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Where's my degree

Discussion in 'Electronic Design' started by Fred Bartoli, Apr 16, 2007.

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  1. Fred Bartoli

    Fred Bartoli Guest

    Ok, this is probably a fraction of a degree.

    Hunting for a supposed thermal offset in a circuit I set up this little
    contraption to get/verify some figures but I fail to see anything...


    +40V--+-------+----.
    | | |
    .-. |/ .-.
    adjust | |<---| R3| |2K
    clamp | | |> | |
    level '-' | '-' 100n
    | Q3 | | ||
    GND '----+-------||---+---->
    | || | 100uV/div
    |/ .-. scope input
    +25.7V >--| Q2 | |
    |> | |1G
    | '-'
    | |
    .-. GND
    Rc| |10K
    | |or short
    '-'
    |
    |
    |/ Q1 = TUT
    +5.7V >--|
    |> 20K
    | ___
    +---|___|-.
    | R2 |
    R1.-. |
    2K| | +-||
    | | ->||
    '-' +-||---< pulse
    | |
    GND GND

    Q1 is the TUT (transistor under test), Q3 is there to clamp the signal
    amplitude and avoid thermal distorsion in the scope stages.

    With Rc=short Q1 dissipates 5mW or 5.5mW depending on the pulse input state.

    Tested Q1 are some TO92 small signal devices, like BC548-9, 2N3904,
    2222, ...

    Pulse has been varied to check for thermal effects.
    Say Tp= 50ms, duty ratio = 20%.
    Looking at some moto datasheet for the BC558 we see the transient
    thermal resistance to be 0.25*200K/W = 50K/W
    500uW*50K/W*2mV/K = 50uV.

    I should see it, but nada. Down to the uV level.
    Strange...
     
  2. john jardine

    john jardine Guest

    "Fred Bartoli"
    How's that clamp transistor work?. I can see the need for one due the
    big -0.5V step but can't get my head round how it's actually clamping.
     
  3. Winfield

    Winfield Guest

    Nice drawing. Can you run some numbers for the lazy reader? We
    see a very small change in power dissipation, from 5.0 to 5.5mW,
    which we imagine won't cause much temperature rise in a device
    that can handle 500mW or more (is it 0.1 or 0.2C?). And we see
    a current sink with 5V across the CS-setting resistor(s), and a
    by-comparison microscopic change in Vbe from the aforesaid small
    change in temperature (about 0.1m/5V = 25ppm?). Just how much
    current change do you expect (about 0.1mV across the load)?

    There's gotta be a better way!

    Thanks for the diversion, hav'ta get back to the taxes!
     
  4. Snip diagram.
    20V and 2.5mA is 50mW?
    I could not find a data sheet for the BC458 (npn) with
    a transient thermal resistance graph, so is this why
    you used the BC558 (matching pnp) data sheet?

    Looking at the Semtech data sheet for the BC558,
    with a tp of 50mS and a tp/T of 50/250mS, rth-A looks
    to be about 180 K/W.
    So it looks more like a 5mW step and 180 K/W.

    Would it be better to step the dissipation by
    stepping the 25.7V on Q2's base (by 2V or so)?
     
  5. Fred Bartoli

    Fred Bartoli Guest

    Tony Williams a écrit :
    Arghh. You're right, and 250uA.20V is 5mW, so it's even worse than I
    thought.
    I used:
    http://www.onsemi.com/pub/Collateral/BC556B-D.PDF

    The way they phrase it isn't very clear, but I think the abscisse time
    is the pulse duration, not the repetition rate (which wouldn't make
    sense for single pulse BTW).

    Also, here we are not interested in peak temperature, rather in
    temperature excursion, so we have to use the single pulse curve, not the
    ones with duty ratio as parameter. This temperature excursion will ride
    on an elevated average temperature.

    From this, I read the 'excursion Rth' to be Rth_ex = 0.25*Rth = 50K/W

    I make it 5mW and 50K/W, so it's now 500uV that are missing.

    Don't know why it should be better, but it'd be less simple.
     
  6. Thanks.... might as well sing off the same songsheet.

    [snip]
    Because you are currently looking for a small slope
    after a relatively large current step (10%), plus
    the change in Q1 current must produce a small change
    in Q1's Vbe,

    Stepping the 25.7V can produce the same change in
    dissipation without having any Q1 current step, maybe
    a small front end transient only.

    It's not that difficult. Put a 2222 ohm resistor
    in series with the base of Q2, and then move the top
    of the 20k to the base. That should step Vb(Q2)
    from 25.7 to 23.13, or by -10%.
     
  7. James Arthur

    James Arthur Guest

    But only during the pulse. The measurement, AIUI, is made
    afterwards. Stepping Vce would be cleaner then, since it doesn't
    directly affect Vbe, but the error from briefly stepping i(e) isn't
    great.
    Since he's trying to measure Vbe(Q1), why not take the 'scope signal
    directly from Q1's emitter? That avoids the need for clamping (not
    knowing the clamp level, Q3 bugs me).

    And, if he then steps Vce as you suggest, Q1's emitter gives the
    error signal directly.

    Best regards,
    James Arthur
     
  8. Buggerit. You've just trashed my next post. :)
    Let's connect Q1's base to +0.7V (using a Vbe at 2.5mA)
    and the bottom end of the 2K to -5V.
     
  9. John  Larkin

    John Larkin Guest

    Yeah, that's clean:

    Vc
    |
    |
    |
    c
    +5---r1---b
    e
    |
    |
    |
    +------- scope
    |
    r2
    |
    gnd


    Step Vc from, say, +10 to +20. That will change power dissipation and
    the delta-Vbe waveform should say a lot about the junction thermal
    time constants. R1 is maybe 33 ohms, just to keep it from oscillating.

    John
     
  10. James Arthur

    James Arthur Guest

     
  11. James Arthur

    James Arthur Guest

    Sounds good. Add a small potentiometer adjustment to that base
    voltage and he can d.c.-couple the 'scope, eliminating more potential
    gremlins (from 'scope bias currents, rectified RFI, blah blah blah).

    Okay Fred, this one's whipped...next please!

    James Arthur
     
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