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Differential amplifier design

  1. Mar 25, 2016 #1
    9rixZEB.png

    The figure shows a differential amplifier with a differential input vid = (vin_d_pos - vin_d_neg) between the bases of Q1 and Q2, and an output (out) at the emitter of Q4.

    I have to Determine R1, R2, R4, R5 and R6 so that the following requirements are met:

    R 1 = R 2
    Av = vout / vid = 100 [V / V]
    Vout = 0V when Vin_d_pos = Vin_d_neg = 0V.
    The output Vout must be able to deliver undistorted peak signal voltage in the range
    from vout-min to vout-max with RL armed.
    Quiescent current IC = 10 * Q3-Q4 IB

    ----------
    I have been given the values:
    Vcc-pos = 21 V
    Vcc-neg = 21 V
    R3 = 2000 Ohm
    RL = 49 Ohm
    I1 = 3,56 mA
    Vout-max = 8,8 V
    Vout-min = -8,8 V
    ----------
    How can I start? For I have difficulty getting started.
     
  2. jcsd
  3. Mar 25, 2016 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    The rules say that you must show some effort before we can offer tutorial help. What have you learned in class about how to analyze differential amplifiers? How are the bias currents set? How is the gain set?
     
  4. Mar 25, 2016 #3

    LvW

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    Where is the current I1 and what is the meaning of "quiescent current iC=10 * Q3-Q4 IB"?
     
  5. Mar 25, 2016 #4
    My English is not so good. I have found this out:

    Vcm = 0 , Vin_D = 0
    VB1 = VB2 = 0 V.

    Ic1 = IE1 = I/2 = 3,65 / 2 = 1,825 mA

    VB3 = 21 V - R3 * Ic1 = 21 V - 2kΩ * 1,825 mA = 17,35 V.

    Ic3 = IE3 = (24 V - VB3 - 0,7 V) / R4

    But I don't know the value of R4?
     
    Last edited: Mar 25, 2016
  6. Mar 25, 2016 #5
    You can see the current I1 under the first set of amplifiers.
     
  7. Mar 25, 2016 #6

    LvW

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    OK - sorry, I have overlooked this current source.
     
  8. Mar 25, 2016 #7

    LvW

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    I think, because you have used nearly all information, as a next step you should use the gain requirement.
    This gives you a relation involving R4 and R5. Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current (you know the DC voltages at the emitter and the collector for Q3)..
     
  9. Mar 25, 2016 #8
    I don't understand this - "Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current".
     
  10. Mar 26, 2016 #9

    LvW

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    VB3=17.35V
    That means: We know VE3=(17.35+0.7)V=21-IC3R4
    We also know: VB4=+0.7V=-21+IC3R5.
    Eliminating IC3 gives a second equation for R4 and R5.
     
  11. Mar 26, 2016 #10
    Oh okay. So we get this by eliminating Ic3:

    18,05 = 21 - R4
    R4 = 2,95

    0,7 + 21 = R5
    R5 = 21,7

    Does it mean, that we have found the values of R4 and R5 now? And now we can find the value of Ic3?
     
  12. Mar 26, 2016 #11

    LvW

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    ???
    Do you substract Ohms and voltages? Why did you neglect IC3 in the equations ?
    "Eliminating IC3" means: Solving for this current in one equation and inserting it into the other one. This is basic for a system of two equations!
     
  13. Mar 26, 2016 #12
    Okay, but when I solve for the current in one equation and inserting it into the other one. I get:

    (17.35+0.7)V=21 V -IC3R4
    (2,95 / R4 ) = IC3

    21,7 V = (2,95 / R4 ) * R5
     
    Last edited: Mar 26, 2016
  14. Mar 26, 2016 #13

    LvW

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    Following my recommendation in post#7 you now have two equations involving R4 and R5.
    Two equations with two unknowns can be solved easily.
     
  15. Mar 26, 2016 #14
    I can solve two equations with two unknowns. But my problem is, what those two equations are? That's what I'm confused about. Are these the two equations?:

    (17.35+0.7)V=21-IC3R4
    +0.7V=-21+IC3R5

    But thats two equations with three unknowns Ic3, R4 and R5
     
  16. Mar 26, 2016 #15

    LvW

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    The first equation is In post#12.
    Question: Did you use already the gain information? NO!
    Because Q4 provides no additional gain, it is Q1, Q2 and Q3 providing the necessary gain.
    Are you able to find a correct gain formula for the first two stages?
    This gives you a second equation.
     
  17. Mar 26, 2016 #16
    I give up. It's too difficult for me to solve. But thanks for your help:frown:
     
  18. Mar 27, 2016 #17

    LvW

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    Too difficukt?
    * You have the DC current and the collector resistor for the first stage. Are you able to find the corresponding gain A1 ?
    * The gain of the second stage A2 is determined (primarily) by the two resitors R4 and R5. Do you know the expression?
    * A=A1*A2=100 . From this you can derive the necessary gain A2 (A1 is known) and the required second equation for R4 and R5.
     
  19. Mar 27, 2016 #18
    ϒe is: VT / IE = 25 Ω
    Av1= Rc / 2* ϒe = 2 / 2* 25 = 40.


    ϒe4= VT/Ie4 = 0,025 / 0,01 = 2,5 Ω
    I have found R6 to: (-21*49)/(R6+49) = -8,8 ⇒ R6 = 67,9.
    Ri4= (β4+1) ( ϒe4+R6) = 401 (2,5Ω+67,9) = 28,2 KΩ
    Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*28,2 KΩ) / (R5+28,2 kΩ)) / (25 + R4)

    Is this correct?
     
    Last edited: Mar 27, 2016
  20. Mar 28, 2016 #19

    LvW

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    What is the meaning of ϒe ?
    What is the gain formula for the first stage?
     
  21. Mar 28, 2016 #20
    ϒe = re

    Gain A1:

    ϒe1 = ϒe2 = VT / IE = 0,025/0,001825 = 13,7
    Av1= (R3 II Rib3) / (ϒe1 + ϒe2 + R1 + R2)

    The gain of the second stage A2:

    ϒe4= VT/Ie4 = 0,025 / 0,34 = 0,339
    Ri4= (β4+1) ( ϒe4+R6) = 401*(0,339+62) = 25 KΩ

    Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*25 KΩ) / (R5+25 kΩ)) / (25 + R4)

    Better?
     
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