Actual gain is very low

Discussion in 'Homework Help' started by kvsingh21, Apr 26, 2008.

  1. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    I have just designed a common emitter transistor circuit with RC 600 ohms and Re 30, and a load resistor of 600ohms. According to calculations, the actual gain should be 20dB. ((600//600)/30)But when i do the simulation, the gain comes out to be only 12dB.
    I am also using a capacitor 26.6uF for upper cut off just before the output.

    What are the possible reasons for a low gain?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Could you sketch your CE amplifier and label all of the values so that we may see all of the details?

    hgmrjr
     
  3. mik3

    Senior Member

    Feb 4, 2008
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    This is not a common emitter configuration but an emitter follower. Can you provide us a schematic of what you have done.
     
  4. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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  5. hgmjr

    Moderator

    Jan 28, 2005
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    At what frequency are you measuring the gain and finding it to be 12dB?

    hgmjr
     
  6. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    Between 100 and 2Mhz. The bode plot shows steep slope on either side, so the cut offs are right, but the mid band gain isnt.
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    The 600 ohm resistance of the signal source Rs is loaded down by the 500 ohm input impedance of the transistor and its bias resistors. The loss is 6.5dB.

    The transistor has an internal emitter resistance of about 1.9 ohms so the gain of the transistor is 300/31.9= 8.4 times which is about 18.5db.

    18.5dB - 6.5dB= an overall gain of 12db.
     
  8. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    Thanks for you reply.

    If i reduce the emitter resistance to increase the gain, what effects will it have on rest of the circuit?
    If i use split emitter resistors (10 ohms and 20 ohms with a decoupling capacitor), why does it effect the output voltage although the total emitter resistance is effectively the same (30 ohms as originally)?
     
  9. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    Anyone ?
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    If you reduce the value of the emitter resistor then the base current becomes higher which makes the collector current higher which might cause the transistor to saturate or to clip the bottom of wavsforms.
    If you reduce the value of an unbypassed emitter resistor then in addition to the above, the input impedance necomes less.

    I don't know what you mean by "split resistors". Maybe you are using an unbypassed emitter resistor to set the AC gain and add a bypassed emitter resistor in series with it to set the total resistance for the DC bias point.
     
  11. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    Thats right mate.
    Thanks
     
  12. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    Can someone please explain this bit, i dont understand it.
     
  13. Caveman

    Active Member

    Apr 15, 2008
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    Look at your bias resistors, they have an AC load of about 500 ohms to ground. If you drive a voltage signal through a 600 ohm source resistance into a 500 ohm load, you will get 0.45*Vin at the transistor because of the loading. This is -6.8dB.

    There is an intrinsice emitter resistance that is usually negligable because it is in the range of 1-3 ohms. It depends on the bias point of the resistor (the DC emitter current). However, since you have only 30 ohms in your emitter path, this should probably be included.

    There is actually a math error here, though. The gain of the circuit is 300/31.9 = 9.4 times. People will typically refer to this in dB, so 19.5dB.

    You need to increase your input resistance. You are giving away gain. There is no reason here to have such a low input resistance.
     
  14. kvsingh21

    Thread Starter Active Member

    Apr 15, 2008
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    Very well explained Caveman.
    Thanks a lot.:D
     
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