555 Hysteretic Oscillator

Discussion in 'Feedback and Suggestions' started by Wendy, Oct 24, 2008.

  1. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Status:
    Ready for proof reading.
    Illustrations Complete.
    Experiments Complete.
    11/22/2009 Minor update.

    ******************************************
    555 HYSTERETIC OSCILLATOR


    PARTS AND MATERIALS
    • One 9V Battery
    • Battery Clip (Radio Shack catalog # 270-325)
    • Mini Hook Clips (soldered to Battery Clip, Radio Shack catalog # 270-372)
    • U1 - 555 timer IC (Radio Shack catalog # 276-1723)
    • D1 - Red light-emitting diode (Radio Shack catalog # 276-041 or equivalent)
    • D2 - Green light-emitting diode (Radio Shack catalog # 276-022 or equivalent)
    • R1,R2 - 1 KΩ ¼W Resistors
    • R3 - 10 KΩ, 15-Turn Potentiometer (Radio Shack catalog # 271-343)
    • R4 – 10 Ω ¼W Resistor
    • C1 – 1 µF Capacitor (Radio Shack catalog 272-1434 or equivalent)
    • C1 – 100 µF Capacitor (Radio Shack catalog 272-1028 or equivalent)

    CROSS-REFERENCES

    Lessons In Electric Circuits, Volume 1, chapter 16: Voltage and current calculations
    Lessons In Electric Circuits, Volume 1, chapter 16: Solving for unknown time
    Lessons In Electric Circuits, Volume 4, chapter 10: Multivibrators


    LEARNING OBJECTIVES
    • Learn how to use a Schmitt Trigger for a simple RC Oscillator
    • Learn a practical application for a RC time constant
    • Learn one of several 555 timer Astable Multivibrator Configurations


    SCHEMATIC DIAGRAM

    Here is one way of drawing the schematic:


    [​IMG]


    As mentioned in the previous experiment, there is also another convention, shown below:


    [​IMG]



    ILLUSTRATION

    [​IMG]



    INSTRUCTIONS

    This is one of the most basic RC oscillators. It is simple and very predictable. Any inverting Schmitt Trigger will work in this design, although the frequency will shift somewhat depending on the hysteresis of the gate.

    This circuit has a lower end frequency of 0.7 Hertz, which means each LED will alternate and be lit for just under a second each. As you turn the potentiometer counterclockwise the frequency will increase, going well into the high end audio range. You can verify this with the Audio Detector (Vol. VI, Chapter 3, Section 12) or a piezoelectric speaker, as you continue to turn the potentiometer the pitch of the sound will rise. You can increase the frequency 100 times by replacing the capacitor with the 1µF capacitor, which will also raise the maximum frequency well into the ultrasonic range, around 70Khz.

    The 555 does not go rail to rail (it doesn’t quite reach the upper supply voltage) because of its output Darlington transistors, and this causes the oscillators square wave to be not quite symmetrical. Can you see this looking at the LEDs? The higher the power supply voltage, the less pronounced this asymmetry is, while it gets worse with lower power supply voltages. If the output were true rail to rail it would be a 50% square wave, which can be attained if one uses the CMOS version of the 555, such as the TLC555 (Radio Shack P/N 276-1718).

    R3 was added to prevent shorting the IC output through C1, as the capacitor shorts the AC portion of the 555 output to ground. On a well used battery it is not noticeable, but with a fresh 9V the 555 IC will get very hot. If you eliminate the resistor and adjust R4 for maximum frequency you can test this, it is not good for the battery or the 555, but they will survive a short test.


    THEORY OF OPERATION

    This is a hysteretic oscillator, which is a type of relaxation oscillator. It is also an astable multivibrator. It is a logical offshoot of the 555 Schmitt Trigger experiment shown earlier.

    The formula to calculate the frequency with this configuration using a 555 is:

    f=\frac{0.7}{RC}

    The 555 hysteresis is dependent on the supply voltage, so the frequency of the oscillator would be relatively independent of the supply voltage if it weren’t for the lack of rail to rail output.

    The output of a 555 either goes to ground, or relatively close to the plus voltage. This allows the resistor and capacitor to charge and discharge through the output pin. Since this is a digital type signal, the LEDs interact very little in its operation. The first pulse generated by the oscillator is a bit longer than the rest. This and the charge/discharge curves are shown in the following illustration, which also shows why the asymmetrical square wave is created.

    [​IMG]
     
    Last edited: Nov 26, 2009
  2. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Stash point
     
    Last edited: Nov 26, 2009
  3. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Stash point
     
    Last edited: Nov 23, 2009
  4. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Now I need to figure out how to make the formula

    f=.7/RC

    Look right. I'll be experimenting with it. I'm also open to input and ideas.

    f=\frac{0.7}{RC}

    Got it. I do think I'm finished except for the wiring and verification stage.
     
    Last edited: Nov 24, 2008
  5. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    Good morning, Bill.

    Great article. I am sure it will find lots of use.

    In a recent thread, the queston of duty cycle came up with this and its CMOS cousin. Would you consider adding a few words about duty cycle? The third paragraph in Instructions might be a good place, as your comment there just touches on the issue.

    John
     
    Last edited: Nov 24, 2008
  6. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    What do you suggest? I thought I had, but maybe a number (like 60/40%)?

    I put a lot of emphasis on the why for and where for's. That's why the extra curves shown in the illustration.

    I've been working on kludges around this for other applications, for what it's worth.
     
  7. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    Suggested revision:
    Notice, I moved one sentence to the end and added an example of a rail-to-rail device.

    John
     
  8. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Changes Input. Better?
     
  9. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    Small change, so I like it just a little bit better. It is excellent -- wish I could write so well.

    John
     
  10. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Thank you for the feedback. Baring problems with the experiments, it should be finished.

    I don't know which is harder, writing or the illustrations. I think I enjoy drawing them more though.

    I started this to get some practice before going for the AAC Book. I haven't heard any of the powers that be take on the Theory of Operation, though I think it is a good feature.
     
    Last edited: Nov 25, 2008
  11. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Finally finished this one, basically had to build an Audio Detector per instructions. It appears Radio Shack is dropping the transformers as an in store item, I drove all all over town to find one.

    It works about as well as can be expected. I only wanted it for consistancy for the rest of the book.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    OK, I know that I'm coming into this discussion WAY late, but had to put in my $0.02's worth.

    A bjt (transistorized) 555 timers' output (pin 3) under very light loading can get down to <100mV, but is limited to approximately Vcc -1.3v due to the Darlington follower configuration in the output circuitry.

    When using the output pin 3 to charge/discharge the timing capacitor (C1) via a single resistor with the control (pin 5) either open or bypassed to ground with a capacitor, the result will be a longer C1 charge time than the discharge time.

    In bjt 555 timers, there is an internal resistive divider consisting of three (nominally) 5k Ohm resistors between Vcc and ground that set the trigger and threshold levels at 1/3 Vcc and 2/3 Vcc. The control (pin 5) input provides a method of adjusting the upper threshold; if the upper threshold level is reduced, the lower (trigger) will be about 1/2 of the control voltage.

    To compensate for the reduced high-level voltage output at pin 3, a resistive current path to ground from the control input will reduce both the upper and lower trip levels. The amount of resistance necessary for the correction will depend upon the output load at both high and low levels, and the actual resistance of the internal divider network.

    The upper 1/3 of the divider network resistance can be measured by removing the 555 timer from the circuit, and measuring from pin 5 to Vcc.

    The lower 2/3 of the divider network can be similarly be measured from pin 5 to ground.

    In practice, it would be easier to simply use a 13k resistor in series with a 10k pot from pin 5 to ground to find the optimal resistance to use for a 50% duty cycle, if so desired.
     
  13. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Actually I was in the process of rewrite when you said this, which is now complete. I now have to pack it and the other articles for the AAC book, another work in progress.

    My personal standards have changed slightly since I started this long term project, I'm going to use designators on all my articles. Since that is a real standard out there I see no reason not to use it for my articles, and any that I update.

    Thank you Wookie.
     
  14. Lightfire

    Well-Known Member

    Oct 5, 2010
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    Hello Bill,

    I want to make the leds 20 leds. not only 2 leds. how? thanks
     
  15. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    I have an article on the basics for LEDs, I'm sure you've seen it.

    LEDs, 555s, Flashers, and Light Chasers

    One of the key concepts taught is the color of the LED dictates how much voltage it drops. This is turn is used to calculate the resistor needed.

    You should try to stay with in the 555 current drives, this is also a main consideration.

    So what color LEDs are you planning on using, the red ones I sent you?
     
  16. Lightfire

    Well-Known Member

    Oct 5, 2010
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    Yes, it is.
     
  17. Wendy

    Thread Starter Moderator

    Mar 24, 2008
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    Did you read the link in post #15? The answer is there. Take a guess after reading it, and I'll help.
     
  18. magnet18

    Senior Member

    Dec 22, 2010
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    If he really wants to drive 20 LED's now might be a good time to introduce him to the transistor.
     
  19. Wendy

    Thread Starter Moderator

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  20. SgtWookie

    Expert

    Jul 17, 2007
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    Gee, Bill - I should've read your initial post a lot more carefully, as the frequency formula isn't correct. :(

    I posted a thread on it here:
    555 HYSTERETIC OSCILLATOR formula for "f" incorrect
    Instead of:
    f = 0.7 / RC
    It should've been:
    f = 1 / 0.7(RC)
     
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