Common emitter bypass capacitor not clear

Discussion in 'General Electronics Chat' started by mishu.daniel, May 28, 2009.

  1. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    Hi all,
    Assuming we have the circuit attached, with Vin amplitude = 240mV, Fin = 6Khz.
    What I don't understand is the behavior of C3 on AC signals.
    Everywhere I searched for an explanation I saw that C3 is calculated to have a very small reactance at the lowest frequency point so that to be a short circuit for the AC signals with the frequency above the lowest point.
    But, what happen if this capacitor is not calculated like this? I want to generalize this problem for all capacitors in this circuit:
    0,01uF means Xc= 1/(2*pi*F*C) = 1/(6.28 * 6000 * 10nF) = ~ 2,7k ohm.
    I didn't find anywhere an explanation for this use case. What I want is to be able to do the math taking into account all capacitor reactances.
    After a spice simulation it is obvious that the output is modifying according to these capacitors values, but I cannot "see" the equations (?!?)

    Thanks for any help.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Gain of this circuit is approximently Rc/Re, Rc includes the load resistance R5 (and it's capacitor) in parallel with the collector resistor R2. Re includes the reactance provided by C3.
     
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  3. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    You mean that:
    Re = β*re + R4 || Xc3 (at 6Khz) and re ~= 26mV/Ie

    and Rc is R2 || (Xc2 + R5)

    so the Gain will be:

    Vout/Vin ~= Rc/Re = (R2 || (Xc2 + R5)) / (β*re + R4 || Xc3) ,
    where Xc are calculated at 6KHz.

    Is it correct this math?
     
  4. Wendy

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    Mar 24, 2008
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    Pretty close, I'm having a little trouble seeing where you got this expression. β does set a maximum gain, but is relatively independent until the maximum gain is hit.
     
  5. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    From ac small signal theory, after building equivalent circuit.
    Maybe I miss something ...
     
  6. Wendy

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    Mar 24, 2008
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    The emitter resistance is a feedback mechanism. As long as the gain of the amp is under the gain of the transistor (which I think is defined as beta. Ic/Ib) then beta doesn't interact too much. Where it gets interesting is when the beta is the defining factor of the gain. This is why well designed amps try to keep beta out of the equation overall.

    I've been wrong before, and will be again, but that is my take on the subject.
     
    Last edited: May 28, 2009
  7. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    Wait, I think I did a mistake in my previous equations: the gain should not include β. It is Vout/Vin ~= Rc/Re = (R2 || (Xc2 + R5)) / (re + R4 || Xc3) ,
    because β disappears due to Ic = β*Ib and Vin = β*re * Ib.

    This means that the gain of this circuit is 640 / 190 ~= 3,3 which is not what graphics shows (?)
     
  8. Wendy

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    Mar 24, 2008
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    If I read your graphics correctly it is showing a gain of 1.9.

    Did you calculate the drop across C2 into the answer? I'll bet you were looking at the collector voltage only. Still that doesn't account for all of it.
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    (re + R4 || Xc3)

    Re = R4 || Xc3

    and

    (R2 || (Xc2 + R5)) Needs vector calculation

    Something else I spotted.
     
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  10. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    I'm trying again to do the math but it seems to be ~3 the gain not ~2. I think that something is not correct here.
     
  11. Wendy

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    Mar 24, 2008
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    So what was the voltage drop across C2?
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You left out C1 in this model.

    If I assume a β for the transistor of 100, solving the circuit (using the nodal method) with all the capacitors taken into account gives me a gain of 2.0278 < -114.86°.
     
  13. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    Hi guys,
    sorry for my late answer.
    I didn't manage to arrive to the same result : gain ~=2.
    Here attached I've put the equivalent circuit for ac signals including all capacitors.
    First I've tried to calculate Zin = Xc1 + (Rp1 || (β*re + (R4 || Xc3))).
    re = 26mV/7.85 mA ~= 3.3 ohm
    Zin = 500 - j*2640 or 2686.9 (< -79,3 degrees)
    This means that input current Iin = 240mV/Zin = 89.3uA (< 79.3 degrees)
    If Beta = 100 => the current gain = 8.93mA which multiplied with R2 = ~6V which is wrong I know. What I did wrong?

    On the other side, from nodal analysis, each time I miss something. If the input impedance is not calculated like above , the input current Iin(or base current Ib) will not be correct. Then, I assume I have to replace Ib in another equation obtained from nodal method at the output side.

    Vout = R5*Io
    Ir2 = Io + β*Ib
    Iin = Vin/Zin

    Can I use these equations to get the result?
    Please help on this topic as I don't manage to "see" the procedure.

    Thanks.
     
  14. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I used a slightly different value for emitter current, but I'll use your value for re which is 3.3Ω.

    Also, in post #12, you don't show ro, but in this model you have ro shown. Do you want to include ro in the gain calculation? If so, what value do you want to use?

    Also, the base input resistor should be (β+1)*re, not β*re. This won't make much difference, but it's what I use in my calculations.

    Instead of Zin = Xc1 + (Rp1 || (β*re + (R4 || Xc3))), you need to use:

    Zin = Xc1 + (Rp1 || ((β+1)*(re + R4 || Xc3)))

    From this I get Zin = 3630.35 < -47.45°

    First off, Zin = 3630.35 < -47.45°, so Iin = 66 uA < 47.45°.

    But the most important thing you have overlooked is that not all of that current enters the base. The impedance looking into the base is (β+1)*(re + R4 || Xc3)) = 20475.2 < -4.24°. This is much higher than Rp1, which means that most of the input current passes into Rp1 (mainly R3). You need to calculate the actual fraction of the input current that enters the base.
     
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  15. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    Yes, yes,yes!
    This was wrong! I didn't multiplied all Re but only re. Honestly I didn't know it. Is a valuable info for me :)
    From the math I knew that to have a 0.5V amplitude on output side, this means that the current gain on colector should be 0,5/675 ~= 0.74 mA, which means that Ib ~= 7.4uA (for β=100). To obtain this current from 0.24V/Zin I saw that Zin should be much greater than my math said, but I didn't know why...
    I saw that Iin is split on 2 branches but from my wrong math Rp=2.78K and the wrong Re was smaller ~=530 ohm which gives almost the same result(almost all Iin current will flow through this second branch).
    And by the way, I've put ro here to have a complete correct equivalent schematic, I know that it is much greater so we can exclude it. The β+1 instead of β was another small mistake, I saw also some discussions about that.
    What about re? the formula is 26/Ie. After calculating the quiescent point it gives me Ie = 7.85mA. What is your value?
    I will do again all math to see if it corresponds with spice.
    And thanks again for your info !
     
  16. The Electrician

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    Oct 9, 2007
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    I ignored the base current when calculating re.

    The 18k/3.3k voltage divider applies 15*(3300)/(3300||18000) = 2.324 volts to the base (ignoring the base current effect on the voltage divider).

    Edit: As mishu.daniel pointed out, the line above should be changed to:

    "The 18k/3.3k voltage divider applies 15*(3300)/(3300 + 18000) = 2.324 volts to the base (ignoring the base current effect on the voltage divider)."

    Then 2.324 - .7 = 1.624 volts are across the 200Ω resistor, which gives Ie = 8.12 ma. Then re = 26 mV/8.12 mA = 3.2Ω.

    But the base current is 1/10 of the current in the bias voltage divider, so Ie will be somewhat smaller if the base current is taken into account.

    If I take the base current (with β = 100) into account, I get Ie = 7.13 mA, which would give an re of 3.65Ω.
     
    Last edited: Jun 4, 2009
  17. Jony130

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    Feb 17, 2009
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    The gain is equal
    Ku=-gmRc*\frac{1-\frac{2gce*Re}{Hfe}}{1+\frac{gm*Re}{(Ic/Ie)}+gceRc*(1+\frac{2gce*Re}{Hfe})}

    where
    gce=h22=Hoe
    Hfe=H21=β
    And all capacitors act like a short circuit.
    http://people.seas.harvard.edu/~jones/es154/lectures/lecture_3/bjt_amps/bjt_amps.html

    In your circuit the gain is equal
    Kusf=Ku1*ku2

    Ku1=[ R2 || (1/Hoe) || (Xc2+R5) ] / [re+(R4 || Xc3)]=3.3[V/V]

    Ku2=Rin/(Xc1 + Rin)
    =(ω*Rin*C1) / √( 1+(ω*Rin*C1)^2 )=0.713[V/V]
    where
    re=Ut/Ic≈25.9mV/Ic≈26mV/Ic at temperature 27°C

    Rin=R1 || R3 || ( (β+1) * [re+(R4||Xc3) ] )=2.7K

    Kusf=3.3*0.713=2.3[V/V]

    for Hfe=570
     
    Last edited: Jun 3, 2009
  18. The Electrician

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    Oct 9, 2007
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    mishu.daniel specifically asked for an analysis where all the capacitors are not acting like short circuits.

    Furthermore, if the capacitors are acting like short circuits, then the voltage gain will be very nearly 675/re ~= 200.

    Your value of 2.3 would seem to be much too small.
     
  19. mishu.daniel

    Thread Starter Member

    May 13, 2009
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    Yes, I've requested the case when the capacitors are not short circuits, but this is due to all tutorials I found where all capacitors where calculated especially to be smaller enough to be considered short circuits. But if the capacitors are not calculated like this and the reactance is big enough to be taken into account ?
    Then, is it really needed to go in that complex formula to obtain voltage and current gain of circuit?
    The Electrician, your voltage divider formula "15*(3300)/(3300||18000)" shouldn't be 15*(3300)/(3300 + 18000) ? And another thing, I don't understand why the base current should be 1/10 from the bias voltage divider. Can you explain this?
    And by the way, I sow in spice model of transistor Vbe = 0.75. I was using 0.65 until now but I tried to follow spice parameters to obtain the same results.

    Thanks for help.
     
  20. The Electrician

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    Oct 9, 2007
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    Well, certainly the addition of capacitor reactance is going to make any gain formula more complicated.

    Yes; I was thinking ahead of my typing, about the equivalent load in parallel with the input of the transistor.

    It's not that it should be, but that it is. Because the base current is about 1/10 of the current in the 3300 Ω resistor, it upsets the operation of the divider a little bit (about 10%), and you can't really ignore it if you want very accurate results. If the base current were only 1/1000 of the divider current (if β were 10000, for example) then ignoring it would be more justifiable.
     
    Last edited: Jun 4, 2009
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