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Discussion in 'Electronic Design' started by Jim Thompson, Dec 30, 2003.

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  1. Jim Thompson

    Jim Thompson Guest

    Posted at....

    Newsgroups: alt.binaries.schematics.electronic
    Subject: Employment Test - EmploymentTest.pdf
    Message-ID: <>

    ...Jim Thompson
     
  2. analog

    analog Guest

    3 0
     
  3. Jim Thompson

    Jim Thompson Guest

    You're supposed to show your supporting equations ;-)

    ...Jim Thompson
     
  4. John Larkin

    John Larkin Guest

    By symmetry. We don't need no stinkin' equations.

    John
     
  5. Jim Thompson

    Jim Thompson Guest

    I happen to live "by symmetry".

    We don't need no education... dum... di... dum ;-)

    "The Wall", one of my all-time favorites... best played at 90+ dBm ;-)

    I also liked the DISCO era... even developed a few sub-bass schemes.

    ...Jim Thompson
     

  6. I heard a Muzak'ed version of it in a shopping mall once. Honest! Still
    chills me to think of it. Sort of a "resistance is futile - you will be
    assimilated" message.
     
  7. John Smith

    John Smith Guest


    I thot the two diodes on the left had a total of -4.4 mV/C while the Vbe on
    the right was -2.2 mV/C giving a net of -2.2 mV/C at the emitter.

    I see by the answers in the threads that I'm wrong. Will someone please
    explain why the TC is 0?

    Thanks,
    John
     
  8. Ummm.. measured from *where* to *where* ?

    Best regards,
    Spehro Pefhany
     
  9. analog

    analog Guest

    Since Jim specified infinite forward beta (current gain) the right
    side transistor does not load the left side circuitry. Therefore,
    by symmetry the center point on the left is always at exactly half
    supply (3 volts) in spite of the TC of the transistors.

    Jim also conveniently specified an Is (reverse bias saturation
    current) for the transistor base emitter junctions that when plugged
    into the diode equation, just happens to yield a 0.7 volt forward
    drop at room temperature (27C) and the 1ma operating current. Note
    that the 3k resistor on the right has exactly 3 volts across it and
    the 2.3k resistors on the left have exactly 2.3 volts across them,
    thus making all currents 1ma. Remember that base currents are all
    zero with the stipulation of infinite beta. By the way, the two
    transistors on the left are hooked as diodes since the bases and
    collectors are shorted.
    It is because the base of the right side transistor is connected to
    the base of the upper left hand transistor. Since both base emitter
    junctions experience identical temperature induced voltage changes,
    the emitters will both be exactly the same voltage below the shorted
    together bases and stay at 3 volts regardless of temperature induced
    changes in the junction voltages.

    Well, almost. There is a second order effect that results in the TC
    only being reduced by a couple of orders of magnitude, so it's not
    actually zero. The left and right hand currents don't stay exactly
    the same when the temperature changes. The right side always
    maintains the 3 volts on its resistor so its current holds at 1ma,
    whereas the left side resistor voltage changes by approximately -2mv
    per degree C due to the transistor TC. This results in a +0.87ua
    per degree C drift in the left hand side current which upsets the
    match in left and right hand side junction drops.

    In a real circuit other effects would probably mask this small
    residual drift, so, for all intents and purposes, the TC can be
    considered to be zero.

    PS to Jim: Does this makes up for my initial terse answer? :)
     
  10. I read in sci.electronics.design that Jim Thompson
    The only way I know to see this is to:

    1. set up my newsreader to subscribe to a.b.s.e in addition to what I
    subscribe to now;

    2. Download about 24 hours of posts to a.b.s.e.;

    3. Search for the PDF.

    Is there an easier way? I have a dial-up connection, if that makes any
    difference.
     
  11. Active8

    Active8 Guest

    I can e-mail it to you. Let me know.
     
  12. Or just download the headers and then the one article you need.
    It's an attachment. It goes wherever you have your newsreader puts
    such things (I'm not familiar with Turnpike).
    Yes, someone could email it to you (it's a 10K .pdf), but you
    specifically ask people not to. ;-)
     
  13. Rick

    Rick Guest

    Current down left-hand branch:
    Ie(Q2) = (3 - Vbe(Q2)) / 2.3K (1)


    CUrrent down right-hand branch:
    Ie(Q1) = (3 + Vbe(Q2) - Vbe(Q1)) / 3k (2)



    If we *assume* that the Vbe's do indeed cancel out to produce
    exactly 3V at vout regardless of the temperature, then
    the above equation simplifies to:

    Ie(Q1) = 3 / 3k (3)


    You can see that the assumption has NOT produced the same equation
    for the currents in Q1 and Q2. If the equations aren't the same
    then neither are the Vbe's, and hence Vout is NOT a nicely cancelled
    3V.
     
  14. John Smith

    John Smith Guest

    Symmetry. Yes. I overlooked that.

    Okay. Now I got it. Had to think about it.

    Thanks, Analog.

    John
     
  15. Jim Thompson

    Jim Thompson Guest

    Correct. Now write a mathematical solution for Ileft =/= Iright to
    demonstrate the variation from ideal.

    ...Jim Thompson
     
  16. Jim Thompson

    Jim Thompson Guest

    You are forgiven, my son ;-)

    ...Jim Thompson
     
  17. Jim Thompson

    Jim Thompson Guest

    In Agent, you simply double-click the Message-ID and you go right to
    it, even if you are not subscribed to that particular group. HOWEVER
    your ISP must carry the group.

    I'll E-mail you a copy.

    ...Jim Thompson
     
  18. Rick

    Rick Guest


    Why stop there? Here's a *full*, algebraic derivation of the currents
    and output voltage, courtesy Maple and Mr Lambert. As a taster,
    I've included the expression for the output voltage just below:

    Vout = 6-k*T/q*LambertW(2300*Is/k/T*q*exp((3+2300*Is)/k/T*q))+2300*Is-
    log(1/Is*k*T/q*LambertW(3000*Is/k/T*q*exp(-(-6+k*T/q*
    LambertW(2300*Is/k/T*q*exp((3+2300*Is)/k/T*q))-5300*Is)/k/T*q))
    /3000)*k*T/q;



    Did I get the job?


    Rick





    # current in left branch:
    # Kirchoff::
    eqn1 := 4600 Ie1 + 2 Vbe1 = 6

    # Let's find out what Vbe1 is by starting with the diode eqn...
    / Vbe1 \
    eqn2 := Ie1 = Is |exp(----) - 1|
    \ Vt /

    # ...and solving it for Vbe1:
    Ie1 + Is
    Vbe1 := ln(--------) Vt
    Is

    # That gives us an expression for Vbe1 in eqn1, so now let's solve eqn1
    # for Ie1:
    3 + 2300 Is
    Is exp(-----------)
    Vt
    Ie1 := 1/2300 Vt LambertW(2300 -------------------) - Is
    Vt

    # And now compute the voltage fed across to Q1's base:
    3 + 2300 Is
    Is exp(-----------)
    Vt
    Vbase := 6 - Vt LambertW(2300 -------------------) + 2300 Is
    Vt

    # compute vout now we know Vbase:
    # Vout is just a diode-drop from Vbase. I've called Q1's Vbe = Vbe2...
    eqn4 := Vout =

    3 + 2300 Is
    Is exp(-----------)
    Vt
    6 - Vt LambertW(2300 -------------------) + 2300 Is - Vbe2
    Vt

    # eqn4 is nice, but we'll need to eliminate Vbe2 from it in order to
    # solve for Vout...
    Ie2 + Is
    Vbe2 := ln(--------) Vt
    Is

    # ...that's just introduced a term in Ie2 (the current in the right-hand
    # branch), but that's easily eliminated...
    Ie2 := 1/3000 Vout

    # Now that we've got an expression for Vout that's independent of Ie2,
    # let's go back to eqn4...
    (3 + 2300 Is) q
    Is q exp(---------------)
    k T
    k T LambertW(2300 -------------------------)
    k T
    6 - -------------------------------------------- + 2300 Is - ln(
    q

    /
    |
    |
    |
    |
    1/3000 k T LambertW(3000 Is q exp(- |
    \

    / (3 + 2300 Is) q \
    | Is q exp(---------------) |
    | k T |
    | k T LambertW(2300 -------------------------) |
    | k T |
    |-6 + -------------------------------------------- - 5300 Is|
    \ q /

    \
    |
    |
    |
    |
    q|/(k T))/(k T))/(Is q)) k T/q
    /
    -14
    Is := .176 10

    # Compute Vout at a couple of temperatures as a sanity check:
    2.997588746
     
  19. Mac

    Mac Guest

    There are a lot of different news readers out there. On mine I can sample
    a specified number of headers for a new group (i.e., one I don't subscribe
    to.) Even so, I don't normally follow any of the links to a.b.s.e, but
    that's just because I am terminally lazy. ;-)

    Mac
     
  20. I read in sci.electronics.design that Active8 <[email protected]
    I've subscribed temporarily to a.b.s.e. and now have 106 more articles
    to scan!

    Thanks for the offer. I am looking for a permanent better method of
    getting stuff from a.b.s.e., if there is one.
     
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