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Eber-Moll Equation --- Transistors

Discussion in 'Electronic Design' started by A.J., May 10, 2004.

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  1. A.J.

    A.J. Guest

    Hello All,

    I have been reading an electronics book that shows the following
    circuit.


    +20V
    -----
    |
    |
    | +10V
    R1 (10k) O
    |_______________|
    |
    |
    Vin /
    --------| Q1
    \
    |
    |
    |
    |
    |
    |
    _____
    ___
    _

    If The base Bias is set so 1mA flows through R1 it will have 10V
    dropped accross. It then goes on to say that an 8 Degree rise in
    temperature will cause the voltage accross R1 to drop to 0.2V. I
    can't say I really unbderstand how this is working! I know that Vbe
    will drop -2.1mV/Degree but how does this affect the current flowing
    through the collector? Does someone with a better undersatnding of
    the Ebers-Moll equation know exactly what is happening here and why
    the 8 degree rise will cause this problem?

    Thanks so much for any assistance offered?

    Regards

    A.J.
     

  2. It looks as though the base bias is achieved by applying a voltage across
    the base-emitter junction. Not terribly stable.


    Graham Holloway
     
  3. A.J. wrote...
    A transistor would not be biased with a fixed temperature-invariant
    base voltage in practice, but it is an interesting student problem.

    Did your book present the Ebers-Moll equation for you to study? You
    can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
    is common (dropping a small term that doesn't matter). We often
    substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

    Using this to find the difference of two base voltages, Vbe1 - Vbe2,
    call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
    simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
    I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
    16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
    in your problem, which creates a new collector voltage of 0.57 volts.

    This analysis may be what your book had in mind, but there's a big
    problem, namely that it assumes that Is remains constant. If this
    was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
    predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
    large positive temperature coefficient, about +18%/deg, so in end we
    get a negative Vbe tempco! So while the above analysis is perfectly
    fine for a 16.8mV change in Vbe at constant temperature, it cannot be
    correct for changing temperatures. But surprisingly, if we painfully
    do the analysis the right way, we'll get roughly the same answer.

    Here's that analysis. The Ebers-Moll equation can also be expressed
    Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
    8-degree rise, if we know how Is changes with temperature. Using a
    spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
    we find that Is must increase by about +18%/deg, as I mentioned above,
    and by about 3.77x for an 8 degree change.

    Noting that VT increases by about 2.7% (from 25.25 to 25.94mV) for an
    8-degree rise (which by itself would lower Ic in the equation), we can
    calculate (using Is => 3.77x higher) that Ic increases to 1.91mA for a
    fixed Vbe bias. Coincidentally, that's pretty close to 1.94mA for the
    "incorrect" method above.

    Hmm, neither matches the book's 1.98mA. If we were to also consider
    the transistor's Early-voltage effect, the discrepancy would be a bit
    larger, and if we were to consider saturation effects, given that Vce
    is nearly zero, it would be larger yet. Oh well. :>)

    Thanks,
    - Win

    (email: use hill_at_rowland-dot-org for now)
     
  4. Active8

    Active8 Guest

    Isn't this the same circuit from AoE where as an excersize you ask
    the reader to show that for an 8 deg C rise in temp the transistor
    saturates? p. 94?
     
  5. Active8 wrote...
    Aha, you mean Exercise 2.9 in our book, page 84? Yes, perhaps so.
    Apparently, according to my calculation above, it almost saturates,
    but not quite. But I certainly didn't mean to work out one of the
    problems in the book! <sigh>

    Thanks,
    - Win

    (email: use hill_at_rowland-dot-org for now)
     
  6. Fred Bloggs

    Fred Bloggs Guest

    I read this a decrease in V(Rc) drop of 0.2V - this does work out at a
    constant Vbe bias of 0.7V to have obtained Ic=1mA, and Ieo nicely
    doubles for each 8oC- so that assuming T=300oC then VT=25.9mV and
    elevated VT=26.6mV- so that [email protected] must be 1.83E-15 making the Ie at
    308oC 2*1.83E-15*exp(0.7/26.6m)=0.98mA for an exact -0.2V change across
    the collector resistor.
     
  7. Fred Bloggs

    Fred Bloggs Guest

    VOOPS- 300oK
     
  8. Rick

    Rick Guest

    The dependence of Is on temperature is well-known, so there's no need
    to compute a value that gives you the dVbe/dT you want.

    Is(T) = Is0*((T/T0)^XTI)*exp(q*EG*(T-T0)/(k*T*T0))
    (where Is0 = saturation current measured at T0)

    plug that into the diode equation:
    Vbe = k*T/q * ln(Ic/Is(T))
    = k*T/q * ln(Ic*(T/T0)^(-XTI)*exp(q*EG*(-T+T0)/(k*T*T0))/Is0)

    Then differentiate with respect to T:
    dVbe/dT = (k*T*ln(Ic*(T/T0)^(-XTI)*exp(-q*EG*(T-T0)/(k*T*T0))/Is0)-XTI*k*T-q*EG)/(T*q)

    substitute Vbe back into that lot:
    dVbe/dT = 1/T * (Vbe - XTI*Vt - EG)

    Job done. XTI = 3 for Silicon, EG=1.11eV or 1.21eV, depending on who you believe.
    This gives about -2mV/C for a junction potential of 0.7V at 300K.

    Cheers,
     
  9. Rick

    Rick Guest

    Your 0.2V drop is due to rounding of your numbers.
    The exact answer by your method boils down to:

    2 * 1mA * exp (0.7 * q/k * (300-308)/(300*308))

    which is .990mA, giving a drop of 0.1V
     
  10. Rick wrote...
    Shheeessh, you guys! Super-accurate wrong answers!

    Thanks,
    - Win

    (email: use hill_at_rowland-dot-org for now)
     
  11. Fred Bloggs

    Fred Bloggs Guest

    Oookay- well just keep adjusting that Vbe until it comes out right.
     
  12. Rick wrote...
    Excellent, Rick, that nicely fills out the subject.

    Thanks,
    - Win

    (email: use hill_at_rowland-dot-org for now)
     
  13. Rick

    Rick Guest

    You're welcome, Win.
    Note that the equation shows that if you make Vbe large enough, say by
    pumping 100,000A through a diode-connected transistor, then the tempco
    goes positive. Spice confirms my equation, but I don't know if reality
    would agree - I'm having trouble finding a transistor beefy enough!
     
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