common emitter design

Discussion in 'Homework Help' started by Musab, May 25, 2010.

  1. Musab

    Thread Starter Member

    Sep 20, 2008
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    Hello,

    I have designed an amplifier that has a gain of 50db (Av=316.23). However, when I simulated the circuit using multisim, the gain was 47.8dB. This is how I calculated the circuit values (the circuit is attached):

    I started off by choosing the transistor. the hfe of the one I chose is 255.9

    DC Analysis:

    let RB = 500k

    Vcc=RBx IB + VBE ==> 30= IBx500k + 0.7 ==> IB=58.7uA

    From IB: IC=15.02mA and IE= 15.09mA

    re = 26m / IE ==> re=1.72Ω

    From AC Analysis:

    Av = Vout/Vin = Rc/r_{e} ==> Rc= 316.23xre = 545Ω


    Note: I chose this value of RB and Vcc randomly. If there is a way to find a good RB and Vcc value, pleases let me know.

    It is the first amplifier that I have ever designed so any comments, suggestions are much appreciated.


    another question: I know how to calculate the badwidth (i.e upper and lower cutt off frequencies) for a single stage and I want to know how to do that for a double stage amplifier.

    Thanks,
     
    Last edited: May 25, 2010
  2. hobbyist

    Distinguished Member

    Aug 10, 2008
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    One way to solve for RB is

    IB = (IC / beta min.)

    RB = {(VCC - Vbe) / IB}
     
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  3. Musab

    Thread Starter Member

    Sep 20, 2008
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    How can I get IB ?

    The only value that I know is Av
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Ib = Ic /β
    β = Hfe = Ic/Ib - current gain of a BJT.
    And voltage gain is equal
    Au = (Rc||rce)/re Rc/(26mV/Ic) (Rc* Ic )/ 26mV ≈ 40 * Ic * Rc
     
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  5. Musab

    Thread Starter Member

    Sep 20, 2008
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    Thanks, but I don't really understand what you are trying to say.

    Do you mean I should find Rc first from the equation Av=40 Ic*Rc , but in this equation I have two unknowns (Ic and Rc).
     
  6. hobbyist

    Distinguished Member

    Aug 10, 2008
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    As your already learning there are several ways to approach any design.

    To find a value for RC, you could choose a value for IC from the transistor data sheet, OR any reasonable value you need,
    then take 1/2 the value of VCC, for this equation.

    RC = {(VCC /2) / (IC chosen value)
    simplified

    RC = { VCC / (2 x IC) }

    That will put a voltage at the collector at around 1/2 the supply value for equal amplitude of both pos. and neg. peeks of the waveform.
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You must assume Ic or Rc and don't forget about rce=H22=hoe
    Au = (Rc||rce)/re

    If you cannot find Hoe in data-sheet we assume hoe = 100V/Ic

    And don't forget that hfe is not a constant but its change with Ic.
    For example Hfe for PN2222 in multisim for Rb=500K and Vcc=30V is equal 213.
    So for Ib = 58uA we get Ic = 12.5mA and Av = (545||8K) * 40 *Ic = 255[V/V] = 48.1dB.
    And don't forget that in simulation temperature is equal 27°C and Vt (26mV) is temperature dependent.

    The lower cutt off frequencies is equal:
    Fd1 = 1/ ( 2 * ∏ * C2 * Rin ) and Fd2 = 1/ ( 2 * ∏ * C1 * (Rout + RL) )

    Fd = √(Fd1^2 +Fd2^2)

    Rout = Rc||1/Hoe
    Rin = Rb|| (hfe+1)*re
    RL - load resistance

    The upper cutt off frequencies is equal

    Fg = 1/ ( 2 * ∏ * Cin * Rin||Rs )

    Cin = Cbe + Ccb*(Au+1)
     
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  8. Musab

    Thread Starter Member

    Sep 20, 2008
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    Thanks guys,

    @Jony130: how can I measure hfe for a given Ic in multisim? and are the frequency equations that you posted for double stage amplifier or single stage ?
     
    Last edited: May 25, 2010
  9. Audioguru

    New Member

    Dec 20, 2007
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    I was taught to never ever bias a transistor amplifier like that.
    Each transistor has a different hFE so one transistor will be saturated and the next transistor will be cutoff.
    Also a hot transistor might be saturated and a cold transistor might be cutoff.

    A transistor neeeds a fairly stable base voltage made with two resistors as a voltage divider. The transistor also needs an emitter resistor to ground as negative feedback for its biasing and to reduce its horrible distortion.

    You can also connect the base resistor to the collector for negative feedback but then the the source resistance controls the gain and distortion.
     
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  10. Musab

    Thread Starter Member

    Sep 20, 2008
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    I have used another transistor, and built the circuit again. Then, I cascaded the circuit with itself and placed a buffer between the two stages. I managed to get the voltage gain that I am looking for (100dB) and the lower cut off frequency (100Hz). But I was not able to figure out how I can modify the upper cut off frequency. any suggestions ?

    The upper cut off frequency for the circuit now is 11MHz, and I want to reduce it 10MHz.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    In school I was taught to never[/b[ simply bias a transistor like that because each transistor (with the same part number) has a different current gain (hFE) and the temperature also affects the biasing.
    Some transistors will be saturated.
    Some transistors will be cutoff.
     
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  12. Musab

    Thread Starter Member

    Sep 20, 2008
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    yeah, I know that this is a very bad biasing if implemented in practice. But I just want to know how to modify the upper cutoff frequency.
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For example you could put a small capacitor between collector and base.
    And your buffer is not properly connect to supply voltage.
    And in real life probably has been already destroyed
     
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