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Diode with very low reverse leakage current?

Discussion in 'Electronic Design' started by smpowell, Jan 10, 2005.

  1. smpowell

    smpowell Guest

    What's a good diode (preferably some jelly bean part) to use
    when one needs an extremely low reverse leakage current?

    The reverse voltage is normally between 1.33 and 1.50 VDC. The
    diode is connected to a high resistance voltage divider that only
    has 3 uA going through it. Its only function is to protect a
    comparator (TS3V393C) from having a negative voltage applied
    to it's inputs under non-operating conditions.

    The diode (D1)is shown in the lower left hand corner of the first
    schematic at:
    http://octopus.freeyellow.com/slaveflash.html




    Stephen Powell
    Electronic Hobby Information
    http://octopus.freeyellow.com/
     
  2. A small-junction NPN transistor, base-emitter junction can
    be used to get leakage down where you appear to need it.
    Tie the collector and base together.
    In that circuit, I see no need for the diode. The current is strictly
    limited, and the comparator surely has its own input protection.
    Why do you believe more is necessary? Is there an exremely
    high voltage on the other end of that 100M resistor?
     
  3. smpowell

    smpowell Guest

    The reverse voltage is normally between 1.33 and 1.50 VDC. The
    The other end is connected to a capacitor at -300 VDC, the other end
    is connected to Vcc which is normally at +4-6 VDC, except if the
    battery is disconnected while the flash's capacitor is still charged,
    in which case, the -300 VDC is going to force the 3 uA into something.

    My guess is that the Vcc rail would go negative with respect to ground;
    as to whether the IC would withstand having 3 uA forced through it the
    wrong way I don't have clue.

    Could some familiar with the inner workings of the TS393C care to
    speculate about this?

    Hmm, would putting a reverse diode between Vcc & ground suffice?

    (Keep in mind that I learned most of my what I know about
    electronics from Forest Mims's books!)
     
  4. Jamie

    Jamie Guest

    i tihnk what you mean is a low cut off voltage.
    most diodes are good for reverse leakage for
    basic things unless your talking about RF?
    in any case i think what your looking for
    with spec's like that is a germanium type
    diode.
     
  5. According to Bob Pease in "Troubleshooting Analog Circuits" (page 66)
    it's the base-collector diode you want to use. He suggests a 2N3707 or
    2N930, although he also mentions 'some 2N3904', saying that there may be
    a problem with some because of gold doping. His claim is 1pA leakage at 7V.

    He also mentions JFET devices, in particular the 2N4117A and PN4117A,
    which apparently have very small junctions, leading to even lower
    leakage of typical 0.1pA, and guaranteed 1pA max.

    Later in that chapter, he also mentions LEDs as having incredibly low
    reverse leakages, but only if you keep them dark, and only reverse-bias
    them up to 1V.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  6. A small switching diode like a 1N4148 (can't get more jelly bean than
    that) typically leaks about 25 na with 20 volts reverse bias.
    http://rocky.digikey.com/WebLib/Diodes_Inc/Web Data/1N4148,1N4448 Diodes.pdf
    At a couple volts, I would expect the leakage to be quite a bit less.
    That would drop something like 12.5 millivolts across your 500k pot.

    If you can use a surface mount and are willing to go with less of a
    bean, you might try the MA2J116:
    http://rocky.digikey.com/WebLib/Panasonic/Web data/MA2J116 (MA116).pdf
    It has a typical reverse current of about 5 pA at 6 volts reverse and
    20 degrees C.
     
  7. With all due respect for Mr. Pease, I'm going to stick my
    neck out and disagree. Most of the leakage of a diode comes
    from surface leakage and thermally generated carriers in the
    depletion region. For a normally built BJT, the B-E junction
    has less exposed surface than the B-C junction and also has
    a smaller volume depletion region, being both narrower and
    occupying somewhat less area. So, unless one requires a
    reverse breakdown higher than the several volts available
    from a B-E junction, it is the better choice. (I am willing to
    be proven wrong on this, but it will take some evidence.)
    I suspect Mr. Pease's advice was directed toward the case
    where more than 3 to 6 Volts of reverse standoff is needed.
    I suppose those might be considered the "jelly bean"
    part the OP requested. I was trying to not overdo
    the low leakage requirement because the circuit in
    question can tolerate a few nA harmlessly.
    I almost mentioned that, but the LED forward drop would
    make it useless in the OP's circuit. There, the diode is
    supposed to conduct so that a "protected" circuit's input
    protection network does not have to source a few uA
    when gently pulled below ground.
     
  8. You can find a few clues in the spec sheet:
    http://www.st.com/stonline/books/pdf/docs/4070.pdf
    No need to speculate. The manufacturer's schematic shows
    protection diodes to each rail from each input. There is no
    harm in asking those diodes to conduct a few uA. They are
    rated to take 50 mA, so your few uA are not abusive.
    Not necessary.
    Never heard of him.
     
  9. Ken Smith

    Ken Smith Guest

    You can get "low leakage diodes" that are in fact JFET die in a two lead
    package.
     
  10. Guest

    Those would be the Vishay (Siliconix) PAD (TO-18b) and JPAD (TO-92)
    parts. Farnell stocks the PAD5 (5pA) and JPAD50 (50pA)- the former
    costs a couple of dollars in small quantities, and the latter around
    $1. There is a PAD1 part.

    http://www.vishay.com/docs/70339/70339.pdf
     
  11. Fred Bloggs

    Fred Bloggs Guest

    The diode is not needed. As you say , the flash capacitor is referenced
    to the BATT(+) rail. When the battery is removed, the capacitor will
    discharge through the BATT(+) rail, R1 , R2, and the 100M. The
    comparator terminals will be no more than 500K/100Mx300V= -1.5V maximum
    relative to the GND terminal. Since the supply voltage across the
    capacitor will be only a few tens of mV, this will not damage the IC,
    any substrate conduction will be limited to 3uA. If you reduce R2 to 50K
    , R5=1M ,R3=27K, R4=1M, then you retain the same adjustment range while
    reducing the maximum reverse discharge voltage to -150mV.
     
  12. Fred Bloggs

    Fred Bloggs Guest

    Well you know that alpha_F x Ieo = alpha_R x Ico, so that right there
    tells you that Ico is an integer multiple or two of Ieo. Pease is
    *never* wrong, so it must be that the Early effect narrowing of the EB
    depletion region causes that junction reverse saturation current to
    surpass the BC diode leakage in some exponential way with voltages
    smaller than breakdown.
     
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