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Questions about npn transistor and operating point of diode

  1. May 3, 2013 #1
    1.jpg
    For question 1, it asks me to find the operating point by different methods.
    But I have no idea to solve it. Can anyone give some hints for me?
    I can just think about the diode can be replaced by an ideal diode, a 0.5V dc source and a 0.2/4=0.05Ω resistor.
    11.jpg

    For question 3,
    3.jpg
    I have solved the (a) part Ic=2.33mA.
    For the (b) part, I have calculated Ib=Ic/β=0.0292mA
    Ie=Ib+Ic=2.3625mA
    However, I don't know how to find the value of Rb without Vbe. I have asked the tutor but he said that I do not need to assume Vbe=0.7V in this question.

    Thanks for reading my questions.
     
  2. jcsd
  3. May 3, 2013 #2

    rude man

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    1(a):

    Write the linear equation relating vD to iD.
    Sum currents to zero at vD. Solve for vD.
    Solve for iD.
    The rest I leave to you.

    3.
    I disagree with your instructor. You do need to assume what Vbe is (0.7V is a good assumption) in order to compute Rb.
     
  4. May 3, 2013 #3
    I do not get the meaning of this sentence...
    Besides, should I ignore the ac power source in (a) and (b)?

    Then I can convert the circuit into Thevenin equivalent circuit and Norton equivalent circuit to get the load line to solve the problem in (b). Is it right?
     
  5. May 4, 2013 #4

    rude man

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    vd is a node. The sum of all currents entering (or leaving) a node = 0.
    No, you should not ignore the ac source. It is a source of current to the node vd, along with Vb.
    I don't know why you would want to convert to thevenin equivalent circuits. For one thing, the id - vd relationship is nonlinear.

    Once you have solved part a) you have an expression for vd. You can take this function of time and plot it on the vd axis, then see the id on the y axis.
     
  6. May 4, 2013 #5

    NascentOxygen

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    Yes, that should work. Thévenin of the source and resistors, you mean? It's best to attempt part (b) first, so you can get a feel for what is really happening.
     
    Last edited: May 4, 2013
  7. May 4, 2013 #6

    NascentOxygen

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    Hard to say. He may have meant that you shouldn't always assume 0.7V specifically. For example, you may first assume 0.7V and then find that the current is so small that 0.55V would be a more realistic figure, and re-work it with that.

    Alternatively, he may have had in mind that since the transistor chemistry is not specified, it may be that it's Germanium where you would assume 0.3V. [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. May 4, 2013 #7
    Is the answer in (a) in terms of cos(ωt) for both vd and id?

    But in this question, I cannot know whether the current is so small as Rb is not given...
     
  9. May 4, 2013 #8

    NascentOxygen

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    Parameters comprise a DC value plus a cos(ωt) term.

    When you see vD, that upper-case 'D' means the complete voltage. If it's a lower-case 'd' then it means just the AC component. (I'm assuming the person who wrote the question adhered to the convention, and that's not always guaranteed.)
     
  10. May 4, 2013 #9

    NascentOxygen

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    You calculate Ib and Rb using the data given. Your results will be accurate to within a few percent, so you really do know the current (to within a few percent).
     
  11. May 4, 2013 #10
    Does it mean that I can only get the result with no cos(ωt) in (b) but not (a)?
    When I use the result in (a) to draw the line, I found that there may be something wrong in my calculation as the my calculated vd is in a small range near 0.5V...
    [itex]i_D=20v_D-10[/itex]
    [itex]\frac{v_s+v_b-v_D}{100}=i_D+\frac{v_D}{100}[/itex]
    [itex]0.1cos(\omega t)+2-v_D=2000v_D-1000+v_D[/itex]
    [itex]0.1cos(\omega t)+1002=2002v_D[/itex]
    [itex]v_D=5.00\times 10^{-5} cos (\omega t)+0.500[/itex]
    [itex]i_D=20v_D-10=9.99\times 10^{-4} cos (\omega t)+9.99\times 10^{-3}[/itex]
    Is there any wrong?

    By the data given, I calculate Ib.
    Then I assume Vbe=0.7V to calculate Rb. But with the Rb, how can I find the actual Vbe?
     
  12. May 4, 2013 #11

    NascentOxygen

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    Yes, the first line.

    The actual precise Vbe would have to be calculated using the exponential relationship for that PN junction. But of course, that accuracy is not required here. If IB is a good fraction of a mA or more, then 0.6V - 0.7V is fine. This isn't a maths subject. :smile:
     
  13. May 4, 2013 #12
    Oh, I found the problem.
    It should be
    [itex]i_D=0.02v_D-0.01[/itex]
    [itex]\frac{v_s+v_b-v_D}{100}=i_D+\frac{v_D}{100}[/itex]
    [itex]0.1cos(\omega t)+2-v_D=2v_D-1+v_D[/itex]
    [itex]0.1cos(\omega t)+3=4v_D[/itex]
    [itex]v_D=0.025cos (\omega t)+0.75[/itex]
    [itex]i_D=20v_D-10=5\times 10^{-4} cos (\omega t)+5\times 10^{-3}[/itex]
    Is it right?
    If yes, then how can I solve it graphically by the results?
     
    Last edited: May 4, 2013
  14. May 4, 2013 #13

    NascentOxygen

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    That's looking better.

    Are you still on (a), or is this going to be (b) now?
     
  15. May 4, 2013 #14
    This is going to be (b) now.
     
  16. May 4, 2013 #15

    NascentOxygen

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    I thought you already outlined your approach, back in what I quoted here.
     
  17. May 4, 2013 #16
    Should I solve (b) with dc load line only? As the ac part of Vth is quite small...
    Vth=0.05cos(ωt)+1V
    Rth=50Ω
    Or I should draw a line with Vth=0.05+1V and a line with Vth=-0.05+1V?
     
    Last edited: May 4, 2013
  18. May 4, 2013 #17

    NascentOxygen

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    I think the question would be answered by drawing the DC load line, then showing a swing around it of 0.05V. So that could be thought of as drawing 3 lines―one solid line and a dotted line either side.
     
  19. May 4, 2013 #18
    Thank you very much and I have solved these 2 questions:smile:
     
  20. May 4, 2013 #19

    NascentOxygen

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    I was waiting to see how you'd do (c)?
     
  21. May 4, 2013 #20
    In (c), I first consider the dc source only and do the similar with (a) and found
    Vo=0.75V, Io=5mA

    In small signal model,
    [itex]r_d=\frac{1}{\frac{d i_d}{d v_d}}=\frac{1}{0.02}=50\Omega[/itex]
    [itex]v_p=0.1\times \frac{100||50}{100+100||50}=0.025[/itex]
    [itex]i_p=0.02v_p-0.01=5\times 10^{-4}[/itex]
    [itex]v_D=v_p cos(ωt)+v_o=0.025cos(ωt)+0.75[/itex]
    [itex]i_D=i_p cos(ωt)+i_o=5\times 10^{-4} cos(ωt)+5\times 10^{-3}[/itex]
     
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