Series-Shunt Feedback Amplifier

Discussion in 'Homework Help' started by jegues, Nov 14, 2011.

  1. jegues

    jegues Thread Starter Active Member

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    I think I managed to get part a), but I'm stuck on part b).

    I wrote all the equations I could think of in the circuit but I'm still a little stuck as to actually solving the DC emitter currents. Is there an assumption I can make about any of the other currents?

    For example,

    Can I assume,

    I_{c1} \approx I_{1}

    Is this acceptable? Or should I be able to solve for everything?

    I feel like I need to solve 1 value or make an assumption in order to get things rolling.

    Thanks again!

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  2. t_n_k

    t_n_k AAC Fanatic!

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    The closed loop gain should be ~100.
  3. t_n_k

    t_n_k AAC Fanatic!

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    For the currents you can do an incremental approach.

    First assume IB1≈0.

    Then calculate the current in R1=0.7/1k=0.7mA. Hence current in R2=0.7mA and hence it follows that IE2=1.7mA.

    Thus IB2=1.7mA/101=16.8uA.

    This makes IC1=100uA-16.8uA=83.2uA. Thus IB1 would be better approximated as 832nA.

    You could redo the process now using IB1=832nA rather than IB1=0A. But I doubt things would change that much.
  4. jegues

    jegues Thread Starter Active Member

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    How did you draw this conclusion?

    I've been reading in the book how to estimate the closed loop gain when we have a large loop gain but obviously I found the wrong relationship.

    My reasoning was as follows,

    A\beta \rightarrow \infty

    A_{f} = \frac{V_{o}}{V_{s}}= \frac{A}{1+A\beta} \Rightarrow \frac{A}{A\beta} = \frac{1}{\beta}
  5. t_n_k

    t_n_k AAC Fanatic!

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    Begging your pardon - it should be ~10 rather than ~100. I slipped an order of magnitude in my original "back of the envelope" estimate.

    Basically, I'm looking at what happens at the output emitter (E2) as the input varies.

    With the input 0V then VB1≈0.7V and hence VE2≈7.7V.

    If the input increases from 0V to +10mV then VB1=0.71 V and hence

    VE2=7.1+0.71=7.81V.

    So the VE2 has changed by (7.81-7.7) 0.11V for a 10mV input change, which equates to a gain of Av=0.11/0.01=11.

    If the input is -10mV then VB1=0.69V and VE2=6.9+0.69=7.59V

    So for a change from 0V to -10mV the E2 voltage changes by 7.7-7.59=0.11V - again indicating a small signal gain of ~11.
  6. t_n_k

    t_n_k AAC Fanatic!

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    I can run a simulation to check the result if you don't have access to simulation software.
  7. t_n_k

    t_n_k AAC Fanatic!

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    Using the idea of feedback you can come up with the gain as follows.

    If the open loop gain is Avo then

    V_o=A_{vo}\[V_s-\frac{R_1}{(R_1+R_2)}V_o\]

    Re-arranging this gives

    V_o\[1+A_{vo}\frac{R_1}{(R_1+R_2)}\]=A_{vo}V_s

    or

    V_o\[\frac{1}{A_{vo}}+\frac{R_1}{(R_1+R_2)}\]=V_s

    And if Avo is very large then

    V_o\[\frac{R_1}{(R_1+R_2)}\]=V_s

    and finally

    A_f=\frac{V_o}{V_s}=\frac{(R_1+R_2)}{R_1}

    So Af=11 with the given values looks consistent.

    Perhaps you have confused the transistor current gain beta values with the feedback beta factor.

    i.e.

    \beta=\frac{R_1}{(R_1+R_2)}
    Last edited: Nov 15, 2011
  8. Jony130

    Jony130 Senior Member

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    I was bored today so I decided to find open loop voltage gain.

    I use this diagram

    [​IMG]

    And I assume Ic1 = 100uA; Ic2 = 1.6mA

    re1 = 26mV/100uA = 260Ω

    re2 = 26mV/1.6mA = 16.25Ω

    So the voltage gain of the first BJT is equal

     A1 = \frac{Rc}{re} = \frac{( hfe+1)*(re2 + R1+R2)}{re1 + \frac{R1||R2}{Hfe+1} } = 4.13K

    And I'm not sure whether this (R1||R2)/(Hfe +1) should be take into account.

    Voltage gain for Q2 is equal


     A2 =\frac{Re}{re+Re} = \frac{R1+R2}{re + R1+R2} = 0.998

    So open loop gain

    Ao = A1 * A2 = 4.12KV/V

    The input resistance

     Rin = {re1 + \frac{R1||R2}{Hfe+1} } = 269\Omega

    So after we close the feedback loop the voltage gain will drop to:

    Af = Ao/ ( 1 + Ao*β ) = 10.97 V/V

    and

    Rin_f = Rin* = 101KΩ

    Rout_f = (R1+R2) /
    (1+ Ao*β) = 29.3Ω But I also not sure about that.

    So what do you think about my simplified analysis?
    t_n_k likes this.
  9. t_n_k

    t_n_k AAC Fanatic!

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    Thanks for your useful analysis Jony130.

    On the matter of which components to include in the open loop AC gain analysis one could perhaps assume that the feedback is 'killed' by shunting Q1 base to ground with a large capacitance. So the DC conditions would be 'preserved' but the AC gain would presumably more closely approach the open loop case.

    Would that mean you then dispense with the (R1||R2)/(Hfe +1) term and replace the R1+R2 combination at Q2 emitter with just R2? Presumably yes.

    In any event the open loop gain is highly dependent on the Q2 hfe value.

    The addition of the load resistance RL to the circuit will also have an effect.
  10. jegues

    jegues Thread Starter Active Member

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    I think this the source of my confusion.

    So the \beta_{1}, \beta_{2} are the transistor current gain beta values NOT the feedback beta factor.

    Thanks again!
    Last edited: Nov 15, 2011
  11. jegues

    jegues Thread Starter Active Member

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    I've started working on part c),

    R_{i} = R_{s} + r_{e1} + \frac{R_{11}}{\beta + 1} \quad ; \text{where} \quad R_{11} = R_{1}//R_{2}

    R_{o} = R_{22}//R_{L} \quad ; \text{where} \quad R_{22} = R_{1}+R_{2}

    Are these expressions correct?

    How does one go about solving for A?

    Thanks again!
  12. Jony130

    Jony130 Senior Member

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    It looks ok to me.


    Simply use classic method.
    Replace BJT with his Hybrid-pi model and find the open loop gain.

    [​IMG]

    And we can use "brutal force" nodal analysis to find the voltage gain.
    I can write equation for Va node.

    \frac{Vout - Va}{r_{\pi 2}} = \frac{ Va - Vin}{ro1} + gm1*(r_{\pi 1} *\frac{0-Vin}{R1||R2+r_{\pi 2}})

    And for Vout node

    \frac{Vout}{R1+R2} +\frac{ Vout}{ro2} +\frac{Vout-Va}{r_{\pi 2}} = gm2*(r_{\pi 2} * - \frac{Vout - Va}{r_{\pi 2}})

    And when I solve this for Vout/Vin this is what I get
    Additional I assume that ro1 = ro2 = ∞

    \frac{Vout}{Vin}=\frac{gm1*(R1+R2)*r_{\pi 1}*(1 + gm2*r_{\pi 2})}{R1||R2 + r_{\pi 1}}

    I also use another method. And I get similar solution.
    The method I use is described here
    http://forum.allaboutcircuits.com/showthread.php?p=165051#post165051 (43)
    http://forum.allaboutcircuits.com/showthread.php?p=289293#post289293

    From all this "methods" my "simplified method" (show in post 9 in thread) is the fastest one.

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    Last edited: Nov 16, 2011
  13. The Electrician

    The Electrician Senior Member

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    Be aware that the output impedance you have given is for the case where the source Vs is not connected to Rs; that is, with the left end of Rs not connected to anything. Assuming Vs is an ideal voltage source with zero output impedance, if Vs is connected to Rs, then Ro will be different. Which do you suppose the problem is asking for?

    Also, if RL is connected, then the expression for Ri will include RL.
  14. jegues

    jegues Thread Starter Active Member

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    How am I supposed to decipher which the problem is asking for?

    Based on the examples in the text, Vs nor Rs are considered when calculating R_{o}.
    Last edited: Nov 16, 2011
  15. The Electrician

    The Electrician Senior Member

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    Since you are the student, ultimately it will be up to you to determine. You could ask your instructor. I was asking if you had any way to know.

    That may be sufficient. However, since Vs would be driving the circuit when the load is connected, the output impedance with a source connected would seem to be relevant.

    Did the other examples in the text you're referring to lack feedback? Perhaps they had output impedances that didn't change when the source was connected.
  16. Jony130

    Jony130 Senior Member

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    We here in Polad use this definition for Rout
    Rout = Vout_open / Iout_short when Vs and Rs is replaced by a short circuit.
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