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How is source-drain voltage defined?

  1. Aug 13, 2012 #1

    mzh

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    Dear Physics Forum users
    Is it true that conventionally, [itex]V_{SD}[/itex] is defined as the difference in potential between the drain and the source. Meaning, [itex]V_{SD} = V_S - V_D[/itex] with the voltage on the source as the reference.
    As an example: if [itex]V_{SD}[/itex] is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source. Right?

    Thanks for hints.
     
  2. jcsd
  3. Aug 14, 2012 #2

    NascentOxygen

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    To adhere to engineering conventions, VSD is the voltage of point S with reference to point D,
    i.e., VSD ≡ VS - VD
    http://img441.imageshack.us/img441/3331/nooo1.gif [Broken]

    The difference [itex]V_{SD} = V_S - V_D[/itex] reveals how many volts S is above D. Suppose S was at 6v and D was at 2v, VS - VD = +4, and you can see that's the voltage at S with reference to D.
    Thanks for asking. :wink:
     
    Last edited by a moderator: May 6, 2017
  4. Aug 14, 2012 #3

    mzh

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    Hi Oxy
    Thanks for your outline. But please explain to me, why my statement is different from yours:

    me: "If V_SD is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source"
    you: "V_SD=V_S − V_D reveals how many volts S is above D."

    I mean, it's the same to ask how many meters Paris is above sea level or how many meters below Paris is the sea level. No?
     
  5. Aug 14, 2012 #4

    NascentOxygen

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    I think you might be confusing yourself with multiple negatives here.

    If VSD is -5v, then S is -5v with reference to D. So S has the lower potential, or, equivalently, you can say D is at the higher potential. So it is equivalent to saying D is +5v with reference to S.
     
  6. Aug 14, 2012 #5

    mzh

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    yes, i think i see it now. If, conventionally and for example, [itex] V_{SD} = V_S - V_D = -5[/itex], then this means [itex]V_S < V_D[/itex], which is what you say.
     
  7. Aug 14, 2012 #6

    NascentOxygen

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    It follows that VSD = –VDS
     
  8. Aug 14, 2012 #7

    mzh

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    still i could not find this conventional definition explicitly nowhere.
     
  9. Aug 14, 2012 #8
    Can anyone help me for preparing mcq of electronics??
     
  10. Aug 14, 2012 #9

    jim hardy

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    GE Transistor Manual , General Electric Company, revised seventh edition 1969, pp 524 & 529:

    https://www.amazon.com/Transistor-Circuits-Applications-Characteristics-Edition/dp/B000U3J62U

    61AIasgbj0L._SL500_AA300_.jpg

    the way i was taught is "Your meter's red lead goes to first letter, black lead to second letter".
    But that was before autoranging digital meters, so we often had to swap our meter leads to get an upscale reading..
     
    Last edited by a moderator: May 6, 2017
  11. Aug 14, 2012 #10
    I could swear [itex] V_{DS}=V_D-V_S[/itex] !!!! Say for a N channel MOSFET, the [itex]V_{DS}[/itex] is always positive. For N FET, Drain is always higher than source. For P-Channel FET, Drain is always lower than the source, so you have [itex]V_{DS}[/itex] is always negative.
     
  12. Aug 14, 2012 #11

    jim hardy

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    for that N channel in a proper circuit:
    red lead on Drain
    black lead on Source
    meter will show VDS positive;

    and for a P channel in a proper circuit, VDS negative.
     
  13. Aug 14, 2012 #12

    NascentOxygen

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    Hi gulu. For help with exam and homework questions go to the homework forum here: https://www.physicsforums.com/forumdisplay.php?f=152 Start a new thread, include details of one question that you are having difficulty with, and show a genuine attempt at answering it. Someone is sure to come by and help you. Guaranteed. :smile:

    uzJdb.gif
     
    Last edited by a moderator: Apr 15, 2017
  14. Aug 15, 2012 #13
    Exactly.
     
  15. Aug 15, 2012 #14
    Except when it's not, e.g. when a NMOS is used as a synchronous rectifier. A MOSFET can conduct in either direction.
     
  16. Aug 15, 2012 #15

    NascentOxygen

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    It's not unknown for an OP to ask a question different from that intended, or at least to be seeming to. If that's the situation here, then mzh should feel free to come back and ask again. :wink:
     
  17. Aug 15, 2012 #16
    I am talking in general. I know you can even reverse the C and E of a BJT and it will work somewhat also.
     
  18. Aug 16, 2012 #17

    mzh

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    Thanks for the feedback on my question.
    No, getting the answer that in V_{SD}, the drain is considered the reference electrode, such that V_{SD} = V_S - V_D was most supportive.
     
  19. Aug 16, 2012 #18
    This is the first time I heard of drain being used as reference voltage!!!! BJT, MOSFET and JFET "usually" act as transconductance devices where the collector/drain is the output of the current source. Voltage is set by the load impedance and the power supply voltage. Control is usually done on [itex]V_{GS}[/itex].

    I put "usually" because there are ways to use transistor in special ways, I am talking the general amplifier working in lineal situation. I know all about that you can even turn the FET around and use the source as drain and it'll still kind of work.
     
  20. Aug 17, 2012 #19

    mzh

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    Ok, so is it [itex]V_{GS} = V_G - V_S[/itex]? My question is really only about the notational convention. Otherwise, correct sign will always be random.
     
  21. Aug 17, 2012 #20
    Yes.

    Sign is not random. For NMOSFET, [itex]V_{DS}[/itex] is "usually" positive. That is,. the Drain is higher than the Source. For PMOSFET, Drain is negative so [itex]V_{DS}[/itex] is negative.
     
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