Series voltage regulator, how is works

Discussion in 'Homework Help' started by bug13, May 8, 2012.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    [​IMG]

    as seen in the schematic, I can understand how Zener diode in series with a resistor to regulate voltage, but I don't understand how the transistor works in this circuit.

    Please explain.

    Thanks
     
  2. BillB3857

    Senior Member

    Feb 28, 2009
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    Look up the properties and operation of an "Emitter Follower"
     
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  3. Woody_Duong

    New Member

    Mar 2, 2012
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    Coz Vb is constant so if Eout increase, Vbe decrease, transitor conducts less => Eout decrease and vice versa, so Eout is regulated.
    I'm newbie here :)
     
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  4. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    I still don't get it:mad:
     
  5. BillB3857

    Senior Member

    Feb 28, 2009
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    Do you understand the theory behind an emitter follower circuit?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The attachment may be of interest.
     
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  7. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Simple answer, no I don't understand it

    Long answer, I seem to understand its operation, but I don't understand why, and what's its used for, and it's properties
     
  8. Jony130

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    Feb 17, 2009
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    To understand how this circuit work first you must understand the limitations of a simply Zener regulator

    [​IMG]

    As you can see the the Zener current is equal to

    Iz = Is - IL

    So as IL varies the Zener current also varies. And this will cause the voltage across the Zener and the load voltage also vary. As IL increase the Iz decreases and zener voltage Vz decreases .
    So the max load current that can flow through that load can not be greater than IL_max = Is - Iz_min

    And the Zener power dissipation is the greatest when the load current is the smallest.

    So to reduce the loading on the Zener we add a current amplifier.

    [​IMG]

    The BJT provides a a current gain equal to

    IL = Ie = Ib + Ic*β = (β+1)*Ib

    So now the change in the load current will cause (β+1) smaller change in the zener current.
    Because now
    Iz = Is - Ib = Is - IL/(β+1)

    And this allows us to increase the max load current that we can drawn from the circuit.
     
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  9. BillB3857

    Senior Member

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    OK then. It looks like we need to back up even further. Do you understand basic transistor theory? Mainly, the effect of base voltage vs emitter voltage and the effect on transistor current? Since this is the HOMEWORK section, we will try to guide as much as we can but really need to know what level of knowledge you have in order to build on it. How long have you been studying electronics? Tell us about your understanding of transistor operation and we can go from there.

    Edit: Jony130 posted while I was typing. Good info Jony130, but looks like it may be over the OP's head at this point.
     
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  10. Jony130

    AAC Fanatic!

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    Perhaps numerical example will help
    Lets as assume that we have a 5.1V zener diode and we pick Rs = 1K.
    And the input voltage is equal to Vin = 10V.
    So for no load we have a this situation
    Is = Iz = (10V - 5.1V)/1K = 4.9mA
    But now if we connect the load resistor RL = 2K. The Zenner current will drop

    Iz = Is - IL = 4.9mA - 5.1V/2K = 2.35mA

    If we further increasing the load current the zener current decreases.
    And when the load current reach 4.9mA there is now current for the zener diode.
    For RL < 1K ---> Iz = 0A
    So the zener diode is OFF and the circuit no longer act as a voltage regulator.

    But now let as add the BJT whose current gain Ic/Ib = 99.
    Because the BJT add Vbe voltage drop we need to increase the zener diode voltage to 5.7V

    [​IMG]

    Now lest as connect RL = 2K
    The Zenner current will only drop to

    Iz = Is - Ib = 4.9mA - (5.1V/2K)/(β+1) = 4.9mA - 22.5μA ≈ 4.87mA

    So now thanks to BJT we can increase the max load current (β+1) times.
    From 4.9mA to 490mA
    So thee minimum load resistance is now equal to
    RL_min = 5.1V/0.49 ≈ 10.5Ω

    Without BJT

    RL_min = 1K

    So by adding the BJT we can now reduce the RL_min = 1K/(β+1) = 10Ω.
     
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  11. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Thanks Jony130, now I can understand why adding a transistor. I like how you start with something I can understand (Zener voltage regulator), and talk about its limitation, and solving it by adding a amplifier (transistor).

    So just check my understanding:

    The Zener is to regulator the voltage on the base of the transistor, and then the transistor is to amplify the current, to increase the usability of this voltage regulator circuit?

     
  12. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Thanks for you input BillB3857, your help is appreciated.

    I have read the DC, AC section of this forum, and I find them very educational, and now I am learning VOL. III - Semiconductors, mainly the diode and transistor, I found they are a bit hard to take on at the moment.

    Understanding of transistor is 1) can be used as switch 2) as amplifer

     
  13. bug13

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    Thanks t_n_k, your explanation gives me an in-deep understand of how this work, I am aware of the saturation and un-saturation of an transistor, but never pay too much attention to it, as I mainly used it as a switch before.

    I will be looking into this from now on.

     
  14. Jony130

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    I don't now it is only me, but the t_n_k explanation is not a key to understand how this voltage regulator work.
    As a matter of fact the "effect" describable by t_n_K (Vbe varies with Ie) is a disadvantage. This "effect" reduces the stiffness our voltage regulator.
    The voltage regulation would be "tighter" if Vbe was independent of the load current.
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hello Jony130,

    Thanks for your comments.

    Clearly the change in VBE in the series regulator output current range is a limitation (albeit small in consequence) and I'm not suggesting that this is advantageous in some mysterious way.

    The point of my observations was that the BJT's forward transfer characteristic (Ic vs VBE) behavior leads to only a very slight change in output voltage [<1.4%] over an operating current range of two orders of magnitude. It is this simple 'virtue' of the BJT pass transistor which transforms its integration with the zener regulator into a relatively inexpensive and useful higher load capacity DC regulator.

    If one has DC regulation design specifications that cannot tolerate even such small variations then an alternative more costly approach is in order.

    In any event why do you seek to single me out for criticism or seemingly invite comment from other forum members regarding the possible deficiency of my contribution? It was a contribution which I thought might be of interest. I'm more than happy to be challenged personally.

    Regards,

    t_n_k
     
  16. bug13

    Thread Starter Well-Known Member

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    Hi Jony130,

    Sorry about my poor communication skills, what I really means is, t_n_k's explanation points me to somewhere I am not aware off, as I mainly use the transistor as a switch. (at this stage of study)

    When I said that, I mainly referring to the properties of a transistor, not the series voltage regulator in in this topic (sorry for being off the topic, I should have focused on the topic)

    Wu


     
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