2N2222A transistor amplifier

Discussion in 'Homework Help' started by stonecrow, Jan 4, 2007.

  1. stonecrow

    Thread Starter New Member

    Jan 4, 2007
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    0
    I have been given the task of designing a one-transistor amplifier to satisfy the following specification:

    Voltage gain: at least 20 dB at mid-band into 600 ohms
    Frequency response:
    Lower cut-off frequency: 40 Hz
    Upper cut-off frequency: 1 MHz
    Input impedance: 600 ohms ± 10%
    Output impedance: 600 ohms ± 10%
    DC supply voltage: 12 V
    DC supply current: not more than 15 mA
    Output power: not less than 2.0 mW into 600 ohms at mid-band.
    The circuit should be based on a 2N2222A bipolar transistor

    The circuit is to be fed from a source that has an output impedance of 600 ohms
    load of 600 ohms

    I have decided on the following setup

    [​IMG]

    I've done a design, but I feel several things are wrong because my resistances are tiny.

    First I assumed a quiescent input current ~ 8mA = IC (are these reasonable and how to improve on them??)
    quiescent collector Voltage ~ 7V = VC

    VCC - DC supply voltage
    VE - quiescent emitter Voltage
    VBE - voltage of base relative to emitter ~ 0.7V
    Rc - collector resistor
    Re - Emitter resistor
    RL = load resistance = 600 ohms
    RS = Source resistance = 600 ohms

    VC = VCC - IC*Rc
    giving me RC = 625 ohms - so i would use RC = 680 ohms
    VC = 6.6v

    then I considered the gain, decided on a gain of 10. Including the RL in the gain calculation:

    voltage gain = -(Rc//RL)/(Re)
    therefore
    Re = (Rc//RL)/10 = ((680^-1)+(600^-1))^-1 /10 = 31.9 ohms (this value seems very small to me, so i can tell i've done something wrong). following the calculation I would use a 12 ohm resistor because 10 is the minimum gain, therefore:

    voltage gain = -(Rc//RL)/(Re) = -(((680^-1)+(600^-1))^-1 /12= 26.6 gain

    next I try and find values for Rb1 and Rb2.

    VE = IE*Re= 0.008*12 = 0.1 V

    Biasing the base: VE + VBE = 0.1+0.7 = 0.8 V

    using the general rule Rb2 ~ 10*Re = 120 ohms

    VCC*Rb2/(Rb1+Rb2) = 0.8

    Rb1 = (Rb2*(VCC-0.8))/0.8 = (120*(12-0.8))/0.8 = 1680 ohms, use 1.5kohm resistor

    I can't help feeling I'm making silly mistakes... how do I check the amplifier meets the specification? I have spent so much time trying to understand this, I really need your help, Thank you...

    Stonecrow
     
  2. stonecrow

    Thread Starter New Member

    Jan 4, 2007
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    note that the values on the diagram are not the values I plan to use, and I only plan to use a capacitor on the input
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    What is the maximum peak-to-peak signal swing you are looking to be able to provide at the output? or put another way, What is the maximum peak-to-peak input signal you expect to occur at the input to your CE amplifier?
     
  4. stonecrow

    Thread Starter New Member

    Jan 4, 2007
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    0

    A maximum output peak-to-peak signal swing ~8V, so that would means it needs to go above and below it's quiescent value by 4V. Can someone please explain how this changes things?
    Thanks once again
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Knowledge of the output signal swing is used to make sure that the quiescent dc collector voltage is chosen so as to optimize this setpoint to minimize any distortion that would take place if the signal swings too positive or too negative. An 8 volt peak to peak signal swing at the collector should be reasonably easy to accommodate with the 12 volt dc power source you have been given.


    hgmjr
     
  6. stonecrow

    Thread Starter New Member

    Jan 4, 2007
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    0
    Ok that makes things clearer. If I have a peak-to-peak output swing of 8V, the output voltage must be able to go above and below the quiescent voltage by 4V. In order to leave a reasonable voltage across the emitter resistor I would choose the quiescent output voltage to be about 4.5V. So:

    VCC = VCC – 4.5 = 12 - 4.5 = 7.5V

    VC = VCC - IC*Rc

    IC*Rc = 4.5V

    To find Rc, I need to choose a value for the collector current (IC), and I am not really sure how. I think I could if I had the output characteristics for the 2N2222A in the form below.

    [​IMG]

    However I can’t find the output characteristics for the 2N2222A!!! Does anyone know where I can find the information, or have it already?? In short I need help deciding on IC..

    Cheers
     
  7. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Following an extensive search of the datasheet sources with which I am familiar, I came up empty handed.

    Unless someone else can locate a source of a plot of Ic versus Vce for the 2N2222A, you will just have to trudge on without that design aide.

    From the calculations that you have already provided, you appear to have a fairly good grip on the general concept.

    hgmjr
     
  8. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    Usually, the quiescent current is chosen so that the output current doesn't affect the q-point that much. As far as I know, many choose the Ic at about 10 times the Io. Larger value is better, but the power draw is bigger. This is why for substantial current output usually an emitter follower is used.
     
  9. stonecrow

    Thread Starter New Member

    Jan 4, 2007
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    0
    Thanks to everyone for the help, I'll post what I come up with from your comments. This place rocks!
     
  10. mrmeval

    Distinguished Member

    Jun 30, 2006
    833
    2
    If there is a specification you need, contact several manufacturers of that part. Then post the one that actually sends you the data so we know which one to buy. :)

    I've had very good luck with National and Dallas, if they have it they'll send it.
     
  11. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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  12. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    Just to chime in - use 2N2222's in the TO-18 package. They always outperform the 2N2222A in TO-92.
     
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