Find the voltage gain of operational amplifier

Discussion in 'Homework Help' started by ADCapacitor, Mar 25, 2016.

  1. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    Find the voltage gain and Io

    Vi=V+=V-
    I1=(V-)/R1
    I1=0.5mA

    Node V-
    Vi-0/R1+Vi-Va/R2=0
    Va=-50mV

    NodeA
    [Va-V-/R2] + Va-0/R3 + Va-Vb/R4 =0
    Vb=-0.2V

    NodeB
    Vb-Va/R4 + Vb-0/R5 +Vb-VO/R6=0
    Vo=-0.55V

    Av= -Vo/Vi
    =-0.55/50m
    = -11V/V

    Io= Vo/RL
    =-0.55/100
    =-55mA

    thank for the help
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    And Vin is ?? Also there is some thing wrong with your answer because voltage gain cannot be negative.
    Also notice that if Vin = 1V the voltage at V- is also equal to 1V and from there we can find Va using only KVL and KCL

    Va
    = VR1 + VR2 = Vin + IR1*R2 = 1V + (1V/100Ω)*200Ω = 3V.

    Vb = VR3 + VR4 = Va + VR4 = Va + (IR3 + IR2)*R4 = 3V + (3V/200Ω + 1/100Ω)*200Ω = 8V

    And if you use the similarly approach we can find that Vo/Vin = 21V/V
     
  3. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    Vin is 50mV

    I use KCL in the above calculation
    current goes into the resistor R2 =current out the resistor R3 +R4
    Vin-Va/R2= VA/R3+Va-VB/R4

    But why
    Va = VR1 + VR2 = Vin + IR1*R2 = 1V + (1V/100Ω)*200Ω = 3V.
    and
    Vb = VR3 + VR4 = Va + VR4 = Va + (IR3 + IR2)*R4 = 3V + (3V/200Ω + 1/100Ω)*200Ω = 8V
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    fir Vin = 50mV the voltage at VA node is equal to:
    Voltage drop across R1 + voltage drop across R2 = VA = VR1 + VR2
    Voltage drop across R1 = Vin because V+ = V-
    Voltage drop across R2 = current through R2 resistor times R2 resistance. And the current through R2 is equal to Vin/R1 because R2 is in series with R1. So, the same current must be flowing through these two resistors.
    Therefore VA = VR1 + VR2 = 50mV + 50mV/100Ω *200Ω = 150mV

    And VB node is equal to VR3 + VR4, Also notice that VR3 = VA and VR4 = I*R4 = (IR2 + IR2)*R4 do you agree with this ?
     
  5. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    0
    sorry, but I still don't understand...
    I find an example which is similar to this question

    short R5
    V+=V-
    0-Vi/R1 + 0-Va/R2=0
    Va=-0.5V

    Then, I follow the same method to find the Va
    Vi-0/R1+Vi-Va/R2=0
    Va=-50mV
    What's problem here?
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The maths is your problem.

    \frac{Vi}{R1}+\frac{(Vi-Va)}{R2}=0

    Try solve this for Va

    \frac{50mV}{100\Omega}+\frac{(50mV-Va)}{200\Omega}=0
     
  7. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    0
    Thank you
    I think I forgot the negative sign , lol
    Va sholud be 150mV
    Then I can use the same way to find Vb,=0.4V ,Vo=1.05V
    then I get the Av=21, Io=10.5mA
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Very good, but you still do not have the right answer for Io. I can assure you that Io is not equal to 10.5mA . By Io I mean the op amp output current. 10.5mA is a current that is flowing through R7 resistor.
     
  9. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    0
    May be Io is equal to I7-I6 ?
    (Io+I6=I7)
    And I6=Vb-Vo/R6
    I know the I6 and I7 ,then I get Io?
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Why you think that Io is a difference between I7 and I6 ?
     
  11. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    Because Io and the current flowing out the resistor R6 are flowing into node Vo?
    So I guess Io which is equal to I7-16
     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Note that for Vin = 50mV we have Vo = 1.05V and Vb = 0.4V. So in which direction IR6 current is flowing ?
     
  13. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    Thank you
    I get it
    So Io is Vo/R7 +Vo-Vb/R6 ?
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, that right but also do not forget about parentheses.
     
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  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You had the same problem with lack of parentheses in post #1. The expressions missing parentheses happen to be such that if they are evaluated left to right without any attention paid to operator precedence, you get the right answer. Some calculators behave this way and perhaps that's what you used when evaluating those expressions, but you may not be lucky enough to have expressions that evaluate properly left to right next time. Better to use parentheses.

    For extra credit, solve the problem for an opamp with a finite gain of 10. :)
     
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