Help Plotting Q-Point

Discussion in 'Homework Help' started by Life617, Nov 18, 2012.

  1. Life617

    Thread Starter Member

    May 23, 2012
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    I am having trouble understanding what I am doing wrong. The question is asking me to plot the Q-Point if Vc=53V.

    It is also asking me if it is possible to amplify a 25Vp-p signal linearly and I am unsure how to go about figuring that out.

    [​IMG]
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    I can't make out any of the text and can't tell exactly what the components are. You've gone too far in reducing the resolution of your attached picture. Try again and you might try making it a PNG file if you can.
     
  3. Life617

    Thread Starter Member

    May 23, 2012
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    Hopefully one of these will be more legible.

    [​IMG]

    [​IMG]
     
  4. WBahn

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    Mar 31, 2012
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    First, thank you for tracking your units in your work -- it is so rare that anyone does that.

    Now, the next thing you need to get in the habit of doing is asking if the answer makes sense. Does a collector-emitter voltage of -8.4V make sense?

    Also, you calculate Ic twice and get different answers, 9.1mA and 6.4mA. Does this make sense? If these two calculations are for different situations, you need to make clear what those situations are. Neither I, nor the grader, are mind readers.

    Next you are making one of the classic mistakes. You know Ohm's Law is I = V/R, but you just grab whatever V is handy. Ohm's Law relates the resistance of a resistor to the current through THAT resistor and the voltage across THAT resistor. On what basis are you claiming that either 75V or 53V (particularly the latter) is the voltage across the resistor. If the collector voltage is 53V, what is the voltage ACROSS the collector resistor?

    If by plotting the Q point it means to plot the operating point on the graph you've provided, then the value on the x-axis is given, namely Vc = 53V. That IS the collector voltage.

    Try to explain why each term in your computation for Vce is there. Keep in mind that Vab means Va - Vb, so Vce means Vc - Ve.
     
  5. Life617

    Thread Starter Member

    May 23, 2012
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    Sorry If I was too vague. The first Ic of 9.1mA is the theoretical maximum current before the saturation cut off point. The second Ic of 5.6mA is the actual current through Rc. The worksheet sates that Vc=53Volts, this means I can use Ohms law to calculate the current through Rc because I have the voltage and resistance across that specific resistor. Am I correct so far?

    I think the problem came when I was calculating Vce which you have made clear. I think the correct answer should be Vce=53V-30.3V = 22.7V.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Why do you think that Vc (i.e., the voltage measured at the transistor collector relative to ground) is the voltage ACROSS the resistor? Is the resistor connected between the transistor collector and ground?

    Yes. Now use that same reasoning to find the voltage ACROSS the collector resistor.
     
  7. Life617

    Thread Starter Member

    May 23, 2012
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    I honestly don't know why I was thinking that, my brain has sort of just assumed that relationship. It is good to have it cleared up now though. So I have done this problem completely wrong then, and the only way for me to calculate Ic or Ie is to rework the problem and not even worry about the voltage at the collector. I will redo the math and repost.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Well, you need to worry about the voltage at the collector because (1) it IS the horizontal value of your operating point, and (2) you need it in order to determine the voltage across the collector resistor, which you need in order to determine the collector current, which is the vertical value of your operating point.
     
  9. Life617

    Thread Starter Member

    May 23, 2012
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    Ok, so first I calculated Vb using the formula (R2/R1+R2)75V. I then used the formula Vbe=Vb-Ve to get Ve=23.9V-.7V . Finally I used the formula Vce=Vc-Ve to get Vce=53v-23.9V. The only problem is that the voltage Vce and Ic Dont line up correctly on the load line so I think i am sitll doing it wrong.

    [​IMG]
     
  10. MrChips

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    Oct 2, 2009
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    What is the maximum Ic?

    What is the operating Ic?
     
  11. Life617

    Thread Starter Member

    May 23, 2012
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    My teacher just informed me that the Vc=53V was not actually supposed to be in the problem. So in order to calculate the voltage drop across Rc I get ... 4.9mA(8.2K Ohms) = 40.18V. My brain is beginning to fail. Ok, so the voltage drop across Vce I can do this 75V = 40.18v + 23.2V + Vce which is 11.62Volts. I can do this because it is a voltage divider problem correct?
     
  12. Life617

    Thread Starter Member

    May 23, 2012
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    Maximum Ic=Vcc/Rc ... 75V/8.2K Ohms
    Operating Ic is what the current is actually at correct? So it would be Ic=Ie?

    Ic=Beta(Ib) ?
     
  13. MrChips

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    Oct 2, 2009
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    No.

    Ic(max) occurs when Vce = 0v.
     
  14. Life617

    Thread Starter Member

    May 23, 2012
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    Maximum Ic = 9.1 mA ... Vcc/Rc ... 75V/8.2K Ohm

    Operating Ic = Ie = Ve/Re
     
  15. MrChips

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    Oct 2, 2009
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    I understand that you did not need to be told Vc.

    Maximum Ic is not Vcc/Rc
     
  16. Life617

    Thread Starter Member

    May 23, 2012
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    Sorry didnt mean to type that twice. Ok so when Vce = 0 that means there is no current flowing. that would mean there is no current flowing through Rc either. So it would be 0mA?
     
  17. MrChips

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    Oct 2, 2009
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    Still wrong. Try again.
     
  18. Life617

    Thread Starter Member

    May 23, 2012
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    The book says Ic= (Vcc-Vce)/Rc . I dont know how else to go about figuring Ic max. What is the difference between Ic saturation, Ic max, and Ic operating?
     
  19. MrChips

    Moderator

    Oct 2, 2009
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    Do your circuit analysis.

    Do you know Kirchoff's voltage and current laws?
     
  20. Life617

    Thread Starter Member

    May 23, 2012
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    hmm.. Vcc/(Rc+Re)? 75/(8.2K+4.7K) = 5.8 mA
     
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