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serial regulator

Discussion in 'General Electronics Discussion' started by pwnstars, Feb 28, 2016.

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  1. pwnstars

    pwnstars

    15
    1
    Oct 6, 2015
    Hi guys... I would appreciate if you could tell me if these calculations are correct or not?

    transistor.png

    D2 is 16V
    HFE in the transistor is to be calculated as 50
    S1 is Connected

    Ur5 = Ucc-Vdz-Ube
    Ur5 = 24V-16V-0.6 = 7.4V

    Ir5= Ur5/r5
    Ir5 = 7.4/1000 = 7.4mA

    Ir5 = Ib = 7.4 mA

    Ic= Ib* HFE
    Ic = 7.4mA*50 = 370mA

    Ie = Ic + Ib = 377.4mA

    Ue = Udz - Ube =
    Ue = 16 - 0,6 = 15,4V

    thanks
     
  2. Alec_t

    Alec_t

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    Jul 7, 2015
    Sadly, it's not all correct.
    This bit is: "Ue = 16 - 0,6 = 15,4V".
    Starting with that, work out the currents in R6 and RL. The total of those two gives Ie.
     
  3. Arouse1973

    Arouse1973 Adam

    5,164
    1,079
    Dec 18, 2013
    NO I don't think its quite right. What is Ur5 please.
    Adam

    Edit: Brain fart moment... sorry.
     
    Last edited: Feb 28, 2016
  4. pwnstars

    pwnstars

    15
    1
    Oct 6, 2015
    Arent Ur5 correct?

    R6 and RL would be 160 ohm
    Ue = 15.4V/160 = 96.3mA
     
  5. pwnstars

    pwnstars

    15
    1
    Oct 6, 2015
    Sorry meant 130 ohm total.

    let me try all over again.

    Ur5 = Ucc - Uzd
    Ur5 = 24V- 16V = 8V

    Ir5 = Ur5 / r5
    Ir5 = 8V / 1000 = 8mA

    Ib = Ir5 = 8mA
    Ub = uzd = 16V

    Ue = Ub - 0.6V
    Ue = 16- 0.6 = 15.4V

    Ir6 = Ue / r6
    Ir6 = 15.4 /1000 = 15.4 mA

    IRL =Ue /RL
    IRL = 15.4mA/150 = 102.6 mA

    Ue = Ir6 + IRL
    Ue = 15.4mA + 102.6mA = 118mA
     
  6. Alec_t

    Alec_t

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    Jul 7, 2015
    Better, but you're not quite there yet ;).
    No. You're forgetting that some of IR5 flows through D2.
     
  7. pwnstars

    pwnstars

    15
    1
    Oct 6, 2015
    So the current through base is calculated as emitter current devided by the Hfe of 50?
    118 mA / 50 = 2.36 mA
    Which makes current through zener diode:
    Ur5 - Ub
    8 mA - 2.36 mA
     
    Last edited: Feb 29, 2016
  8. Alec_t

    Alec_t

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    Jul 7, 2015
    Got it!
     
  9. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    I read the circuit incorrectly and I have deleted my comments.
     
    Last edited: Mar 3, 2016
  10. Alec_t

    Alec_t

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    Jul 7, 2015
    Not so, Colin.
    Spice confirms them (as do members on another forum where the same topic is raised).
    RegulatorTest.PNG
    Eh? Most of the current goes through the 150Ω resistor, not through the zener.
    So the rest of the world is wrong :). Hmmm.
     
  11. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    I read the circuit incorrectly and I have deleted my comments.
     
    Last edited: Mar 3, 2016
  12. Alec_t

    Alec_t

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    Jul 7, 2015
    I certainly can, but this thread is all about encouraging the OP to do the calculations.
    How do you get that figure? R1 is the main current-limiter for the 16V zener. The only way 200mA could flow would be if there were reverse-voltage breakdown of the transistor collector-base junction.
     
  13. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    What current flows through the zener when the switch is open?
     
  14. Alec_t

    Alec_t

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    Jul 7, 2015
    That's a calculation I can do in my own head :). It's a bit more than when the switch is closed, but not the ~200mA you allege ;). Are you asking for an answer for yourself, Colin, or asking the OP to do the math?
     
  15. BobK

    BobK

    7,671
    1,681
    Jan 5, 2010
    So can I. I get approximately 9mA.

    Here is my math:

    Voltage at the top of the Zener: 15V.
    Voltage at the top of the resistor, 24V

    Voltage across R5 24 - 15 = 9V

    Current through R5 = 9 / 1000 = 9mA

    Current into base is negligible, 15mA / beta.

    Colin, why don't you show us your calculation that comes out to 200mA, so we can point out where you are wrong.

    Bob
     
  16. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    You are right. I read the circuit incorrectly.

    The only comment I can make is the fact that the transistor will deliver a maximum of about 300mA due to the gain of the transistor.
     
    Last edited: Mar 3, 2016
  17. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    The figures should be:
    base current = 24 - 16 = 8v current = 8mA.
    All this means is this: the maximum emitter current can be as high as 50 x 8 = 400mA.
    but the actual emitter current is determined by the load.
    Doing the calculations in your head:
    The 1k resistor will take 15.4mA
    and the 150R will take about 16/150 = 106mA
    making a total of about 120mA.
     
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