Transistor Linear Region for CC Amplifier

Discussion in 'General Electronics Chat' started by ELECTRONERD, Aug 20, 2009.

  1. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Hey Fellows,

    I am trying to design a CC audio amplifier and wanted some help. My goal is to put the 2N3904 transistor into it's linear region and away from its saturation region. I want it to be as efficient as possible. In the attachment you'll find the basic schematic and so I'm trying to find the resistor values. R3 and R1 form a voltage divider and I'm wondering if it would be better to take out R1. Vcc=9V and the maximum gain I can provide is 30 for a current of 22mA. So based on the 2N3904 datasheet, how to I accomplish my goal?

    I'd appreciate the help!
     
    Last edited: Aug 21, 2009
  2. Audioguru

    New Member

    Dec 20, 2007
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    Your transistor is not a common-collector type and does not have a power supply connected anywhere.

    A common collector transistor is an emitter-follower that has a voltage gain of only 1 when its load resistance is fairly high. It is not a mic preamp circuit.

    You need to use a common-emitter transistor circuit. You also need to bias the electret mic.
     
  3. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Fixed. It was getting late in the night...:D
     
  4. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Audioguru,

    I chose CC due to impedance transformation but I'll take your word for CE. How do I bias the electret mic? Also, since I am now doing the CE configuration, do I take Re out or should I have Re?
     
    Last edited: Aug 21, 2009
  5. Audioguru

    New Member

    Dec 20, 2007
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    Re is needed to help set the DC operating point of your Ce transistor circuit. Re needs to be bypassed with a capacitor to avoid signal loss.

    the circuit needs a supply bypass capacitor so that the transistor does not oscillate.

    An electret mic needs about 0.5mA through a resistor (10k) fed from 3V to 9V. The voltage feeding the resistor should be decoupled with a 1k resistor and 47uF capacitor.
     
  6. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Thanks Audioguru,

    What value should this bypass capacitor be? Is there a way to calculate it or do you just pick a standard value (such as 10uF)? Could you show me the things you just explained on a schematic please (especially with the electret mic)?

    Thanks!
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    I think you will be sadly disappointed by your circuit because the input impedance of your amplifier will be an order of magnitude too low, compared to the impedance of the microphone.

    With a ceramic or crystal mike and a BJT input you need to go for some impedance raising techniques such as bootstrapping, cascode connection.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Calculation may look like this:
    Au=30V/V; Vcc=9V; Vbe=0.65V
    I pick Rc1=3.3K and Re1=RC/Au=110=100Ω
    Ic_opt=0.5*Vcc/(Rc1+Re1)=4.5V/3.4K=1.32mA
    VRe1=Ic*Re=0.13V a bit too small.
    So I pick Rc1=1K and change the schematics (add re2).
    Ic≈4.5V/1K=4.5mA
    Re1=1V/4.5mA=220Ω
    I assume that Hfe_min=100
    Ib_max=Ic/hfe=45uA
    So current that is flow through R1 and R2 mus be large then 5..30*Ib_max.
    R2=(VRe1+Vbe)/( 5..30*Ib_max)=1.64V/220uA=7.2K=10KΩ
    Idz=1.64V/10K=164uA
    R1=(Vcc-Vb)/ (Idz+Ib)=7.36V/210uA=36KΩ

    re+(re2||Re1)=Rc/Au=33Ω
    re=26mV/Ic=5.8Ω
    re2||Re1=27Ω
    re1=1 /( 1/27-1/100)=36Ω
    Ce=0.16/(F*re2)=220uF (F=20Hz)
    C1=0.16/(F*Rin)=3.3uF
    Rin=R1||R2|| [ (Hfe+1)*(re+re2||Re1)]
    CL=0.16/(F*RL=)=1uF

    [​IMG]

    I did not take into account source impedance.
     
  9. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Thanks for the help! I suppose I could use a transformer to match the impedance...
     
  10. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Jony,

    I'm not saying that anything is wrong with your design, but I could stick any transistor in place of the 2N3904 as long as it meets the requirements. You see, I want to take a look at the specs of a transistor (the 2N3904 in this case) and look at the graphs and be able to see where to put that sucker in its linear region. Could someone take me through the specs and take out what information I need and then use that to design a linear audio amplifier? I just want to be able to make that transistor as efficient as possible using external components.

    My goal is to learn how to do this and with practice I can design with transistors not only using class A amplifiers but the others as well; and not just using one transistor but more than one transistor.
     
  11. Jony130

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  12. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Hey Jony,

    I know what hFE is and Vbe but what is the other abbreviations you listed? I am not familiar with them.
     
  13. Jony130

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  14. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Well Jony, sounds like you know your stuff. Thanks for the help.

    Could someone tell me why the common-emitter configuration is best for this? I suppose it's for impedance reasons?
     
  15. Audioguru

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    Dec 20, 2007
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    The common-emitter transistor circuit with its emitter resistor bypassed has high voltage gain for use as a mic preamp and has a medium input impedance.
    But it has high distortion at high output levels and its max gain might not be high enough. An audio opamp is much better.
     
  16. Jony130

    AAC Fanatic!

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    You as a young designer don't need to know all this stuff.
    Design your circuits by a help of a Ohms law and set the voltage on collector (emitter for CC amplifier) equal 1/2*Vcc.
    And remember that is always nice to Rin>>Rs and Rout<<RL
    For example for source impedance 10K it will be good to have amplifier with 100K input impedance or higher.
    Similarly situation is with Rout. If our load impedance is 10K then output impedance of our amplifier should be 1K or lower.
    And the size of a capacitors is selected that Xc is much smaller then resistance of a cooperating resistor at the lowest frequency of operation Xc=0.1*R.
    Xc=1/(2*PI*F*C)=0.16/(F*C)
    C=0.16/(F*R)
     
    Last edited: Aug 22, 2009
  17. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Yeah, but I want to know it! :rolleyes: Thanks for your help!
     
  18. ELECTRONERD

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    May 26, 2009
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    Ok, so the capacitor in parallel with Re increases gain, correct? I believe you calculated this value with the input frequency from a function generator but what If I am using a microphone? Do I use the average voice frequency for man?

    Thanks!
     
  19. Jony130

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    Yes. Capacitor eliminates negative feedback. So
    Au=Rc/re≈ 40*Ic*Rc
    Yes, I I typically choose between 10Hz..20Hz
     
  20. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Ok, thanks for that.

    Alright, that sounds good.

    I don't seem to be thinking clearly today, my mind seems a bit slow (I think it's that darn coffee! :D) Anyway, what is Au again? I know I've used it before but I can't remember it right now! Arrrgh!
     
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