Finding relevant voltages and current in a BJT

Discussion in 'Homework Help' started by l46kok, Jan 23, 2016.

  1. l46kok

    Thread Starter New Member

    Apr 7, 2011
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    [​IMG]


    I'm supposed to find Vb, Ve, Ic, Vc and Vce for this circuit but I have never seen anything like this, so I'm lost on how to even approach it.

    Do I start from KVL and write two loops? But then there's way too many unknowns. Here's a very poor attempt at writing one:

    Top loop: -VR1 - Vbc - VRc + 20 = 0 Bottom loop: -VR2 - VRe - Veb = 0

    Do I assume the NPN transistor is in active mode and set Vbe = 0.7v? We don't know that until we find the actual relevant voltages around the transistor.

    This somewhat resembles a voltage divider, but it doesn't seem like one.

    Rough guideline is fine, I'd like to know a starting point on how to tackle this problem.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    First, do they allow a more analytic approach like using Vbe for the base emitter diode drop or do they want a numerical result like using 0.7v for the base emitter voltage, and if they do want a numerical result then is 0.7v a good enough approximation or not?

    If you use Vbe then you can just write equations, without having to actually know what Vbe is right now.
    For example, then Vb=Ve+Vbe and that's it.
    If you use 0.7, then you probably have to calculate everything numerically.
     
  3. l46kok

    Thread Starter New Member

    Apr 7, 2011
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    They want a numerical result. 0.7v is a good enough approximation.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    For your loop equation you can't just say "Vr1" since the voltage drop across a component has a polarity.

    Also, where does the "20 V" come from? There is nothing along that loop the drops 20 V. Just because the voltage of one of the nodes relative to some unrelated reference node is 20 V does not mean that anything around any particular loop drops 20 V.

    As with most diode/transistor circuits, you have to assume which state the decides are in, analyze the circuit under that assumptions, and then evaluated the results to determine if the assumption was valid.

    All of this you should have seen and become comfortable with long before being asked to analyze this circuit.

    To get an initial estimate of the answers, assume that the transistor has infinite beta. This means that no current enters the base and that the collector and emitter currents are equal. That will likely get you a result that is within a few percent of the actual results.
     
  5. dannyf

    Well-Known Member

    Sep 13, 2015
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    Assume sufficiently large beta for Q1: any reasonable transistor + Re will overwhelm R2 so we are good here.

    R1/R2 divides Vcc to determine Vb. Ve = Vb - 0.7v.

    Ve / Re => Ie ~= Ic.

    Vc = Vcc - Ic * Rc.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Wbahn:
    Not sure what you mean here. Dont we usually use the passive sign convention?

    I46kok:
    Your top loop is not formed correctly. Note that the 20v source is not in series with those other components.
    Try that again and see what you can make of it.
    Transistor circuits like this can be handled just like any other circuit with a dependent source that is reckoned to be a current controlled current source. If you have never dealt with dependent sources you will need to study those a little more.
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    WBahn is right about assuming the state of the transistor. That is the only way to define Vce.
    If I recall correctly, we usually assumed that transistor was on and saturated, and used Vce(sat) from datasheet. My textbook used 2N2222 for all these problems and the datasheet was included in the end of the book.
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Also. It is not obvious, but you actually have 3 loops.


    bias.jpg
     
    Last edited: Jan 23, 2016
  9. WBahn

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    Mar 31, 2012
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    That might be possible, except that he doesn't define the current direction. You have to define one or the other. You also have to define it consistently, so you can't just use the direction of a mesh current because the voltage needs to be defined with the same polarity even though shared meshes have opposite polarity.
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I guess for me i automatically assume a current direction based on the equation i would write, even without showing it on the schematic.
    For example, if i had a 10v source in series with three resistors R1, R2, and R3 i would probably just write:
    10=vR1+vR2+vR3

    and because it is a positive source and because i use conventional current flow the direction of current must be obvious or it would not work out to anything reasonable. So the current direction is implied using the passive sign convention. I thought everybody did it this way.

    Of course his circuit doesnt work this way because the source in not in series :)
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    If you have three resistors in series, as you describe, then you can get away with that. You could also get away with that in this circuit because the direction of current flow in each resistor is obvious -- though notice that the TS is subtracting VR1 and VRc in their first loop equation, so either they are using a mesh current or they have messed up. It's pretty much impossible to tell without being a mind reader. Had they bothered to indicate their terms on the diagram we could tell.

    Also, there are many times when it isn't obvious what the direction of the current is going to be in a branch (think of bridge circuits) and when you are doing loop or mesh equations it is important, so if you don't annotate it then you are much more likely to mess up and anyone reading your work had to become a mind reader to tell if you are doing it right.
     
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