Zener Diodes Problems

Discussion in 'Homework Help' started by slodnulius, Feb 11, 2013.

  1. slodnulius

    Thread Starter New Member

    Feb 11, 2013
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    Hello. I finished my Electronics homework… but I’m not so 100% sure if I did it right. The problems that I have doubts about are number 5 and 6. Respectively a zener voltage regulator with fixed power supply and load, and a zener voltage regulated with fixed power supply and variable load.
    In both problems, I need to find the voltage and the current in each resistor. In problem number 5, I also need to find the power in each resistor and the zener.

    Anyways, this is pretty much what I did.
    [​IMG]

    I also simulated the circuit, just to compare if the voltages are the same. I realized that they are not equal, but are close from the simulation’s voltage.
    [​IMG]
     
    Last edited: Feb 11, 2013
  2. slodnulius

    Thread Starter New Member

    Feb 11, 2013
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    Here is what I did for problem number 6.
    [​IMG]

    This time the simulation gives me pretty much the same result (or very close) to the analytical solution.
    [​IMG]


    Well, if you guys can point out what I did wrong, I’ll highly appreciate it. Thanks! :)
     
    Last edited: Feb 11, 2013
  3. WBahn

    Moderator

    Mar 31, 2012
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    Welcome to AAC. I'd be happy to look over your work, but the contrast on your scans isn't good enough for me to be able to read, particularly in the bottom half of some of them. Could you try to improve the contrast (sometimes a black and white scan helps, sometimes it makes it worse) and repost.

    It sounds like yoiu probably have the right answers. In gneral, you can almost always verify the answers to circuit problems by using the results to calculate things you didn't use in the answer, such as the total voltage around some loop or the currents at some junction or the total power in the circuit, and see if the results are consistent with reality.
     
  4. slodnulius

    Thread Starter New Member

    Feb 11, 2013
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    Thanks for the reply. Well... I don't know why, but it seems that I can't edit my posts anymore. Anyways, I edited the scans and improved the contrast:

    Problem #5
    http://i49.tinypic.com/k3ak9g.jpg

    Problem #6
    http://i48.tinypic.com/219dg5u.jpg
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,716
    4,788
    Those are much better. Thanks.

    It looks like you are making some kind of effort at tracking your units through your work, which I very much commend and appreciate, but you are doing so inconsistently. Try to get in the habit of tracking units at every single step of the way. It will pay off.

    It looks like you've got problems right from the get-go, but the numbers are such that the simulator is probably producing correct values that are reasonably close to your incorrect ones. Choosing Vt>>Vz really hides the impact of the Zener. A value of 24V for Vt would have been much better.

    You don't indicated what voltage Vs is referring to, but your first few lines make it pretty clear that it is the node shared by R1, R2, and R3. If that's the case, then it is incorrect to say that the voltage at that node is going to be Vt-Vz. The voltage at the top of the Zener is going to be Vz (assuming it is reverse conducting, which it is in the first circuit). So if Vs is also Vz, then R2 has no voltage across it and, hence, no current through it.

    To analyze a circuit with a zener diode in it, you do the same thing you do with a circuit that has a normal diode. You make an assumption regarding the state of the diode and then analyze a modified circuit that is consistent with that state. Once the analysis is done, you then check if the resulting current in the diode is consistent with that state.

    For a normal diode, you can either assume that it is conducting or not conducting. If you assume it is conducting, you replace it with a short (or a battery with a voltage equal to the forward voltage drop of the diode) and check for a forward current that is greater than zero. If you assume it is not conducting, you replace it with an open and check for a voltage that is less than the forward voltage drop of the diode.

    For a Zener diode you have three states, forward conducting, not conducting, or reverse conducting. The analysis for the forward conducting case is the same as for a normal diode. The not conducting case is almost the same, except that you check to make sure that the voltage is less than the forward diode drop but greater than -Vz. For the reverse conducting case, you do something very similar to the forward conducting case, you replace it with a battery of -Vz (or Vz with the positive terminal on the cathode end) and check to be sure that you have a current that is less than zero (or greater than zero going into the cathode).

    It looks like you might be mixing these up because your modified circuit shows an open for the Zener. Yet you seem to be allowing current to flow through the open, so perhaps it is just a strange way of depicting things graphically.

    Now, remember the part where I was saying that you can usually check your work starting with your answer? Let's do that.

    You have an answer for the current in R1 of 114.5mA. But you also have the current in R2 as 114.5mA. That means that all of the current in R1 must be going down R2. So where does the 56.29mA (check your math, by the way) that is supposedly in R3 come from?

    Always track your units, and always ask if the answer makes sense.

    Oh, let's see if the teacher was correct when they said the voltage across R3 was 68V. That means that the current in R3 would be 68V/1.2kΩ which is 56.67mA. The voltage across the 500Ω resistor would be 57V yielding a current in it of 114mA. This would make the current in R1 equal to 170.67mA with a corresponding voltage drop of 170.67mA*1.5kΩ which is 256V. Thus the value of the supply voltage that would be needed would be 256V+68V or 324V.

    Methinks that's a tad high.

    The actual voltage on that node is somewhere between 50V and 55V.
     
  6. slodnulius

    Thread Starter New Member

    Feb 11, 2013
    5
    0
    Thanks a lot! Your answer really did help me. :)
     
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