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    IC 4060 is an excellent integrated circuit for timing applications. Its ten active high outputs can give time delay from few seconds to hours. With a few components, it is easy to construct a simple but reliable time delay circuit.

    IC 4060 is an Oscillator binary counter cum frequency divider. Its inbuilt oscillator is based on three inverters. The basic frequency of the internal oscillator is determined by the value of the capacitor connected to its pin 9 and that of the resistor in its pin 10. By increasing or decreasing the value of capacitor / resistor, time delay can be changed. Each output goes high after the completion of the timing cycle. To get maximum time period, output Q10 is omitted in the IC itself so that double time is available between Q9 and Q11.

    Inside the IC, there is an oscillator and 14 series connected bistables (Ripple cascade arrangement). Internally the oscillator signal is applied to the first bistable which drives the second bistable and so on. Since each bistable divides its input signal by two, a total of fifteen signals are available, each of half the frequency of the previous one.

    Ten of these fifteen signals are available on the output pins Q3- Q13.HEF 4060 is CMOS version which can operate at 3 volts while CD 4060 is high voltage type that can operate between 5 to 15 volts. It is necessary to add a capacitor close to pin 16 of IC so that minute voltage changes will not affect the timing cycle. Reset pin 12 resets the timing cycle once it is grounded. Outputs can give almost full supply voltage to drive light loads. Heavy loads such as relay can be operated through a driver transistor. When the high output is connected to the pin 11(clock input) through a diode, oscillation stops and IC remains latched in high state till it resets. Pin 11 can be used to give clock pulses from an external source.

    4060 Timing cycle calculation

    Time t = 2 n / f osc = Seconds

    • n is the selected Q output number
    • 2 n = Q output number = 2 x Q no times Eg. Q3 output = 2x2x2 = 8
    • f osc = 1 / 2.5 (R1xC1) = in Hertz

    R1 is the resistor at pin 10 in Ohms and C1, the capacitor at pin 9 in Farads.

    For example if R1 is 1M and C1 0.22, the basic frequency f osc is

    1 / 2.5(1,000,000 x 0.000,000 22) = 1.8 Hz

    If the selected output is Q3 then 2 n is 2 x 2 x 2 = 8

    Therefore time period (in seconds) is

    t = 2 n / 1.8 Hz = 8 / 1.8 = 4.4 seconds

    A ready reckoner for selecting the output to get a required time delay is given along the diagram

    attentionThe author D Mohankumar is not an active member anymore. Please take into consideration that the presented information might not be correct.
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    26 Responses to "IC 4060 Design Note"

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    1. vineela mallina says: on July 9, 2010 at 4:12 pm

      sir.iam from ece.
      iam doing miniproject on midnight security light.in that the capacitor and resistor values are taken as u mentioned above but iam not getting requried output.i need time period adjustment .plz can u give me a suggistion

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    2. The oscillation frequency can be changed using an 1M preset in the place of the resistor at pin10.A number of circuits using CD4060 are in the site. Search and see the designs

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      • Hearty congratulations for your work. I am from Pune, India. seeking your help: I want to know whether we can ‘pause’ ( hold ) oscillations, remember the instnt of ‘pause’ and resume to ocillations from the point where it had paused.The circuit should remember the instant whre it paused and start again from that point. I look forward to hearing from you a solution. Thanks in advance.
        – Ramesh Divekar.

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      • sir . .can u send the complete detail about the midnight security light for my educational project . .thanks. . .

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    3. vineela mallina says: on July 21, 2010 at 9:15 pm

      sir i saw the design notes i the site.but i want small time period to choose the out put. in my circuit there is a 100k preset is avaliable.sir plz can u provide me rc values for time period less thab 30 min..in other sites i saw the formula as (2^n)/(2*rc) but u given as (2n)/(2.5*rc).i also want the clarification about the formula sir.

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    4. Sir, I want to make a big stroboscope using 100 white LEDs of 1 watt each. The flash rate should be adjustable from 10 to 100 flashes per second. The flash on time should be from 10 to 30 micro second. The most important part is to how to get the maximum light from an LED even at the lowest flash on time of 10 micro sec. I shall be obliged if you can help me out

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    5. D Mohankumar D.Mohankumar says: on July 31, 2010 at 3:16 pm
        member

      The lowest flash rate of 10 microseconds cannot be perceived by the eye as flashes.The LED remains lit even if it is flashing. So exact timing can be determined only by measuring the pulse width of the oscillator circuit. You can use a 555 based astable circuit. Select the RC network by calculating the on time/ off time. A Medium power transistor is necessary to drive the LED. Brightness can be controlled by adjusting the value of the series resistor of LED from 10 ohm to 470 ohms 1 watt

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    6. yidnekachew says: on August 14, 2010 at 11:45 am

      sir, It is good Explanation about the 4060 IC.
      But, I have also saw that the timing formula is t=1÷(f ÷16) ,Note that 16 is 2^4 for Q3, for Q4 2^5 and so on,for Q13 2^14 because there is no Q10.
      And the frequency of 1÷(2.3*R1*C1)
      How do u see that ?The IC is HCF4060B

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    7. D Mohankumar D.Mohankumar says: on August 15, 2010 at 6:54 am
        member

      Two versions of timing. This slightly varies depending on the make. Anyway exact timing cannot be obtained due to many facts

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      • i am not satisfiedJagjit Mahal
        July 31, 2010 at 8:01 am

        Sir, I want to make a big stroboscope using 100 white LEDs of 1 watt each. The flash rate should be adjustable from 10 to 100 flashes per second. The flash on time should be from 10 to 30 micro second. The most important part is to how to get the maximum light from an LED even at the lowest flash on time of 10 micro sec. I shall be obliged if you can help me out
        Repl

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    8. kec(dhapakhel) says: on January 13, 2011 at 7:55 am

      how can it be used in inverter as oscillator(12 volt dc to 220 volt ac 100w?)

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    9. Rajarshi says: on May 13, 2011 at 10:51 am

      Sir can u please explain about the working of 4060 refering to logic diagram, how time delay is set in the flipflops.

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    10. Rajsekhar Nag says: on May 30, 2011 at 10:17 pm

      According to datasheet Q4 to Q10 and Q12,Q13,Q14 are assigned.
      Please see http://pdf1.alldatasheet.com/datasheet-pdf/view/66436/INTERSIL/CD4060.html

      But you have used Q3 to Q9 and Q11,Q12,Q13.
      Again according to your calculation, if basic frequency is 1.8Hz (time period .56 Sec)then pin 7(described as Q3 as in the above diagram) will be 8 (no unit mentioned here) and time period will be 4.4 second!
      But according to the datasheet, if basic frequency is 1.8Hz (time period .56 Sec) then pin 7 will be (divided by 16) 0.1125Hz time period will be 8.8Sec.
      Please explain if I am missing something. I have seen many article of yours in Net and in EFY. I always read your writings with great interest.

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    11. sir does the midnight security ckt works properly..?? as the DC ckt is clubbed wid AC .. so wil it make any trouble to the users?

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    12. sir can u help me in designing a circuit that should provide 7 microseconds of delay with respect to input pulse.
      which ic would u prefer and asssist me to take up and the rc values

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    13. sir . .how is the function of midnight security light, because if plug to ac, the light will be automatically on . . .what is the error of my project.

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    14. Teufel964 says: on November 29, 2012 at 6:32 pm

      Hi,

      I have a problem with the value

      C1 is 0.1 what Farad, pikoFarad or nanoFarad

      C2 is 0.22 what Farad, piko or nano.

      who kan tell me, what it is.

      Thanks

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    15. HI mohan,
      The circuit working fine.Excellent job.But i have problem.That The above circuit working in fine in BREADBOARD.But if it not working in the general purpose board.I checked all the connection.Its not working can U Just

      Regard
      m

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    16. sir,This circuit give continuous time delay .i need a circuit when the time completed the circuit will totally turn off please sir.

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      • Rajsekhar Nag says: on October 11, 2013 at 7:41 pm

        @Dinesh,
        You can use a diode from the output pin to one of the oscillator pin #11. This diode will act as a latching component, which will latch the IC once the set time lapses and the output of the IC goes high.

        Connect the cathode (band side) of a 1N4148 diode to the pin 11 and anode to the selected output pin.

        If this diode is not inserted, the output would go freewheeling from logic high to logic low and keep repeating the time delays.

        Also you can use a reset switch between pin 12 and 16.

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    17. Thank u sir to giving information abt this project says: on April 22, 2014 at 6:10 pm

      i have used my r1=150k and c= 0.1 micro n my freq is after calculation is 26.66 hz
      and time period is 0.3 sec howz it possible

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    18. chaitanya says: on April 23, 2014 at 6:15 pm

      What about using IC 4010 instead of IC 4060

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