Emitter Follower

Discussion in 'Homework Help' started by nubelube, Feb 9, 2010.

  1. nubelube

    Thread Starter New Member

    Jan 3, 2010
    16
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    I'm reading "Practical Electronics" by Ralph Morrison. On p.99, he shows an Emitter-Follower circuit. See attached image for my rendition of it -- both voltage sources are 25V; the resistors are 10k and 2.7k; the transistor has Beta=100. The resistor is supposed to be a 2N3904 (NPN) transistor, although I used the simulator's default NPN transistor. (The graph shows the output voltage, according to the simulator. The "V=-862.05mV" refers to the simulator's calculation of voltage at Base of the transistor.)

    Here is what he says: "Assume there is no input signal connected. The emitter current is about 10mA. It is determined by the emitter resistor and the minus power supply voltage. If the Beta of the transistor is 100, the base current is 100uA. This current flows in the 10kOhm resistor and the base voltage rises to 1V. This means that the emitter voltage is actually about 0.4V."

    I do not understand why "the base voltage rises to 1V". By my calculations, it should be about -0.860V, which is reasonably close to the -0.862V that I found with the simulator. Specifically,
    (0-v_Base)/10k = i_Base = (1/101)*[(v_Base - 0.7)-(-25)]/2.7k
    This gives v_Base = -24.3/28.27 = -0.860V

    What am I missing? (Is it possible that the author mistakenly concluded that the Base goes to 1V rather than -1V?)
     
    Last edited: Feb 9, 2010
  2. jlcstrat

    Active Member

    Jun 19, 2009
    58
    3
    I think the math works more like this:

    25v/2.7k=9.593mA

    9.593mA/100(beta)=92.593uA

    92.593uA*10k=.92593V

    But, I believe you're correct regarding the sign. Just switch the probes on the meter around and you'll have -v ;)
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Actually, his math is correct. Yours is approximate.:)
     
  4. jlcstrat

    Active Member

    Jun 19, 2009
    58
    3
    I'm a student, so I'm trying to get this straight. I still can't see how I was off so much. I just realized I used 100 instead of 101, but that still doesn't lead me to .8. I've attached my solution using the original math. I still come up with roughly my original answer. Any help would be great.

    Vb/10k=1/101*([(Vb-0.7)-(-25)])/2.7k

    Vb/10k=((Vb+24.3))/272.7k

    272.2kVb=10kVb+243K

    262.2kVb=243K

    Vb=0.92677
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You missed a minus sign:

    Ib= - Vb/10k
     
  6. jlcstrat

    Active Member

    Jun 19, 2009
    58
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    Why do you have a negative Vb on one side and not the other?
     
    Last edited: Feb 12, 2010
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    656
    On the left side:
    Base current is defined as flowing into the base.
    Ib=(0-Vb)/Rb

    On the right side:
    Ie=(Ve-(-25V))/Re
    Ve=Vb-0.7
    Ib=Ie/101
    Therefore,
    Ib=(Vb-0.7-(-25))/(101*Re)

    We have 2 equations for Ib. Setting them equal to each other,

    -Vb/Rb=(Vb+24.3)/(101*Re)
    cross-multiplying,
    -101VbRe=VbRb+24.3Rb
    Vb(Rb+101Re)=-24.3Rb
    Vb=-24.3Rb/(Rb+101Re)
    Rb=10k, Re=2.7k
    Vb=-860mV
     
  8. jlcstrat

    Active Member

    Jun 19, 2009
    58
    3
    Got it...I forgot about the 0-Vb. Thanks!
     
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