Having a really hard time with Diodes and transistors...

Discussion in 'Homework Help' started by domandrzejczuk, Nov 19, 2007.

  1. domandrzejczuk

    Thread Starter New Member

    Nov 19, 2007
    9
    0
    .l.l.l.l.l.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Your diode conducts whenever the cathode is .7 volts less positive than the anode. That should make 1b & c a snap.
     
  3. domandrzejczuk

    Thread Starter New Member

    Nov 19, 2007
    9
    0
    how would i go about solving for vout then if i dont know the current through the diode, and assumming the diode has no resistance?
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Once it's conducting, it has effectively no resistance. Treat it as a short with a .7 volt drop.
     
  5. domandrzejczuk

    Thread Starter New Member

    Nov 19, 2007
    9
    0
    so in that case, if the right most node has a v of 8 volts, then vout is simply 7.3 volts?
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
    beenthere, I think we've "been had". This is the same problem electo101 asked about last Friday. And this doesn't look like a "practice" test to me. At the top of the page it says it's due tomorrow. The first paragraph says "You may use any non-human and non-internet references to help you solve the problems" and "Naturally, you are on your honor in these respects; I am certain none of you will let me down in that department".
     
  7. mogadeet

    Member

    May 1, 2007
    19
    0
    man that's sad. you know, talking electronics might cheer us up. for instance i've got a couple of questions regarding the foregoing dialogue:

    *) "Your diode conducts whenever the cathode is .7 volts less positive than the anode."

    ok so clearly the diode conducts when anode-cathode is exactly 0.7 but what about these cases?:

    1) anode - cathode < 0.7
    2) anode - cathode > 0.7

    *) "Once it's conducting, it has effectively no resistance. Treat it as a short with a .7 volt drop."

    as i understand it resistance is (volt drop)/current. we're told the diode is conducting so presumably current must be nonzero. we're also told the volt drop is 0.7 volts which is likewise nonzero. how does this square with the device being "a short" with "no resistance"?

    peace
    stm
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
    You're not taking this same test, are you? :)

    Go read this thread:
    http://forum.allaboutcircuits.com/showthread.php?t=7922

    Does that help?

    I'm relatively new here, and I've been helping a number of people. I thought I was helping with homework, not exams. Will one of the moderators tell me, is this a big problem, namely people representing exams as homework?

    I think what I would prefer to do in such a case, is invent a new circuit, not on the exam, but similar. Then discuss that circuit, with the goal of teaching principles, not solving an exam problem.
     
  9. chesart1

    Senior Member

    Jan 23, 2006
    269
    1
    We don't know the resistance of the forward biased diode. But if you include a 0.7 volt drop across the diode in your calculations, your implicit assumption is that it has some finite resistance. The relationship between diode voltage and diode current is not linear. Therefore, the diode's resistance is not directly proportional to the current through it.

    John
     
  10. domandrzejczuk

    Thread Starter New Member

    Nov 19, 2007
    9
    0
    That doesn't really tell me much, other than the answer.
     
  11. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    Electrician,

    That is a judgement call on behalf of those responding.

    Yes, we could be responding to test questions or homework questions. One never knows.

    We could be offering design advise to those you normally wouldn't offer design advise.

    It's always a tough call, after all, this is the internet.
     
  12. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    145
    If you know the characteristics of the diode you can use the Ebers-Moll Equation to plot the I-V characteristic. The non-linear behaviour arises out of the exponential function implicit in Ebers-Moll. For a given characteristic you can calculate the conductance for a given voltage by taking the derivative of the Ebers-Moll Equation - the resistance is the reciprocal.

    Beyond 0.7V for silicon, and 0.3V for Germanium \partialI/\partialV \rightarrow \infty, therefore the resistance tends to 0.

    Dave
     
  13. GS3

    Senior Member

    Sep 21, 2007
    408
    35
    My practice is to never give the short answers but to give information which helps understand what the answer should be. I do not care whether it is homework or an exam, I do not want to do anybody's homework for them. I am willing to help them with their study and understanding but I am not directly going to do their homework (well, I can discuss a fair price ;) ) as that serves no purpose.
     
  14. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    GS3,

    At that philosphy is congruent with the posted Rules.
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
    In this case, I think we knew. Apparently you didn't see the OP's original question and attachment, and he edited his original post to delete his question and attachment. As it happens, I downloaded his attachment, and here it is. It's an exam, due today.

    As for the actual question he wanted answered, it was dealt with a couple of days ago, in this thread: http://forum.allaboutcircuits.com/showthread.php?t=7922

    The question said to take the diode voltage as .6 volts and assume it has zero resistance.
     
Loading...