transistor amplifier help

Discussion in 'Homework Help' started by manish1991, May 7, 2014.

  1. manish1991

    Thread Starter New Member

    Jun 17, 2013
    11
    0
    hello, I am trying to design a transistor amplifier with the following characteristics:

    Vcc=9V
    C1 = 100nF
    C2 = 100nF
    hfe = 350
    ic (quesent) = 18mA

    I have to find the resistor values of the circuit, however cannot seem to find the correct value for RL. Its probably easy but i can't seem to remember for the life of me.

    I have managed to calculate R1 and R2 values along with RE.

    My calculations:

    IB = (5.14X10^-5)A
    IC = 18.03mA

    Firstly:
    R2=(VRE+ VBE)/(10 X IB )
    R2=(1+ 0.6)/(10 X (5.14X10^-5)

    R2=3.11 KΩ

    Secondly:

    R1=(VCC-(VRE+VBE))/(11 X IB )
    R1=(9-(1+0.6))/(11 X (5.14X10^-5)

    R1=13.09 KΩ

    RE=VRE/IE
    RE=1/(18.08X10^-3)

    RE=55.31Ω

    Can anybody show me how to calculate RL as I am stumped??

    I have uploaded a circuit diagram

    Thank you

    http://i695.photobucket.com/albums/vv313/patel5557600/ScreenShot2014-05-07at132542_zpsf1242ee7.png
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,440
    3,361
  3. manish1991

    Thread Starter New Member

    Jun 17, 2013
    11
    0
    Not quite sure how to use the load line

    I used the following...

    RL = VCC/IC
    RL = 9V/18X10^-3
    RL = 500ohms

    Would I be correct?
     
  4. #12

    Expert

    Nov 30, 2010
    16,283
    6,793
    No. You've calculated, "What amount of resistance will use up the entire 9 volts?".
     
  5. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    You have given us an incomplete specification. You need to include the required gain. Otherwise there is a large number of values RL could be. At the collector we have (Vcc - Vc)/Ic = RL. Notice that without the gain specification RL is a variable.
     
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