_rank: Contributor
timestamp: July 15, 2014, 10:23:45 am
content: QuoteWhat is ln(3)?

author: bwat
author_rank: Frequent Contributor
timestamp: July 15, 2014, 10:28:40 am
content: Quote from: Mad Mart on July 15, 2014, 10:23:45 amThenatural logarithmof 3.

author: rolycat
author_rank: Super Contributor
timestamp: July 15, 2014, 10:33:59 am
content: Quote from: Mad Mart on July 15, 2014, 10:23:45 amThe natural logarithm of 3 (its logarithm to the base e). Enter 3 on a scientific calculator and press the ln button and you will get a more exact value.The formula is derived from the one which determines the charge curve of a capacitor, which I referred to earlier.If you aren't too comfortable with the maths then you can ignore all that and just use the approximation, which is close enough for most practical purposes.

author: Mad Mart
author_rank: Contributor
timestamp: July 15, 2014, 10:47:04 am
content: Ah! yes, I remember all that stuff (not really). It's been quite a number of years since I last used logs. But I am learning a lot on here.

author: Mad Mart
author_rank: Contributor
timestamp: July 15, 2014, 06:44:02 pm
content: This circuit is obviously not going to work..for me!!! Could I have fried all 6 of my ICs? I've tried all your suggestions but that pin 6 won't start timing. Is there a way to check the IC, apart from build a 555 tester circuit?

author: Electronics-Repairman
author_rank: Contributor
timestamp: July 15, 2014, 07:04:42 pm
content: I have one of these and it works well on many chips including the 555http://www.ebay.co.uk/itm/Multi-Function-IC-LED-Optocoupler-LM399-DIP-CHIP-TESTER-Model-Number-Detector-/251549975540?pt=LH_DefaultDomain_0&hash=item3a918c03f4

author: rolycat
author_rank: Super Contributor
timestamp: July 15, 2014, 07:13:44 pm
content: Quote from: Mad Mart on July 15, 2014, 06:44:02 pmWell, you could send me one and I'll test it for you. You can buy a new one for less than the cost of a letter, so it's scarcely worth returning it (I will if you want), but you would have some assurance that you haven't blown them all. PM me if you want my address.Another option is to buy a couple of the PCBs likethisavailable on EBay for two or three pounds and solder up a permanent circuit with an IC socket. They do complete kits for a few pounds more. This might provide a more reliable test bed (if your soldering is up to scratch)

author: Electronics-Repairman
author_rank: Contributor
timestamp: July 15, 2014, 07:36:58 pm
content: I just may get one myself they look fun, could be done in a dinner break "Yeah" there's a plan

author: Mad Mart
author_rank: Contributor
timestamp: July 15, 2014, 08:43:06 pm
content: Thanks rolycat, I think I will just buy another batch for now. I've made three different 555 tester circuits now and none of my ICs work on any of them. The LEDs glow but they should flash alternately. I will definitely buy one of those testers though. It probably would have saved me many hours of head scratching.I'll let you know the outcome.

author: rolycat
author_rank: Super Contributor
timestamp: July 15, 2014, 09:31:17 pm
content: No worries. To paraphrase one of Dave's sayings, you learn a lot more when your circuits don't work.P.S.Make sure you don't get the CMOS version for general experimentation - they are a lot more vulnerable to accidental damage.

author: Mad Mart
author_rank: Contributor
timestamp: July 15, 2014, 09:39:49 pm
content: Quote from: rolycatThat is so true. I've only been "playing" with electronics for just over a week and I now know the pinouts of the 555 timer and can build a simple circuit without too much thought.

author: dentaku
author_rank: Frequent Contributor
timestamp: July 16, 2014, 12:32:54 am
content: You could always try the good old 555 blinker to see if they've been fried. I've found that the average 7555 chip is ridiculously tough.http://www.555-timer-circuits.com/flashing-led.html

author: Mad Mart
author_rank: Contributor
timestamp: July 17, 2014, 04:51:46 pm
content: Great!all that time spent messing around (read, learning) and it's the ICs that were buggered. Thanks for your help guys. Onwards & upwards.Note to self - Be more careful with your components!

author: Electronics-Repairman
author_rank: Contributor
timestamp: July 17, 2014, 05:04:08 pm
content: Quote from: Mad Mart on July 17, 2014, 04:51:46 pmAnd where you buy them from Simon on here sellshttp://www.ebay.co.uk/usr/sparkylabs-co-uk?_trksid=p2053788.m1543.l2754, this is who I mainly buy from and at reasonable prices, I think you get a discount as a eev blog subscriber, not that I've ever mentioned it.May beo if Simon reads this he'll put you right, funny thing is I just bought some 555's from him.

author: rolycat
author_rank: Super Contributor
timestamp: July 17, 2014, 05:16:05 pm
content: Quote from: Mad Mart on July 17, 2014, 04:51:46 pmAh, that explains the extreme non-cooperation of your circuits.Happy to help - I'm glad you haven't been put off.QuoteAny idea how you managed to nuke them? Reversing the supply polarity is a leading cause of magic smoke.

author: Mad Mart
author_rank: Contributor
timestamp: July 17, 2014, 05:32:07 pm
content: Well it seems that it's probably down to a naff power supply, as when I tested my old ICs, 4 out of the 6 were fine. I had a feeling that it could possibly be down to the mains adaptor but sort of dismissed it. So when I got my new batch today I made sure I tested them using a 9V battery.Why the adaptor? Well it's one I've had knocking around for years (along with about a dozen others). The output is given as 6V DC 300mA. When I checked it it was outputting around 9V but I thought that would be ok. Maybe a regulator or something not working properly.

author: rolycat
author_rank: Super Contributor
timestamp: July 17, 2014, 05:50:39 pm
content: Quote from: Mad Mart on July 17, 2014, 05:32:07 pmIf it's one of the old linear type it may not have any regulation, in which case that 9V is likely to be the no-load voltage.Try putting 100 ohms or so across it and measure the voltage then.

author: Mad Mart
author_rank: Contributor
timestamp: July 17, 2014, 06:50:56 pm
content: QuoteOf course it is. doh!QuoteToo late. I ripped the lead out and chucked the adaptor.At least I'm now able to move on.