Decoupling capacitor in common base

Discussion in 'Homework Help' started by anhnha, Oct 21, 2013.

  1. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    I have just seen the circuit in another thread. What I am confused is about decoupling capacitor at base. I think the main purpose of this capacitor is to ground base in small signal model.
    If the capacitor is moved, can we say it is common base topology?

    [​IMG]
     
  2. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Hi anhnha - really a good question.
    Think of a common emitter amplifier with a emitter resistor Re.
    Is it still common emitter? Yes - with degeneration (negative feedback).
    Something similar applies to the common base circuit.
    Without the capaitor Cdec you still have common base - however, with a little negative feedback effect for signals.
     
    anhnha likes this.
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes you will still have a common base topology. But without Cdec capacitor the base voltage divider resistors will be seen at the emitter as a (R1||R2)/(Hfe + 1).
    And this resistance will add to re resistor.
     
    anhnha likes this.
  4. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Thanks, I thought about common emitter degeneration too. I think it is still called common emitter because the characteristic of the circuit basically same as common emitter without emitter resistor. There is only one different is voltage gain is lower.

    I can't figure out the negative feedback here. Could you explain? To me it seems that there are no connections between R1, R2, Cdec and input, output points. I don't see how Cdec creates a negative feedback.
     
  5. LvW

    Active Member

    Jun 13, 2013
    674
    100
    No, of course it is NOT common emitter.
    And the voltage gain is NOT lower.
    The only difference is: Common emitter produces phase shift (180 deg) and common base NOT.

    Regarding negative feedback:
    Compare the effects of the degenerating resistances for common emittter and common base. In both cases, there is a voltage "follow" effect - however, just very small for common base (because of the small base current). That means: Without the base capacitor the voltage gain remains nearly unchanged. It reduces only to a small amount.
     
  6. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Yes - it`s correct.
    For clarification: The input resistance at the emitter node is the sum of
    re=1/gm (gm=Ic/Vt=transconductance) and the contribution of the base circuitry.
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Cdec - remove "negative feedback".
    Try see this in terms of Vbe changes. Without Cdec capacitor, change in emitter current causes change in base current. This increase base voltage, because of a voltage drop across RB resistor. So any change in Vin voltage causes smaller changes in Vbe voltage than we have in the circuit with Cdec capacitor. Rin is now greater and voltage gain is smaller.
    Also note that negative feedback also increase Rin and decreases voltage gain.
     
    Last edited: Oct 21, 2013
    anhnha likes this.
  8. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Jony, please help me check this.

    Let's call the resistance seen at emitter Rin.
    + Without Cdec: Rin1 = re + (R1//R2)/(Hfe + 1)
    + With Cdec: Rin2 = re

    The total input resistance seen at emitter Rtot (now Re is included):
    + Without Cdec: Rtot1 = Rin1// Re
    + With Cdec: Rtot2 = Rin2//Re

    Rin1 > Rin2 => Rtot1 > Rtot2.

    We want high input resistance, right? If so, why not remove Cdec or am I missing something?
     
  9. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Well, with common emitter I didn't talk about the circuit I post in this thread. I was talking about two kinds of common emitter.
    + Common emitter
    + Common emitter degradation
    I meant common emitter degradation has the same characteristic with common emitter but the voltage gain is lower. And therefore, common emitter degradation is still called "common emitter" but "degradation" is now added.
    I hope this is correct!
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes, it`s correct but in this case we use CB amplifier in RF circuit, And in RF circuit low Zin (50Ω /75Ω) is not a disadvantage.
     
    anhnha likes this.
  11. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Amazing! I see it now. Just one thing I am not sure. What is the resistance RB? Is it the resistance between base and emitter?

    PS. Now I think you meant RB = R1//R2?
    But how can you know that Vbe increase because it seems that the voltages at two ends of base-emitter increase?
     
    Last edited: Oct 21, 2013
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes,

    Without Cdec capacitor Vbe voltage decrease, when Vin voltage increase and base voltage also increase
     
  13. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Jony, I still don't get it completely.

    Please help me explain it with the picture.

    [​IMG]

    It is the small signal model for the circuit without Cdec.

    I see this through the relation: Ie = (Hfe + 1)Ib
    And from the figure, Vbe = Ie*re.

    => As Ie changes, Ib and Vbe change simultaneously, right?

    The voltage at base (relative to ground), VB:

    VB = (R1||R2)*Ib

    Also, this voltage changes simultaneously with Ib and Vbe?

    Summary: Ie, Vbe and Ib change simultaneously.

    Now assuming that Cdec is added. Then in small signal model, base is connected directly to ground.
    Similar to above, as Ie changes, Ib and Vbe change simultaneously, right?

    Summary: Ie, Vbe and Ib change simultaneously.

    In both cases, Vbe changes according to Ie (or Vin). The only difference I can see is that in the first case VB ≠ 0 and in the second case VB = 0.

    Vbe = Ie*re and in both cases Vbe changes with the same amount. What I am wrong here?
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Without Cdec

    Vin = Ie*re + Ib * R1||R2 So Vbe = Vin - Ib*R1||R2

    Now if you add Cdec Vin = Vbe

    So for the same change in Vin we will have a smaller change in Vbe in circuit without Cdec.
     
    anhnha likes this.
  15. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Well, great. So simple as read but I have never thought about this. :D
    Now I can see it directly with the circuit in the first post. With Cdec two ends of R2 are connected to ground in small signal model and therfore there is no voltage drop across it.
     
    Last edited: Oct 21, 2013
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    I hope that now you see the feedback. Vin wants to increase Vbe voltage but
    VB = (R1||R2)*Ib factor wants counter back and return to the previous state.
     
    anhnha likes this.
  17. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,039
    287
    Howdy:

    The capacitor is there to assure the base stays at "signal ground." It's still a common base amplifier, except it's often awkward to build a common base amp with the base at DC ground (this usually requires two separate supplies). So as far as the signal goes, this IS grounded base. :)

    Eric
     
    anhnha likes this.
  18. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Thanks, Eric.
     
  19. LvW

    Active Member

    Jun 13, 2013
    674
    100
    anhnha, in case you want to see the gain of a CB amplifier with feedback, here comes the formula:

    Vout/Vin=gm*Rc/(1+RB/h11)=gm*Rc/(1+gm*RB/h21).

    RB=R1||R2
    h11=rbe: small signal input resistance at the base (measured in CE configuration)
    h21=beta: small signal current gain (measured in CE configuration)
    gm=Ic/Vt: small signal transconductance
     
    anhnha likes this.
  20. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    773
    45
    Thank you.
    Here is what I get. I think you made a small mistake in the final formula.

    [​IMG]

    Collector current, Ic:

    I_{c} =  g_{m}  v_{be}

    Base current:

    I_{b} =  \frac{ I_{c} }{ h_{21} } =  \frac{g_{m}  V_{be}}{ h_{21}}

    Let's call the voltage across RB= R1||R2 is VB:

    V_{B}  =  -I_{b}  R_{B}=  -\frac{g_{m}  V_{be}}{ h_{21}} R_{B}

    And from the picture:

    v_{be} =  V_{B} -  V_{in} = -\frac{g_{m}  v_{be}}{ h_{21}} R_{B} - V_{in}

    Therefore:

    v_{be} =  -\frac{ V_{in}}{ \frac{g_{m} R_{B}}{h_{21}} + 1 }

    Output voltage:

    V_{out} =  - I_{c}  R_{c} =  g_{m}  v_{be}  R_{c} = -g_{m}R_{c}. \frac{ V_{in}}{ \frac{g_{m} R_{B}}{h_{21}} + 1 }

    Voltage gain:

    A_{v} =  \frac{ V_{out} }{ V_{in} } = \frac{ g_{m}R_{c}}{ \frac{g_{m} R_{B}}{h_{21}} + 1 } =  \frac{ g_{m}R_{c}}{1 + \frac{g_{m} R_{B}}{h_{21}}  }
     
    Last edited: Oct 22, 2013
Loading...