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555 Beginner problem

Discussion in 'Homework Help' started by NLT, Apr 10, 2016.

  1. NLT

    NLT New Member

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    Hi

    I have this assignment, and I have absolutely no idea what I’m doing.

    I need to design a circuit that produces the output as attached.

    T1=1ms

    Duty Cycle1 = 0.75

    Duty Cycle2=0.75

    T2 can be calculated from T1

    tpw3 can be calculated from T2


    I have to use monostable and astable multivabrators.

    It seems there are 3 circuits that need to be interconnected.

    I have the following:

    2 (possibly 3) 555 timers.

    A variety of caps.

    A variety of resistors.

    Can anyone please help!? Assume I know nothing.

    I’m supposed to simulate in LTspice. If you could provide a schematic, and explain what it does, that would be awesome!

    Thank you very much!
     

    Attached Files:

  2. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    Have you downloaded and read the 555 data sheet?
     
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  3. Ian Rogers

    Ian Rogers Super Moderator Most Helpful Member

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    Eric did some fabulous tools to make 555 timer circuits go to the tools menu above and check them out..
     
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  4. dave

    Dave New Member

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  5. JoeJester

    JoeJester Active Member

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    What's with pulse 1 through pulse 3
     
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  6. NLT

    NLT New Member

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    OK. So, this is my attempt. It doesn’t work.

    From left to right:

    Monostable for pulse 3 - > Astable for pulse 2 -> Astable for pulse 1.

    Calcs:

    For pulse 1:

    f= 1/T = 1/1m = 1k

    Choose

    C3=100n

    R4=3.6k

    R5=7.2k

    Then f= 1.44/((R5+2*R4)*C3)

    = 1.44/((7.2k+2*3.6k)*100n)

    =1k as required

    Duty cycle = (R5+R4)/(R5+2*R4)

    =(7.2k+3.6k)/(7.2k+2*3.6k)

    =0.75 as required


    For pulse 2:

    I reasoned that (I don’t know if this is right) the up time for pulse 2 is equal to (trailing edge of last cycle of pulse 1) – (leading edge of first cycle of pulse 1)

    =4*1m+1m*0.75

    =4.75m

    Then the period is

    T=4.75m/0.75=6.33m

    Then f= 1/6.33m

    =0.158k

    Choose

    C2=100n

    R2=22.8k

    R1=45.6k


    Then

    f= 1.44/((R1+2*R2)*C3)

    = 1.44/((45.6k+2*22.8k)*100n)

    = 0.158k as required

    Duty cycle = (R1+R2)/(R1+2*R2)

    =(45.6k+22.8k)/(45.6k+2*22.8k)

    =0.75 as required


    For pulse 3:

    I reasoned that (I don’t know if this is right) the up time for pulse 3 is equal to (trailing edge of last cycle of pulse 2) – (leading edge of first cycle of pulse 2)

    Then pulse width is

    tpw3=4*6.33m+4.75m

    =30.07m

    Choose

    C4=100n

    R5=273.48k

    Then

    tpw = 1.1*R5*C4

    =1.1*273.48k*100n

    =30.08m as required (nearly)


    The circuits for pulses 1 and 2, when run individually – not connected to other circuits – do produce the required values. If I run the circuit for pulse 3 on its own, output just stays high, and does not return to low.

    The measurement I’m interested in is at OUT2. As I said, it’s not correct. The up time for the 1st and last pulse in a cluster is either too long, or too short.


    I have tried connecting the 555s in a million different ways, but have had no luck. Any help will be much appreciated.

    THANKS!
     

    Attached Files:

  7. JoeJester

    JoeJester Active Member

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    I take it there are NO constraints on the values you choose. However, should you build it in the lab, you will be restricted to an EIA standard value.

    Using the EIA E12 series for capacitors and the EIA E24 series for resistors, you can get the 1000 Hz within a couple of Hz and a duty cycle of 76 percent.

    As far as combining the signals to produce the output as displayed, you are working with digital levels, assuming the professor wanted you to use the +5 Volt supply.
     
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  8. NLT

    NLT New Member

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    No. No restrictions. I will rework everything to use standard values, but that still doesn't solve my problem.

    Voltage wasn't specified, but we have been working with 5V thus far, so I kept it like that.
     
  9. Colin

    Colin Member

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    To start with, 555's don't like 5v.
     
  10. JimB

    JimB Super Moderator Most Helpful Member

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    Well, the Fairchild datasheet for the 555 says that the minimum supply voltage is 4.5v.
    upload_2016-4-13_2-25-2.png

    JimB
     
  11. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Out3 does not work because its 555 is not triggered.
     
  12. alec_t

    alec_t Well-Known Member Most Helpful Member

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    I expect Colin reads that as "5.4V" in Australia :D.
     
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  13. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The LMC555 is a Cmos 555 and its datasheet says, "1.5V supply operating voltage guaranteed". The TLC555 and ILM555 work from a supply voltage down to 2V.
     
  14. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Maybe the new math says that 5.4 is less than 5?
     
  15. JoeJester

    JoeJester Active Member

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    There have been a few changes since the 555 was introduced.
     
  16. JoeJester

    JoeJester Active Member

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    NLT

    You are not synchronizing the three timers so the sequence is set the way you want it.

    NLT-1.png

    NLT-2.png

    Review the datasheets and application notes carefully.
     

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