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Differential Amplifier find currents, and Gain

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data
    For the differential amplifier circuit given above, for all transistors except Q1 and Q2 we have:
    β=100, VA=80 V, RC =20 k, VT=0.026 (Thermal voltage), VBE(on)=0.7 V

    a-Determine the collector currents IC1, IC2, IC3 and IC6
    b-Calculate the Differential mode voltage gain Vo2/Vd.
    c-What is the output resistance of the three transistor current source that is biasing the differential pair? Find the numerical value. Does this effect the common mode voltage gain? Explain briefly.
    d-calculate the overall Gain

    2. Relevant equations



    3. The attempt at a solution
    So the first thing I did was find I1 = V-Vbe-V(-)/R1 = 1 mA and Ic3=I1=1 mA then Ic1=Ic2=Ic3/2=0.5 mA and I have V02=V-Ic2(Rc) = 15-0.5(2) = 5 Volts then Ic6 = 15-5-.7/6 =1.55 mA. Next I found differntial mode gain rpi1= Vtβ/Ic = 0.026*100/0.5 = 5.2 kΩ and Rib6= rpi6+(1+β)Re = ?? rpi6= 100*0.026/1.55 = 1.67 mA so Rib6=1.67+(101)(4.3) = 435.97 and gain is β(R||Rib6)/2(rpi1+Rb) = 132.8. For part c Ro= Va/Ic3 = 80/1 = 80 kΩ and yes because if Ro changes so does Av when Av=something/(something)*(Ro). Calculate the overall gain well I would first need to find A2 and A3 and then plug into the formula Ad=Ad1*A2*A3 = ?? However I am unsure of how to proceed what should I do now
     

    Attached Files:

  2. jcsd
  3. Apr 18, 2014 #2
    Ok I uploaded a pdf of my work to make it cleared and I also made an attempt at part d.
     

    Attached Files:

  4. Apr 20, 2014 #3
    Hey is there anything I can do to make the question more clear? ? I know I tend to write poorly worded questions, so just let me know if it is too confusing.
     
  5. Apr 22, 2014 #4

    rude man

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    DodgeV, I think one reason for no-shows is your attempt description is too run-in. Split the sections up into paragraphs would help a lot. On top of that, there's a lot of detail here that I suppose many of us just don't have the time/energy/patience/inclination/whatever to sift thru all your work.

    It would be good if you concentrated on one of the parts you're least sure about. We're here to give you hints if you're stuck, not necessarily proofread all your work (if I'm out of line here I'm sure I'll hear about it soon ....).

    Anyway, for part (d) the overall gain is the gain from either input to Q2-c and then from that to Q6-c. I would consider Q2-c unloaded but maybe your teach wouldn't agree.

    How are you modeling the Early voltage? By a resistor = Va/Ic between emitter and collector?
     
    Last edited: Apr 22, 2014
  6. Apr 22, 2014 #5
    Ok man yah you are probably right well I found my error on part (a) but I am stuck on the differentail mode gain I had one of my friends assist me Adm = -gmRC/2 = -IC1RC/2VT = -(0.5)(20)/2(.026) = -192.3075 is that ok?
     
  7. Apr 22, 2014 #6

    rude man

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    Do you define the differential gain as Vo2/(v1-v2)? If so, why the 'divide by 2'? You're already figuring the Q2 current as 0.5 mA.

    You're ignoring the 2K base resistors but frankly I would too.
     
  8. Apr 23, 2014 #7
    Ok so in your opinion which looks better because I got that equation -Ic1RC/2VT from a friend and im defining it as Vo2/Vd. Honestly as far as dividing by 2 I am not sure what does the division by 2 mean? I get lost on the gain stuff I can handle the DC analysis.
    This is my equation:
    Beta(R||Rib6)/2(rpi1+RB)
     
  9. Apr 23, 2014 #8

    rude man

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    The differential gain IS a dc computation.
    I don't see any Vd on the schematic. I guess Vd = v1 - v2?
    You "equation" is the differential gain? I don't think so. What is Rib6? What is R? Try to stick to symbols on the schematic where possible. You were a lot closer before. I don't have time right now but I think I goofed and I believe the divide by 2 is correct. Right now looks like the differential gain is indeed IC1*RC/2VT but i want to double-check.
    More later.
     
  10. Apr 24, 2014 #9
    Because we have 2K resistors at differential pair input the voltage gain will not be equal to RC/2re = IC1*RC/2VT = 192.
    The gain will drop to

    [itex]\Large Av = \frac{Rc}{2re + \frac{2*2K}{\beta+1}} = 139V/V[/itex]

    And if we include Early voltage rce ≈ 80V/500μA = 160kΩ and Q6 input impedance we have

    rce = 160kΩ and RinQ6 ≈ (β + 1)*5.3kΩ = 535kΩ. The gain will drop to

    [itex]Av =\Large \frac{Rc||rce||RinQ6}{2re + \frac{2*2K}{\beta+1}} = 120V/V[/itex]
     
  11. Apr 24, 2014 #10

    The Electrician

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    Q

    Hi, Jony130,

    I noticed that in post #1, the OP gives us:

    "For the differential amplifier circuit given above, for all transistors except Q1 and Q2 we have: β=100, VA=80 V, RC =20 k, VT=0.026 (Thermal voltage), VBE(on)=0.7 V"

    You have done what I would, which is to assume that Q1 and Q2 have the same specs, but I wonder why the problem excepted Q1 and Q2? Does the problem want the student to use some other parameters for them?
     
  12. Apr 24, 2014 #11
    Hmmm just read that I think it means they have different parameters. would that affect Ic1 and 2 I guess not because I never used parameters to find them. On another note if I where to try to find the gain like jony how would I do that
     
  13. Apr 24, 2014 #12
    Also im in an area where I cant really give you Rib6 or the other values requested but I will later on
     
  14. Apr 24, 2014 #13

    The Electrician

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    It won't affect Ic1 and Ic2, but if the Early voltage and β is different for Q1 and Q2, the gain will be affected.
     
  15. Apr 24, 2014 #14
    ok so how would I go about starting to find gain? I mean I know what I see in the book and what my buddies show me but I really dont get it.
     
  16. Apr 24, 2014 #15
    Rib6=rpi6+(1+beta)Re rpi6=beta*Vt/Ic6 rpi1=beta*Vt/Ic1 and Rc= 20 k
     
  17. Apr 24, 2014 #16

    The Electrician

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    It would make it easier for those of us who are helping you if you wouldn't put everything on one line; it's hard to read. Do it like this instead:

    Rib6=rpi6+(1+beta)Re
    rpi6=beta*Vt/Ic6
    rpi1=beta*Vt/Ic1
    Rc= 20 k

    To get the voltage gain of the differential pair you have take into account that the load on Q1 and Q2 is not just Rc.

    The Early effect creates an apparent resistance (sometimes denoted "ro"; Jony130 called it rce) from collector to emitter of a transistor. The ro for Q1 and Q2 is 160000 ohms; you have to include that in the load on Q2, as well as Rpi6.

    Look at what Jony130 did in post #9. That should give you some guidance.
     
  18. Apr 24, 2014 #17
    Ok, I am starting to get it. Where exactly is rpi located? I now know ro is over the collector to emitter. Also would it help for me to draw the small-signal equivalent? Furthermore what is K and re is his formulation I assume re means Re which is 5.3 K
     
    Last edited: Apr 24, 2014
  19. Apr 24, 2014 #18

    The Electrician

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    Where is rpi located? I guess you could say that it's inside the transistor, just like ro and re are. Yes, it might help to draw the small-signal model, because then Rpi, ro and re will be shown explicitly.

    K means 1000. Where Jony130 has 2*2K, that means 4000 ohms.

    re means the internal emitter resistance, given by .026/Ic for each transistor.

    Your schematic shows a resistor RE of 5300 ohms connected from the emitter of Q6 to ground; that's not the same as Re.
     
  20. Apr 25, 2014 #19
    Last edited: Apr 25, 2014
  21. Apr 26, 2014 #20
    I get how to do small-signal equivalents for single transistor circuits, but multistage circuits are giving me problems. Are there any example multistages, that break it doen step by step? I have uploaded my first attempt.
     

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