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MOSFET use

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by malc9141, Sep 17, 2014.

  1. alec_t

    alec_t Well-Known Member Most Helpful Member

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    AG, the OP has tried what you suggest as the alternative to use of a 4093 and it doesn't work with the optosensor :(. Simulation suggests this is due to insufficiently sharp-edged input pulses: hence the use of the Schmitt-action 4093. See post #110 et seq. Thanks for adding the "internals" to the 4093 symbol.
     
  2. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Sorry, I did not read all the previous pages of this long thread.
     
  3. malc9141

    malc9141 Member

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    Hello:
    Just want to check before diving in:
    (my ignorance here will be esp about duration of events)

    at "neutral", pin 1 and pin 2 are High. So pin 3 is Low, and connected via RC to 0 v.

    at "Pulse", pin 1 is still High but pin 2 goes Low. So pin 3 goes High (for time determined by C7 and R13 ?).

    Does that put pin 5, 6 High; and thus pin 4 Low?

    So at this time pin 2 is still Low, and pin 4 has gone Low. So pin 2 at 555 gets a Low pulse??

    But pin 2 is Low (we have just said) and the slow Trigger response keeps pin 1 Low, so pin 3 is simultaneously High...........................(a)

    I can see the Trigger will give the 555 a Low, but it seems that because of (a) this Low will copy the duration at the Trigger, not give a short pulse. I suppose thats OK. But am I arguing correctly?

    And did you say I don't need R12?

    Thanx
     
  4. dave

    Dave New Member

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  5. alec_t

    alec_t Well-Known Member Most Helpful Member

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    No. After the time defined by C7/R13 the voltage at the top of R13 will have fallen close to 0V so pin 4 of the 4093, hence the Trig input of the 555, will go high. Thus the 555 Trig input is low only for the defined time, regardless of how long pin 3 of the 4093 stays high. The two gates of the 4093 are here configured as a conventional monostable circuit.
    R12 is intended to limit current through the input protection diodes inside the 4093. Since C7 is only 1nF it probably doesn't store enough energy to damage the IC. It is just a precaution in case (a) you have a particularly sensitive IC or (b) we need to increase the value of C7 at any time. It was AG who suggested R12 is unnecessary :). He may well be right, but I prefer to play safe.
     
  6. malc9141

    malc9141 Member

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    That helps. I will have a go on Monday> Toes crossed.

    On the original circuit, the equivalent of C7 was 33nf and 150k R. That was likewise to clean up the pulse before it went to another subcircuit, for final adjustable timing.
     
  7. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Toes crossed here, too :).
     
  8. malc9141

    malc9141 Member

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    OK. But we must remember that the pulse Trig (Low) lasts for a non-specified time and thats what we are trying to get away from.

    So at pulse Trig, pin 1 goes Low , pin 2 held High by 555 pin2.

    So pin 3 High for time, t sec. During t, pin 4 is switched to o v (via the pin 7 in the 4093?). 555 pin 2 goes Low. All good.

    But the pulse Trig is still Low so pin 1 is Low. But so is pin 2 at this moment. So pin 3 goes High. That starts the whole thing off again till the pulse Trig passes.

    ********************************************************************

    Do we need a long time delay RC between pin 2 and pin 4?
     
    Last edited: Dec 3, 2016
  9. alec_t

    alec_t Well-Known Member Most Helpful Member

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    No, it doesn't. It's already high, so the monostable doesn't get re-triggered.
    Perhaps the waveforms below will help:
    InjectorFETDriver5Waveforms.PNG
     
    Last edited: Dec 3, 2016
  10. audioguru

    audioguru Well-Known Member Most Helpful Member

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    R12 is not needed because a gate on the CD4093 drives C7 with a current that is much less than its maximum input current of 10mA.
     
  11. malc9141

    malc9141 Member

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    1) Thank you. There must be something I don't get, which isn't surprising.
    I thought pin 4 and also pin 2-555 (direct to pin 2) would go high and (while pin 1 is still low) keep 3 high for the duration of the drop from the sensor at pin 1, as in the schema.
    But then I dont see how the C7, R13 is helping (see the picture). If pin 3 stays high, what is C7, R13 achieving?

    It still looks to me as if the pin 3 duration is controlling (via pin 4) the 555 output, even tho your schema is different..

    ************************************
    2) If the 555 gets a persisting drop at its pin2, the 555 output stays On till that drop ends. That has been the problem. (Your schema is not like this, I accept)
    So keeping the 4093 pin 3 high, will keep the signal to the 555 low and 555 output stays on. ?????

    **************************************
    3) Is the problem of light "staying on" at the 555? Should pins 4 and 8 be getting power without a resistance as in the original diagram (page 6 here)? Could it be damaged? (I replaced it but that wouldn't help if it blows at once)
     
    Last edited: Dec 4, 2016

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