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Request help with 555 timer circuit

  1. Nov 21, 2011 #1
    Hi all,

    I would like to make a 555 timer circuit that when 12v is applied to the circuit an LED will light for 3 - 5 seconds and then go out. and remain out until the 12v is removed and reapplied.

    Any help greatly appreciated.

  2. jcsd
  3. Nov 21, 2011 #2


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    Yes, you can do this.

    It is usually an unwanted side effect, but if you use a monostable circuit and include a bypass capacitor on pin 5 of the 555 (as is always shown in 555 circuits) then you will get one cycle of output when you turn the circuit on.
  4. Nov 21, 2011 #3
    Thanks for the reply, I found this schematic at theis webpage (http://www.eleinmec.com/article.asp?4)

    [PLAIN]http://www.kipper2k.com/monostable.gif [Broken]

    Based on the formulas provided i can make a 5 second (approx) LED display using

    R = 47K
    C= 100uf

    I am assuming that i can just bridge the trigger and when 12v is actually supplied to the circuit that the device at pin 3 (which would be my LED) would light up for approx 5 seconds. Using a 5v through hole LED, should i feed the cathode to a 150ohm resistor to ground to limit current or would it be fine going directly to ground

    Thanks again

    edit, i think i forgot that output is probably equal to the input so in order to use a 5v LED then i would probably need to use a 7805 regulator on the input to reduce output or reduce the output of the 555 to 5v by voltage divider or other means ?
    Last edited by a moderator: May 5, 2017
  5. Nov 21, 2011 #4


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    Yes, that looks OK.

    To exaggerate the effect, you could make the capacitor on pin 5 a 0.1 μF

    It is better to use a tantalum capacitor for timing as normal electros have a lot of leakage.
    The price of tantalum capacitors depends on the capacity (how many μF) so you could make the capacitor 10μF and the resistor 470 K if you like.

    For this circuit, you could use a CMOS 555. They cost about the same as a normal 555 but have a high input impedance and hence give better charging of the capacitor. They also draw less current than a normal 555.
    LMC555 and 7555 are some CMOS 555 chips.
    You can read about this here:
  6. Nov 24, 2011 #5
    I have the circuit almost working, the only problem is that it requires the button to be pushed to complete the circuit and if the button is held then it will not turn off the led.

    What i am trying to do is to use an external 12v power supply so that when the power is turned on i would receive a visual 3 - 5 second lighting of the LED which would then go off. If the external power is turned off and then turned on the cycle will repeat.

    Basically what i am saying is the 555 timer circuit should not need to be touched once installed and the trigger button removed

    Any ideas ?

    Thanks for the replies
  7. Nov 24, 2011 #6


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    Putting a .1 uF cap across the switch will make the circuit activate every time the power is applied. You should probably increase the 10K to 100K or larger. The idea is that every time the circuit powers up you want to hold pin 2 low very briefly. The cap on pin 5 is usually put there since doing this is good engineering practice to help prevent noise on the pin. However, in this case I would make sure it is absolutely no larger than the .01 uF and if it were my design I would change it to .001 uF since it needs to be established at 2/3 supply voltage or higher before pin 2 gets above 1/3 supply voltage. I would also put a small signal diode across the 10K so the cap placed across the switch has a place to discharge through upon power down instead of through the internal protection diode on pin 2. Cathode towards the supply. Some may consider this unnecessary but I always lean towards cautionary. One last thing, if the power supply comes up too slowly you may have to increase the size of the cap placed across the switch.
    Last edited: Nov 24, 2011
  8. Nov 25, 2011 #7


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    You can get a similar effect with just a CMOS logic inverter.

    I used a 74C00 NAND gate with the two inputs of one gate tied together to produce an inverter.

    http://dl.dropbox.com/u/4222062/turn%20on%20pulse.PNG [Broken]

    When power is first applied, the 10 μF capacitor charges via the 100 K resistor and produces a high output for about a second.

    When power is removed, the capacitor can discharge through the diode and the 22K resistor, so that it is ready to be used again in about 250 mS.
    Last edited by a moderator: May 5, 2017
  9. Nov 25, 2011 #8

    Thanks for the help. Circuit is working, the discharge through the signal diode is slow, can take about 5 - 7 seconds, i haven't tried too many different values to judge the response.


    Thanks for the different option. Looking at the circuit and trying to understand what you did...

    The Nand truth tables (2 input), Any time there is a 0 input then the output will be 1. So when power is applied the top input on your diagram instantly goes to high, the lower input tied to the capacitor will remain low until the capacitor charges. The LED tied to the output will instantly come on and remain on until both inputs are high (caused by combination of 100k resistor and the tantalum capacitor).

    So, heres where i am... guessing, in order to increase the on time of the LED i need to slow down the charging time of the capacitor to allow it to remain on for at least 3 seconds. Can this be done by altering the RC of the input or maybe even feeding the output to another logic gate/RC circuit to increase the on time

    Thanks again.
  10. Nov 25, 2011 #9


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    If the discharge through the signal diode is slow it is probably because your power supply voltage is not dropping out very quickly. To increase the on time I would adjust the RC value on the input. I have used 555 timers in monostable mode stretched out to almost 10 minutes. That is pushing it. Usually I would prefer to increase R instead of C as long as the R doesn't get above several meg. The reason is that I like to avoid discharging a large capacitance into pin 7 when the thing times out. Look at a block diagram and you will know what I mean by this. I recall using a CMOS version of the 555 with a 10 meg pot and 100 uF or more (tantalum caps) to get out to about 10 minutes. This type of setup tests the 555 for sure. Any damage done by ESD will probably show up when using the 555 like this. As far as I know this unit is still in operation sitting at the bottom of a water tower running an amateur radio repeater controlling when the transmitter sends a morse code ID. It's been over 15 years since I built it.
  11. Nov 25, 2011 #10


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    kipper2k, Yes, you have that NAND stuff OK.

    I like to just keep NAND gates in stock and use them exclusively for oscillators and buffers. You can get dedicated inverters that have 6 inverters to a package, but I find the layout a bit crowded.

    You could also tie one input of a NAND gate high, so the output would only go low only if the second input became high, which it does eventually after the capacitor charges up enough.

    Anyway, yes, you can change either the R or the C to get longer times.
    You just multiply the value of R (in Megohms) by the value of C (in μF). to get the approximate time.
    eg if you wanted a 5 second delay, you could use a 10 μF Tantalum capacitor and a 470 K resistor.
    Don't use electros for timing. They have too much leakage current.

    You can leave the 22 k resistor unchanged as this only affects the discharge time.

    The diode should be a silicon type. I measured the reverse resistance of some Schottky diodes and it was around 1 Megohm and very sensitive to temperature, so these probably would not be suitable.

    CMOS logic works OK up to about 15 volts, so look for the 74C..... number.
    These are becoming rare now as the 74HC.. devices (which have to have a 5 volt supply) take over.
    These are great, but not as flexible as the 74C devices.
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