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Calculating time when SCR starts up

  1. May 5, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    Given a circuit to control the volume of an electric bell P. Potentiometer T is set to 600K ohms. The breakout voltage of the SCR (at the SCR gate) is 20 micro-amperes. And the max gate that will start up the SCR is 0.7 Volts


    http://img41.imageshack.us/img41/4322/yest2.jpg [Broken]



    Calculate the time when the SCR starts up.


    3. The attempt at a solution

    http://img205.imageshack.us/img205/1268/yest1.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 5, 2012 #2

    I like Serena

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    Hmm, how did you get U(diac)?
    It does not look right.

    Btw, the quotient would not be in degrees.
    You would need to take the inverse sine to find alpha.
    Or did you already do so?
    Either way, that does not look like the right angle.
     
  4. May 12, 2012 #3

    Femme_physics

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    I thought Udiac was given as 0.7V, but perhaps I misunderstood.

    And you're right, it should be a radians.

    I admit, I glanced at another person's solution in my class. Problem is that it confuses me even more!

    http://img703.imageshack.us/img703/2155/vinvm.jpg [Broken]

    To my understanding, V(t) is Vin's voltage during the set-off of the SCR.

    I failed to understand the equation: Vt(t) = Vin(t) - Vd - Vp

    Because if it uses Kirchhoff's law on the loop there should be 5 voltages. 1 source and 4 drops.

    Source: Vin

    Drops:
    Vt (potentiometer), Vd, Vp and the SCR.

    Can this be clarified for me?
     
    Last edited by a moderator: May 6, 2017
  5. May 12, 2012 #4

    I like Serena

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    I you take the loop through the potentiometer, your drops are:
    Vt, Vd, Vs (SCR gate).

    Apparently the drop over the gate has been marked as Vp, but I find that confusing, since I'd expect Vp to be the drop over the electric bell.


    If I understand your problem statement correctly, the SCR is triggered when Vs (SCR gate) is 0.7 V.
    And when the current at the SCR gate is at least 20 μA.
    Btw, the current at the SCR gate is the current that flows through the potentiometer, the diode and the SCR gate.

    Can you calculate what Vin has to be, if the current through the potentiometer is 20 μA, and if the voltage drop over the SCR gate is 0.7 V?


    Btw, in the other person's calcution, you first get an expression for the voltage across the potentio meter, and then an expression for the current through the potentiometer.
    If this current reaches 20 μA, the gate opens.
    You can solve this equation mathematically to find the corresponding time.

    However, there is a mistake in it.
    I'm missing a factor 2 in the 3rd and 4th lines.
     
  6. May 19, 2012 #5

    Femme_physics

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    Ah..true

    How so? After the firing angel been set-off I see 3 different currents

    http://img96.imageshack.us/img96/1683/i1i2i3.jpg [Broken]


    I can't see it, but I rather try my own calculations anyway
     
    Last edited by a moderator: May 6, 2017
  7. May 19, 2012 #6

    I like Serena

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    Yep.
    I3 is the current through the resistor, the diode, and the SCR gate.
    I2 is the current through the SCR anode.
    I1 is the current through the SCR cathode.
     
  8. May 19, 2012 #7

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  9. May 19, 2012 #8

    I like Serena

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    If only it were that simple... ;)

    Your equation applies to a straight line, but your voltage is a sine and not a straight line.
    A different formula is needed here.

    You're alos leaving some stuff out, in particular the voltage drop over the resistor and the voltage drop over the diode...
     
  10. May 19, 2012 #9

    Femme_physics

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    I had a feeling it's not that easy, but give me some points for general sneakiness ;)

    At any rate, I suppose that by "different formula" you mean the

    Vin(t)=Vmax sine omega t

    I get that. It relates the voltage to the angle and the time... That's logical.

    Why need the voltage drop over the resistor and diode is a bit beyond me because the same voltage drops on all the components between the point where the current split to when it reunites.
     
  11. May 19, 2012 #10

    I like Serena

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    I'll give you 0.4125 point for sneakiness! ;D

    And yes, I meant that formula.

    In your problem statement it says that you need 20 micro amperes flowing through the SCR gate (I3=20 μA).
    If I3=20 μA, then you get a voltage drop over the resistor.
    If Vin is not high enough, this will mean the the voltage at the SCR gate is not high enough to trigger it.

    You need to find the Vin that is high enough to trigger the SCR gate.
     
  12. May 19, 2012 #11

    Femme_physics

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    Hehe

    But the SCR is not in series to the potentiometer. We need 20 micoamperes through the bell and the SCR. I guess the potentiometer and the diode are for... Actually, I'm not sure entirely what is the use of them..
     
  13. May 19, 2012 #12

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    It says "The breakout voltage of the SCR (at the SCR gate) is 20 micro-amperes."

    Since it says "at the SCR gate", I interpret that to mean I3.

    Here's a picture of an SCR:
    120px-Thyristor_circuit_symbol.svg.png

    "At the gate" should mean the connection on the left side.
     
  14. May 20, 2012 #13

    Femme_physics

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    Ah, true :)

    http://img18.imageshack.us/img18/4859/scragain.jpg [Broken]

    OK, so it means Vin is 12.6 [V] because:

    Sigma V = Vin - Vp - Vscr = 0
    Vin = 12.6 [V]

    So that's the value of Vin during SCR's set-off.

    Now do I use the

    Vin (t) = Vmax Sin ωt

    Formula? Whereas ω is 2pi, Vin is the value I found at set-off, and Vmax is just 12√2

    Is that right? And the T I find is the T during SCR's set-off that they ask me to find.
     
    Last edited by a moderator: May 6, 2017
  15. May 20, 2012 #14

    NascentOxygen

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    Hi FP! Are you challenging the laws of physics again? :wink:

    The triggering current flows gate→cathode, and is largely independent of the SCR's main anode→cathode current, so you are not at this stage involving current through the bell.

    For a gate current of ≥20uA, and the variable resistor set at 600kΩ, the required instantaneous level of the AC supply voltage level is going to be:

    ≥ 0.7 + 0.6 + 12v

    Correct.
     
  16. May 20, 2012 #15

    I like Serena

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    Close.
    The Vscr voltage is between the gate and the cathode, and not between the gate and the anode (that's also what NascO said).


    And yes, the "t" you find is the t during SCR's set-off.
    But ω is not 2pi, but ω = 2 pi f, where f is the frequency 50 Hz.
     
  17. May 22, 2012 #16

    Femme_physics

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    Why my favorite pastime uncle Nascent! :)
    Oh
    Ohx2 :)

    In this case:

    -Vp +Vscr +Vd + Vt

    -Vp +0.7 + 0.6 +12

    Vp = 13.3 [V]

    And then,

    Vin (t) = Vmax Sin ωt

    13.3 = 12√2 x Sin 2pif x t

    Sin2pif = sin(2pi x 50 )= -0.717


    13.3 = -12.174t

    t = 1.09 [secs] (since time can't be negative it's swapped to positive)

    The result doesn't make much sense to me...these things suppose to take mili if not micro secs to set off
     
  18. May 22, 2012 #17

    NascentOxygen

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    13.3 = 12√2 sin(250t)
    Parentheses and brackets are cheap; I recommend that you make good use of them.
     
  19. May 22, 2012 #18

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  20. May 22, 2012 #19

    NascentOxygen

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    Okay, so you included some parentheses but then did something wrong.

    Actually, I can't figure out what you are doing, but it doesn't look right.
    Following on from what I wrote above, the next step is:

    13.3 / 12√2 = sin(250t)

    sin⁻¹ (13.3 / (12√2)) = 250t

    ∴ t = https://www.physicsforums.com/images/icons/icon5.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  21. May 22, 2012 #20

    Femme_physics

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    Last edited by a moderator: May 6, 2017
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