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Input/Output Impedance

  1. Mar 1, 2014 #1
    The Meaning Of An Active Bandpass Filter

    1. The problem statement, all variables and given/known data

    Calculate the node voltages V1 and V2. Assume that the 0 ohm resistor is an ammeter, and that the zero current element is a voltmeter.

    https://scontent-b.xx.fbcdn.net/hphotos-ash3/t1/1898156_10151912573775919_1335607656_n.jpg

    2. Relevant equations

    KVL and KCL calculation format.

    3. The attempt at a solution


    Can you please tell me if my calculations are correct.

    Node 1: V1/0ohm -V1/30ohm + 100V/10ohm = 4

    -V1 + 300V = 120V

    V1 = 180V


    Node 2: V2 = (4A)(7.5ohm)

    V2 = 180V


    Thank You
     
    Last edited: Mar 1, 2014
  2. jcsd
  3. Mar 2, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    That's an interesting choice of title for your thread, Duave. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken]

    I'm thinking you have not drawn the circuit correctly. You show the ammeter shorted out by a connecting wire. That wire also connects node 1 to node 2, making them one node. Please check and correct.
     
    Last edited by a moderator: May 6, 2017
  4. Mar 2, 2014 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    At node 1, two of the terms you will probably have are V1/10 and (V1 + 100)/30 plus some more.
     
  5. Mar 17, 2014 #4
    Thank you for any help that you can offer.

    Did I answer all of the questions correctly and thoroughly? Can you please find any errors, and point them out to me so that I can fix them?

    1. The problem statement, all variables and given/known data


    Calculate the input impedance looking directly into the base of the BJT
    Calculate the output impedance including the 3.3k resistor



    https://scontent-a-pao.xx.fbcdn.net/hphotos-prn2/t1.0-9/1900003_10151939940590919_952631427_n.jpg

    2. Relevant equations

    Zin = (hFE + 1){ZLoad}
    ...................................................
    Zout = {(R10k)(R3.3k)/(R10k) + (R3.3k)}/(hFE+1)
    .................................................................................................................


    3. The attempt at a solution

    1(a)

    Calculate the input impedance looking directly into the base of the BJT

    Zin = (hFE + 1){ZLoad}
    ...................................................
    Zin = (hFE + 1){RL
    ...................................................
    Zin = (100 + 1){10 x 10^3(ohms)}
    ...................................................
    Zin = (101){10 x 10^3(ohms)}
    ...................................................
    Zin = {10.1 x 10^5(ohms)}
    .....................................
    Zin = {1010k(ohms)}
    .....................................


    1(b)

    Calculate the output impedance including the 3.3k resistor

    Zout = (R10k)(R3.3k)/(R10k) + (R3.3k)}/(hFE+1)
    .................................................................................................................
    Zout = {(10 x 10^3(ohms))(3.3 x 10^3(ohms))/(10 x 10^3(ohms)) + (3.3 x 10^3(ohms)}/(100 + 1)
    .................................................................................................................
    Zout = {2.481 x 10^3(ohms)}/(100 + 1)
    .................................................
    Zout = 24.56(ohms)
    ..............................

    Are there any errors?

    Thank you.
     
  6. Mar 17, 2014 #5
    Are you sure about Zout ?? Also check your numbers in Zin; ZL = ?
     
    Last edited: Mar 17, 2014
  7. Mar 17, 2014 #6
    Jony130,

    Okay, no, I am not sure about Zout. Seeing that my answers are not right. What assumptions can I make to improve my next attempt? What kind of questions does one ask to tackle a problem like this?
     
  8. Mar 17, 2014 #7
    For this simple circuit with CCCS

    attachment.php?attachmentid=67738&stc=1&d=1395073484.png

    try to find Zout = Vin/Iin for Vin = 1V and β = 100;
     

    Attached Files:

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  9. Mar 17, 2014 #8
    Answer

    This is my answer to the question

    Iin = IB + (hFE)(IB)
    ...................................................................
    Iin = (hFE + 1)(Ib)
    ...................................................................
    Ib = (Vin)/(Rb)
    ...................................................................
    Iin = (hFE + 1){(Vin)/(Rb)}
    .........................................................................
    Zout = (Vin)/(Iin)
    .........................................................................
    Zout = (Vin)/[(hFE + 1){(Vin)/(Rb)}]
    .............................................................................................
    Zout = (Rb)/(hFE + 1)
    ........................................................
    Zout = (10k)/(100 + 1)
    ........................................................
    Zout = 99(ohms)
    ..................................................
     
  10. Mar 17, 2014 #9
    Very good, so Zout for your emitter follower circuit will be equal to Zout = ?
     

    Attached Files:

  11. Mar 17, 2014 #10
    So Zout = {R10k/(100 + 1)}||{R3.3k}
    ............................................................................................................
     
  12. Mar 17, 2014 #11
    Excellent work
     
  13. Mar 17, 2014 #12
    Okay this is a revision of the whole problem. Can you please look at all eight questions and answers? Thanks Jony130.

    Thank you for any help that you can offer.

    Did I answer all of the questions correctly and thoroughly? Can you please find any errors, and point them out to me so that I can fix them?

    1. The problem statement, all variables and given/known data

    Calculate the input impedance looking directly into the base of the BJT
    Calculate the output impedance including the 3.3k resistor
    Calculate VB neglecting loading of the bias network by the BJT
    Calculate VE neglecting loading of the bias network by the BJT
    Calculate IE neglecting loading of the bias network by the BJT
    Calculate VB including loading of the bias network by the BJT
    Calculate VE including loading of the bias network by the BJT
    Calculate IE including loading of the bias network by the BJT


    https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-frc1/t1.0-9/1932404_10151940102460919_638617745_n.jpg


    1(a): Calculate the input impedance looking directly into the base of the BJT

    Zin = (hFE + 1){ZLoad}
    ...................................................
    Zin = (hFE + 1){R10k
    ...................................................
    Zin = (100 + 1){10 x 10^3(ohms)}
    ...................................................
    Zin = (101){10 x 10^3(ohms)}
    ...................................................
    Zin = {10.1 x 10^5(ohms)}
    .....................................
    Zin = {1010k(ohms)}
    .....................................


    1(b): Calculate the output impedance including the 3.3k resistor


    Iin = [IB + (hFE)(IB)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ...................................................................
    Iin = [(hFE + 1)(Ib)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ...................................................................
    Ib = [(Vin)/(Rb)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ...................................................................
    Iin = [(hFE + 1){(Vin)/(Rb)}]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    .........................................................................
    Zout = {(Vin)/(Iin)}{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    .........................................................................
    Zout = (Vin)/[(hFE + 1){(Vin)/(Rb)}]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    .............................................................................................
    Zout = {(Rb)/(hFE + 1)}{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ..................................................................................................................................
    Zout = {(10k)/(100 + 1)}{3.3k}/{(10k)/(100 + 1)} + {3.3k}
    .....................................................................
    Zout = 96(ohms)
    ..................................................


    https://scontent-a-pao.xx.fbcdn.net/hphotos-prn1/t1.0-9/1964785_10151940672820919_881396491_n.jpg


    2(a): Calculate VB including loading of the bias network by the BJT

    rin = (hFE + 1){R}
    ...................................................
    rin = (hFE + 1){R7.5k
    ...................................................
    rin = (100 + 1){7.5 x 10^3(ohms)}
    ...................................................
    rin = (101){7.5 x 10^3(ohms)}
    ...................................................
    rin = {7.575 x 10^5(ohms)}
    .....................................
    rin = {757.5k(ohms)}
    .....................................


    IB = IE/(hFE + 1)
    ...................................................
    IB = {(VCC/R7.5k)/(hFE + 1)}
    .................................................................................
    IB = {(15V/7.5k/(100 + 1)}
    ..........................................
    IB = {(15V/7.5k/(100 + 1)}
    ..........................................
    IB = 1.98 x 10^-5
    ..........................................
    IB = 19.8uA
    ..........................................
    VB = (rin)(IB)
    ................................................
    VB = (757.5k)(19.8uA)
    ..................................
    VB = (757.5k)(19.8uA)
    ..................................
    VB = 15V
    .............................


    2(b): Calculate VE including loading of the bias network by the BJT

    VE = VB - VBE
    ................................................
    VE = 15V - 0.6V
    ............................
    VE = 14.4V
    ............................


    2(C): Calculate IE including loading of the bias network by the BJT

    IE = IE/R7.5k
    ................................................
    IE = 14.4V/7.5k
    ................................................
    IE = 1.92mA
    ................................................


    2(D): Calculate VB neglecting loading of the bias network by the BJT

    {R150k/(R150k + R130k)} x VCC = VB
    .........................................................................................
    150k/(150k + 130k) x 15V = VB
    .......................................
    8.03V = VB
    ........................................


    2(E): Calculate VE neglecting loading of the bias network by the BJT

    VE = VB - VBE
    ................................................
    VE = 8.03V - 0.6V
    ............................
    VE = 7.43V
    ............................

    2(F): Calculate IE neglecting loading of the bias network by the BJT

    IE = IE/R7.5k
    ................................................
    IE = 7.43V/7.5k
    ................................................
    IE = 0.99mA
    ................................................

    Are there any errors?

    Thanks again for your help.
     
  14. Mar 17, 2014 #13
    NascentOxygen,

    This is a whole problem. Can you please look at all eight questions and answers and tell me if you see errors?

    Thank you for any help that you can offer.

    Did I answer all of the questions correctly and thoroughly? Can you please find any errors, and point them out to me so that I can fix them?

    1. The problem statement, all variables and given/known data

    Calculate the input impedance looking directly into the base of the BJT
    Calculate the output impedance including the 3.3k resistor
    Calculate VB neglecting loading of the bias network by the BJT
    Calculate VE neglecting loading of the bias network by the BJT
    Calculate IE neglecting loading of the bias network by the BJT
    Calculate VB including loading of the bias network by the BJT
    Calculate VE including loading of the bias network by the BJT
    Calculate IE including loading of the bias network by the BJT


    https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-frc1/t1.0-9/1932404_10151940102460919_638617745_n.jpg


    1(a): Calculate the input impedance looking directly into the base of the BJT

    Zin = (hFE + 1){ZLoad}
    ...................................................
    Zin = (hFE + 1){R10k
    ...................................................
    Zin = (100 + 1){10 x 10^3(ohms)}
    ...................................................
    Zin = (101){10 x 10^3(ohms)}
    ...................................................
    Zin = {10.1 x 10^5(ohms)}
    .....................................
    Zin = {1010k(ohms)}
    .....................................


    1(b): Calculate the output impedance including the 3.3k resistor


    Iin = [IB + (hFE)(IB)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ...................................................................
    Iin = [(hFE + 1)(Ib)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ...................................................................
    Ib = [(Vin)/(Rb)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ...................................................................
    Iin = [(hFE + 1){(Vin)/(Rb)}]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    .........................................................................
    Zout = {(Vin)/(Iin)}{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    .........................................................................
    Zout = (Vin)/[(hFE + 1){(Vin)/(Rb)}]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    .............................................................................................
    Zout = {(Rb)/(hFE + 1)}{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
    ..................................................................................................................................
    Zout = {(10k)/(100 + 1)}{3.3k}/{(10k)/(100 + 1)} + {3.3k}
    .....................................................................
    Zout = 96(ohms)
    ..................................................


    https://scontent-a-pao.xx.fbcdn.net/hphotos-prn1/t1.0-9/1964785_10151940672820919_881396491_n.jpg


    2(a): Calculate VB including loading of the bias network by the BJT

    rin = (hFE + 1){R}
    ...................................................
    rin = (hFE + 1){R7.5k
    ...................................................
    rin = (100 + 1){7.5 x 10^3(ohms)}
    ...................................................
    rin = (101){7.5 x 10^3(ohms)}
    ...................................................
    rin = {7.575 x 10^5(ohms)}
    .....................................
    rin = {757.5k(ohms)}
    .....................................


    IB = IE/(hFE + 1)
    ...................................................
    IB = {(VCC/R7.5k)/(hFE + 1)}
    .................................................................................
    IB = {(15V/7.5k/(100 + 1)}
    ..........................................
    IB = {(15V/7.5k/(100 + 1)}
    ..........................................
    IB = 1.98 x 10^-5
    ..........................................
    IB = 19.8uA
    ..........................................
    VB = (rin)(IB)
    ................................................
    VB = (757.5k)(19.8uA)
    ..................................
    VB = (757.5k)(19.8uA)
    ..................................
    VB = 15V
    .............................


    2(b): Calculate VE including loading of the bias network by the BJT

    VE = VB - VBE
    ................................................
    VE = 15V - 0.6V
    ............................
    VE = 14.4V
    ............................


    2(C): Calculate IE including loading of the bias network by the BJT

    IE = IE/R7.5k
    ................................................
    IE = 14.4V/7.5k
    ................................................
    IE = 1.92mA
    ................................................


    2(D): Calculate VB neglecting loading of the bias network by the BJT

    {R150k/(R150k + R130k)} x VCC = VB
    .........................................................................................
    150k/(150k + 130k) x 15V = VB
    .......................................
    8.03V = VB
    ........................................


    2(E): Calculate VE neglecting loading of the bias network by the BJT

    VE = VB - VBE
    ................................................
    VE = 8.03V - 0.6V
    ............................
    VE = 7.43V
    ............................

    2(F): Calculate IE neglecting loading of the bias network by the BJT

    IE = IE/R7.5k
    ................................................
    IE = 7.43V/7.5k
    ................................................
    IE = 0.99mA
    ................................................

    Are there any errors?

    Thanks again for your help.
     
  15. Mar 17, 2014 #14

    berkeman

    User Avatar

    Staff: Mentor

    The ammeter is shorted out, as NO says. There is likely an error in your original post. Can you please check with the instructor?
     
  16. Mar 17, 2014 #15

    berkeman

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    Staff: Mentor

    Thread closed while we sort out multiple posts of the same question...
     
  17. Mar 18, 2014 #16

    berkeman

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    Staff: Mentor

    Thread re-opened after 2 threads merged.
     
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