BJT Transistor HELP!!

Discussion in 'Homework Help' started by de1337ed, Mar 6, 2011.

  1. de1337ed

    Thread Starter New Member

    Mar 6, 2011
    23
    0
    Hey Guys, I've been stuck on a BJT problem for a very long time, and I can't figure out what I am doing wrong. The Image is attached. In the image, I personally added in the Vo1 and the Vo2. Also, β=100
    For the problem, I need to find the operating point, and I need to find the region that the transistor is in.

    My work:
    I first found an expression for Ic and Ib, and combined them together through Ic = βIb.
    I found that Ic = (15-Vo1)/Rc
    I found that Ib = (Vs-0.6)/Rb
    Then, (15-Vo1)/Rc = β(Vs-0.6)/Rb
    And I solved for an equation for Vo1 in terms of Vs.
    I then did the same for Ie. Where Ie = (β+1)Ib.
    And I came up with another equation for Vo2
    I then took the difference Vo = Vo1 - Vo2, to find a Vo in terms of Vs. And the operation point (which i think is the same as the quiescent point), would have a Vo at (15-0.2)/2 which is 7.4V. At this point, I solved for a corresponding Vs.
    I then showed that this transistor is saturated (which i believe is what they mean when they ask for the region), because it is passed the Vs point where Vce = 0.2.

    Am I going about this correctly? Please help. When i do this, I find that the FA curve is between a very small range of values which doesnt seem correct to me. Thank you!
     
  2. Vahe

    Member

    Mar 3, 2011
    75
    9
    Yes, I think you are doing it right. Assuming forward active mode, if you calculated your base current you will get
    <br />
I_b = \frac{V_{cc}-V_{be,on}}{R_B + (\beta +1) R_E}=\frac{15-0.6}{1\text{k}\Omega + 101 \text{k} \Omega} = 141\mu\text{A}\\<br />
I_c = \beta I_b =14.1\text{mA}<br />
    So the voltage V_{o1} is
    <br />
V_{o1} = V_{cc} - I_c R_C = 15 - (14.1)(2.2) = -16 \text{V}<br />
    So the transistor cannot be in forward active region. So, it is probably in saturation. For this mode, assume V_{be,on}=0.6\text{V}, V_{bc,on}=0.4\text{V} since both junctions are forward biased. This will make V_{ce}=V_{ce,sat}=0.2\text{V}.
    You should be able to write some KVL expressions to figure out the currents and voltages in the circuit now.

    Cheers,
    Vahe
     
  3. Hagen

    Active Member

    May 8, 2010
    30
    1
    <br />
I_b = \frac{V_{cc}-V_{be,on}}{R_B + (\beta +1) R_E}=\frac{15-0.6}{1\text{k}\Omega + 101 \text{k} \Omega} = 141\mu\text{A}\\<br />
I_c = \beta I_b =14.1\text{mA}<br />

    Where did you get the 101k res?
     
  4. mjhilger

    Member

    Feb 28, 2011
    119
    16
    Vahe did some really good work for you. The 101k comes from the current gain of the base current through the emitter circuit to cause the base to the the resistor on the emitter multiplied by the Beta you supplied. It has a one added because of the original base current. (100 * 1K) + 1K.
     
  5. Vahe

    Member

    Mar 3, 2011
    75
    9
    The 101\text{k}\Omega is not a resistor in the circuit. It is (\beta + 1) times the resistor in the emitter. Sometimes (\beta + 1) R_E is referred to as the emitter resistance as seen from the base. This term occurs because if you do a KVL equation around the loop involving V_{cc}, R_B, V_{be,on} \, \text{and} \, R_E you get
    <br />
Vcc = R_B I_b + V_{be,on} + R_E I_e<br />
    and the emitter current I_e=I_b+I_c=I_b+\beta I_b=(\beta+1) I_b. Solving for I_b leads to the expression I had previously posted for I_b. I hope that clears it up.

    Cheers,
    Vahe
     
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