Hello, I need help with BJT scheme.

Discussion in 'Homework Help' started by krisio, Sep 2, 2014.

  1. krisio

    Thread Starter New Member

    Sep 2, 2014
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    I've attached an image of what my problem is. The resistors are 10k, 1k, 300 ohms, and Vcc is 5volts. I am being asked what will be Uout if Uin is as shown on the characteristic under the scheme.
    Any help will be appreciated. Thanks in advance. bjt.JPG
     
  2. Jony130

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    Feb 17, 2009
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    First you need to show us your work.
     
  3. dalam

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    Aug 9, 2014
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    The output will have a 180 degree phase shift as it is Common emitter configuration. Value of beta(Common emitter DC current gain) or hfe is required to find current Ic. Using Ic we can find gm(transconductance). And gain is~= -gm*RL(Load resistance). - sign represents 180 degree phase shift.
     
  4. ScottWang

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    Aug 23, 2012
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    The circuit is an bjt inverting circuit, so the waveform of Vout(not Uout, V is equal to Volt) will be from +5V to 0V as the square wave.
     
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  5. dalam

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    Aug 9, 2014
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    How did you determine Vo without Beta ??
    I got Vb=2.273 V and Ib=0.1573mA. Without Beta how did you find Ic?
     
  6. Jony130

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    Are you sure about that result? And we don't need to know Beta here. Because the BTJ will be saturated.
    You can prove this by assuming that Beta is larger than 10.
     
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  7. dalam

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    Aug 9, 2014
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    Yes I am sure. Vb(max) = Vin(Max)*10/11 and Ib =Vb/10k.
    Even if the BJT is saturated then Vo=5-(300*Ic).
    And using KVL : -5+300Ic+0.2(Vce sat)=0
    This results in Ic= 16mA.
    And Vo= Vce sat =0.2V.
    Also BJT does not act as amplifier in sat it is just an switch.
    Am I right here?
     
  8. Jony130

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  9. dalam

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    Aug 9, 2014
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  10. Jony130

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    Yes, if we assume Vce_sat = 0.2V. Then your answer is right. Vo = Vce_sat for Vin = 2.5V.
    But what about the case when Vin = 0V.
     
  11. dalam

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    Aug 9, 2014
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    Well in that case the BJT enters into Cut-off mode and it is an open circuit which implies Vo= 5V.
     
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