Question about BJT biasing

Discussion in 'Homework Help' started by Heidi L, Mar 8, 2013.

  1. Heidi L

    Thread Starter New Member

    Nov 12, 2012
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    In the figures attached below, figure (b) is the Thevenin equivalent of (a).

    It is said that for a given value of the supply voltage V_{CC}, the higher the value we use for V_{BB}, the lower will be the sum of voltages across R_C and the collector-base junction (V_{CB}). Can you please clarify why the sum of the two voltages will be lower?

    My thought is as follows:

    For a fixed V_{CC}, I guess we can use a higher V_{BB} by fixing the value of R_2 while reducing the value of R_1, then R_B will be smaller,

    R_B = \left( \frac{R_1R_2}{R_1+R_2} \right)

    For the loop containing R_C abd R_B,

    V_{CB} + I_CR_C - I_BR_B = V_{CC} - V_{BB}

    Why can we tell that V_{CB}+I_CR_C will be smaller when (V_{CC}-V_{BB}) is smaller?
     
  2. WBahn

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    Mar 31, 2012
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    When that claim is made, what constraints are being placed on R1 and R2?
     
  3. Heidi L

    Thread Starter New Member

    Nov 12, 2012
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    It seems no constrants on R_1 or R_2.

    The original statements are attached below. Would you please tell me why (I_CR_C+V_{CE}) will be lower when using a higher V_{BB} and a fixed V_{CC}? Thank you!
     
  4. WBahn

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    Now you are saying something different. Originally you said that the sum of the voltage across the collector resistor and the voltage across the collector-base junction. Now you are talking about the voltage across the collector resistor and the voltage across the collector-emitter junction. Which is it?

    Derive an expression for the voltage you are interested in as a function of Vbb and Rb and then see if it is higher for higher Vbb and whether or not there is any constraint of Rb.
     
  5. Heidi L

    Thread Starter New Member

    Nov 12, 2012
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    Sorry, it should be the voltage of (V_{CB}\ +\ I_CR_C).

    The equation around the loop containing Rc and Rb seems to be

    -V_{CC}+I_CR_C+V_{CB}-I_BR_B+V_{BB}=0

    or, substituting \frac{I_E}{\beta+1} \ for I_B

    I_CR_C+V_{CB}=V_{CC}-V_{BB}+\left(\frac{R_{B}}{\beta+1}\right) \left(\frac{V_{BB}-V_{BE}}{R_E+ \frac{R_B}{ \beta+1 } } \right)

    The problem is, for a given V_{CC}, V_{BB} would be increased if we reduced R1 while keeping R2 constant,

    V_{BB} = V_{CC} \left( \frac{R_2}{R_1+R_2} \right)

    but then R_B = \frac {R_1R_2} {R_1+R_2} = \frac {R_2} {1+\frac{R_2}{R_1} } will be smaller.

    So how can we tell that I_CR_C+V_{CB} will be lower if we use a higher V_{BB} for a given V_{CC}?

    More importantly, what effects will it have on the biasing scheme if we make V_{BB} higher?
     
    Last edited: Mar 9, 2013
  6. Heidi L

    Thread Starter New Member

    Nov 12, 2012
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    I am so comfused about this problem. Your help will be greatly appreciated.
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    If we increase Vbb voltage the base current Ib also increases.
    Increase in base current result a corresponding increase in Ic and Ie current. So voltage drop across Rc and Re is also increase. So Vcb and Vce decreasing.
     
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  8. WBahn

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    If Vbb increases and Rb stays the same, this is true. But if you are increasing Vbb by changing one or both of the voltage divider resistors, then that may not always be the case.

    Saying that the sum of the drop across the collector resistor and Vcb will decrease is just another way of saying that Vb will increase if Vbb increases. But if there is no constraint on Rb, then this is not guaranteed. After all, I can tie Vbb to Vcc and use a base resistor sized such that the base current will decrease and get the base voltage down all the way to the 0.7V range.

    If the question truly imposes no constraint on Rb, then I think it is a poorly framed question since, to increase Vbb, I can either decrease the upper resistor or increase the lower resistor or I could increase both resistors but tilt the ratio in favor of higher Vbb.

    A useful exercise would be to come up with the relationship between Vbb and Rb in which the base voltage remains the same and then plot on that same graph the relationship between Vbb and Rb if just the top resistor is changed and if just the bottom resistor is changed.
     
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  9. Heidi L

    Thread Starter New Member

    Nov 12, 2012
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    Would you please explain it a little bit for me? I don't understand because I only know that
    -Vcc + IcRc + Vcb -IbRb +Vbb = 0, assuming that the transistor is operated in forward active mode.

    (1)--- V_{BB} = V_{CC} \left( \frac{R_2}{R_1+R_2} \right) = V_{CC} \left( \frac{1}{\frac{R_1}{R_2}+1} \right)

    (2)---R_B= \frac{R_1R_2}{R_1+R_2} = \frac{R_1}{ \frac{R_1}{R_2} +1} = \frac{R_2}{1+\frac{R_2}{R_1}}

    (3)---I_B= \frac{V_{BB}-V_{BE}}{R_B+(\beta+1)R_E}

    For a given Vcc, to increase Vbb, I choose to
    (A) increase R2 and keep R1 not changed, resulting in the increase of Rb,
    or (B) decrease R1 and keep R2 not changed, but Rb will be decreased.

    To see if it's possible that Ib will decrease when Vbb is increased, I need to check only option A by finding the derivative of Ib with respect to R2.

     \frac{dI_B}{dR_2}= \frac {(\beta+1)R_ER_1(V_{CC}-V_{BE}) + V_{BE}R_1[R_1+(\beta+1)R_E]}{ [[R_1+(\beta+1)R_E]R_2 + (\beta+1)R_ER_1 ] ^2 }

    As long as Vcc > Vbe and fordward biasing the B-E junction, the value of this derivative is positive, so I think Ib will increase if Vbb is increased by increasing R2 while keeping R1 not changed.

    Please do tell me if I made any mistakes. Thank you very much!
     
    Last edited: Mar 11, 2013
  10. WBahn

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    Mar 31, 2012
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    The voltage at the base of the transistor, relative to your ground, can be determined by starting at ANY point with a known voltage with respect to ground and then moving from their to the transistor base. So let's start from Vcc.

    Vb = Vcc - IcRc - Vcb = Vcc - (IcRc+Vcb)

    Hence, if the sum of IcRc and Vcb decreases, Vb increases.

    Your work trying to determine if it is possible for Ib to decrease if Vbb is increased is a bit off the mark because a decrease in Ib does not correspond directly to what you are looking for. Let's say that Vbb is increases and Ib decreases -- does that tell us whether Vb increased or decreased? No, because we don't know about the relationship imposed by the change in Rb. If Rb went up by just a little, then the voltage dropped by the lower Ib might be less than the previous voltage dropped across the old Rb and the Vb will increase. At some point, the increase in Vb would be enough that, even at a lower Ib, we could get a higher total voltage drop across it to compensate for the increase in Vbb and end up with the same Vb. If the increase in Rb was even more, then the lower Ib could result in a sufficiently higher voltage drop across Rb to result in a Vb that was lower than the original, despite the higher Vbb.

    Now, if you are coming at it from the other direction and saying that

    Vb = IeRe + Vbe = (1+β)IbRe + Vbe

    NOW you have sufficient justification to say that an increase in Vb requires an increase in Ib (assuming that β and Vbe do not change enough to matter).

    Assuming that your derivative is correct, then I would say your conclusion is also correct. Note that you can significantly simplify your numerator.

    You've identified two ways to increase Rb, but they aren't the only two. The third one is to increase BOTH R1 and R2 but just increasing R2 more.

    I don't know if this is turning into overkill, but I think it is a useful exercise for you. Hopefully you are getting a better appreciation for how to think about these kinds of questions and to go about find ways to answer them.
     
  11. Heidi L

    Thread Starter New Member

    Nov 12, 2012
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    Thank you so much for your time, WBahn, your explanations are helpful.
    Although I've improved my proof based on what you explained, hopefully, I still need some help with another condition regarding resistors R1 R1. English is not my native language, hope you don't have to guess what I'm writing about. ^_^
    Since Vb=Vbb-IbRb=Vcc-(IcRc+Vcb),
    so the problem of proving why we get lower (IcRc+Vcb) for a given Vcc when using a
    higher Vbb now becomes the problem about how to prove Vb will increase when Vbb is increased.

    Here's what I thought:
    Because Vbb and Rb are determined by R1, R2 and a fixed Vcc,

    V_{bb}=\left( \frac{R2}{R1+R2} \right)V_{cc} = \left( \frac{1}{ \frac{R1}{R2}+1 }\right)V_{cc}

    Rb=\frac{R_1R_2}{R_1+R_2}

    to increase Vbb, we may
    a) keep R2 not changed and use a lower R1,
    b) Keep R1 not changed and use a higher R2,
    c)
    If we find that the value of Vb after increasing Vbb is larger than the previous one, we solve the problem. I use differentiation to accomplish it, but I can't figure out how to proceed with Condition #c, could you give me a hint?

    Rearranging the loop equation
    Vbb-IbRb-Vbe-IeRe=0,
    we have

    I_b= \frac{ V_{bb}-V_{be} }{ R_b+(\beta+1)R_e}
    V_b= V_{bb} - I_bR_b= V_{be}+I_eR_e= V_{be}+(\beta+1)R_e \left( \frac{V_{bb}-V_{be}}{R_b+(\beta+1)R_e}\right) = \frac{V_{be}R_1R_2+(\beta+1)R_eV_{cc}R_2}{ R_1R_2+(\beta+1)R_e(R_1+R_2) }

    #a, differentiating Vb with respect to R1, we get

     \frac{dV_b}{dR_1}= \frac{ (\beta+1)R_eR_2[ R_2(V_{be}-V_{cc}) -(\beta+1)R_eV_{cc} ] }{ [\ [R_2+(\beta+1)R_e]R_1 + (\beta+1)R_eR_2 \ ]^2 } \ < \ 0 \ if V_{cc}>V_{be}

    d(Vb)=(dVb/dR1)(dR1)>0, which means that if we reduce R1, leaving R2 not changed, we get a larger Vbb and a larger Vb, and it's independent of the value of positive beta.

    #b, in a similar way but this time we differentiate Vb with respect to R2, we have

    \frac {dV_b}{dR_2}= \frac{(\beta+1)R_eR_1^2V_{be}+(\beta+1)^2R_e^2R_1V_{cc} }{[\ \ \ \ ]^2} \ >\ 0

    That is, if we increase R2 (hence dR2>0) while keeping R1 fixed, we get a higher Vbb and a higher Vb.

    #c, ????
     
    Last edited: Mar 13, 2013
  12. WBahn

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    Mar 31, 2012
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    I can follow your english well enough, so that's not much of an issue.

    I think you are getting a good handle on the notion and are now only missing a small piece that gets at the heart of the matter and that is that, without putting constraints are Rb, the assertion that increasing Vbb will result in an increase in Vb simply is not true.

    If you have no constraint on Rb1 and/or Rb2, then you can simply model the bias circuit as a voltage source, Vbb, in series with the Thevenin equivalent resistance, Rb. That way you only have one resistance to deal with. If you don't have any constraint on Rb1 or Rb2, you can make Vbb and Rb whatever you want them to be (subject to Vbb having to be between 0V and Vcc).

    I think what is tripping you up is that in the first two cases you know how to change your variable in such a way that you know that Vbb is going up and can then see whether the derivative of the value of Vb is positive or negative. But when you can change both Rb1 and Rb2 (or just Rb in the case of the Thevenin equivalent) you don't know whether Vbb is going up or down. You are just stepped into the realm of multivariable calculus and you need to take the total derivative of the function Vb(Vbb,Rb). What you will find, and which you can also conclude based on the chain of reasoning I went through many posts back, is that if someone tells you how much they are going to raise Vbb, you can turn around and tell them how much they need to increase Rb by in order to make Vb actually go down.
     
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