Solving Circuits

Discussion in 'Homework Help' started by davidand, Jun 3, 2005.

  1. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    [attachmentid=685][attachmentid=686][attachmentid=687]Can someone point me in the right direction,
    I attached a few homework problems I'm looking for some advice maybe someone can pass along or recommend a good book or web site to help me learn more about electronics.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    If you haven't already done so, take a look at the material contained in the tutorials on this website. There is some very good basic information available. Well worth the read.

    hgmjr
     
  3. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    I've been going through the tutorials,
    I tried to solve HW 3 and I come up with R=0.564 ohms.
    using 11.3/20.
    Am I going in the right direction?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    You are very close. The only thing you overlooked is that the current is 0.020 Amps rather than 20 amps. If you make that adjustment you will be right on the money.

    hgmjr
     
  5. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    Thanks
    I'm tryng HW#2 now
    so far I have V
    B = 7.9V
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    I believe you need to take another look at your work in calculating Vb.

    HW#2(hint)
    The three diodes play a key role in the setting the value of Vb.

    One important parameter appears to be absent from the definition for HW#2 problem. I am surprised that the transistor's beta is not part of the set of givens for the problem.

    HW#1(observation)
    Is it possible that the circuit node where the 430 ohm emitter resistor and the 5V zener diode's anode are connected is also tied to a source of negative voltage that is not shown?

    hgmjr
     
  7. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    Your correct for HW #1 I went online with instructor and he gave me the corrected
    HW#1.

    For HW#2 I did subtract 0.7 for each diode. I'm researching the chapter on diodes now.

    Thanks for you input.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    Hi davidand,

    I think that there may be some information missing on the figure for HW#2. Is it possible that the circuit node associated with the cathode of the bottom diode and the bottom leg of the 15 ohm emitter resistor is attached to a negative power source?

    I made the assumption that the node was connected to ground based on the figure. That may be why I did not get the same answer that you did.

    Once this question is cleared up I think you are on the right track.

    I am guessing that the node in question is tied to -10V. Am I right?

    You're doing pretty good so far.

    hgmjr
     
  9. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    Yes it is connected to -10V
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    Now that I have the missing piece of information from HW#2, I can see where you got your answer.

    Suggest you take a second look at the sign of your result. Keep in mind that the diodes are referenced to the -10V.

    hgmjr
     
  11. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    thanks hgmij
    The book I'm using id a reprint for Bura of Navy,
    I'm taking a distance learning course and this is a reference book, it's pretty basic
    I'm trying to find a book on solving circuits.
    I'm still trying.
    again thanks
     
  12. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    I will get that info to you soon, I left book at work.
    I'll post info soon.
     
  13. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    For HW#2
    I find
    V
    B = -7.9

    V
    C = 0

    V
    E =0

    I use the following

    V
    C = 10V - (IR)

    It does'nt look right can someone check this and explain it to me?
     
  14. hgmjr

    Moderator

    Jan 28, 2005
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    Your answer to Vb agrees with mine.

    You need to take a second look at your Vc and Ve calculations.

    HINT: The voltage at the emitter will be more negative than the voltage at the base. Keep in mind that you are dealing with an NPN silicon transistor.

    Once you have Ve then you will have most of what you need to calculate Vc. Since you have no value for the beta of the transistor we will be forced to assume a value of say 100. That is unless you know the actual value posed by the problem. Or we can assume that Ie is equal to Ic and proceed on that basis.

    Hang in there.
    hgmjr
     
  15. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    NAVPERS 10087-C
    Basic Electronics

    I bought it at Barnes +Noble

    I'm in Army All the manuals for electronics courses are made advailable for me
    but no hard copy I must download them and print it if I want hard copy.
    Cheaper to buy the book for $14.95
     
  16. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    I worked HW#2 some more

    V
    B = -7.9V

    V
    E = -8.6V

    V
    C = 24.4V
     
  17. hgmjr

    Moderator

    Jan 28, 2005
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    Your value for Ve is now correct.

    Your value for Vc is off. I think I can understand where you made the error.

    HINT: The value of the emitter current is computed by the voltage "drop" across the emitter resistor.

    I think what you may have done is used the value of Ve directly in your calculation of the current flowing in the emitter resistor.

    HINT: In the transistor stage of HW#2, any computed voltage that is more positive than the positive power supply or more negative than your negative supply should be view with suspicion.

    Still you are making progress.

    hgmjr
     
  18. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    I calculate it to be 9.96V now.
     
  19. hgmjr

    Moderator

    Jan 28, 2005
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    That is still not quite it.

    What value do you get for the emitter current?

    hgmjr
     
  20. davidand

    Thread Starter Active Member

    Jun 2, 2005
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    -8.6V/15 Ω =0.57mA

    10V-(0.57)(60 Ω )=9.49V
     
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