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Help understanding Darlington CB leakage current and its ramifications

Discussion in 'Electronic Basics' started by [email protected], Jul 21, 2005.

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  1. Guest

    Just when I think I'm advancing as an engineer I come across something
    that is simple in theory but completely stumps me. I am going through
    a product design review. I am stuck on a comment a senior design
    engineer made.

    Sorry for the lack of ASCII artwork depicting my circuit but I don't
    know how to do that, nor can I read circuits others have posted in
    their threads. They look garbled. I'll try to describe this verbally.

    I have a circuit that operates off a 13.5V supply. I am switching a
    relay coil on and off with a FMMT614 Darlington NPN transistor acting
    as a low side driver. One side of the relay coil connects to the 12V
    supply, the other side connects to the transistors collector. The
    emitter is grounded. The base is pulled to ground through a 100K ohm
    resistor. Base current is supplied by a microcontroller output port
    through a 10K ohm resistor. I think that's about it for the time
    being. Simple I hope.

    Oh yeah, I have a back to back zener/rectifier diode clamping network
    across the relay coil to suppress transients to a safe level protecting
    the transistor.

    This circuit must operate from -40C to 85C. It must survive
    environments as high as 125C without damage.

    I thought this was a no-brainer. That's when my mentor suggested I
    check out the collector to base leakage current of the transistor. He
    did not elaborate any further.

    I have two problems:

    1.) Why did he say this?
    2.) This data is not furnished in the data sheet nor in any
    application note I found so far.

    I assume he asked me to check this since the Darlington is a high
    current gain device. A small amount of leakage C->B can turn into a
    substantial current flow C->E. Could the device turn itself on and
    activate the load!

    I've read a few threads posted here through the years that argue where
    C-B leakage current goes. Into the base load (pulldown), into the B->E
    junction, both? That I'm not clear on. I don't mean to open that can
    of worms again but feel I must. Can any amount of C->B leakage current
    flow into the B->E junction? My thoughts are that enough leakage
    current has to flow to develop a proper bias voltage across the base
    pull down resistor. If this voltage approaches/exceeds the turn on
    requirement of 1.5V for this transistor base current will flow. If the
    amount of C->B leakage current does not create a high enough voltage
    drop across the base pull down transistor (arbitrarily say 300mV) then
    no current will flow B->E. Is this hypothesis flawed? Can C->B
    leakage current sneak right through to the B->E junction regardless
    without voltage drop developed across the base pull down resistor?

    Another problem. How do I characterize the leakage current? It is not
    published in the data sheet. I wrote the manufacture (ZETEX) regarding
    this issue but their support team in India (being sarcastic here) has
    not responded.

    I hear leakage current increases with temperature and probably varies
    depending on the potential across VCE when the device is off. But,
    what are the hard numbers? I would set up an experiment to check this
    out but do not have the necessary equipment to measure currents this
    low in magnatude.

    One might ask: Why not go to your mentor and ask him to elaborate.
    Well, a little bit of this has to do with pride. He typically asks
    such questions hoping you will go out, perform the research, and come
    back to him with the answer. If you need to go to him again without
    the answer it it like admitting defeat. Even if I set pride aside,
    there is the fact that he is extremely busy. He probably has 20 design
    reviews going at one time. I don't want to be a nuisance. I'd rather
    try and iron things out here first.

    That is if you don't mind helping.

    Thank you

    George W. Marutz II
     
  2. John Larkin

    John Larkin Guest

    Those resistor values sound a little high to me, for no extremely
    rational reason.
    All sounds right to me.
    Well, if the leakage is there, it's real collector current, but it's
    likely small. If your base-ground resistor diverts most of the leakage
    to ground, the collector leakage will *not* be multiplied by beta,
    which could make it a lot bigger.


    Measure a few parts?

    Assume it doubles every 10 degrees C.

    Most any handheld DVM can measure nanoamps... on its *voltage* range.
    Just measure the voltage drop across a 1M resistor (1 mv per na) or
    even across the meter's internal resistance, usually 10 megs. A Fluke
    and a 9-volt battery and a couple of clip leads will quickly measure
    Iceo, Icbo, and Icex.


    He sounds OK to me. And he's likely aware that some darlingtons,
    especially the bigger ones, can be very leaky.

    John
     
  3. PeteS

    PeteS Guest

    Some quick thoughts, without going in to any real depth.

    I pulled the data sheet, and it's a little sparse - I assume (but do
    not know) that there is no internal resistor (pull from input device
    base to lower emitter) ; Assuming that to be so, then it won't affect
    other things.

    The base in a darlington, is the base of the actual top collector
    device. As (when off, anyway) a reverse biased PN junction, there will
    be leakage current which increases linearly with temperature, and can
    usually be assumed to be roughly equal to that of an ordinary bipolar
    device (unless other guidance is given). (When on, it's more like a low
    quality current sink towards the emitter).

    For typical PN reverse leakage, I suggest you look at some PN diode
    data sheets (they usually specify such things).

    I would be wary of a 100k pull down on *any* bipolar device - you
    wouldn't need much leakage to give you significant problems. I usually
    reserve such values for FETs.

    You may want to reduce that to more like 10k (or even less). That will
    still be the vast majority of the current drawn from your processor
    output pin (the minimum current transfer is 15,000 for the device).

    Others here will no doubt weigh in :)

    Cheers

    PeteS
     
  4. Reading ASCII drawings requires that you set your newsreader to a
    fixed width (per character) font, like Courier. This puts the
    characters on a Cartesian grid.
    Clear enough.
    He probably got bit on the ass for neglecting this effect at one time.
    If you have no data on the input transistor's collector to base
    leakage versus temperature, how do you know that your 100k resistor
    will drain the leakage without the voltage building up high enough to
    keep the darlington from being turned on by it?
    There are two transistors in a darlington. The first one can be turned
    on by its collector to base leakage, and then the second one is driven
    by that leakage current times the beta of the first one. The second
    device can also be driven directly by its own collector to base
    leakage. Many darlingtons contain an internal resistor base to
    emitter of the second transistor to handle the second case, since you
    rarely have access to this base.
    Almost all of the CB leakage will pass through the resistor, till the
    base emitter voltage of that device gets high enough to start pushing
    current through that junction. That voltage may be as low as .3 volts
    at high temperature. Then that base current is amplified by the
    transistor beta and is applied either to the second base emitter
    junction or to the parallel combination of that junction and any
    internal base emitter resistor.
    Yes, it is the voltage that determines how much current passes through
    the junction.
    I don't think so.
    (snip)

    I would perform and experiment on several devices from different
    batches with no input resistor and a higher temperature than you
    expect the circuit to ever see (up to the rated temperature for the
    device) and the collector voltage at least equal to the supply voltage
    plus the zener voltage, to get an idea how bad the problem might be.
    If there is an internal base resistor on the second transistor, the
    problem may be slight. Then find out by experiment how low a base
    resistor is needed to bring the collector current down to an
    essentially fixed value (no amplification of leakage). Then use one
    1/10th of that value.

    Of course, if your logic signal pulls down to essentially zero volts
    (and there is no way the logic can become unpowered while the
    darlington is powered), the effective pull down resistance is the
    parallel combination of the base resistor and the series resistor to
    the logic signal. So by the time you get the series resistor low
    enough for good saturation at lowest temperature and device gain, the
    effective low state resistance may be well below that needed to keep
    the leakage under control under all circumstances with no shunt
    resistor across the darlington.
     
  5. Guest

    OK, took your advice and looked at diode leakage on a variety of
    diodes. Leakage ranges from the low single digit nA range clear up
    into the double digit uA range. I was dead in the water without
    knowing how to characterize this specific transistor in comparison to
    one of the numerous diode models.

    So, I decided to run a experiment. I took a transistor and connected
    its collector to 13.50VDC. I connected a 979K (at 25C) resistor from
    the base to ground. I connected the emitter to ground. Measured
    leakage current at room temperature was 0.2nA. I next moved the
    experiment over to an environmental chamber that is running a test at
    70C. I placed the test sample inside (wiring all connected). The
    resistor dropped in resistance due to the rise in temperature so I had
    to compensate for this. The end result was 4nA of leakage current.
    This seems awefully low but makes sense if you look at the levels at
    room temperature. If you keep halving leakage current for every 10C
    drop in temperature from 70C it comes out to roughly 0.18nA at room
    temperature. Close to what I measured. The un certainty of my
    measurement equipment could have had a role to play in this small
    discrepancy or perhaps a perfect doubling does not occur at exactly 10C
    rise. Whatever...

    Do the levels of leakage I mention make sense?

    At 120C leakage should increase to approximately 128nA. Across my 100K
    pulldown this results in a 12.8mV drop across the load resistor. IMO
    this is nothing to be concerned about (assuming my findings are valid).
     
  6. Guest

    He sounds OK to me. And he's likely aware that some darlingtons,
    You are absolutely right. He is OK. As a matter of fact, I have
    nothing but respect for the man. He challenges me to push myself to be
    better.

    I'm sure you have all been there at one point or another. Trying to
    impress your mentors. You know, show them that you're approaching
    their skill base and soon may surpass them,,, At least that's how I
    feel.

    George W. Marutz II
     
  7. Guest

    Of course, if your logic signal pulls down to essentially zero volts
    Unfortunately the Darlington always has voltage across it. The micro
    can turn on and off freely. So, I cannot count on the base resistor to
    be a stiffer pull down at all times.

    George W. Marutz II
     
  8. John - KD5YI

    John - KD5YI Guest


    Hi, George -

    The data sheet gives Icbo of 10 nA max at 25C. At 125C that would increase
    by 2^10 or to 10 uA. (Probably won't be this high since the spec is at 100V)

    The Vbe(on) is given as .8V typical at 50 mA Ic and about 125C (Vbe vs Ic at
    temperature graph on page 2)

    So, Icbo is about 10 uA and 8 uA goes through the 100k leaving 2 uA for Ib.
    At high temperature, your gain is over 65,000. This could lead to a
    collector current of 130 mA (if the relay coil permits). If it was my
    design, I would go as low as I could with the b-e resistance as others have
    recommended. It may help with noise as well.

    I realize you effectively have a 10k shunt via the logic input signal, but
    what is the microcontroller low output voltage at 125C? And, as John Poplish
    pointed out, what if, somehow, your micro output tristates to floating? Go
    for the robustness.

    Just some thoughts. Good luck.

    John
     
  9. Guest

    All three of you made mention that my base pull down resistor is too
    high of a value. Funny, so did the senior design engineer.

    Why did I do this? Well, for one thing, I did not realize that using a
    high resistance value pull down could cause potential problems. I know
    better now.

    My train of thought was to keep current being sourced by the micro pin
    as low as possible (around 500uA as an arbitrary target). Higher
    source current can cause other problems in my system that I will not go
    into now (A/D accuracy). I decided that the high current gain of the
    darlington over the temperature extremes I will be working with would
    aid in this effort. I first chose the base resistor from the parts
    list already on my BOM. I chose another resistor that was already on
    the BOM for the base pull down. I wanted this value to be greater than
    the base resistor so I would ensure I would be turning the transistor
    on. The 100K part fit the bill.

    I might have to break down and add a new component to the BOM. Perhaps
    a 4.7K base resistor and 10K pull down?

    Thank you for your valuable input.

    Ge0rge
     
  10. Guest

    John - KD5YI (so many John's around here:) )

    So what you are saying is that cut off current is the same as leakage
    current?

    Hmm... I guess I had a misunderstanding of what cut-off current meant.
    THANKS. If I had known this I could have saved myself a lot of
    anguish. I learned a few valuable leasoons here today.

    Ge0rge
     
  11. John - KD5YI

    John - KD5YI Guest

    Actually, George, I ASSUMED it was the same as leakage. If I'm wrong, I need
    to know it and I hope one of the regular gurus will come along and correct me.

    John
     
  12. Guest

    Actually, George, I ASSUMED it was the same as leakage. >If I'm wrong, I need to know it and I hope one of the regular >gurus will come along and correct me.

    Sounds right to me. I checked your hypothesis before I responded the
    first time. Go to:

    http://www.electronicdefinitions.com/definition.php?defid=362

    Ge0rge
     
  13. John  Larkin

    John Larkin Guest

    You might consider using a fet instead of a darlington, especially if
    the micro is a 5-volt part. A 2N7002 costs a few cents, saves one
    resistor, and will sink over half an amp with +5 on the gate.

    John
     
  14. John - KD5YI

    John - KD5YI Guest

    Even so, I would support John Larkin's suggestion that you use a FET. No
    drain on the micro and none of these other problems. It is a match made in
    heaven.

    Cheers,
    John
     
  15. Guest

    Even so, I would support John Larkin's suggestion that you use a >FET. No drain on the micro and none of these other problems. It is >a match made in heaven.

    The original design used a FET. That's why the resistor values are the
    way they are. The darlington was supposed to be a drop in replacement.
    FETs for this purpose cost 5x as much as the Darlington in my circuit
    now. Their are only a hand full of options for a N-Channel MOSFET with
    a 100V breakdown rating in a SOT-23 package. The limited number of
    competitive products on the market dictate higher prices. When you
    manufacture 2 million+ products a year this sort of thing adds up.

    Thanks

    Ge0rge
     
  16. Guest

    You might consider using a fet instead of a darlington, >especially if the micro is a 5-volt part. A 2N7002 costs a few >cents, saves one resistor, and will sink over half an amp with >+5 on the gate.

    Thanks for the input John. However, read my comment to
    John - KD5YI. I need a part with a higher VDSS rating. The 2N7002,
    and many other comparable devices, just would not cut it. I desire as
    small of package as is possible. The SOT-23 fits the bill for now.
    Unfortunately you don't find many SOT-23 N-Channel FET offerings with a
    VDSS of 100V or greater. The price takes a considerable jump over the
    60V counterparts.

    Oh, one last question. What resistor would you eliminate? The Gate
    pull down? Why?

    Ge0rge
     
  17. John - KD5YI

    John - KD5YI Guest

    From the Mouser catalog:

    Fairchild BSS123, 100V, 6 Ohms, SOT-23, $.08/250.
    Supertex VN2110K1, 100V, 4 Ohms, SOT-23-3, $.24/1000.
    Siliconix SI2328DS-T1, 100V, 0.25 Ohms, SOT-23, $.55/500

    I didn't look any further. I don't understand why you would need a 100V
    device to operate a 12V relay, but I'm sure you do. If you find that you
    could use a 60V device, you'll have a much wider selection. But, you knew
    that already.

    Good luck.

    John
     
  18. John Larkin

    John Larkin Guest


    Assuming your uP starts up with its output port pin tri-stated, you
    would still need a pulldown to ensure the fet is off. But you wouldn't
    need a series resistor.

    You can, incidentally, leave the gate lead of a 2N7000/7002
    disconnected, and turn it on/off by briefly charging or discharging
    the gate by touching V+ or ground with one finger and then tapping the
    gate lead with the other. Once on or off, it will stay that way for a
    while... about a week is the longest I've verified.

    John
     
  19. Guest

    Snip:
    Excellent. Never came across this one. I was using the $0.55
    Siliconix part.
    Unsuppressed battery feed / automotive load dump.

    Ge0rge
     
  20. John - KD5YI

    John - KD5YI Guest


    In that case, the Rds of the Fairchild part may be too high for you.


    Ah! As I recall from SAE J1211, you can have transients well over 100V at
    times. But, I don't know if that is applicable to your particular case.

    Cheers,
    John
     
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