design of ce amplifier

Discussion in 'Homework Help' started by Renu B, Nov 29, 2008.

  1. Renu B

    Thread Starter New Member

    Nov 29, 2008
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    hey...im new around here..actually i rly rly need sum help with this...ive gotta find the general process for designing a ce amplifier....ohk...here's wat the question states
    Design a single state RC coupled CE amplifier using the transistor type BC147 to meet the following specifications:
    A >= 100, S = 10, frequency (lower cut off) <= 20 and V(out) = 3.5 V rms. Assume suitable Vcc.

    Honestly i hav no clue how to proceed with this, can someone please help me out...ive got a test next week and this is sure to be in there...please help!!!
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    I'm not too sure about the variables you stated, you might redefine them a bit using the full names. Here is a schematic to start off with, take your best guess where you think you need to start and we'll talk you through it.

    [​IMG]

    You got one of them, the low freq cutoff. What is the output impedance it will be feeding? What is the amplitude of the input?
     
  3. Renu B

    Thread Starter New Member

    Nov 29, 2008
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    well...there's no input given...its said dat the voltage gain is >=100, the stability factor is 10 and the output is supposed to be 4.5 V rms.
    Now ive got the value of Rc = 2.2 K. What i can't figure out is how to find a suitable value for Vcc....
    btw, thanks for replying...
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    First, you have to choose the bias base, emitter and collector currents. These currents affect the AC gain of the amplifier. Just to give you an idea, the greater the bias currents the greater the AC gain.

    For example, gm=Ic/25mV (approximately)

    The greater the Ic the greater will be the gm and thus the greater the AC gain (of course you cant set the bias currents too high because other problems will arise, you trade off between gain and other parameters).

    Then you have to decide what will be the quiescent emitter voltage and set the collector quiescent voltage 8 V above the emitter quiescent voltage. This is because you need an output voltage swing of 6.36 V (4.5 rms times 1.4142) plus a significant voltage drop across the transistor as not to drive it into saturation and clip the signal.

    For the choice of the power supply voltage, think that the output signal will swing 6.36V above the quiescent collector voltage plus the voltage drop across Rc. Thus you need to choose a power supply voltage greater than the maximum value of the voltages mention before. I think a 20V power supply will be enough.
     
  5. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    3.5 RMS = 10 V P-P, assuming a sine wave. This tells you 12VDC probably isn't going to be enough for the power supply, though if you had exact values it could come close. 15VDC perhaps?
     
  6. Renu B

    Thread Starter New Member

    Nov 29, 2008
    3
    0
    hey guys...i think im finally starting to understand this now....thanks a lot....uve been a great help...
     
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