i need help to get the Inputs and Output

Discussion in 'Homework Help' started by luna, Oct 19, 2008.

  1. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    hi how are you i have a question and i tryed to solve it but i couldnt i hope you can help me
    i have a design common emitter amplifier with emitter resister and emitter bypass capacitor for the follwoing specifications :
    i have got Voltage gain is : 70
    and Lower cut off frequency: 135 Hz
    and WE take RL as 10k and quiescent operating points ,Vce=6v And Ic = 2mA

    i need to get the Input and Output impedances of the Amplifier
    Those are my questions i need some one to help me
    i tryed to take this rule :
    Vcc-RcIc-Vce=0
    Vce=Vcc-IcRc
    but we have got already the Vce i couldnt know how to figuer out
    :( PLEAS HELP ME
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Can you post a schematic of your transistor circuit so that we can better answer your question?

    This sounds like a homework question. If so, it should be in the homework section.

    hgmjr
     
  3. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    i just got those theres no schematic of the
    transistor .... yes its ahomework i put it here by mistake :S
     
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    This looks like a homework problem, so I don't want to give a full answer. However, just to point you in the right direction, consider the following.

    Input and output impedances are AC specifications, and they will not depend directly on DC variables like Vcc or Vbe (but DC variables show up indirectly as constant parameters for the Q-point). What you need to do is make an equivalent AC linearized circuit model that describes how small AC signals behave around a DC operating point (Q-point). Then you solve for the ratio of input AC voltage to input AC current to get input impedance. And, solve for the ratio of output AC voltage to output AC current to get output impedance.

    Note, this is not trivial to do, and AC impedance calculations always confuse people more than AC gain (i.e. output/input ratio) calculations, but the principles are the same.
     
    Last edited: Oct 19, 2008
  5. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    Thank you so much but can u help me in getting the ruls like what i should make step1??
    coz what u wirtten is a information but i need how i can findthe ruls like are they VBB-IBRB?
     
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Can you tell us what level class you are taking. The procedure I'm describing is typically done in a college junior level electronics design course in electrical engineering. If you are at this level, your book should describe how to determine Q-point (DC conditions) and set up an AC equivalent circuit. If you are not at this level, then perhaps you are expected just to look up some formula.
     
  7. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    am in level two i have book called (electronic device) and i tryed to search but i couldnt ok can u give me example ? for this question?
     
  8. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    any help??
     
  9. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    In the equations above You keep forgetting to add the IcRe

    Vcc-IcRc-Vce-IcRe=0
    Vce=Vcc-IcRc-IcRe

    Have you learned yet about the inherent emitter resistance "re" where re=26/Ic (ma)
    then Zin=(Beta+1) x (Re+re) ?

    And with the emitter bypass cap. depending on the Xc can sometimes eleiminate Re out of the equation?

    These are things you need to look at in solving this problem.

    If you didn't study these yet, then I'm steering you in the wrong direction. Whoops..
     
    Last edited: Oct 19, 2008
  10. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    Ops nonon we didnt study those still :D
    OMG its too hard what u written :confused: re=26/Ic (ma)
    then Zin=(Beta+1) x (Re+re) :confused: i didnt study notyet...
     
  11. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    That my be a bit more advanced, But to give you an idea of what that means, is Ib and Ic flow through the same resistor Re. That is the physical emittor resistor used to bias the emitter of the transistor.

    Because this resistor is in both the input and output currents, then because (Ic is Beta x Ib) then any signal current flowing into the base of the trans. will see two resistances, Re and re.

    First re is a inherent resistance in a diode, it is not a stable value, but it changes with respect to the current flowing through it. So the equ. to solve for this value is approx. 26 divided by Ic where Ic is given in (ma)
    Known as "shockleys relation"

    Now the (Beta+1) x all this,

    Youll learn more about this in depth, but for now juist know that a signal current into the base of a transistor will see Re as a resistance of Beta x Re. so if Beta =100 and Re=68 ohms then the signal current (ac) will not see 68 ohms but rather around 6.8K ohms for Re and even more than this when you include re as well, but at least 6.8k ohms. BUT>>>
    this may be shunted by your bias base rersistors as well as any capacitive reactance.

    When you learn this it will become more clear to you.
     
  12. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Why doesn't your teacher teach you anything??
     
  13. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    Thank you so much
     
  14. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    i have got those answers but i dont know what els ihave to get
    vcc=vce*2=6*2=12
    vre=10% of vcc =vre=1
    VRC=VCC-VRE-VCE
    12-1-6=5
    Rc=5\2=2.5
    IE=Ib=1/2.5=0.5
    Rc=26mv/2=13k
    Ro=RL//Ro
    10*2.5/10+2.5=2k
    26/2=13
    re1+RE1=RO/AV=
    2*10^3/70=28.57
    RE1=2*10^3/70-13=15.57

    THIS WHAT I GOT FROM THIS EQUATION I HOPE ANY BODY CAN HELP ME TO KNOW ELS WHAT I SHOULD DO..THANK YOU SO MUCH
     
  15. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Quoting "and WE take RL as 10k and quiescent operating points ,Vce=6v And Ic = 2mA"


    When you say Q points I assume being an amplifier it is the half supply voltage at Vc which is respect to ground.. When you have a emitter resistor Re then Vc is taken as the Q point, because it couples the output signal to the load, with respect to ground. ... without Re then Vce could then be the Q point, because the emitter is grounded.

    If by chance your looking at the emitter capacitor as coupling the emitter to ground that is for calculating Ac gain and impedances,

    But all Q point bias voltages are DC.

    This make sense????:rolleyes:

    Just trying to help clarify your question in the first post.
     
  16. luna

    Thread Starter Member

    Oct 19, 2008
    12
    0
    Thank you all for helping me :D
     
  17. fouadalnoor

    Active Member

    Oct 20, 2008
    30
    0
    sorry to point this out... though it has nothing to do with the question, but you should seriously type better, you don't seem to be able to speak or write proper English, so at least when you type something, make it clear....
     
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