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PNP transistor and argument with my teacher

  1. Jun 17, 2012 #1

    Femme_physics

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    I just had arguments with my teacher about my test answer in class. I'm a little upset, but maybe I'm acting like a spoilt brat (I'll be the first to admit it) I went out and I'm using a hallway computer in my college so I can't post images. I still disagree with him-- But I'll try to explain it as best as I can.

    Say I have a PNP transistor. I have a source of 6 volts. I have an unkown resistor RE, and a currect Ie = 1.2 mA. I am also told that VBE = -0.7 volts. VBE is connected to the ground.

    THE FOLLOWIG IS A DRAWING OF ONLY THE LEFT SIDE OF THE CIRCUIT


    6 volts ----/\/\/\/\/\-------(EMITTER--->BASE)----GROUND

    (the jagged lines describe the resistor RE)

    Consider the right side of it (with the collector) irrelevent. Both me and my teacher did.



    So, in the solution, can I just do:

    Sum of all voltages= 6 - ReIe - 0.7 = 0
    5.3 = 0.0012 x Re
    Re = 4416.6 ohms


    BUT my teacher say I should write VBE in plus in my equation! That since it's been defined as -0.7 volts, it's a mistake to write it in minus in the equation


    But that doesn't make sense to me. Vbe is supposedly an electrical component, not a voltage source. There must be a voltage drop there, it can't act as a votlage source!
    Please help me see the light!


    PS If anyone could tell me with explanation who is wrong and who is right (his explanation that it can act as a voltage source is weird to me!) he should still be in class in the next 2 hours....otherwise I'll talk to him next week
     
    Last edited: Jun 17, 2012
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  3. Jun 17, 2012 #2

    I like Serena

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    Hey Fp! :)

    Looks to me like a misunderstanding.

    What you write looks correct to me.
    Indeed the transistor is not a voltage source and you'll have voltage drop from E to B.


    Apparently VBE is the voltage from E to B, which is negative.
    If you write it down as a formula, you should have:
    ƩV = +6 - ReIe + VBE​
    Note the plus in front of VBE.
    Is that perhaps what your teacher meant?
    Did you perhaps write this down with a minus in front of VBE?


    If you fill in the number for VBE, you get:
    ƩV = +6 - ReIe - 0.7​
    which is what you wrote down and which is correct as far as I can tell from this.


    (Btw, VBE in a similar situation with an NPN transistor would be +0.7 V.)
     
  4. Jun 17, 2012 #3
    Do you mean diagram 1 or 2 in my sketch?

    As you have described it the transistor would burn out.
     

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  5. Jun 17, 2012 #4

    Femme_physics

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    Let's stick to just PNP - how come you set it as plus, but plug it in minus? And how come in NPN it is plus?
     
    Last edited by a moderator: Jun 18, 2012
  6. Jun 17, 2012 #5
    A pn junction will conduct for Vnp > 0
    A np junction will conduct for Vpn < 0
    Hint: You will also have to think about the polarity of your power supply and about sign conventions for currents.
     
  7. Jun 17, 2012 #6

    Femme_physics

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    Diagram B
     
  8. Jun 17, 2012 #7

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    As I understand your description, you have the following diagram:

    attachment.php?attachmentid=48436&stc=1&d=1339953097.png

    Is that what you intended?
     

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    Last edited: Jun 17, 2012
  9. Jun 17, 2012 #8

    Femme_physics

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    Correct ILS, plus the the fact that current Ie flows downwards from the resistor ( as it should be ) and equals 1.2 mA
     
  10. Jun 17, 2012 #9

    Femme_physics

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    Could you elaborate please?
     
  11. Jun 17, 2012 #10

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    Okay, okay. I added the current. :)
    What about my earlier post?
     
  12. Jun 17, 2012 #11

    Femme_physics

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    I read your earlier post wholly through and replied to it. Its right under Studiot's first post
     
  13. Jun 17, 2012 #12
    OK, taking ILS diagram now makes sense

    Vbe = -O.7 means The Vb - Ve = -0.7

    that is (0) - (+0.7) = -0.7

    Which is as it should be.
     
  14. Jun 17, 2012 #13

    Femme_physics

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    @ Studio - So in KVL you plug it counts as addition to the 6 volts like my teacher proposed?
     
  15. Jun 17, 2012 #14

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    Oops. I didn't recognize it as an answer to my post. :blushing:

    This is about sign conventions.

    VBE could be either VB - VE or VE - VB.
    From your context I deduce that the sign convention your teacher is using is that:
    VBE = VB - VE

    With VB = 0 [V] and VE = +0.7 [V], it follows that VBE = -0.7 [V].

    With this sign convention, since you already know that in your formula you need -0.7, you need to plug in VBE as "+".


    Using the same sign convention, in a similar NPN circuit you have:
    VBE = VB - VE = 0.7 [V] - 0 [V] = +0.7 [V]​

    That is, Vb must be higher than Ve (contrary to a PNP transistor).
     
  16. Jun 17, 2012 #15
    If you got that, good, because my next statement may make sense and may not. Tell me if it doesn't.

    You are mixing up voltages measuresd across a component with voltages appearing at particular points in the circuit.

    The voltage across the base emitter is -0.7 volts (because the transistor is pnp)

    but the voltage at the emitter is +0.7 volts in relation to the circuit reference.

    You have to decide if you are going to use voltage drops or voltages at circuit points and then use them in a consistent manner.

    Obviously the voltage drop = difference between the voltages at two points.

    However we normally count this positive going up the circuit and negative going down.

    Are you still with me?
     
  17. Jun 17, 2012 #16
    I was referring to the fact that pnp circuits often have negative supply, so ground is the highest potential.
    And, as the others have pointed out, Vbe = -Veb
     
  18. Jun 17, 2012 #17

    Femme_physics

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    ILS - in ur first post u said I was correct, now from what I read out you say I'm wrong... Originally you said I write it in plus in KVL but plug as minus... I get your voltage explanations of VBE = vb - ve ... What i don't get is how I plug it in the formula for a KVL loop where i denote the voltage drops... Up till this day, all my voltage drops were minused UNLESS it was an opposite voltage source or the current was flowing the opposite direction. I'm still confused of whether I did right or not...or whether my teacher was correct. Ill get home in 15 mins and i could use my scanner... I wanna get to the bottom of this so i will be more thorough and post the original exercise with my full solution. Im on the bus with 15% iphone battery so it might die any moment. Talk to u guys soon from my headquarters
     
  19. Jun 17, 2012 #18

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    What you wrote in your opening post is all correct.

    I already said it might be a misunderstanding with your teacher.
    What formula exactly did your teacher disagree on?
     
  20. Jun 17, 2012 #19
    I would be interested, where have you seen the second 'convention'.

    I have only ever seen 'first minus second' in electrical circuits.

    This is then consistent with the zero referenced node voltages

    thus Vb = V at b minus zero and is positive if Vb is more positive than zero.

    I would imagine the other convention to be a nightmare to work with.
     
  21. Jun 17, 2012 #20

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    As you know, I'm not an electrical engineer.
    So I can't say for sure if the second convention is used in electrical engineering.
    I'd have to research it to find out.
    (But I find it hard to believe that all electrical engineers world wide use the same convention.)

    I do know that in different fields of science different conventions are used and also that different conventions are used in the same field.
    I have learned to always check which convention is used and not assume anything.
    Often the convention must be deduced from the context.

    As for other scientific fields with a different convention, the first that springs to mind is vector algebra:
    ##\vec{AB} = \vec B - \vec A##.
     
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