Negative voltage and Zener diode

Discussion in 'Homework Help' started by zork, Nov 9, 2013.

  1. zork

    Thread Starter New Member

    Oct 12, 2013
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    Hi everybody,

    In the circuit joined to this post :
    Used diodes are ideal diodes.
    R1 = R2 = 100 Ω
    U = -20 V
    Uz = 6V

    I am asked to find the value of current passing through the zener diode.
    Here's how I did :

    I've omit the diode D1 because no current flow though it (reverse biased). I've also omit the zener diode because it acts as a short circuit (forward biased).


    Then,
    U - R2.I - R1.I = 0
    I = U/(R1+R2)
    I = 0.1 A

    Is it right ? Because I've a doubt if I've to reverse the "arrow" of the voltage source or not.

    I am a total novice in this field.
    Thanks in advance.
     
    Last edited by a moderator: Nov 9, 2013
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The post is moved to its own thread.

    Bertus
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    You have two arrows next to the two resistors. What do they indicate? Is the I in your equations refer to the I next to the left-to-right arrow in the bottom branch?

    If so, then you got the correct answer, but it appears that luck and happenstance played a role.

    Your U is given as -20V, but then you drop the sign when you plug in the value for U when evaluating your final equation. Why?
     
  4. zork

    Thread Starter New Member

    Oct 12, 2013
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    0
    @bertus : Thanks.

    @WBahn : The two arrow near resistors refers to their voltage (I use them to apply Kirchhoff's 2nd law). And the I in the equation refers to the current across the circuit (it is the same because it is a series circuit).

    I've drop the negative sign of U because I have reverse it's arrow. And this is what disturb me… What's the difference between 20V and -20V ?!
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    The difference between 20V and -20V is 40V.

    The problem is that you are being a bit sloppy with your notation and conventions. You say that the resistors refer to the voltages across the resistors and that you use them to apply Kirchhoff's 2nd law (which I'm using you are referring to Kirchhoff's Voltage Law), but there's no indication that you have actually used them for any such purpose. Since you have defined them in opposite directions relative to a path taken through both resistors, the signs associated with them should opposite.

    The voltage U is defined by the arrow in the original diagram as being the voltage on the top node relative to the voltage on the bottom node. You put a different voltage in the second diagram that is the voltage on the bottom node relative to the top node. This voltage is NOT the voltage U, it is a different voltage. Call it J. They are related by J=-U.

    For the two resistors, your chosen voltage arrows (and using + and - symbols is generally better than using arrows because arrows are more commonly assocated with current directions) you have that both voltages are positive on the left side relative to the right side. Let's call the voltages across R1 and R2 V1 and V2, respectively.

    Going around the loop as you've set it up, in the direction of the current I, and summing up the voltage gains you have:

    -V2 + V1 + J = 0

    We can replace V2 and V1 with the current and resistance, per Ohm's Law.

    V1 = I1*R1
    V2 = I2*R2

    -(I2*R2) + (I1*R1) + J = 0

    By the passive sign convention, the current in each resistor, I1 and I2, has the polarity such that it flows through the resistor from the positive terminal to the negative terminal.

    Thus, I1 = -I and I2 = I

    We then get

    -((I)*R2) + ((-I)*R1) + J = 0
    -I*R2 - I*R1 + J = 0
    -I(R2+R1) = -J

    I = J/(R1+R2)

    We've already established that J = -U, so we have

    I = -U/(R1+R2)

    Now we plug in our values and we get

    I = -(-20V)/(100Ω+100Ω) = (20/200)(V/Ω) = 0.1A
     
  6. zork

    Thread Starter New Member

    Oct 12, 2013
    7
    0
    Thanks for your answer.
    I guess I was completely wrong it when I set up the arrow that indicate the voltage, I have corrected them now (attached file) the equation should correspond to the diagram now.

    Did I do wrong when using 20V instead of -20V ? Is there any other way to find the current through that Zener diode ?
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    When you draw an arrow on a diagram, you need to include a designation that idicates what that arrow means and provides a name for use in equations. With just an arrow, there's no way to even know if you are talking about a current or a voltage.

    The 20V corresponds to the voltage across the terminals as you defined it with your arrow, but that voltage is NOT the voltage indicated by the parameter U. Don't redefine parameters (without being very, very explicit). It will get you in trouble. You have a line in one place that states U=-20V. You then have an equation toward the end that has a U in it, but you substitute +20V into it. It's correct, but only because the two instances of U are not the same. Very confusing and looks like an error and, at some point, you are going to put in the wrong value because you yourself will be confused. If you want to change a parameter, define a new parameter (like I did with J) and make the connection between the two explicit and clear.
     
  8. screen1988

    Member

    Mar 7, 2013
    310
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    Zork, you missed the voltage across Zener diode; it is not zero.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    He said he is using ideal diodes, so that generally means 0V when forward biased. This is the case even for an ideal Zener diode, which then also has exactly -Vz across it when sufficiently reverse biased.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Yes. So?

    Uz is the zener voltage, which is the voltage it would have across it if it were conducting in the reverse direction.

    What direction is the diode conducting in if U=-20V?
     
  11. screen1988

    Member

    Mar 7, 2013
    310
    3
    Well, sorry. I recognized my mistake and deleted that post while you was typing.
    (that is not what I want)
    Yes, the diode is forward biased. My apologies.
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,751
    4,797
    Cool. Glad we are in agreement.
     
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