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Effect of and on ambient temperature?

Discussion in 'Electronic Design' started by The little lost angel, Dec 17, 2003.

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  1. On reading spec docs, most parts appear to be specified for X ratings
    at ambient temperature of 25C. Then they have a thermal derate to zero
    at some temperature like 275C.

    The problem now is, ambient temperature right next to the part will
    get hotter too right? So does the spec mean, as long as I start with
    25C ambient, it's ok to run the part up to 275C.

    Or do I have to figure out the temperature rise of the ambient air and
    at what temperature it will stabilize given the heat dissipated by the
    part. Then see how much temperature rise am I allowed?

    In addition, if I'm constantly pushing 25C air through using forced
    air cooling. Does this means I can assume 25C constant temp or does it
    still mean nothing since the air next to the part will always be
    hotter than 25C?

    Thanks!

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  2. N. Thornton

    N. Thornton Guest


    Hi

    If you mount the part in the way specified, in an unenclosed space,
    then the mfr's deratings are as stated, and you use the room temp, not
    the temp 0.01mm from the part. If its in an enclosed box you use the
    temp of the air in the box.

    Once you start fan cooling a part that is only specced for non-forced
    cooling, the spec sheet info wont be valid for that. Consider the heat
    path: the die has the max temp of 250 or whatever the derate to zero
    figure is. Heat has to travel from die to part case, from part case to
    heatsink, and from heatsink to air. With unforced cooling the heatsink
    to air barrier is likely to be the least thermally conductive barrier.
    Fan cooling reduces that resistance considerably.

    Ultimately it is the temp of the die that counts.

    Regards, NT
     
  3. Without reading the spec for the particular part, I don't think anyone
    will be able to help you with your exact situation. Every vendor
    seems to state their temperature dependancy (and derating)
    differently. Some spec their parts with respect to ambient temps
    while others use case or junction.

    I have to say that I've never seen 275C even mentioned for an
    allowable temperature though (but I'm not in the miltary world, so
    that may be why). Make sure you aren't reading the soldering
    temperature. Max junction temperature of the parts I'm used to
    working on range from 105C to 150C.

    But to answer your question, yes, the vendor is generally referring to
    the temperature of the airflow in the general area of the part, not
    the temperature of the air just a fraction off the body of the IC.

    However, that still doesn't necessarily mean you're off the hook. It
    all depends on how much heat the part generates, and what its thermal
    resistance is:

    Let's assume the part has a ThetaJA (thermal resistance from junction
    to ambient) of 50 deg C per Watt when airflow is moving at 100 lfm
    over the part, and that it's max safe operating junction temperature
    is 125C.

    Assuming you provide that much airflow, the part can give off 2 Watts
    [ (125C - 25C)/50] without exceeding its specs. Any more heat than
    that, and you risk the long-term reliability of the part. The only
    way to get more heat out of it is to lower the ambient temperature or
    lower the thermal resistance (bigger package or add a heat sink or
    increase airflow).

    And don't forget that if you have more than one part in a row
    (parallel to the airflow), each part will heat the air for the next
    part. If you generate lots of heat, your 25C air could easily become
    40C air (or more) by the time it reaches the last chip in the row - so
    you have to redo the above calculation with higher ambient.

    Have fun,

    Marc
     
  4. Mac

    Mac Guest

    Yes. You might just have to guess. Calculating how much an enclosure will
    heat up based on the dissipation inside is not easy.

    No, if you have 25C air blowing on the part, you can use 25C ambient. But
    unless you are in a climate controlled room, you should use a higher
    temperature to allow for hot days. Maybe 35 or 40 deg, depending on what
    seems reasonable. For example, a piece of outdoor equipment which isn't in
    the shade can experience very high ambient temperatures.
    Yeah, I'm with you on that. Some power MOSFET's are spec'd at 175C
    junction temperature. That's the highest I've ever seen.
    Power MOSFET's give power ratings with a case temperature of 25C. This has
    to be one of the most unrealistic ratings in the world! If the part is
    dissipating power, it will be nearly impossible to keep the case at 25C
    unless you use liquid cooling or something.

    Anyway, the typical thermal data in a datasheet are ThetaJA, as mentioned
    above, and ThetaJCase (thermal resistance from junction to case), and
    sometimes Theta from the case to the heatsink. Note that Theta is a greek
    letter. We are spelling out the letter, but on the datasheet they have a
    font which supports the real theta, which looks a bit like an 'O' with a
    horizontal line through the middle of it. Actually, they might use
    Rtheta. But the real clue is that the units are degrees C per Watt. You
    probably already know this, but it's hard to tell, so please don't be
    offended. ;-)

    You can put several Rtheta's in series. For example, R theta junction to
    case (Rtjc) and R theta case to heatsink (Rtch) and R theta heatsink to
    ambient (Rtha). This last one would be from the heatsink manufacturer and
    it will depend on the airflow. The resistances add up just like electrical
    resistances. So let's say all the above resistances add up to 40 degrees C
    per Watt. If you estimate that ambient will be around 60 degrees C, and
    you are dissipating 2 Watts, then you can work backwards and see that the
    junction will be at 60 degrees + 2 Watts * 40 deg. C/Watt = 140C. If the
    maximum operating junction temperature is 175 deg. C, then you have 35
    deg. of margin, which should be plenty.

    Good luck.

    Mac
     
  5. Well, I guess I should explain that I was looking at the parts spec
    for power resistors. Such as

    http://www.ohmite.com/catalog/pdf/metalohm.pdf
    http://www.ohmite.com/catalog/pdf/hs_hsn_series.pdf
    http://www.isotekcorp.com/dataSheets/PDF/ulh.pdf
    I'm not quite sure how to obtain this thermal resistance values from
    those documents.

    But using the isotek as an example, at the very last page, a chart
    gives % of rated power vs ambient temp. Using line 2 for 200W~300W.
    Assuming I want to dissipate 200W.

    Assuming that I have a standalone piece and ambient goes to 40C, I am
    looking at a derate to around 66% or 130W.

    Now refering back to the Ohmite estimates here,
    http://www.ohmite.com/catalog/pdf/appnotes_res_select.pdf (pg 7)

    if I put a 700 FPM airflow on this part, the rating factor is 0.4.
    Therefore I will be able to dissipate 130W using a 53W equivalent
    part, or using the same 200W part, dissipate up to 325W.

    However, the isotek part seems to have some kind of operating range of
    -55C to 200C. So the limited temp rise factor from the Ohmite guide is
    around 2.8 for 160C limit. Which brings that figure down to 116W.

    Since I am planning to use it for 200W, I would either have to use a
    pair of them or improve my cooling say by adding a heatsink or
    increasing the airflow.

    Would this method of estimate be sufficiently close to that arrived
    with by using thermal junctions (which for most parts I don't see as a
    given value in most docs)

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  6. Roy McCammon

    Roy McCammon Guest

    unless clearly stated otherwise, assume that the blurb
    has rated it in the most optimistic fashion (bolted to
    a battleship sitting in the arctic ocean). You have to
    hold the ambient at 25C to get the rated dissipation.
    In practice, that means the ambient is going to heat up
    and you must derate the part. Or something exotic like
    liquid cooling or a thermoelectric cooler.
     
  7. Mac

    Mac Guest

    You should have said you were using resistors in your original post. They
    are a bit different. The resistors you are looking at are specifically
    designed to dissipate lots of power and are made from heat resistant
    materials. Materials being what they are, it is possible to design
    resistors for higher temperatures than it is for semiconductors. It looks
    like the manufacturers, at least Ohmite, use other guidelines, but you
    seem to have figured it all out.

    Yeah, sounds right. Forced air would help, though. The combination of
    forced air and a heat sink will help tremendously.
    I can't fault your logic. Note that 700 feet per minute is a lot of air
    flow.
    You'll have enough trouble getting to 700 feet per minute. I would go for
    the heat sink and/or multiple resistors. If you use more than 1, try to
    keep some distance between them.
    You seem to be following the worksheet on page 7 of the Ohmite document.
    That seems reasonable to me.

    BTW, it's not thermal junctions. It's semiconductor junctions. Resistors
    don't have semiconductor junctions. But the principal is the same. There
    is a max temperature and a thermal resistance which can be affected by air
    flow and heat sinking. In their own way, it seems like the documents try
    to address this.

    It seems like you are on the right track. If you haven't already, make
    sure you read that whole Ohmite document. It seemed to have lots of good
    information, although I only skimmed it. I believe that the worksheet on
    page 7 should apply more or less equally to resistors from other
    companies.

    The only other thing I should point out is that 200 Watts is a lot of
    power. You will definitely need to put a fan on the enclosure, and if the
    enclosure is in a small (closet sized) space, you may need to ventillate
    the space, too. Also, since the resistor or resistors will be so hot, make
    sure that they are downstream from any electronics that may also be in the
    enclosure.

    Good luck.

    Mac
     
  8. I must have gotten something wrong then. I've got easy access to 80mm
    fans rated for 40+ CFM each. I arrived at the 700 FPM by assuming I'll
    put a 2x2 array of these fan at one end, to get around 160 CFM. Which
    according to some online documents I've found for converting CFM/FPM
    would result in a 700FPM airflow.

    Roughly, 80mm radius gives 0.216 sq feet. 160 CFM / 0.216 sq ft would
    give 700 fpm.
    Hehee, I was hoping to keep costs down as far as possible i.e. low
    component count for this current sink / load bank. My friend said it
    would cost upwards of $20~30 USD per high wattage resistor and I'm
    looking at a binary ladder with values up to 32A 12V

    Was hoping to use power mosfets or something but apparently I cannot
    control the current sunk as their characteristics will be constantly
    in flux due to the incoming voltage/temperature and I haven't figured
    out how to control them using opamp feedback... few of my simulations
    on EWB demo gives me what I expect when I tried :(
    Yesh, I've been through that doc a few times, just not 100% I got it
    right that's all :)
    The enclosure is probably going to be the heatsink/shroud itself and
    in an open room. Though 200W is just one of the piece of the whole
    gadget, I'm supposed to be cooking up the design for something to
    dissipate 1 to 63A at 12V, 5V and 3.3V
    Yup, definitely, nothing's expected to be downstream of them except
    cables to ground. Though one of my friend is talking about putting the
    whole resistor/heatsink part into a water tank with a water pump :ppPp

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  9. I read in sci.electronics.design that The little lost angel <[email protected]
    overgirl.lrigrevol.moc.com> wrote (in <
    t.sg>) about 'Effect of and on ambient temperature?', on Fri, 19 Dec
    2003:
    I very much doubt it. Even at Digikey prices. Check for yourself.
     
  10. N. Thornton

    N. Thornton Guest

    My first thought is why buy the resistors when you can use resistance
    wire for the cost of about 2 peanuts.

    Regards, NT
     
  11. Well, that idea was thought of. But the following question arises

    1) How do we know we've got exactly 0.11 ohm worth of resistance when
    the best multimeter we have, has a base probe resistance of 1.5 ohm
    apparently.

    2) What kind of material to use for the wire so that the resistance
    won't change drastically as temperature goes up?

    3) Assuming we found the materials, and wound the wire... how do we
    cool it since we can't attach a heatsink to a coil of wire... or can't
    see how yet.

    The problems won't be so bad if we could figure out how to produce a
    compensating circuit that monitors the current draw and switches
    on/off another load. But none of my experiments on an opamp circuit is
    working out on EWB :(

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  12. You use a combination of an amp meter and a millivolt meter, apply a
    large current ot the resistor and measure both the applied voltage and
    the current.
    Manganin. Nichrome is not as stable but fairly common and cheap. A
    coating of silver solder makes it possible to soldr copper wires on.
    Copper wire has a higher thermal coefficient but is really cheap and
    available. Iron and brass wire are also usable and people hang
    pictures on them.
    Weave the wire across a frame and blow the air through the web of
    wire.
    Is this a question?
     
  13. But isn't the current determined by the resistor itself? The closest
    to this I've thought of was using the 1mOhm current sense resistor in
    series with the load resistor and measuring the millivolt drop across
    the sense resistor. But again, the accuracy of the method was a big
    question mark in my own head.
    Hmm, what would be the thermal characteristics of such a setup? As in,
    how high can the temperature be allowed to go up to and such?
    heehee, sorta, I did post a schematic a few days back. Only one person
    replied and said it wouldn't do what I expect but when I asked why,
    there was no response.

    The circuit I did had a current sense resistor in line with the load
    bank, e.g.

    +ve ---+-Rcs-+----+--/ -R1---+
    | | | |
    | | +--/ -R2---+--- Gnd
    | | | |
    | | +--/ -R3---+
    OA+ OA-

    The idea was to feed OA+ into the input of the OpAmp and the OA- into
    the -Vcc of the opAmp, thinking that that should make the OpAmp read
    the drop over the current sense Rcs. Then I would apply a reference mV
    to the +input of the OpAmp. Except of course, it didn't work.

    There was also the issue of somehow tying the reference voltage to the
    switches for R1 to R3. I thought a voltage adder might be the idea but
    I'm stumped on exactly how to do it. EWB doesn't work the way I
    expect, I can't use it as a test and try thing since I have no idea
    why it's producing the voltages it is (generally full +Vcc or -Vcc)

    Would greatly appreciate it if you could give me some hints about it
    too! Thanks!!!

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  14. The current will certainly be influenced by the resistance, unless the
    current comes from a current regulated supply (which is a fine thing
    to use). The point is that you do not assume some current, you
    measure it by putting an amp meter in series with the resistor. At
    the same time to put a separate volt meter across the resistor, and
    you calculate its resistance (under those conditions, it you don't
    want to make assumptions about temperature coefficient) by dividing
    the voltage across the resistor by the current through it. You may
    need to limit the current to a safe value for the source by also
    having some other resistance in the circuit, like a large incandescent
    lamp or a toaster.
    That depends on the melting temperature of the wire, and how much
    thermal change in the resistance you can tolerate. But you can
    measure the thermal effect with the same resistance measurement method
    I spoke of above.

    The big advantage of this method is that there is very little air drag
    compared ot an aluminum heat sink, so you can move a lot of air with a
    small, axial fan. The wire is just a spider web across the air flow.
    They carry a thousand watts out of an ordinary hair dryer, this way.
    And you know how small those are.
    It didn't capture my attention, and I thought you had a conversation
    going on with someone, so didn't butt in.
    Linear opamp circuit make the assumption that the two inputs stay at
    essentially the same voltage for the whole range of output voltage.
    The input signal and the feedback have to make that happen when you
    are getting the desired output.

    A subtractor is one configuration that amplifies a voltage difference
    and outputs that amplified difference on an arbitrary reference
    voltage (like zero volts). There are also current mirror circuits
    that rereference a high side current (or small voltage difference) to
    another voltage rail.

    Section 3 of this circuit collection shows several ways to do this
    from a full, 3 opamp instrumentation amplifier (top of page 14), a two
    opamp version that can have inputs outside the amplifier input voltage
    range (top of page 15), and a two opamp version that has to have the
    input signals within the opamp input voltage range. But since both
    your voltages come from a low impedance source you can just use the 1
    opamp subtractor that is all the resistors and the right amplifier out
    of the I.A. at the top of page 14.
    http://www.national.com/an/AN/AN-31.pdf

    Here is an example of moving the current shunt signal to a different
    reference with a current mirror:
    http://www.zetex.com/3.0/pdf/ZDS1009.pdf
    Are you trying to select the load resistor based on a measurement of
    the load current, or step the load resistors to hold the current to
    some setpoint (as close as the switching will allow)?
    If the load resistors are all equal there is a different method than
    if they have a binary multiple relationship. In the first case, any
    resistor is equivalent to any other so you need a circuit that just
    switches them on or off in a straight sequence. In the second case,
    you need ot switch them on and off with the equivalent of a binary up
    down counter. But in both cases, this selection circuit has to be
    connected to a decision making function that does this selection in a
    stable and orderly fashion (at a speed that does not exceed the the
    decision making process's ability to evaluate its previous decision
    before making another one). This can all be done with opamps and
    comparators and discrete logic (fast but large), or with a program
    running in a microprocessor (slower but smaller and more feature
    rich).
     
  15. Thanks for the pointers, though I'm running a flu now and it's making
    my slow brain even slower, so I'll go through those documents later.
    I'm stepping the load resistors to hold the current to a set point.
    Essentially to be able to put a load on the voltage(current?) source
    at say 1,2,4,8,16,32 amps settings to see how well the power supply is
    holding up. Since the power supply is supposed to remain stable with
    the load changing up to 20Khz, having the load oscillate a little
    around the set point shouldn't cause it problems... if it does, the
    supply's crap and I'll like to know that!
    I'm still not quite sure which is the best method. The
    buy-fixed-resistors-in-binary-steps-without-current-control seems to
    be the easiest but most expensive and there's only that close I can
    get to what I want.

    I think I've mentioned (during my last vacation) of using mosfets to
    control the resistance to a bank of same value resistors, except at
    this point I realize (maybe wrongly) that it would be almost
    impossible to control the mosfet Rds to the exact level I want. Once
    the load bank is started up, temperature will kill the preset
    calibration and again the same problem of I don't know how to control
    it electronically comes in.


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  16. N. Thornton

    N. Thornton Guest

    1. fix the multimeter
    2. look up the resitance per metre of the wire type/gauge you have,
    and cut to length - easiest option.
    3. apply a known V and measure the i. V=IR.

    I'd suggest googling s.e.d for "resistance wire", or looking for
    constantan's thermal characteristics. There are many wire types to
    choose from.
    Look at how a fan heater is made. Wire strung across an open ceramic
    frame. Look at the amount of fanning, the amount of wire and the power
    dissipation to get a rough idea. 200 or 300w is easily done.

    I dont really know what youre trying to achieve, so cant add much
    useful.


    Regards, NT
     
  17. It takes fewer resistors for a given step size, with the binary
    method, but the worst case switching spike is bigger. you can combine
    a single stage of continuously variable load (linear mosfet) with the
    switched blocks to be able to use pretty coarse steps and still settle
    to a completely specified current, as long as the setpoint does not
    end up right on the boundary between blocks.
    Lets take the simplest (to understand) case and see if that is good
    enough.

    Lets say you have 4 equal blocks of resistance selected by 4
    comparators that have switching thresholds set in a nicely spaced
    sequence of voltages by a small divider of 5 resistors with switching
    thresholds of, say, 1,2,3,4 volts. There is an opamp error amplifier
    that evaluates the match between the continuously measured current
    (via one of the subtractor circuits we talked about last time) and a
    voltage that represents the current setpoint.

    The error amplifier has a capacitor feedback, so that it ramps up and
    down for positive and negative errors. The output is what drives the
    4 comparators that switch the 4 equal resistors on and off, in
    sequence. If the desired current is half way between that drawn by
    two of the combinations, the circuit will ramp back and forth between
    those two number of blocks being on, producing a square wave load that
    swings back and forth between the current that is a little too high
    and the current that is a little too low, with the time in each state
    dependent on the size of the feedback capacitor around the error
    amplifier. This constitutes the coarse load selection part of the
    system.

    Are you with me so far?
     
  18. That would be everything to the right of the word "balance" in the
    middle, IOW, R2, R3, R4, R5, one LM107? I hope so because I'm not sure
    how to fit R1 in except by connecting it in series to R2//R3.
    Does this thing work without external power? I got confused (ok now
    everybody realizes why I've got lost in my nick), because on page one,
    the right hand side schematic seems to indicate that no external Vcc
    needs to be applied to the IC, or rather nothing is connected to the
    X1 Y1 pins. However in pg2 the spec table says max of 120V between X1
    and Y1.

    Also, the spec seems to indicate that it can only be used on a circuit
    with no more than 10V, since Max Voltage (E1-E2, E3-E4) is 10V and no
    more than 1 amp can be measured Continuous Current (E1-E4, E2-E3).

    Or did you mean that I should take the IC schematic as a reference and
    build one using four transistors and resistors?

    Thanks!!!

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  19. No R1. R2 and R3 would connect to the ends of your current shunt.
    The main limitation of this circuit is that the opamp inputs can
    respond only inside a certain voltage range, called the common mode
    range (a range of voltage that they have in common). For the LM107
    that would be a couple volts inside the supply rails. For an LM324,
    that would be from 1.5 volts below the positive rail to .3 volts below
    the negative rail.
    so you could use this to subtract the voltages across a current shunt
    that had voltage about 1.5 volts below the positive rail of the opamp
    (or a little higher because of the voltage dividers) and still keep
    both inputs inside the common mode range. So if you build the
    subtractor with low gain (say, 1) the dividers increase the common
    mode range on the inputs to almost double the positive rail.

    At the moment, I don't know what opamp supply voltage you are
    planning, and what the positive supply rail voltage will be at the
    current shunt.
    The only power it needs is that the outputs be held at a negative
    voltage (about 2 to 120 volts) compared to its inputs at the current
    shunt end.
    Normally, the voltage from E1 to E2 is the low voltage produced by the
    current shunt, and the voltage between E3 and E4 is the output
    voltage. The voltage between the E1 E2 pair and the E3 to E4 pair is
    how far from the shunt rail the error amplifier will be.
    The current mirror requires well matched resistors to have a
    predictable accuracy. However, you can easily copy it with separate
    small transistors and get it to work, as long as you don't expect it
    to have exactly the specifications on this data sheet. So, either one
    might work for a one off.

    The purpose of this current mirror is to take a small voltage as an
    input (the drop across a current shunt, R2) and turn it into a current
    difference, scaled way down (by the resistor R1) to something in the
    milliamp range. Then that current difference can be changed back to a
    voltage difference (and enlarged) by the load resistors R3 and R4. So
    you get voltage amplification by the ratio of R4/R1 and relocation of
    the signal from a positive rail to a negative one.
     
  20. David Dong

    David Dong Guest

    The final temperature is power multiply the thermic resistor which can
    be found in the devices datasheet.
     
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