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Reference Voltage Schematic

Discussion in 'Electronic Design' started by Marte Schwarz, Feb 16, 2007.

  1. Hi,
    I found a schematic for a low voltage reference in LM311 datasheets from
    National, TI and ST. It looks like:

    VCC
    |
    \
    / R1
    \ 3.9k
    /
    |
    ,-----+
    | |
    | |
    | |/c Q1
    +---| 2N3708 (Si), 2N797 (Ge) or 2N1304 (Ge)
    | |>e
    | |
    | |
    | +-------Uref
    | |
    | |
    | |/c Q2
    '---| 2N2222
    |>e
    |
    |
    |
    gnd

    followed by a LM311 as a noninverting operational amplifier (!?!)

    I want to understand this and need a little help. As I play a few with spice
    it seems to give a voltage about a few ten mV but not really like a
    reference source. What is the benefit of this schematic and how does it
    work?

    Marte
     
  2. Hello Marte,

    This circuit may be as bad as it looks nowadays.

    Prerequisite:
    A Si-transistor has a Vbe of about 0.7V with a tempco of
    about -1.8mV/degree.
    A Ge-transistor has a Vbe of about 0.36V* with a tempco of -1.xmV/degree**.

    The concept of this circuit:

    Vbe of Q2(2N2222) is at about 0.7V with a tempco of -1.8mV/degree
    regardless of Q1.
    The base of Q1 is 0.36V* below this 0.7V at 0.34V (0.7-0.36).
    The net effect is that the voltage Ve_Q1 at the emitter of Q1 is 0.34V with
    a tempco of 0mV/degree if the Vbe_tempco of both transistors is the same.

    Ve_Q1=Vbe_Q2-Vbe_Q1=0.7-0.36=0.34V

    Ve_tempco=Vbe_tempcoQ2-Vbe_tempco_Q1=-1.8mV-(-1.xmV)=-0.ymV.

    The opamp circuit around requires a potentiometer to adjust the output
    voltage because of the wide variation of Vbe.

    My advice: Don't use this circuit for a commercial application because
    Ge-transistors are dead since 30 years.
    I also have never seen Ge-transistors in SMT.

    Best regards,
    Helmut

    * Just a guess. The real value may have a wide variation in the datasheets.
    ** -1.xmV means -1.5mV or something like that. Ideally it should be equal
    the tempco of the Si-transistor.
     
  3. Hi Helmut,
    Right, all clear. With Ge transistors it is clear. With Si transistors the
    Voltage will be very low and sensible to Vcc.
    May be they wanted to tell, that LM311 may also be useable as a opamp and
    lead to its ability of common mode input below gnd.
    What's about Schotthy Diode instead of Germanium transistor? Or LED instead
    of 2N2222?

    Marte
     
  4. Fred Bloggs

    Fred Bloggs Guest

    That's getting close but not quite there. Using your notation for Q1 and
    Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance.
    Q2 must saturate and Q1 is diode connected with its Vce=Vbe,Q2 for
    practical purposes. The Vce,sat is inherently the difference between two
    forward biased PN junctions so that on that basis alone you would expect
    the tempco to be an order of magnitude below that of single diode. Also,
    when Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with
    Q1 providing negative feedback by shunting the supply current from
    driving the base of Q2 into its CE circuit. The net effect is to produce
    a Vce,sat with regulated base drive. It is not necessary for Q1 to be
    Germanium, an Si will work too but the Vout is of lower magnitude.
     
  5. ----- Original Message -----
    From: "Fred Bloggs" <>
    Newsgroups: sci.electronics.design
    Sent: Saturday, February 17, 2007 2:41 PM
    Subject: Re: Reference Voltage Schematic

    Hello Fred,

    Sorry, your assumption is wrong. The basic idea is to compensate the tempco
    of Vbe1 wth the tempco of Vbe2 by using Q2 in its active region with Vce >
    Vce_sat.
    This can only be achieved with a Germanium transistor for Q1 with a much
    lower
    Vbe than the Vbe of Q2.

    Best regards,
    Helmut
     
  6. Gibbo

    Gibbo Guest

    Why does the National data sheet show Q1 as being either Si or Ge then?
     

  7. Hello Gibbo,

    Don't believe all what's printed.

    Best regards,
    Helmut
     
  8. Gibbo

    Gibbo Guest

    I believe very little of what's printed until I see it with my own eyes.
    But I do believe the circuit does indeed operate as FB says it does.
     
  9. ----- Original Message -----
    From: "Marte Schwarz" <>
    Newsgroups: sci.electronics.design
    Sent: Saturday, February 17, 2007 12:27 PM
    Subject: Re: Reference Voltage Schematic

    Hello Marte,

    Both methods may be at least as good as the combination of a Ge-transistor
    and a Si-transistor.
    The ladder may be better because it gives a higher voltage. You have to
    "play" with LEDs
    of different color and from different manufacturers of course.
    Just try it in an oven.

    Best regards,
    Helmut
     
  10. Hi Fred,
    right here. But guess Vce,sat with a few mV in case of two Si-transistors?
    But I get the tempco with about .x mV/K at a voltage of a few ten mV instead
    of 500 to 600 mV from a simple si-diode. I would say that wouldn't be a
    real benefit then.
    At least here the simulations with LTSPICE disagrees with you. Making a DC
    sweep with V1 creates a nearly linear function of Ib(Q2)
    In my simulation it looks like that changing the more powerfol transistor to
    the upper side (Q1) and the "smaller" one (I took 2N3904 for simulations
    now) as Q2 produce more stable results. I think this stabilases Helmuts
    theorie, isn't ist?

    Marte
     
  11. James Arthur

    James Arthur Guest

    It doesn't, at least not in my copy. See http://www.national.com/ds/LM/LM111.pdf

    The circuit only makes sense if Q1 is germanium (e.g. 2n797), in
    which case Q2 robs its own base drive via diode-connected Q1. The
    output, then, is Vbe(Q2) - Vbe(Q1).

    The datasheet's "Precision Squarer" (pg. 14), and "Low Voltage
    Adjustable Reference Supply" (pg. 15) support this interpretation,
    each clearly needing a reference voltage on the order of 300mV.

    The "Precision Photodiode Comparator" (pg. 16) appears to be in
    error when it states "at comparison, the photodiode has less than 5mV
    across it..."

    As for a modern-day version, yes, you might make do with an LED and
    a silicon transistor:

    ..
    .. Vcc >-+-------,
    .. | |
    .. .-. |
    .. | | |
    .. R1 | | |
    .. '-' |
    .. | |
    .. | |/c
    .. o-----|
    .. | |>e
    .. --- +--------> U(out)
    .. LED \ / .-.
    .. --- | |
    .. | | | R2
    .. | '-'
    .. | |
    .. === ===

    Keep in mind that d(Vbe)/dT is not constant; you can adjust it by
    changing i(c), possibly achieving better cancellation / temperature
    compensation. Also, Vce(sat) has a small positive tempco, which can
    be handy too.

    Cheers,
    James Arthur
     
  12. Gibbo

    Gibbo Guest

    You're right it doesn't (I was working from memory). It't the TI one
    that shows both Qs as Si.

    http://focus.ti.com/lit/ds/symlink/lm111.pdf

    And the circuit does indeed work with both Qs as Si but with (obviously)
    a much lower voltage as FB details.
     
  13. Hi James,
    look at Page 15 on the top.
    or with a schottky as Q1 as I mentioned before.

    Marte
     
  14. It doesn't, at least not in my copy. See
    Sorry, i mised the sentence before :) But look at ST's LM311 datasheet
    there is a silicon transistor used.

    Marte
     
  15. James Arthur

    James Arthur Guest

    We agree the circuit works per Fred's description when Q1 is
    silicon, and we agree the output voltages are quite different for
    silicon versus germanium Q1. I measure Uo = 18.18mV with Q2=PN2222a,
    Q1=KTC3198, with drift of roughly 0.3% / C, line regulation = 0.9% /
    V. With Q1=2n5772, Uo = 6.88mV.

    However, the application circuits then make no sense, as this
    massive a change in reference voltage would obviously have a huge
    impact on the "Low Voltage Adjustable Reference Supply" and the
    precision waveform squarer I identified earlier. The reference
    voltage would then be only a few times the comparator's offset
    voltage, for one thing, and massively dependent on Vbe matching, for
    another.

    Since the LM311 was originally a National Semiconductor design
    ("LMxxx," after all) I conclude T.I. copied National's datasheet, that
    some helpful T.I. app engineer noticed the obsolete transistor, and
    erred in recommending a substitute Q1. Perhaps he's the same fellow
    who left the "dot" off the precision squarer circuit?, which, as
    drawn, produces no useful output!

    Cheers,
    James Arthur
     
  16. James Arthur

    James Arthur Guest


    No, at lest in this datasheet ST correctly shows the voltage
    reference application circuit (pg. 9) with a 2n1304, which is
    germanium:

    http://www.st.com/stonline/products/literature/ds/4848/lm311.pdf

    T.I. is alone in showing a silicon transistor, which doesn't work in
    their circuits for the reasons I described. What good is a "precision
    squarer" circuit with an unpredictable output?

    Best wishes,
    James Arthur
     
  17. James Arthur

    James Arthur Guest

    ............^^^^

    --James
     
  18. Gibbo

    Gibbo Guest

    You could well be right. I never said it was a good circuit, just ageed
    about it *could* (indeed does) work with silicon Qs. Well spotted re
    the "dot"!
     
  19. Sorry, i mised the sentence before :) But look at ST's LM311 datasheet
    Sorry that I did blame the wrong publisher. You're right.

    Marte
     
  20. Hi James,
    I don't see, why Helmut should be wrong here.
    did you really measurements or simulations?
    Try Q1 the more powerful transistor and Q2 the smaller ones. So Uo will be a
    little better.
    I agree, one look on the offset voltage and all the dreams are gone :)
    This may be the next topic: What is a "precision squarer" for?
    Right, I saw so many application hints in datasheets, with obvious mistakes
    ;-)

    Marte
     
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