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what is the difference/s if i use a thyristor instead of the transistor in this circuit?

Discussion in 'General Electronics Discussion' started by Name..., Feb 16, 2017.

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  1. Name...

    Name...

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    Apr 6, 2016
    upload_2017-2-16_20-38-20.png
     
  2. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    If you can supply enough current to trigger the gate then the Thyristor will latch on and stay on until the current through the device falls below its holding current. This type of device is also called an SCR.
    Thanks
    Adam
     
  3. Name...

    Name...

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    Apr 6, 2016
    O thanks!

    But my question is more like, what are the advantages and disadvantages of using them in this situation?
     
  4. AnalogKid

    AnalogKid

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    Jun 10, 2015
    A thyristor will change the operation of the circuit in a significant way. Once the buzzer is activated, it will not turn off when SW2 is pressed again. It will stay on until power is removed by SW1. Are you sure you want the circuit to behave that way?

    A thyristor will dissipate slightly more power than a saturated transistor in the same low power application.

    ak
     
  5. Name...

    Name...

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    Apr 6, 2016
    Oh, i understand now.

    THANKS!
     
  6. Name...

    Name...

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    Apr 6, 2016
    Can someone please explain in detail what happens when the reset switch (SW3) is pressed to reset the output to 0( in the 4017b) and turn off the flashing LED and buzzer.
     
  7. AnalogKid

    AnalogKid

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    Jun 10, 2015
    Internally, a 4017 isn't a counter in the traditional sense. It is a 5 stage shift register with a bunch of decoding gates around it. When the Reset input goes high, the shift register is set to all 0's by resetting each of the five flipflops.

    Forcing the 4017 to the 0 state means the 0 output is high (turning on D2) and all other outputs are off (turning off Q1).

    ak
     
  8. Name...

    Name...

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    Apr 6, 2016
    Ok thanks. This is what i had, does this sound correct?

    So when the reset switch is pressed pin3 is taken to high voltage so the output is reset back to output 0 by the clock (pin 14). This happens with the help of the enable pin which is held low and is in operation when voltage is high. Also when this Is pressed, there is high resistance between the collector and the emitter; and the change of voltage above 1.5V allows the collector and emitter to fall and so current can flow out of the transistor through the base and the transistor is switched off and the buzzer and flashing LED are turned off.
     
  9. AnalogKid

    AnalogKid

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    Jun 10, 2015
    No. If you look at the internal schematic on the 4017 datasheet, you will see that the Reset input is not gated by either the Clock or Enable inputs; it goes directly to the flipflop reset inputs. This is called an asynchronous reset, and happens no matter what the state of the clock input is.
    No. The change of base voltage *below* 0.6 V turns off the transistor. As the transistor turns off, the voltage on the collector *rises* because it is pulled up by the impedance of the buzzer. Also, there is no flow of current out through the base other than the base-emitter junction charge which is extremely small and not related to the collector current.

    ak
     
    Last edited: Feb 17, 2017
  10. Name...

    Name...

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    Apr 6, 2016
    OK thanks so much!

    Also, in terms of D7 i was advised to put it there. Is that there to prevent electromotive force(emf)? I dont think it is but if you can help me with that, it would be great!
     
  11. Chemelec

    Chemelec

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    Jul 12, 2016
    D7 should Only be needed if its a Mechanical Buzzer that has a Coil in it.
    Yes to prevent BACK EMF from a Coil.
     
  12. Name...

    Name...

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    Apr 6, 2016
    Ok thanks!

    Can you please explain to me in detail how this diode prevents back EMF from a coil?
     
  13. AnalogKid

    AnalogKid

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    It doesn't prevent it. When current is removed rapidly from an inductor, the magnetic field collapses. This causes *induces* a current into the wiring of the inductor (hence the name). That current produces a potential difference (voltage) between the two ends of the inductor. Depending on the current before the turn-off and the size and construction of the inductor, the voltage can be in the thousands very briefly. This is part of what makes an automotive ignition coil work. Every engine with spark plugs has a switching power supply in it, based on a design worked out over 100 years ago, before the invention of the vacuum tube. But I digress.

    The voltage produced by the collapsing inductor field is of the opposite polarity to the voltage that was across the inductor before being turned off (due to a minus sign in Faraday's law of Induction). So while the diode was reverse-biased with the operating DC voltage, it now forward conducts the back emf and tries to short it out. This reduces the tens, hundreds, or thousands of volts t0 around 1 V, or whatever is the Vf rating of the diode.

    ak
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The diode doesn't prevent back EMF, it prevents the back EMF from doing damage.

    "A coil" in this sense is referring to a device with inductance.

    Inductance means that it has the property of resisting the change of current flowing through it. (If you understand inertia, it is somewhat analogous). Initially, when power is applied, this means that the current ramps up -- this is not a problem for your circuit. However, when the transistor turns off, the inductance wants to keep the current flowing.

    Keeping the current flowing when the transistor is turned off doesn't sound possible. However, what happens is that the voltage across the inductor rises to allow the current to flow. This voltage will rise until sufficient current flows to expend the energy that has been stored in the inductor (incidentally, as a magnetic field).

    In your circuit, the most likely thing is that the voltage will rise until the transistor breakdown voltage is reached, and energy will flow through the "off" transistor. This avalanche breakdown of the transistor is not a normal mode for most transistors, and can easily cause damage.

    So how does the diode help? Simply speaking, it provides an easy path for the current to flow, minimizing the magnitude of the voltage spike that is created.

    If you imagine the current flowing down from the +ve rail, through the buzzer, through the transistor, and finally to the ground (or -ve) rail. When the transistor turns off, the current continues to flow through the buzzer. Without the diode it has to find a path through the transistors and your power source. With a diode, it simply goes back up the diode to the other side of the buzzer.

    If you are thinking of conventional current (which does from +ve to -ve), then the Arrowhead of the diode must point the other way (toward positive).
     
    Last edited: Feb 18, 2017
  15. Chemelec

    Chemelec

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    Jul 12, 2016
    Your right, So I should have Defined it a bit better.
    Yes to prevent BACK EMF from a Coil in doing damage to the Transistor.
     
  16. Name...

    Name...

    40
    1
    Apr 6, 2016
    Ok Thanks guys!
     
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