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BJT and MOSFET complete analysis

  1. Mar 24, 2017 #1
    1. The problem statement, all variables and given/known data
    I have been given the world's longest transistor problem as an assignment :wink: Here is the circuit:
    upload_2017-3-25_0-15-54.png
    I am asked to find:
    a) V1, V2 and V3 using DC analysis
    b) AC equivalent circuit
    c) AC tension gain: Ava=Vx/Vsig
    d) AC tension gain: Avb=Vo/Vx
    e) Total AC tension gain: Avt=Vo/Vsig
    f) Rin
    g) Rout
    h) AC current gain: Aia=ix/is
    i) AC current gain: Aib=io/ix
    j) Total AC current gain: Ait=io/is

    Now, I realize this is a long question... I'm just completely lost. I figured I would post here step by step what I do and try to get help.

    3. The attempt at a solution
    I think I can analyze the BJT and the MOSFET separately?

    So far, I have started with the BJT DC analysis:
    upload_2017-3-25_0-21-36.png
    upload_2017-3-25_0-22-9.png
    I tried to write as many equations as I could think of. I didn't want to take all the time to solve this for 6 eqns and 6 unknowns before having someone give me their opinion.

    For the MOSFET:
    upload_2017-3-25_0-29-20.png
    with Vt=1, then here I could solve for VD and Vs
     
  2. jcsd
  3. Mar 25, 2017 #2

    gneill

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    Staff: Mentor

    For the BJT DC analysis I'd suggest replacing the base bias network with its Thevenin equivalent to start. Also remember to take into account the base-emitter diode potential drop!

    Take advantage of how base, collector, and emitter currents are interrelated; you should know how ##I_E## is related to ##I_B## through the transistor's ##\beta## (which is in addition to how ##I_C## is related to ##I_B## of course).

    With all that you should be able to write a single KVL equation to solve for ##I_B##. Once you have that all the other currents and potentials should fall into place.
     
  4. Mar 25, 2017 #3
    This is what I did
    upload_2017-3-25_12-49-25.png
    upload_2017-3-25_12-49-58.png
    upload_2017-3-25_12-50-38.png

    a lot easier than my 6 equations from before !

    For the MOSFET, was I on the right path or completely lost ?
     
  5. Mar 25, 2017 #4

    gneill

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    :smile: Yup! Well done!
    I admit that I haven't looked at the mosfet part. I'll do so when I find time... sorry about that.
     
  6. Mar 25, 2017 #5
    No problem, I know this problem is super long, and I always post a lot for my homework and such. I really do appreciate the help, otherwise I am completely lost all the time !
     
  7. Mar 25, 2017 #6

    gneill

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    I had a minute so I took a quick look at what you've done with the mosfet stage. It looks like you've assumed that the mosfet is operating in saturation mode by your choice of equation for the drain current:
    upload_2017-3-25_14-21-28.png
    Can you justify this choice? The circuit would appear to be an amplifier. Would saturation mode be a suitable guess to begin with?
     
  8. Mar 25, 2017 #7
    Well, VDS > VGS-Vt since VGS-Vt would be negative ?
     
  9. Mar 25, 2017 #8

    gneill

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    I think you'll have to do the analysis with your assumption in mind, then check that the results you find tally with your assumption.
     
  10. Mar 25, 2017 #9
    Hmm. When I solve my two equations I get either
    VD = -1.152 and VS = -1.149 or VD = -0.857 and VS = -0.853

    So then VDS isn't bigger than VGS-Vt... so it's not in saturation. Lovely, we have never done an example in class that wasn't in saturation. :confused:
     
  11. Mar 25, 2017 #10
    Also, I tried to draw the AC small signal equivalent circuits
    upload_2017-3-25_21-3-13.png
    upload_2017-3-25_21-4-37.png
     
  12. Mar 25, 2017 #11

    gneill

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    Fortunately there are web resources that you can tap into. A bit of googling quickly turns up such gems as:

    http://www.ittc.ku.edu/~jstiles/312/handouts/Steps for DC Analysis of MOSFET Circuits.pdf

    http://whites.sdsmt.edu/classes/ee320/notes/320Lecture27.pdf
     
  13. Mar 25, 2017 #12
    I'm doing something wrong. When I use the saturation equations I end up with it not satisfying the conditions and then when I use the linear equations I end up with it not satisfying the linear equations...
    upload_2017-3-25_21-31-31.png
    I used wolfram to solve for VD and Vs
     
  14. Mar 26, 2017 #13

    gneill

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    Let's take another look at your calculations for saturated mode.

    First note that the potential at the FET gate is a result coming out of the BJT section analysis (V2). What value are you bringing forward from that analysis? How will you relate this value to VGS?
     
  15. Mar 26, 2017 #14
    Ok so
    upload_2017-3-26_17-11-18.png
    This is the formula I can use and my only unknown will be ID. I know V2=6v from the last part
    upload_2017-3-26_17-12-12.png
    But when I solve this, I get that ID=4.86x10^-3 or ID=5.14x10^-3

    When I solve for VS I get VS=4.86V or VS=5.14V
    and then for VD = 0.28V or VD=-0.28, respectively.
    Neither of these sets will give me in saturation, and my prof told us it was for sure in saturation.
     
  16. Mar 26, 2017 #15

    gneill

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    I suspect that you're running into problems in the handling of k'. That 0.5 value has units of mA/V2. If you're going to work with terms like ##I_D R_S## and retain the current as milliamps, then you'll have to scale the resistor value accordingly. Either that or convert k' to A/V2.
     
  17. Mar 26, 2017 #16
    I see what you mean. The worst part is, I sent an email to my prof and he finally responded with that formula and it has an error o_Oo_O
     
  18. Mar 26, 2017 #17

    gneill

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    Well, I suppose the formula would be fine so long as it's stated that RS is specified in kΩ.
     
  19. Mar 26, 2017 #18
    It still doesn't work... where am I going wrong
    upload_2017-3-26_17-50-54.png
     
  20. Mar 26, 2017 #19

    gneill

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    ##\frac{1}{2} 0.5~mA/V^2 = \frac{1}{2} \frac{1}{2000} ~A/V^2 = \frac{1}{4000}~A/V^2##
     
  21. Mar 26, 2017 #20
    Finally i figured it out, Id=2.1mA and then VD=5.8V
    Now, for the first gain. WIll Vx = V2? Do I need to short circuit the capacitor to find Vo? Do I use the AC equivalent circuits ?
     
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