555 timer 40khz 50%duty cycle

Discussion in 'The Projects Forum' started by SNVHKC, Nov 15, 2012.

  1. SNVHKC

    Thread Starter New Member

    Jun 28, 2012
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    Could someone help me to generate a 40khz square pulses(Astable multivibrator) using 555timer IC,with 50% duty cycle.
     
  2. JDT

    Well-Known Member

    Feb 12, 2009
    658
    85
    This is not the correct forum for this!

    A 555 cannot generate 50% duty cycle. Also I'm thinking that a 555 is not the best device for operation at this frequency.

    If a 50% duty cycle is important I would have an oscillator that runs at twice the required frequency and follow it with a divide by 2 flip-flop. Easily done in 4000 series CMOS.

    Edit: In fact have a look at the datasheet for a HEF4047B. It's all in there!
     
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  3. BillO

    Well-Known Member

    Nov 24, 2008
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    Here is link to the data sheet. Page 10 shows one method of doing 50% duty cycle.

    You can also google "555 50% duty cycle" and you'll get a few hundred thousand hits.
     
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  4. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    In the attached schematic, make R1 = 910 Ω, C1 = .001 μF, and R2 = 17545 Ω. (You may use a 15 or 17 kΩ fixed resistor and a series trimpot for R2.) Mathematically, this should produce a 40 kHz output with a 51.26% duty cycle. Of course, component tolerances will cause variation.
     
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  5. spark8217

    Member

    Aug 29, 2011
    64
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    Formula: T = 0.7 * (R1 + (2 * R2)) * C1 (without diode)
    f = 1.4 / ((R1 + (2 * R2)) * C1) (without diode)

    T = 0.7 * (R1 + R2) * C1 (with diode)
    f = 1.4 / ((R1 + R2) * C1) (with diode)

    Symbols: T = Time period (Units: Seconds)
    f = Frequency (Units: Hertz)
    Mark = High time of output waveform (Units: Seconds)
    Space = Low time of output waveform (Units: Seconds)
    Duty Cycle = % of period T in which output is high (mark)

    Data: Mode: Astable
    Calculate component values
    Frequency = 40 Kilohertz
    Duty Cycle = 50 %

    Results: Period = 25 Microseconds
    Mark = 12,5 Microseconds
    Space = 12,5 Microseconds
    C1 = 10 Nanofarads
    R1 = 1 Kilohms
    R1 nearest preferred value = 1 Kilohms (E24 / 5%)
    R2 = 1,786 Kilohms
    R2 nearest preferred value = 1,8 Kilohms (E24 / 5%)

    Diode D1 should not be fitted
     
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  6. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Don't use pin7, drive the cap from pin3 which is bipolar (through one resistor), and you just need 1 cap and one resistor and it automatically makes a 50% duty oscillator.

    I have not used pin7 in a while, unless I specifically need a 555 with a duty that is NOT 50%.
     
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  7. claudewang

    New Member

    Oct 23, 2013
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    Hi, there. I have been looking for a way of using ne555 as a 50% duty oscillator with only one resistor tuning for frequency. Your method sounds awesome, but it's still not very clear to me how to implement it. Could you please do me a favor to show me an example circuit that can do exactly what you said? Many thanks.
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is an example of using the 555 to generate a square wave. If doing this at 5V Vcc or lower, use the CMOS version of the 555. Take the output as shown so as not to screw up the timing. Note that V(rc) turns around at 1/3Vcc and 2/3Vcc.
     
    Last edited: Nov 10, 2013
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  9. sakar

    New Member

    May 2, 2014
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    spark8217 can you please post the circuit diagram relative to which you are providing the values.
     
  10. sakar

    New Member

    May 2, 2014
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    Can you please post the circuit diagram relative to which you are provoding the values.
     
  11. MrChips

    Moderator

    Oct 2, 2009
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    Note the date of the post and number of posts of member spark8217.
    Member probably not hanging around AAC too often.
     
  12. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    50/50 is the easiest 555 circuit of all, and 40 KHz is easily within the range of both bipolar and CMOS parts.

    http://www.555-timer-circuits.com/simplest-555-oscillator.html

    The circuit will run slightly slower than the calculations for the standard oscillator because the pin 3 output does not swing all the way up to Vcc.

    ak
     
  13. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    Mathematically, with R1 and R2 at 120 ohms, and C1 at .1uf, you should get 40kHz from the attached circuit.
     
  14. Johann

    Senior Member

    Nov 27, 2006
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    Another resurrected thread!;)
     
  15. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    And you helped keep it breathing. :)
     
  16. ScottWang

    Moderator

    Aug 23, 2012
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    It's very strange, although I like that way, but when I tried it on one month ago and it didn't work, I did that again on yesterday, but it still can't reach to 50%/50%, I was used LM555CN and one resistor, one capacitor.

    So, could you tell me how did you do that?
    Thank you.
     
  17. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    What duty cycle did you get?

    It's possibly a bit off 50% due to the old TTL 555 having slightly different sink/source characteristics of the output pin 3. Some brands are worse than others.

    I'm generally using CMOS 555s these days like the 7555. Faster, less power required etc.
     
  18. ScottWang

    Moderator

    Aug 23, 2012
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    I just thought maybe you were used the CMOS type, but do they really can reach to 50%/50% or just close to.

    I used two TTL type, one is LM555, another one is NE555(TI), but they just approaching to 50%/50%, I'm not sure how much different for the high and low duty cycle, but I can see from the scale of O'scope, they are not equal.

    I can use another way to reach to 50%/50%, but I still hope the easy way that you mentioned could work.
     
  19. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    Scott,

    Have you looked at the output from the circuit I attached to post #13? My scope is not sufficient to see that much detail.
     
  20. ScottWang

    Moderator

    Aug 23, 2012
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    I was used IWATSU 100Hhz O'scope to measured your circuit, it shows as below:
    left side high level 5uS*2.1, right side Low level 5uS*1.4.
    R1=R2=10K, C1=103, 57Khz.
    1000000Hz/(5uS*(2.1+1.4))=1000000Hz/17.5uS=57.142 Khz
    So, the high level and low level are different, I was used NE555(TI), I don't have CMOS type to try, so I dont' know the result.
     
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