serial regulator

Discussion in 'Homework Help' started by pwnstars, Feb 29, 2016.

  1. pwnstars

    Thread Starter Member

    Feb 29, 2016
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    1
    Hi guys, im a new electronic student, but im stucked in the basics :)
    So would appreciate some help correcting my calculations in this curcuit.
    transistor.png

    D2 is 16V
    S1 is connected

    here is my calculations

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

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

    Ir5 = IB
    8mA = 8mA

    Ue = Uzd - Ube
    Ue = 16V - 0.6V = 15.4V

    Ir6 = Ue/R6 =
    Ir6 = 15.4V/1000 = 15.4mA

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

    Ie = Ir6+ IRL
    Ie = 15,4mA + 102,6mA = 118mA

    Ic = Ie-Ib
    Ic = 118mA - 8mA

    Is there some truth in these calculations?

    Best, Peter
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    You need to annotate your diagram with the definitions of the variables you use. When you say Ur5, that is ambiguous because you are not only requiring the reader to assume that it refers to the voltage across R5, but you are requiring them to guess what the polarity of that voltage drop is. Similarly, when you talk about Ir5 you are requiring the reader to guess what direction you are defining that current to be in. Engineering is not about guessing -- make your definitions clear.

    This would appear (again, you are forcing people to guess) to say that all of the current that flows through R5 must then flow into the base of Q4. Does this seem reasonable?
     
  3. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    When you say that Ir5 is Ib. How does the zener diode regulate?
    A zener diode will need a minimal amount of current to be able to regulate.

    Bertus
     
  4. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    Thanks guys.
    I have subtraced the .5mA zenior current from the calcs and added my values to the picture..
    transistor2.png
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    What is your reasoning for concluding that the zener current is 0.5 mA?

    What would the zener current be if the wire going to the base of the transistor were cut?
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Have a look at the datasheet of the zener for the minimal current.

    Bertus
     
  7. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    I meant 5 mA, sorry. and i've looked it up in the datasheet.

    If i cut the base of the transistor then the zener would pass all the current delivered to it.
    And with that knowledge i have no explaination for why the current wouldn't pass more than 5mA.. (most likely because it's wrong)

    I cant seem to find the solution at this point
     
  8. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    Is it right to say that the current through R5 is:
    (Ucc - Udz)/R5
    (24V-16V)/1000ohm = 8mA ?
     
  9. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Yes, the current in R5 is 8 mA.
    This current is split into the zener current and the base current of the transistor.
    The currents will be different in the case the switch is open and closed.
    You will have to make calculations for both cases.

    What is the gain of the transistor?
    Have a look in the datasheet.
    For the calculations I would use the minimum gain.

    Bertus
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,775
    4,804
    You are on the right track and making good observations. We can explore this a bit more, but your next posts provide a better opportunity.
     
  11. WBahn

    Moderator

    Mar 31, 2012
    17,775
    4,804
    Yes, this is correct. And you correctly tracked your units -- bravo!!

    Like a normal diode, a zener diode will hold a constant voltage across it (if possible) while allowing whatever current through it that is needed to hold that voltage.

    So if the base of the transistor is cut off, then all of the current will go down through the zener. With the transistor in the circuit, some of it will go into the base and the rest will go through the zener. The amount that will go into the base will be dictated by the circuit to the right of the transistor combined with the current gain (beta) of the transistor.
     
  12. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    Thanks for your patience :)
    According to the data sheet the minimum gain is 40.

    I assume the next stage is to calculate the currents after the emitter. Is that right?

    Uemitter = Uzener - Ubasis/emitter diode
    Uemitter = 16V - .6 = 15.4V

    current through the resistances would be calculated as the emitter voltage devided by the resistance values.

    15.4V/1000 = 15.4mA
    15.4/150 = 102.6mA
    total current ( current-emitter) = 118mA

    am i on the right track? :b
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,775
    4,804
    The typical gain is probably in the 100 to 300 range, but using 40 will make most of your calculations more conservative, which is usually (not always) good.

    Watch those units. 16 V is a voltage, while 0.6 is just a number. You can't subtract a number from a voltage. It should read

    16 V - 0.6 V = 15.4 V

    What those units! 15.4 V / 1000 is 15.4 mV. What you meant was

    15.4 V / 1000 Ω = 15.4 mA
    15.4 V / 150 Ω = 102.6 mA

    This is only IF the switch is closed -- something that you haven't indicated one way or the other.

    Yes, you are.

    So what would the base current in the transistor be, assuming a beta of 40?
     
  14. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    Sorry, I'll make sure to add units :)

    Assuming the gain is 40.
    Base current is: emitter devided by 40
    118 mA / 40 = 2.95 mA

    And as far as i know the collecter current is approximately the same as emitter.

    Doesn't the base current determine if the transistor is closed or open?
     
  15. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    When there is 2.95 mA going to the base of the transistor, how much current is going to the zener diode?

    Also calculate the currents for the open switch.

    Bertus
     
  16. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    The remaining current from the resistance (R5) will go through the zener diode.
    8 mA - 2.95 mA = 5.05 mA

    Im not sure how to calculate the open switch.
     
  17. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You already have calculated the current through the 1k resistor.
    This is the only one connected to the regulator when the switch is open.

    Bertus
     
  18. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    Bertus, Im thick headed.

    I dont really have a clue how to go from here.

    Am i supposed to change resistance values?
    Everything is constants to me.
     
  19. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Think of it like the switch and the RL do not exist, like this:

    pwn_transistor2_open_switch.png

    This would be the situation when the switch is open.
    Now calculate all currents.

    Bertus
     
  20. pwnstars

    Thread Starter Member

    Feb 29, 2016
    44
    1
    Ohh.. of course :)

    Current at emitter is calculated by voltage at emitter devided by R6
    15.4V / 1000 Ω = 15.4 mA

    Base current is emitter current devided by the Hfe of 40

    15.4 mA / 40 = 0.385 mA

    Current through zener is current in r5 subtract with current at the base
    8 mA - 0.385 mA = 7.615 mA



    Current at collecter is current at emitter subtract current at base
    15.4 mA - 0.385mA = 15.015 mA
     
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