Am I insane? v.BJT biasing

Discussion in 'Homework Help' started by Six_Shooter, Mar 11, 2013.

  1. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    In lab this week, we need to calculate several parameters of a biased BJT amplifier circuit.

    Before going any farther I'm going to preface this by saying that the lab is titled as a voltage divider design, and the related text book reading is also about voltage divider biased BJT amplifier. BUT in last week's lab the emitter biased circuit was labeled as a voltage divider biased circuit, and in this week's lab there is only a value of Rb, not Rb1 and Rb2. There is also no value of β given, which makes it impossible (from what I can find) to calculate Rb1 and Rb2, anyway. Formula I have for Rb2 is ((β+1)Re)/10, then Rb1 calculated from there.

    Now, here's the greater confusion, each of us has been given our own values to use in this lab, and mine are these:

    Vcc: 18 V
    Vc: 8.07 V
    Ic: 14.6 mA

    Now assuming Vce is Vcc/2 = 18V/2 = 9V

    Vrc (voltage across Rc) Vcc-Vc = 18V - 8.07V = 9.93 V

    Ve = Vcc - Vrc - Vce = 18V - 9.93V - 9V = -.93V

    Now obviously Ve cannot be negative, so is there something I am missing, or I haven't found yet?

    I have sent an e-mail to my teacher, but I don't expect to hear back until morning at the earliest, and want to get this done before I need to show up in lab.

    Thanks
     
    Last edited: Mar 11, 2013
  2. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    Why can''t Ve be negative? You haven't provided any schematic for us to look at, so for all we know there is an emitter resistor going to a negative supply.

    We are NOT mind readers!

    I.m guessing you have a single supply circuit (Vcc and Gnd), but why is it that people that you are asking for help have to guess at what the circuit is that you are using?

    You are "assuming" (*your* word) a particular Vce and then don't like the resulting value for Ve. Well, why did you assume that particular value for Vce?
     
  3. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    You've clearly made a bad assumption about VCE. If VC is already down to 8.07V, then you don't have enough volts left for VCE=9V.
     
  4. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    The closest I can find for a bias circuit design is this:
    [​IMG]

    The problem is that we have not been provided a schematic, just some values and a title of the lab called voltage divider that would include Rb1 and Rb2 values, which have not been said to be values for this lab, only Rb.

    Single voltage source, 18V Vcc, no negative sources, nothing special. One transistor (I am assuming 2N4404 since that's what we used last week), 3 resistors (Rc, Rb and Re), again, because of other aspects of the lab write up, and naming that has been used over the last couple of labs.

    Our pre-lab calculations are supposed to be to find the values of Rb, Rc, Re, Ve, Ib, to be able to design this circuit using E12 standard resistor values.

    So given a single source, the negative value of Ve is an impossibility, at least from the standpoint of what we have been taught so far, and I can't find any information that says otherwise. This would put the transistor in a reverse bias mode and would not act as an amplifier then, and would require another power source.

    Since I have no β and no specified value for Vce, we were told in theory class to use a Vce of Vcc/2.

    If there is something I'm missing, please tell me, but this seems to be a circuit that will not work. I can get values that make sense if I reduce VCC to 16, but that's not my assigned values.

    On a sorta related note, is there a way to have subscript show up when posting messages?
     
  5. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    But you know good and well you can't get 9V across the transistor when you only have 8V margin. This rule you have to break.
     
    Six_Shooter likes this.
  6. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    Right...

    I hadn't looked at it that way, but yes that makes sense.

    I have no other way of determining values in this circuit. Other than the exact reading from the text book, all value and information has been posted.

    Is there some other way to go about this, other than to assume 0V accross Re (no resistor), and assume Vbe of the 8.07V? This would make this a fixed bias circuit.

    This would also give a value of Vb of .7V, and a Vb of 1.4V...
     
    Last edited: Mar 13, 2013
  7. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    Texting a couple of my classmates, they are equally confused.
     
  8. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
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    When given values as you showed above, you made a wrong assumption on the VCE value, in a practical transistor amp stage, you would like to assume a value of VCC/2 at the collector terminal, (VC), since you were given a VC value, you can design your circuit with those parameters, which will give you a value of VCE depending on what bias resistors you choose to use.

    Your challenge in this is to calculate values for your bias resistors, that will allow you to have the voltage VC, and the current IC working linearly in the amp stage.

    You will find that there are many different design values for each resistor for any given voltage and current parameters.

    next post I'll give an example of choosing bias resistors.
     
  9. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    Yeah, I see that now, and should have seen that long before. lol

    I'm looking forward to your next post.
     
  10. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Given your values for VC and IC as well as VCC, you can solve for RC using your formulas for solving it.


    you can choose a value for RE by trying voltage values and comparing them to stage gain values.
     
  11. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    Yes, I already have Rc, but after that, I tried to find Ve, in the ways I've done in previous examples and that's where the confusion set in. I have Rc of 630Ω, E12 value being 680Ω.

    So you're saying to basically arbitrarily try values until I get some that work? I'm unfamiliar with what you mean by "stage gain values". Are you referring to Av (Voltage gain between input and output)?
     
  12. hobbyist

    Distinguished Member

    Aug 10, 2008
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    now choose a value for VE, or choose a stage gain, example,

    Av= 10, that woiuld make RE = to 68 ohms.
    which makes VE = approx. 1v.
     
  13. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    VE ~= IC*RE < 8.07V - VCE
    => RE < (8.07 -VCE)/IC

    VCE must be > .2V for active.

    Upper limit RE = (8.07 - .2)/IC
    Lower limit RE = 0

    Pick your poison. Personally, I'd decide what I need for gain, and choose RE = RC/AV and go from there.
     
  14. hobbyist

    Distinguished Member

    Aug 10, 2008
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    or choose a value for VE, example, make VE = 3v.
    RE now becomes to 200 ohm resistor, nom. value being used.
     
  15. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Then solve for VE + Vbe, to get your voltage at the base VB.
     
  16. hobbyist

    Distinguished Member

    Aug 10, 2008
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    than choose a value for Rb1 base to ground resistor, to be any where from 10 to 20 times RE for a starter, then rework it for input impedance requirements and such as needed later on in the design.
     
  17. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Rb2 is solved using the equations (VCC - VB) / (VB / RB1)

    Simply divide the VB by your choice of RB1 and that will give the divider current,
    then subtract the VB value from VCC to give the voltage across RB2, then use that value to divide by divider current to solve for RB2 value resistor.
     
  18. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    Hmm, in any case of just choosing a value for Ve or Re, that makes the Q-point move away from center, which it has to anyway, looking at the parameters given.

    I'll try that and see what I come up with.

    It seems I'm not insane, just didn't want to assume that selecting my own value for Vce (by selecting a value for Ve), would be acceptable practice.
     
  19. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Then you have some bias going on wioth your stage, now you need to rework values to match input and output impedances and any other parameters that come into the picture.
     
  20. Six_Shooter

    Thread Starter Member

    Nov 10, 2012
    33
    1
    There have been no parameters of input or output impedance given, only the values and parameters that I have posted in the first couple posts of this thread. The biggest restriction on resistors is that they are E12 standard.

    Thanks
    :)
     
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