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Emitter follower help.

  1. Aug 26, 2012 #1

    perplexabot

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    Hi all. So I have been trying to understand how a emitter follower (common collector) works. But I can't seem to grasp the physics behind it. I know it is used as a buffer, it has high input impedance and low output impedance. I don't know what properties cause the output voltage to follow the input voltage. Also if you have "The Art of Electronics," on page 67 where it says "Important points about followers," I can't understand why clipping occurs for the signal. For some reason this whole thing is kind of vague to me. Please help. Thank you.
    http://upload.wikimedia.org/wikipedia/commons/b/b8/NPN_emitter_follower.svg
     
  2. jcsd
  3. Aug 26, 2012 #2

    sophiecentaur

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    Hi
    Without any Maths, you can say that current will pour into the base and, hence, through the transistor (current amplification factor of at least several tens) from C to E, until Vbe is around 0.7V. When that happens, Vout is 0.7V below Vin. If you vary Re, the current through the transistor will change to re-establish the 0.7V Vbe. There is so-called Voltage Feedback because any small change in Vout will cause a change in Ib and hence a large change in Ie - and this will maintain Vout to be what it was. This gives an effectively low output impedance - as the Volts are independent of the load resistance. If Vin is raised, the current that will be drawn through the base will not change appreciably because Ve 'chases Vb', remaining 0.7V below. So the current into the base of the Emitter follower will not change as Vin increases - which is what you would get with a very high resistance - hence the high input resistance.
    Why does it 'limit'? If you try to take Vb too high then the transistor will saturate and no more current can pass through Re, so the volts cannot increase. The feedback mechanism runs out. Vbe need only increase by a miniscule amount beyond this and the forward biased be diode will pass loads of current. The input impedance will drop as Vb increases until it approaches that of a forward biased diode in series with Re.
    There you are - no Maths and plenty of arm waving. Does it help?
     
  4. Aug 26, 2012 #3
    perplexabot,

    The physics are the same as any BJT. Do you mean the topology of the common collector?

    Its use won't help you understand how it works.

    The property of any transistor in the active region. The emitter terminal voltage is around 0.7 volts different than the base terminal. So the emitter follows the base or vice versa voltage wise.

    I don't, but clipping occurs whenever the input is overdriven or the operating point is chosen badly.

    Ratch
     
  5. Aug 26, 2012 #4

    sophiecentaur

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    It isn't clear at what level he's asking the question. I think the confusing phrase is "the physics behind it". Is it a real Physics question or a simple application question? I interpreted it as the latter. Very little 'Physics' is needed beyond V=IR, in a first stab at transistors. (Plus remembering the rules of how they behave).
     
  6. Aug 26, 2012 #5
    Emitter follower is a very simply circuit. The output voltage is always 0.6V lower the the input voltage.
    See some examples
    attachment.php?attachmentid=50210&stc=1&d=1346004990.png
    Also notice that the base current is (β+1) smaller then emitter current (load current).
    So our base current source B1 see our load ( Re resistor) not as 100Ω resistor. But B1 see (β+1)*Re load.
    Here you have anther example.
    Spouse we have a 1K resistor voltage divider supply from 10V battery.
    Without any load connect to the output terminal of our voltage divider the output voltage is equal 5V. Now we connect a 100Ω load resistor across the output terminal.
    And now our voltage divider output voltage drops from 5V to 0.83V. So we ruin our circuit.
    To fix this issue we add a buffer (emitter follower) see the diagram
    attachment.php?attachmentid=50209&stc=1&d=1346004805.png
    Now I hope that you see why we say that emitter follower has a high input impedance and low output impedance. It's all thanks to BJT and his current gain.
     

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  7. Aug 26, 2012 #6

    perplexabot

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    Thank you all for your great replies.

    Why will a change in Vout cause a change in Ib? Why would Vout change in the first place? Isn't Ib the one that is being "controlled" and so is subject to change?


    May you please clarify this phrase? I don't understand what you mean by "overdriven or the operating point is chosen badly." Do you mean if Ib is low enough to cause Vbe to be lower than .7V then clipping shall occur?

    For the first image, second figure, the current through Re, should it be 54mA?
     
  8. Aug 26, 2012 #7

    perplexabot

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    I still don't understand page 67 of "The Art of Electronics" under "Important points about followers," number 1. If anyone can help that would be great.
     
  9. Aug 26, 2012 #8
    Yes, my mistake.

    As for the clipping. The clipping can occur when transistor change his state eg. conduction to cut-off. See the example form The Art of Electronics
    I slightly change the diagram
    attachment.php?attachmentid=50211&stc=1&d=1346008102.png
    And ask yourself one question. What voltage you need to provide to open BJT in this circuit. Also look at figure 2.9 in Art of Electronics.
    Vout can change for example if you change load resistance.
     

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  10. Aug 26, 2012 #9

    perplexabot

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    Thank you for your help.

    Cut-off happens at voltages less than .7v?

    Is the answer .7V ?

    That was what I was thinking, but I wasn't sure. Thanks.

    On page 67 of "The Art of Electronics" it says:
    The output can swing to within a transistor saturation voltage drop of Vcc, but it cannot go more negative than -5 volts. That is because on the extreme negative swing, the transistor can do no more than turn off, which it does at -4.4 volts input (-5V output).

    I don't understand why clipping occurs at -5V (or -4.4V input) when cut-off happens at .7V. In other words shouldn't clipping occur at .7V?
     
  11. Aug 26, 2012 #10
    No 0.7V Is a Vbe voltage not the base voltage or load voltage.

    attachment.php?attachmentid=50211&stc=1&d=1346008102.png

    In this imagine without BJT the Ve voltage is equal 5V so to open the BJT Vb voltage must be higher then 5.6V. So If input voltage is greater than 5.6V the BJT is active mode and the voltage across the load resistor is Vout = Vin - Vbe. But if Vb is smaller than the 5.6V the BJT is in cut-off stage. And the Ve voltage is constant 5V. Input voltage don't affect the VE voltage any more because BJT is cut-off.
     
    Last edited: Aug 26, 2012
  12. Aug 26, 2012 #11

    perplexabot

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    Thank you so much that makes so much sense. Continuing on your example (assuming Ve is 5v when BJT is off), if Vin is 7v then Ve will be 6.3V, is that correct? Also does it saturate at 20V?
     
    Last edited: Aug 26, 2012
  13. Aug 26, 2012 #12
    Yes for Vb= 7V The output voltage Ve ≈ 6.3V
    As for the saturation for Vb = 20V bjt is almost enter the saturation region.
    To enter saturation Vb must be higher the Vcc.

    Ok So now if we change the GND position as you have in Art of Electronics
    attachment.php?attachmentid=50212&stc=1&d=1346016381.png
    I hope that now you see that without BJT we have VE = -5V. And now if you want to open the BJT the Vb voltage must we greater (lees negative) than -4.4V. If Vb is lower (more negative) than -4.4V the BJT is in cut-off mode.
     

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  14. Aug 26, 2012 #13

    perplexabot

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    Truly amazing explanation. I finally understand. Thank you so much.
    Sorry for my many questions. With the BJT in cut-off mode, can it be treated as an open circuit (or infinite resistance)?
     
    Last edited: Aug 26, 2012
  15. Aug 26, 2012 #14
    You can even delete BJT from the diagram.
     
  16. Aug 26, 2012 #15

    perplexabot

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    That simplifies a lot.

    Once again thank you so much, you have been a great deal of help.
     
  17. Aug 27, 2012 #16

    perplexabot

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    How did you solve for Vb or Ib in the second circuit?
     
  18. Aug 27, 2012 #17
    I use a circuit theory and solve the circuit.

    attachment.php?attachmentid=50231&stc=1&d=1346083224.png

    From the II Kirchhoff's law we can write

    Vcc = I1*R1 + I2*R2 (1)

    I1 = Ib + I2 (2)

    I2*R2 = Vbe + Ie*Re (3)

    And Ib = Ie/(β+1) (4)

    And we can solve this for Ib.

    [itex]\LARGE {I_b = \frac{R2Vcc - Vbe(R1+R2)}{(\beta +1)Re(R1+R2) + (R1R2)}}= 419.047619\mu A[/itex] I assume β = 99 and Vbe = 0.6V

    Knowing Ib we can easy solve for Ie. Ve and Vb

    But there is also a simpler way to solve this circuit by using thevenin's theorem.
    We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit


    Vth = Vcc * R2/(R1+R2)

    Rth = R1||R2 = (R1*R2)/(R1+R2)


    attachment.php?attachmentid=50232&stc=1&d=1346084910.png

    And now we can solve for Ib

    Vth - Ib*Rth - Vbe - Ie*Re = 0


    And we also know that

    Ie = Ib*(β +1)

    so we end up with


    Vth - Ib*Rth - Vbe - Ib*(β +1)*Re = 0


    [itex]\LARGE Ib = \frac{Vth-Vbe}{Rth+(\beta +1)*Re}[/itex]

    And this is the end.
     

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  19. Aug 27, 2012 #18

    perplexabot

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    Most of what you wrote made sense but I do have 3 questions.

    How did you get equation (1)? Is Vcc = 10V, if Yes, then I think I get it.

    I am familiar with thevenin's theorem, but I don't know how you got Vth = Vcc * R2/(R1+R2)? Once again if Vcc = 10V, then I think I understand.

    Isn't this supposed to be: Vth - Ib*Rth - Vbe - Ie*Re = 0 ?
    Or are you assuming the emitter current going from ground to transistor?

    I am sorry if I am asking for too much. You are helping me a lot.
     
  20. Aug 27, 2012 #19
    Vcc is a supply voltage Vcc = 10V in our example.

    No I made a typing error. Point for you.
    I edit my post and correct the error.
     
  21. Aug 27, 2012 #20

    perplexabot

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    Thank you.

    I have one last question for you (I hope). For the image above (top figure), did you just "copy/paste" Vin from collector to ground?
     
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