Single Stage BJT Common Emitter Amplifier

Discussion in 'The Projects Forum' started by NiCeBoY, Aug 20, 2008.

  1. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    hello,

    can anyone help me in the design of a simple amplifier.

    The specification of the amplifier is:

    1. should operante from a single 15v supply
    2. have a small voltage gain between 10dB to 20 dB.
    3. uses voltage divider bias
    4. has a load resistance of 10k-ohms.

    only transistors/capacitors and resistors should be used in this design..

    Thanks in advance for your kind help..

    regards,
    niceboy..
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Freq response? Output impedance? Is this homework?
     
  3. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    dont bother abt the freq. and impedance as long as it meets the specifications..

    I know the basic circuit diagrams.... but i dont know how to work this out to get those values and all.....
     
  4. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    its a project
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Hmmm, a common emitter design leaves a lot of room. You figure the Base impedance is the gain of the transistor at the freq times the emitter resistor. This interacts with the two base bias resistors, which are treated as if they are parallel to each other as far as input impedance goes.

    Have you read this section of the AAC eBook?

    http://www.allaboutcircuits.com/vol_3/chpt_4/5.html

    Can you use more than one transistor? I would be tempted to use a common collector to boost the input impedance followed by a common emitter for gain.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    By load resistance do you mean input impedance, or loading to the power supply?
     
  7. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    The load resistance means loading the power supply

    will a transistor 2n222 do the job man?
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    Yep. It has a gain of about 50 average, I'm not sure where the freq rolloff is, but I believe around 200Mhz (unity).

    Just to have something to reference to I drew up this schematic on the fly, it will give us something to use when calculating values.

    [​IMG]

    If you leave off C3 the gain will be R3/R4, within the limits of the transistor. Figure R4 is 1KΩ, R3 is 10KΩ, and R1 and R2 are something like 47KΩ and 62KΩ respectively. Adding the C3 will max out the AC gain to the transistors value if it is large enough.

    All of this is off the cuff, you really ought to go through the math on it. We'll help.

    Close enough?
     
  9. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    yes yes this help me a lot man...
    but can you help me with the transistor selection to get the gain between 10dB to 20 dB not more or less.. i think if it gonna be 15 should be great..

    sorry its the first time am trying to design something :p

    If i end up with some calculation problems will let u know,,
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    Transistors don't work like that, their gain even within a family varies. If you want a specific gain you use R3 and R4 in the formula I listed. Just to restate it...
    Gain = R3/R4.

    The input impedance is approx. R1 and R2 in parallel with each other, this lets you calculate the value for C1, it should 1/10 the value of the parallel resistance for your minimum frequency. A similar rule applies for C2. For precision gain control eliminate C3 all together.

    If you need more gain you can with either a Darlington configuration or multiple stages.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    One other thing before I call it a night. What the amp is feeding (impedance) will affect the gain somewhat. For accuracy I should have included load resistors for the input and output.
     
  12. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    man check this screenshot out for me please..is it a proper way of doing the test..
    i have not yet calculated anything..
    am just trying..:p
     
  13. thingmaker3

    Retired Moderator

    May 16, 2005
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    Looks like there is a short across R1...
     
  14. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    oh yeah thankx...
    here is another circuit without the short circuit,,
     
  15. Audioguru

    New Member

    Dec 20, 2007
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    A voltage gain of 20dB is only 10 times. A voltage gain of 10dB is only 3.16 times.
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    You've dropped your emitter resistor a lot, which will increase loading considerably. You'll also need to drop your base resistors proportionally, which will also increase loading. Try changing the ratio on R1 and R2 to drop the voltage, say 2V or less.
     
    Last edited: Aug 20, 2008
  17. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    Hello,
    still i am not able to get the good values for the resistor..
    can anyone help me out with the values to get an output between 10dB-20dB please?
    below is my cricuit..
    [​IMG]


    Thanks
     
  18. Wendy

    Moderator

    Mar 24, 2008
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    Reread my post (#10). Reverse the values for the Collector Resistor and Emitter Resistor. Why did you choose to loose my designations on the schematic I drew on post #8?

    R1 and R2 are also backwards (again, reread post #10 and #16).

    The emitter needs to be a low voltage, because the collector will be dependent on this value. Remember, gain is Rc/Re. As drawn your schematic will have a gain of 1/10, or .1, a negative gain number. Also, this amp is inverting, doesn't matter, but you need to be aware of it.

    Whatever the current you have flowing through the emitter is also the collector current. You use this to determine the voltages. On your current drawing you have a base voltage of 10.9 Volts (straight voltage divider). You can generally ignore the transistor base load, which is beta of the transistor times Re (β X Re).

    This means the emitter voltage is 10.9V - .7V, or 10.2V. The emitter current is 10.2V/12KΩ, or 0.85ma. This translates as 1.02V across Rc (1.2KΩX0.85ma), so the collector voltage is 12V-1.02, or approx. 11V.

    Let me restate the following using your values.

    [​IMG]

    R1 = 1.2KΩ, R2 = 12KΩ, R3 = 1KΩ, R4 = 10KΩ, Vcc = 12V

    Base Voltage = 10.91V
    Emitter Voltage = 10.21V
    Emitter Current = 1.02ma
    Collector Current = 1.02ma
    R3 Drop =1.02V
    Collector Voltage = 10.9V
    Gain = .1
    Max AC on Collector around 1V P, or 2V P-P



    OK, lets try following the advice I've given for biasing.

    [​IMG]

    R1 = 12KΩ, R2 = 1.2KΩ, R3 = 10KΩ, R4 = 1KΩ, Vcc = 12V

    Base Voltage = 1.091V
    Emitter Voltage = 0.39V
    Emitter Current = 0.39ma
    Collector Current = 0.39ma
    R3 drop = 3.9V
    Collector Voltage = 8.1V
    Gain = 10
    Max AC signal on Collector = approx 3.9V P, rough estimate 7.8V P-P
    Max AC signal on Base is 10X less minus a little, around .76V P-P.


    If you don't understand where the numbers came from study it a while, then ask. Since this is an AC amp (audio, RF, whatever) after you feel comfortable with the DC numbers we can go through the AC analysis.
     
    Last edited: Sep 2, 2008
  19. Audioguru

    New Member

    Dec 20, 2007
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    You must first understand how a transistor works then understand how a single transistor amplifier is biased and how to determine the values of its resistors.

    Your circuit now has a voltage gain of about 0.09 (it has a loss) and the transistor is almost saturated.
     
  20. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
    59
    0
    ah ok ;)

    I have modified the values now.. i order to get an output of approximately 5v..

    I have attached the circuit..

    I will continue to work onthis tonite..

    thanks.
     
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