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Transistor NPN BJT 2N2222

  1. Nov 11, 2015 #1
    Hello people,

    I am working on a BJT NPN amplifier the model that I've choosen is 2N2222 : http://www.e-voron.dp.ua/files/pdf/tranzistor/2N2222.pdf

    I am having 3 problems here:

    1) am not having any AC voltage at output Vce seems like I have set wrong operating point?

    2) I am trying to figure out how to determine hoe resistance of this transistor from the graph hybrid output resistance hoe=dIc/dUce at operating point Q

    3)Is my schematics for the small signal analysis wrong?

    I hope pictures are not blur.
     

    Attached Files:

  2. jcsd
  3. Nov 11, 2015 #2
    Okay they are blur BJT_NPN_Shema.jpg
    DSC_0792.jpg
    http://postimage.org/ [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Nov 11, 2015 #3
     
    Last edited by a moderator: May 7, 2017
  5. Nov 12, 2015 #4

    rude man

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    What's your frequency and amplitude? Your RC and RT are both very low.
    forget about hoe. It's negligible.
    No idea. Depends on your software. Your schematic in post 2 does not show a signal source ...
     
  6. Nov 12, 2015 #5
    Okay forgotten to say about voltage source it's sinus with amplitude 100 mV and frequency 50Hz
     
  7. Nov 12, 2015 #6
    I know that hoe is so small but I am trying to do things atleast as correct as I can I do this for my soul.There is Ic-Uce curve in the link that I've given but it's in logaritmic paper how to I convert it from dB to Ampers and Volts?
     
  8. Nov 12, 2015 #7

    rude man

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    Due to C1 being so large you should be seeing a horribly distorted sine wave at the collector of about 200 mV.
    At the output there is very little ac since the time constant formed by RC, RT and CV2 is very low (cutoff freq. about 2 KHz).

    This is a really bad design so worrying about hoe is a total waste of time.
    What does your simulator say?







    .
     
  9. Nov 13, 2015 #8

    I have taken schematics from the internet,voltmeter shows no voltage at output and oscilloscope shows no signal at output
     
  10. Nov 13, 2015 #9

    NascentOxygen

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    For a signal frequency of 50Hz your coupling capacitors are way too small; they are attenuating the signal more than the 2222 is amplifying it. What is the reactance of a 1uF capacitor at 50Hz?
     
  11. Nov 13, 2015 #10
    it's 3813 ohms.Okay then make it 10kHz so the reactance goes down by 200
     
  12. Nov 13, 2015 #11
    Forget about this task it has very bad resistances.

    I got a new task from my professor:

    Compile common emmiter amplifier(NPN) with the following default parameters:

    VCC = 12V, power transistor 50 mW max. output signal between 3-5V, VRE = 1V.

    Determine Computing and measuring AV, AI, Rin, Rout, the upper and lower limit frequency, THD.

    Then, instead of RL connect the speaker and try something heard on the speaker.

    I have choosen transistor BC547.

    I assumed that the voltage at RL is 5V and from PRL=VRL^2 / RT I get that RL=250 ohms.I measured that Vbe=0.68V.

    I am trying to figure out what should be resitances R1 and R2



    transistor.jpg
    image upload no limit
     
  13. Nov 13, 2015 #12

    rude man

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    R1 and R2 are chosen to give you the required 1V at the emitter. Consider the base dc current also.
    The professor meant to limit your dissipation to 50 mW, not that the transistor cannot sustain more than 50 mW, probably.
    EDIT:
    I didn't see the 1V VRE requirement. So this will not be a real common-emitter circuit.
    What are your upper and lower frequencis?
    You can probaly connect the speaker in series with the collector resistor.
     
    Last edited: Nov 13, 2015
  14. Nov 13, 2015 #13

    NascentOxygen

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    The specs say Vcc = 12V, yet your calculations use Vcc of 15V. Why? :oldconfused:
     
  15. Nov 14, 2015 #14
    My bad I was really tired at evening when doing this that I didn't even notice it was 15 V.
     
  16. Nov 14, 2015 #15
    How do you mean it won't be real common emitter circuit when it has a resistor RC connected directly to the collector electrode?R1 and R2 are voltage dividers I know that they are used to set Quiescent Point.If R1 and R2 are voltage dividers of Vcc and if I want 13.35 V at R1 and 1.65 V at R2,R1 should be 8.09 times larger than R2.

    Edit(after drinking a coffee :D):
    Btw I am using BC547 for this circuit https://www.fairchildsemi.com/datasheets/BC/BC547.pdf maybe I should take some data from the datasheet and try calculating resistances.For example there is a Ic-Vce curve and I tried to make load line and then it's a piece of cake for finding the resistances.

    Load line equation is:
    Ic=-Uce/Rc+Re + Ucc/Rc+Re (Asumming that Ic=Ie)the maximum Vce voltage is Vcc which is 12 volts ,maximum current Ic will flow when Ucc/Rc+Re and therefore it's y-intercept of the curve.But still I dont know Rc and Re so therefore I can't find y-intercept...maybe I should consider a worst case scenario and take values for Vce ,Ic and Ib from the datasheet?
     
    Last edited: Nov 14, 2015
  17. Nov 15, 2015 #16

    NascentOxygen

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    If you're aiming for 50mW dissipation, and you bias for VCE of around 6-7V, then what will be the DC value of IC?
     
  18. Nov 16, 2015 #17
    Well I don't know since I am trying to figure out how to find R1 and R2 since they determine the bias IB and then IC.If I assume that Vce is around 6 V then VRC=VCC-VCE-VRE=12-1-6= 5 V but still I don't know the value of the resistance RC?
     
  19. Nov 16, 2015 #18

    NascentOxygen

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    power dissipation = voltage x current

    - which voltage?
    - which current?

    Remember that power into the base is a negligible fraction of total power, for these small signal amplifiers.
     
  20. Nov 16, 2015 #19
    I think he meant power dissipated at the output load resistance
     
  21. Nov 16, 2015 #20
    Yes power into the base is negligible but I didnt understand first part of ur post?
     
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