Need help with gain of a BJT amplifier, circuit.

Discussion in 'Homework Help' started by Andreas Hultberg, Dec 7, 2018.

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  1. Andreas Hultberg

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    Dec 7, 2018
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    Hello!
    i have tried for some time to get the right gain of this amplifier-transistor Circuit.
    Im now trying to get help or a push in the right direction to let me solved this, there is probably something i have missed but i cant seem to figure it out.

    I need to decide the gain of V(out)/V(g).
    From the text (Swedish) i got:
    hfe=B(ac)=300
    (Ohm)
    Rg=5k
    RL=5k
    Rc=2.7k
    R1=68k
    R2=15k
    Re=100
    RE=470
    Vcc=10V

    * i did take 1/hoe out of account because i cant solve for it and its most likely to big to take in account for the 4-pole

    First i got ICQ from (DC). IC=ICQ
    By tumb rules of constructing a DC curcuit i take 45% of the VCC over RC = 45% of 10 = 4.5v
    By ohm it gives me about 4.5/2.7k=1.7mA (I have also calculated this the long way and got 1.7mA) - it checks out.

    With ICQ i can calculate gM. = Gm=ICQ/VT, i guess VT is aproxx 0.025v
    by standars transistors. So:
    1.7mA/0.025~=0.067 =gm

    Now i can get hie=r(pi) by gm=hfe/hie
    300/~0.067=4500 ohm.

    To save time i write the final eqvation and how i use them: ( i used 4-pol system to get this), also the equation
    V(out)/V(g)=(-(Rin)/(Rin+Rg)) * (hfe*(RC))/(hie+Re(hfe+1) - Is know since before
    V(out)/V(g)=(-(Rin)/(Rin+Rg)) * (hfe*(RC//RL))/(hie+Re(hfe+1). - Added RL (Load) in 4-pole equation.

    Rin = R1//R2//(hie+Re(hfe+1) this gives: ~9068.3 ohm
    so:
    V(out)/V(g)=~-0.65*~15.2, this give me = ~ -9.88 gain.

    ----------------------------------------------------------------------------
    Now i use voltage diver to calculate:
    *equation shorten

    (-hfe*Rc//Rl)/(hie+Re(1+hfe)) = ~15.2
    The voltage devider to take Rg in calculations
    Rg/(Rg+Rx)*Vg=V(out)
    Rx=(R1//R2)+Rg = ~ 17300 ohm
    this gives: (since gain is absolut value so the negative don´t matter in this case, just for the signal)

    Rg/(Rg+Rx)*Vg --> ~0.22*vg=V(out)
    Set voltage to 1v --->
    15.2*0.22 = 3.3 ---> so i remove 3.3 from 15.2 gain = -11.9 gain

    ----------------------------------------------------------------------------
    The answer by the book is a gain on: -11.3.

    If you want me to show more calculations/drawing i can post, but i hope this is enough to maybe help me see my error
     
    Last edited: Dec 7, 2018
  2. Jony130

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    Feb 17, 2009
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    I recalculate your circuit. And this is what I got:

    Ic = (Hfe *(Vth - Vbe) )/(Rth + (Hfe + 1)*(Re1 + Re2)) = 1.88823mA and gm = 25mV/1.88823mA = 13.2399125Ω

    Where:
    Vth = Vcc * R2/(R1 +R2)
    and
    Rth = R1||R2

    And the Transistor voltage gain is :

    Av1 = - (Rc||RL)/(Re1 + re) * Hfe/(Hfe + 1) = - 15.4312V/V

    and the whole amplifier gain is :

    Av = Uout/Ug = Rin/(Rg + Rin) * Av1 = - 9.9363V/V

    Where:

    Rin = Rth||(hfe +1)*(Re1 + re) = 9.03161kΩ

    re = Vt/Ie = 25mV/1.89453mA = 13.195Ω




    Why did not include Rin in your calculations ? As you did in the first time?
     
  3. MrAl

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    Jun 17, 2014
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    Hello there,

    I am getting -11.1 which is close to -11.3 but not exact of course.

    Since Jony got -9.94 and i got -11.1 and the supposed answer is -11.3 i suspect that something is missing or incorrect in the problem statement.
    I'll try to go over my results though to make sure i did everything right as soon as i can.
    Also note i used a different technique than Jony did that's why there is a difference in numerical results.

    [LATER]
    Ok i went over this analytically and got the result:
    -11.10902

    This is using Nodal analysis, first doing a DC analysis and then perturbing the input to get the gain. This is without using the quantity 're'.
    I suspect that adding the effect of 're' will just reduce the gain, but if someone else wants to try this that's cool too.
    The capacitors were considered to be of infinite value as per the original document with the given information.
     
    Last edited: Dec 8, 2018
  4. LvW

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    Jun 13, 2013
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    Hello to all, here are my results - derived by hand calculation using classical methods.
    In addition I have performed a Spice-analysis (BC548).
    For my opinion, during calculation of the DC bias pont the influence of the BC548 current gain (other than B=300) plays no great role because of the negative feedback (RE=570 ohms).
    I think, the most "critical" point is finding the correct DC collector current and the corresponding transconductace gm.
    All other steps are - more or less - standard procedures.

    Results (Simulation in brackets):

    IE=1.675 mA (Sim: 1.785mA)
    IC=1.67mA (Sim: 1.776 mA)
    Transconductance gm=0.0642 A/V
    Vbasis=1.605 V (Sim: 1.696 V)
    Input resistance at the base node (incl R1||R2): 9.1 kOhm
    Total gain: A=-9.78 (Sim: -9.355).

    So - I am very close to Jonys results. At the moment, I don`t know the reason for the small difference.
    Question to Jony: Is it correct that your value for Rth does not yet contain the small base current IB=IC/300 ??

    That means: It seems to be clear that the gain magnitude is below A=10.
    Hence, the "book answer" with "-11.3" seems to be wrong (no surprise...each book contains errors).
     
    Last edited: Dec 8, 2018
  5. Jony130

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    Did you assume Vbe = 0.65V? As this value was given in the problem.
     
  6. LvW

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    Jun 13, 2013
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    Jony - yes, I did of course.
    What about IB and its influence on Vth?

    Your value: Vth(=Vbasis)=1.807V
    My value: Vbasis=1.605 V.
    Simulation: Vbasis=1.696 V
     
    Last edited: Dec 8, 2018
  7. Jony130

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    But Vth is not the voltage at the transistor base.

    Vbasis = Vth - Ib*Rth = 1.807V - 6.294μA*12.289kΩ = 1.729V
     
  8. LvW

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    Jun 13, 2013
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    OK - thank you.
    Yes, as I have expected - the "crux" is the calculation of the Bias point.
    Nevertheless, does your expression for Vbasis appear in your equation for Ic ?

    However, our results are close enough - I think.
     
  9. Jony130

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    I was using a Thevenin's Theorem.

    Ib = (Vth - Vbe) /(Rth + (Hfe + 1)*(Re1 + Re2)) = (1.807V - 0.65V)/(12.289kΩ + 301*0.57kΩ) = 6.293μA

    And the collector current is Ic = β*Ib = 300*6.293μA = 1.88mA
     
  10. LvW

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    Jun 13, 2013
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    Well, at the moment I cannot reveal the source of the small disagreement (I have used another method for finding the bias point).
    Perhaps not too important since we are very close to each other ?
     
  11. MrAl

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    Jun 17, 2014
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    Hello again,

    Oh, did you change any of the original resistor values?
    You are saying that RE=570 but i am not sure where you are using that and if you eliminated any capacitors.
    I am following the original schematic with the top 'RE' resistor 100 Ohms and the bottom 470. The combined value however is NOT 570 so i am hoping you did not do that or else you'll never get the right answer. That's unless you changed the problem itself, but then you wont get the same as the book anyway.
    This is a very important point BTW, although i cant see how you could use 570 anyway.
     
  12. LvW

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    Jun 13, 2013
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    Is not?
    100+470=570 ohms.
    I think, for bias point calculation this sum is correct.
    Why do you think it is not?
     
  13. Jony130

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    But 570Ohms is only used for finding the DC condition. And to find AC gain we use RE = 100.
     
  14. MrAl

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    Bingo :)
     
  15. MrAl

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    See Jony's post.
     
  16. LvW

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    Jun 13, 2013
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    Hello again,
    Surprisingly - even a rough calculation (without explicitely calculating the bias point) can show that the overall gain will be below A=10.

    A reasonable DC bias point may exist if we allow app. 4 Volt across the collector resistor Rc (60% of Vcc for the C-E path and the emitter path) - equivalent to a collector current of app. Ic=1.5mA and a transconductance of app. gm=0.06 A/V. Hence, 1/gm=16ohm.
    This results in h11=rbe=300/0.06=5k and a total input resistance (at the base) rin=68||15||(5+300*0.1)=9k.
    The gain (ref. to the base) is A= - (2.7k||5k)/(100+16)= - 15
    and the total gain is Ao= - 15*[9/(5+9]= - 9.64.

    I think, this is a good example to show that - in case of sufficient negative feedback - the overall gain value becomes more and more independent on the (unknown, uncertain) properties of the active element.
     
  17. The Electrician

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    Oct 9, 2007
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    Undoubtedly the Spice simulation includes the effect of hre and hoe. That alone is enough to account for the difference between -9.78 and -9.355

    And as was pointed out earlier, failing to take into account re can boost the gain to around -11.1
     
  18. MrAl

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    Hello,

    First, thanks to Electrician for joining the discussion.

    Here is the circuit simplification for AC analysis.
    Note the lack of power supplies and capacitors and lumped elements. Some of the components are not lumped so that the currents though those sections can be looked at separately if needed. The AC equations can be obtained readily from this.

    As we look at the reduced schematic, we see right off that any 'Re' (which is 're') greater than zero will decrease the output magnitude, not increase it, and with Re=0 we get:
    |Vout|=11.1

    We are looking for 11.3 here, and the only way to get that is to make Re=-1.8896, which is obviously not possible at room temperature.

    If i make Re=26/2, i get a gain of:
    -9.9519

    Could the original paper just have a simple typo?
    That happens all too much.
    It is also interesting though that with even a small change in gain (like maybe 305 instead of 300) we might see the right value, which shows the practicality of looking at this in such a detailed manner. Academically however it could be important.

    For what it is worth, when we make Re=26/i with 'i' the emitter current in milliamps, the general solution for Re (which is 're') is:
    Re=
    -(13*(B*R2*RE2+R2*RE2+B*R1*RE2+R1*RE2+B*R2*RE1+R2*RE1+B*R1*RE1+R1*RE1+R1*R2))
    /((B+1)*(500*E3*R2-500*E2*R2+13*R2+500*E3*R1+13*R1))

    where E2=Vcc, E3=Vbe, B=Beta.
    This result depends only on the circuit element values and not any specific current.
     
    Last edited: Dec 8, 2018
  19. The Electrician

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    There are several possibilities. The textbook author could just have made an arithmetic mistake.

    Making re zero gets the gain to the neighborhood of -11; then making β equal to 355 makes it -11.3

    If re = 13, including hre in the calculation increases the gain. With β=300, re=13 and hre = .00806 I get a gain of -11.3, but hre is typically less that that.

    The last two don't seem very likely. Leaving out re and them making an arithmetic mistake seems to me the likely explanation.
     
  20. LvW

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    Jun 13, 2013
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    There was a slight misunderstanding (caused by my unclear words).
    Of course, there will be always a difference between hand calculations and Spice simulations - I think, it is not necessary to explain why.
    But with my comment regarding the "small difference" I was referring to the small deviation between Jonys and my hand calculations (both of us did not take into account the finite output resistance of the BJT, which will further decrease the gain)
     
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