Help with 555 formula

Discussion in 'The Projects Forum' started by campeck, Sep 14, 2009.

  1. campeck

    Thread Starter Active Member

    Sep 5, 2009
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    I basically want to get a quick formula for frequency but am not too good with math. I'm surprised I figured out t1 and t2....Unless they are wrong.

    Could someone one show me how to get it?
    Thanks!:cool:
     
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  2. Wendy

    Moderator

    Mar 24, 2008
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    That is a new schematic to me. The .6V diode drop will screw up your equations though. Basically it will charge one way, and discharge through the other path. The total resisance of both paths add up to the total pot resistance. So you are going to have to add two completely separate equations, one side includes R1, the diode (no designator), R2, and the cap for ½ the waveform, the other equation will only have the other diode, R3, and the cap. Pin 7 shorts to ground during the latter cycle, so R1 is not involve, though it is a passive drain on the power supply.

    It would be a good variable PWM, but the frequency isn't even a little stable as the PWM ratio is changed.

    That data sheet I drew depends on sharing the current paths charging and discharging the capacitor to simplify the math, this one is going to be slightly convolted and pretty long.
     
  3. campeck

    Thread Starter Active Member

    Sep 5, 2009
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    yup.
    Its a PWM circuit I pulled offline somewhere.

    And yes. Since I can calculate f using t1+t2.

    frequency at R2 = 50k and R3 = 50K is 2.6kHz
    and with R2 at one extreme or the other it is around 26kHz

    (all true only if I didn't mess up. which.....yeah. probable!)
     
  4. campeck

    Thread Starter Active Member

    Sep 5, 2009
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  5. campeck

    Thread Starter Active Member

    Sep 5, 2009
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    oh. And is there a 555 configuration that can change duty cycle without a frequency change?

    Also. Removing R1 from this circuit would result in what?

    And Diodes are 1N4148
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    With this one, the frequency change is minimized, but I understand from another post not totally eliminated...

    [​IMG]

    I plan on running a set of experiments and documenting this design after I get caught up on my other articles. I think I've shown them to ya before.

    The 555 Projects

    There is another variation that address several issues at once, stable frequency (but can be varied at whim) and much greater range of duty cycle.

    [​IMG]

    If you removed R1 (leaving a short) Pin 7 goes to ground. The magic smoke would promptly escape that 555.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    That's better than what he was using.
    He had 1N34 diodes in his schematic, which are ancient point-contact germanium diodes. He probably thought that would give a lower Vf. However, 1N34's have a Vf of 1v at around 5mA.

    If you want to improve things, replace them with Schottky diodes, like 1N5817 thru 1N5819; they have a lower Vf.

    Smoke and flames from the 555. :eek: Pin 7 is an open-collector current sink that is used in this circuit to drain the charge from C1 through R2 and the diode next to pin 7 to ground. Without R1, the power supply would be shorted to ground. As it is, with Vcc=12v and R1=1k, there will be 12mA flowing through R1 to pin 7. This alone may increase the voltage on pin 7 to 0.2v, causing it to take longer to discharge C1.

    Not easy to do. Replacing the diodes with Schottky versions will certainly help.
     
  8. campeck

    Thread Starter Active Member

    Sep 5, 2009
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    alright so the Diodes affect the circuit calculations. Can anyone show me how? I bet it has something to do with that plot...
    But the parallel-ness and the diodes facing different ways and the capacitor are throwing me around.
     
  9. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
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    Last edited: Sep 14, 2009
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, it does have something to do with "that plot". ;)
    (Current vs Vf plot for a 1N4148 diode)

    If you want the calculations to come out closer, you'll need to adjust the control voltage input (pin 5). There are three 5k resistors in a voltage divider network that set the trigger and threshold levels to 1/3 and 2/3 of the supply voltage. Since the diodes are dropping some voltage, pulling pin 5 lower by using resistance to ground will lower those 1/3 and 2/3 thresholds to compensate.

    It's needed even more when using pin 3 to source/sink current. Basically, if you disconnected pin 7, then disconnected R1 from Vcc and connected it to pin 3, you'd have Bill Marsden's version of a PWM circuit. But the 555 output pin 3 never goes much higher than Vcc-1.3v, so you have to adjust pin 5 to compensate by adding resistance to ground. About 27.4k works if Vcc=12. It's too late to re-calculate for other voltages right now.

    It's just like eating an elephant - just take one bite at a time. ;)

    If current can't flow through a diode, that part of the circuit doesn't carry current.

    As I've already mentioned, the pre-set threshold levels are about 1/3 and 2/3 Vcc. Start with C1 charged to 1/3 Vcc, and pin 7 high. What will be the current flow through the appropriate circuit path? Don't forget the diode Vf at the current.

    Then what will the current flow be right when it reaches 2/3 Vcc?

    Then pin 7 goes low, down to around 0.2v. What will the discharge current be initially when C1 is charged to 2/3 Vcc?

    What will the current flow be when C1 is nearly down to 1/3 Vcc?

    Don't forget, R1 is effectively in the circuit only when charging C1. When pin 7 goes low, the discharge path is right to pin 7.
     
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