high freq cutoff beta

Discussion in 'Homework Help' started by stupid, Mar 25, 2010.

  1. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi,
    could someone provide guidance or idea for a start?

    the question is to find an expression for β cutoff frequency Fb which is simply a high freq current gain cutoff point of the transistor with collector & emitter terminals shorted as seen in the attached.

    is the expression Fb= 1/2\pir(C_{}\pi + C_{}\mu)

    is that correct?


    thanks
    stupid
     
    Last edited: Mar 25, 2010
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You're trying to find h21 for this network, right?

    As you learned in your other thread about h-parameters, RB will play a part.

    If you apply an input voltage and calculate the input current, you can then find the ratio of Ib to Iin.
     
  3. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi the electrician,
    the question askd for β's cut off freq.
    would h21 help?

    attached are more to give some references

    thanks
    stupid

     
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    • 4.pdf
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  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Apply a 1 amp current source at B, and calculate the voltage that appears at B'. The input circuit is the only thing that will limit the frequency response. The output circuit is just the transconductance source, and it has no capacitors associated with it as in your references.

    So just recognize that you have a divider circuit. Calculate the input impedance at B and then you have the voltage there when you apply a 1 amp source at B.

    Once you have the voltage at B, then you have a voltage divider giving the voltage at B'. The expression for the voltage at B' will involve jω because there's capacitance involved.
     
  5. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi, i got the expression,
    (1-Ib)Rb-Ibr\pi=Vb'

    is that right?
    what can i do with that?

    we know no dc current flows thru capacitance.

    btw, how to do subscript? i have had trouble to make it happen.

    thanks
    stupid

     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    First, see if you can derive an expression for the input impedance at the B node. It will have to involve jω and the capacitors.
     
  7. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    (Vb'/R∏) +(Vb'/Xc)=Ib

    (Vb'Xc+Vb'r∏)/r∏*Xc=Ib

    Xc=(Ib(r∏*Xc)-(Vb'r∏))/Vb'

    1/2∏fc=(Ib(r∏*Xc)-(Vb'r∏))/Vb'

    2∏fc=Vb'/Ib(r∏*Xc)-(Vb'r∏)

    f=Vb'*2∏c/Ib(r∏*Xc)-(Vb'r∏)

    is that right??

    thanks
    stupid

     
  8. The Electrician

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    Oct 9, 2007
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    Upon thinking about this problem, it's not clear to me whether they want the cutoff frequency for the whole circuit, or just the part to the right of RB. I'm going to assume they want the whole thing. If they don't, it's easier to do the it without RB, and having done it with RB, you can go back and make the necessary deletions.

    You want to find the ratio h21 = Iout/Iin. Recognize that Ib is not Iin, because Rb absorbs some current. Here h21 means for the whole circuit, not just the transistor.

    h21 = Iout/Iin = (Vb'e*gm)/Iin

    But, Iin = Vin/Zin, so (Vb'e*gm)/Iin = (Vb'e*gm)/(Vin/Zin)

    Rearranging, h12 = Vb'e/Vin * Zin * gm

    The first part, Vb'e/Vin, which is the same as VB'/VB, is found by applying the voltage divider rule to find VB' when you know VB.

    I'll give you the result for the first part. See if you can do the derivation. it's VB'/VB = 1/(2 + 2*∏*f*C*R∏)

    The second part, finding Zin, is just a bunch of algebra, but it's easy to make mistakes, so be careful.

    Once you have the first part and the second part, both of which will involve expressions like 2*∏*f*C, you have to multiply the first part times the second part times gm; that will be h21. Then solve for the value of f that will make the expression for h21 equal to 1.
     
  9. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    the electrician,
    i cant follow yur explanation.

    is Zin=Rb//r∏//r∏ ?

    & how do u get the VB'/VB = 1/(2 + 2*∏*f*C*R∏)?

    thanks
    stupid
     
  10. The Electrician

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    Oct 9, 2007
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    I'm sure you know how to work out the impedance of various series and parallel combinations of resistances/reactances, don't you? You've done in the other threads where you derived h11 and h22. Look at the circuit again.

    Zin = RB || (R∏ + (R∏ || XC))
     
  11. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    use admitance 1/Z,

    r∏//C=1/R1=1/r∏+jωC
    1/R1=(1+jωCr∏)/r∏

    R2=r∏//R1
    1/R2=1/r∏ + (1+jωCr∏)/r∏
    =(2+jωCr∏)/r∏

    1/Zin=1/Rb + 1/R2
    =1/Rb + (2+jωCr∏)/r∏
    =(r∏ + Rb(2+jωCr∏))/Rb*r∏
    Zin=Rb*r∏/(r∏ + Rb(2+jωCr∏))

    h21=Vb'e/Vin*Gm*Rb r∏/(r∏ + Rb(2+jωCr∏))

    it has become too complicated to compute Freq!!!

    any better way around this?

    thanks
    stupid
     
  12. PRS

    Well-Known Member

    Aug 24, 2008
    989
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    ---------------------------------------------------------------------------------

    Hi stupid. First, you don't sound stupid to me. I think your answer is correct.

    To derive this, short the collector to the emitter as in both of our drawings. (They are equivalent). First we get an expression for

    h_{fe} = \frac{I_c}{I_b}\

    By inspection,  Ic = (g_m - sC_\mu)V_\pi

    and V_\pi = I_b(r_\pi||C_\pi||C_\mu)

    Therefore h_{fe} = \frac{I_c}{I_b} = \frac{g_m - sC_\mu}{\frac{1}{r_\pi} + s(C_\pi + C_\mu)}

    At frequencies for which this model is valid, gm is much greater than \omega C_\mu resulting in:

    h_{fe} = \frac{g_m r_\pi}{1 + s(C_\pi + C_\mu)r_\pi]

    So h_{fe} has a single pole response given by \omega_b = \frac{1}{(C_\pi +C_\mu)r_\pi}

    And \omega_t = \frac{gm}{(C_\pi +C_\mu)}

    Now that I compare this with your answer in the original post they are the same. This evaluation comes from MicroElectronic Circuits by Sedra and Smith.
     
    Last edited: Mar 27, 2010
  13. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    R2 should be r∏ in series with R1, not in parallel with R1.

    There's no way around it, and this is only the computation of Zin; you still have to do Vb'e/Vin in terms of the circuit R's and C's.

    The final expression has some cancellation of terms so it's not all that bad.

    So, fix the error I pointed out, and do the Vb'e/Vin computation.
     
  14. PRS

    Well-Known Member

    Aug 24, 2008
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    35
    This is just a bump so that you look at my post above this. Your original answer is correct.
     
  15. The Electrician

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    I said in post #8:

    "it's not clear to me whether they want the cutoff frequency for the whole circuit, or just the part to the right of RB. I'm going to assume they want the whole thing. If they don't, it's easier to do the it without RB, and having done it with RB, you can go back and make the necessary deletions."

    The circuit he posted in post #1 has an extra resistor, Rb, which I assume is an equivalent to the biasing network. Ordinarily, this wouldn't be considered if one wanted the intrinsic Ft of the transistor. But the circuit he posted includes it.

    If he finds Ft for the complete circuit he posted, then to delete the effect of Rb will be easy. He needs practice solving networks like this, and I decided to lead him through the steps for solving the complete circuit.

    This is the homework help forum, after all.
     
  16. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    assume Zin is to the right of Rb,

    Rin = r∏//Xc
    =r∏/(1+jωCr∏)

    Zin = r∏ + Rin
    =[r∏ + r∏(1+jωCr∏)]/(1+jωCr∏)

    Vb'e = Vin*Rin/(Rin + r∏)
    Vb'e/Vin = Rin/(Rin + r∏)
    =[r∏/(1+jωCr∏)] / [r∏/(1+jωCr∏) + r∏]
    =r∏/[r∏ + r∏(1+jωCr∏)]
    =1/(2+jωCr∏)

    hfe=Gm/(2+jωCr∏) * [r∏ + r∏(1+jωCr∏)]/(1+jωCr∏)

    [Gm/(2+jωCr∏)]*(2+jωCr∏)/[(1/r∏) + jωC]
    r∏Gm / (1+jωCr∏)

    is that right???
    pls help!

    stupid
     
    Last edited: Mar 28, 2010
  17. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    hi PRS,
    mind showing how the following arrived?

    h_{fe} = \frac{g_m r_\pi}{1 + s(C_\pi + C_\mu)r_\pi]

    So h_{fe} has a single pole response given by \omega_b = \frac{1}{(C_\pi +C_\mu)r_\pi}

    And \omega_t = \frac{gm}{(C_\pi +C_\mu)}

    btw, how to make my working arrives at those shown above?

    thanks
    stupid
     
  18. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    I showed in the post above that h_{fe} = \frac{I_c}{I_b}

    I_c is taken straight of the hybid pi model by inspection.
    I_b is obtained by noting that V_\pi is the parallel combination of r_\pi , C_\pi , and C_\mu times I_b

    Substitute the second relationship into the first and solve for h_\{fe}

    The second term in the numerator (see my original post) can be thrown away since it is so small. This simplifying assumption results in the formula given above. The pole then is just the s term in the denominator.
     
    Last edited: Mar 28, 2010
  19. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It's not the same as in his original post. What he had in his original post is:

    "is the expression Fb= 1/2\pir(C_{}\pi + C_{}\mu)...correct?"

    Which may or may not be correct depending on what r he's referring to.

    He posted in post #3 some pdf files (3.pdf, 4.pdf) that give this derivation:

    f_\beta=\frac{f_\tau}{\beta_{mid}}=\frac{1}{2\pi r_\pi(C_\pi C\mu)}

    so

    f_\tau=\frac{1}{2\pi r_{e}(C_\pi C\mu)}

    These expressions are equivalent to yours, but since the OP didn't specify which r he is referring to in his original post, we don't know which of your expressions might be equivalent to his. As his expression stands, none of yours are equivalent to his.

    Also, he is somewhat confused over what Fb is. He said in post #1:

    "the question is to find an expression for β cutoff frequency Fb which is simply a high freq current gain cutoff point of the transistor with collector & emitter terminals shorted as seen in the attached"

    The file 4.pdf he attached says "The transit or cut-off frequency, Ft,...is defined as the frequency where the current gain falls to 1."

    They've used the phrase "cutoff frequency" to mean "the frequency where the current gain falls to 1"

    On the other hand, his reference material (3.pdf) does refer to a frequency f_\beta, which is apparently the 3dB down frequency.

    So, my best guess is that he wants:

    f_\beta=\frac{1}{2\pi r_\pi(C_\pi C\mu)}

    The schematic he posted has a resistor R∏ as a series element, which is not correct; R∏ is the shunt element. The series element is Rx in Sedra and Smith. Having R∏ as a series element has no effect on Ft when calculated for a transistor driven by a current source, but it will affect the response when driven from a more typical source.

    Sedra and Smith don't include the shunt element RB, and the fact that the OP included it led me to think that perhaps his instructor intended it to be a part of the circuit for which Ft is to be calculated.
     
  20. The Electrician

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    Oct 9, 2007
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    This is correct for the expression which doesn't include the effect of RB. Is the circuit you posted in post #1 something which is in your text book, or did your instructor give it to you?
     
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