Circuit with Zener Diode

Discussion in 'Homework Help' started by AD633, Jun 23, 2013.

  1. AD633

    Thread Starter Member

    Jun 22, 2013
    96
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    Determine the range of values ​​of the voltage VE ensures a constant value of
    Vo = 9.1V.

    I think that for the diode to operate in the zener current must be between [100 micro A, 10 milli A]

    Writing the equations of the mesh for both cases:

          VE=Rs*Izmin+Rl*Izmin<br />
             VE=(1 kOhm)*(100 micro A)+(4 kOhm)*(100 micro A)=0,5 V<br />
<br />
              VE=Rs*Izmax + Rl*Izmax<br />
            <br />
              VE=(1 kOhm)*(10 mili A)+(4 k Ohm)*(10 mili A)=50 V

    Is this correct?

    Thanks
     
    Last edited: Jun 23, 2013
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, how Ve = 0.5V can be smaller than the Vz voltage?
     
  3. AD633

    Thread Starter Member

    Jun 22, 2013
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    It can't.So is the equation wrong?Have i missed something?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes you equations are wrong. This is not the right answer. You simply forget about Zener voltage and Kiechhoff''s law. Because as we can see on the diagram the Zener diode is connected in parallel with RL resistor. So voltage across RL is equal to Vz voltage.
     
  5. bertus

    Administrator

    Apr 5, 2008
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  6. AD633

    Thread Starter Member

    Jun 22, 2013
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    Yes you are right the voltage across RL will be equal to 9,1 V if the diode is in the zener mode.I did not remember how the zener diode worked.

    So VE will have to be between [9,2;19,1] V for the diode to operate in the zener region.What if i used VE greater than 19,1 V.This would damage the zener because the current across it would be to large right ?

    Thaks
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Why you skip RL resistance in you calculations? Also if Iz > 10mA the power dissipation will finally destroyed the diode.
     
  8. AD633

    Thread Starter Member

    Jun 22, 2013
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    Because i already know that voltage drop across RL will be 9,1V since it is in parallel with the zener diode.Since i am trying to determine VE ,what i want to calculate is the voltage drop across the resistors in the circuit.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    But if we have RL in the circuit Ve voltage can be greater than 19.1V. Why? Simply because RL "steals" some current from the Zener diode.
    Is = Iz + IL
     
  10. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    No.
    Why are you multiplying the series resistance by the current flowing through the diode instead of the current flowing through the series resistance? Again the load resistance by the current flowing through the diode instead of the current flowing through the load resistance?
     
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