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Homework Help: Zener diode help! (self-learning)

  1. Aug 16, 2012 #1
    sorry just found out this is the wrong thread
     

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  2. jcsd
  3. Aug 16, 2012 #2
    What is your question?
     
  4. Aug 16, 2012 #3
    1. The problem statement, all variables and given/known data

    Hello, i'm trying to teach myself the basics of EE to kind of get an idea of what i'd do in college. I'm finding this question particularly troubling. Is this due to my low intelligence, or is it normal?

    I've attached the photo of the problem. It is asking to find the power rating that the zener diode should have. Also, I included the answer that the book provides; the most confusing this is that it says the assume a zener current of .5 amps which doesn't make sense to me because I don't know how they came up with .5 amps. If any more info is needed let me know, thanks!

    2. Relevant equations

    P=IV V=IR



    3. The attempt at a solution

    I know that the zener diode must get the same voltage as the lamp because it is in parallel so it has 15V. For the resistor on top, I figured that it must have a voltage drop of 9 to 45 volts. However, even with this I am unable to find the current or resistance through the resistor. This is where I get stuck. The solution in gray says to assume a zener current of .5 amps, but I have no idea where this number comes from. Please help!
     
    Last edited by a moderator: Aug 16, 2012
  5. Aug 16, 2012 #4
    Don't know where you got this question but it is ridiculous.

    The 15 volt zener needs to draw about 0.02A to be considered 'on'.

    Additionally it needs to be able to cope with the lamp current, when the lamp is off or broken.

    This is stated to be 0.075 amps

    So the zener needs a standing current of 0.02 + 0.075 = 0.1 amps.

    Power = volts x amps = 0.1 x 15 =1.5 watts say 2 watts for safety.

    The idea of a zener is that it is (as you say) connected in parallel with the load (lamp) and it adjusts its own current draw so that the result is always 15 volts across it.
     
  6. Aug 16, 2012 #5
    Thanks for the reply studiot. The answer says 42 watts so im confused, why do you say it is ridiculous?
     
  7. Aug 16, 2012 #6
    It's wrong.

    It's wrong to choose 0.5 amps, that's way too much for the job.

    But let us say that we did choose this standing current.

    Then the wattage dissipated by the zener = 0.5 x 15 = 7.5 watts, not 42.
     
  8. Aug 16, 2012 #7
    I am incredibly confused, do you have any external resources I could use to learn more about zener diodes?
     
  9. Aug 16, 2012 #8
    Do you understand what a potential (voltage) divider is?
     
  10. Aug 16, 2012 #9
    Not quite sure... I'm fairly new to all this.
     
  11. Aug 16, 2012 #10

    berkeman

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    Staff: Mentor

  12. Aug 16, 2012 #11
    OK this is very important to learn as it is the basis of many circuit arrangements.

    A voltage divider divides the input voltage in a defined ratio.

    It is very simple - at its simplest it is just two resistors in series with the input voltage connected across them.

    In my sketch I have shown this with an input of 1 volt the drop across the larger resistor is 90% of the voltage ie 0.9 volts.

    In your circuit that is what is used.
    Your input resistor is my R1
    The parallel combination of the zener and the lamp is my R2

    The object is to waste as little power as possible and the only component we really want to supply is the lamp.

    Are you with me so far?
     

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  13. Aug 16, 2012 #12
    Why are the resistors in series? How is that going to divide voltage?
     
  14. Aug 16, 2012 #13
    What happens to to the voltage if you place it across one single resistor?
     
  15. Aug 16, 2012 #14
    It drops while current stays the same.
     
  16. Aug 16, 2012 #15
    Not really. I think you need to revise the basics.

    Remember this is for circuit theory, not physics. The physics is confusing at first until you have got hold of simple circuit theory.

    The voltage is what I measure with a voltmeter.

    Let us say we have a 12 volt auto battery.

    If I connect it across a 12 ohm resistor 1 amp flows

    If I connect it across a 24 ohm resistor 0.5 amps flow

    If I connect it across a 48 ohm resistor 0.25 amps flow

    If I connect it across a 1 ohm resistor 12 amps flow

    But the voltage is always 12 volts.

    Now what happens if my 12 ohm resistor is actually two 6 ohms resistors in series?

    I have 12 volts across the series combination, but 6 volts across each resistor.

    Are we doing any better?
     
  17. Aug 16, 2012 #16
    Wait what? How is that wrong? "The voltage dropped across the resistor in a circuit consisting of a single resistor and a voltage source is the total voltage across the circuit and is equal to the applied voltage."


    http://www.tpub.com/neets/book1/chapter3/1-12.htm
     
  18. Aug 16, 2012 #17
    Don't you think that is a bit of a mouthful?

    It's not actually wrong, but what does it mean?

    What is the 'total voltage across the circuit' and how would you recognise it walking down the street?

    Seriously - don't go there - that way makes a simple thing very complicated.
     
  19. Aug 16, 2012 #18

    berkeman

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    Staff: Mentor

    Your link says exactly what Studiot was saying. What is the confusion?
     
  20. Aug 16, 2012 #19
    A voltage is what I measure with a voltmeter.

    A voltmeter has two leads: you connect one to each of two different points in your circuit.

    Here is a (slightly) more complicated arrangement to show what you would read on a voltmeter, connected at various locations.
     

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  21. Aug 16, 2012 #20
    I thought the link was easier to understand... :(
     
  22. Aug 16, 2012 #21
    So how did your link lead you to think that the current remains the same but the voltage drops in your post 14?
     
  23. Aug 16, 2012 #22
    Well in a series circuit the same current is used by all the components. And all the components use a certain voltage which leaves less for the rest. Am I wrong?
     
  24. Aug 16, 2012 #23
    Yes that is correct.

    This bit is another way of saying they divide the available total voltage.
    Or another way of saying that they form a potential divider.

    Which brings us back to post12
     
  25. Aug 16, 2012 #24
    Go on. Do you have a EE degree? How do you know this stuff?
     
  26. Aug 16, 2012 #25
    My first degree was applied maths, but I have been doing this for quite a few years.

    Since you are self learning you might like to investigate this site

    I have placed the link in vol 3 of their online textbook.
    Look at chap 3 for zener diodes

    http://www.allaboutcircuits.com/vol_3/index.html

    I need to go to bed now but keep asking questions
     
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