Common-emitter amplifier question

Discussion in 'Homework Help' started by beijim, May 11, 2012.

  1. beijim

    Thread Starter New Member

    May 11, 2012
    3
    0
    First, sorry for my poor English.

    The question is shown below.

    diagram4c.jpg

    Here are my calculations, is that right?

    Ci) Rc:

    Because Vce=Vcc/2 , so Vcc=Vce*2

    Vcc=12V

    Rc=(Vcc-Vce)/Ic

    =(12-6)/15*10^-3

    =400Ω

    IB:

    I know IB=Ic/βDC but i don't know how to find βDC :eek: !

    Because I know RB, so I find another way :

    RB=(Vcc-VBE)/IB

    This time, I find that I don't know how to find VBE :confused: ...

    Because I cannot find IB ...I also cannot find IE and
    βDC .

    Can anyone help me?

    Finally, thanks for you help. :)
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    First of all your Vce must be equal to 6V and Vcc = 10V
    As for the Vbe simply assume Vbe = 0.6V
    And are you sure that R1 is 10K ohm ? If this is the case I'm very strange surprised.
    Since most of a BJT as a current gain greater than 100.
    And for condition as in you diagram the BJT will by in saturation.
     
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  3. MrChips

    Moderator

    Oct 2, 2009
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    You are told VCC = 10V.

    Why do you think VCE= VCC/2 ?
    This is a wrong assumption.

    You are also told that R1 = 910KΩ

    You do not need to make any assumptions about VBE. If this is small it will not affect the calculation for IB.

    Assuming VBE= 0.6V will give a better estimate of IB.

     
    Last edited: May 12, 2012
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  4. beijim

    Thread Starter New Member

    May 11, 2012
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    Thanks for your reply. :D

    I would like to know why Vce is 6V? (Sorry for my silly question, because I am a novice in Electricity. :p)

    And here are my corrections, is that right?

    Ci) Rc:

    Given : Vcc=10V, Vce=6V, Ic=15mA

    Rc=(Vcc-Vce)/Ic

    =(10-6)/15*10^-3

    =266.67Ω

    Ib:

    Rc=(Vcc-Vbe)/Ib

    910k=(10-0.6)/Ib

    Ib=10.33μA

    Cii) Ie:

    Ie=Ic+Ib

    =15m+10.33μ

    =15.01mA

    βdc:

    βdc=Ic+Ib

    =15m/10.33μ

    =1452.08
     
    Last edited: May 12, 2012
  5. Rick Martin

    Active Member

    Jun 14, 2009
    31
    2
    It all looks correct now although BETA is extremely high.

    Also Vce = 6.0V is given, why are you questioning this?
     
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  6. Audioguru

    New Member

    Dec 20, 2007
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    A transistor is NEVER biased like that because Beta is a range of numbers. Its beta might be high or it might be low and it changes when the temperature changes.

    Usually the base of a common-emitter amplifier transistor is biased from a voltage divider and the transistor has an emitter resistor.
     
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  7. beijim

    Thread Starter New Member

    May 11, 2012
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    Oh...I would like to know why Vbe=0.6V but typed Vce=6V...sorry.
     
  8. MrChips

    Moderator

    Oct 2, 2009
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    beijim likes this.
  9. Rick Martin

    Active Member

    Jun 14, 2009
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    0.6V is the voltage lost across a silicon P-N junction. Vbe = 0.6 - 0.7 is a standard assumption made.
     
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  10. mlog

    Member

    Feb 11, 2012
    276
    36
    Chips and Martin pretty much answered the question. The assumption is that the silicon transistor begins to turn on when V_{be}=0.6 and is fully saturated when V_{be}=0.8. So the active region is a voltage between those two points.
     
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  11. Audioguru

    New Member

    Dec 20, 2007
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    No.
    That is an assumption that might be true for only a few transistors.
    Vbe changes with the input current and with temperature.
    The maximum Vbe of a 2N3055 power transistor with a 4A collector current is 1.5V, not 0.8V and is more when it is cold.
    A transistor input is a current, not a voltage. Its input voltage requirement increases a little when its input current increases.
     
  12. mlog

    Member

    Feb 11, 2012
    276
    36
    Of course it is a rule of thumb and is typical of a small signal transistor. There are always exceptions to the "rule," which doesn't make what I said wrong. Thank you for your clarification.
     
  13. Rick Martin

    Active Member

    Jun 14, 2009
    31
    2
    For what Beijim is doing Vbe=0.6-0.8V is the value, there are ALWAYS variations however when learning and doing entry level small signal circuits Vbe=0.6-0.8V is accepted.
     
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