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TPS61040 Intrinsically Safe Application - HELP

Ralph Gable
Prodigy 115 points
Other Parts Discussed in Thread: TPS61040

Single Lithium cell input: 2.5 to 4 Volts in

40mA @ 6V out

Intrinsically safe => three zeners on either end of the inductor.

7.5V zeners being used. Have also tried 8.2V Zeners, but the 7.5 is preferred.

Zeners on the input, no problem.  Zeners installed on the output side of the inductor kills performance.

Is there a way to change the basic design to accommodate the addition of these zeners?

  • 0
    TI__Mastermind 40080 points
    Hi Ralph,Could you attach the draft schematic for this application?BR,Helen
  • 0 in reply to Ralph Gable
    TI__Mastermind 40080 points
    Ralph,
    Why choose 7.5V zener diode? Can the zeners installed on the output side of the inductor be higher than 7.5V? Because 7.5V is too near to the SW normal operating voltage.
    Another question is this is a very low voltage application, why need these extra zener diode? Could you please help to explain the exact reason to me, thanks!
    BR,Helen
  • 0 in reply to Helen chen
    Prodigy 115 points

    Intrinsic safety is all about energy available to create an ignition spark. Open circuit voltage, short circuit current, and stored energy in capacitors and inductors. In I.S. testing, they are allowed up to two countable faults. That means they can eliminate two of the three zeners. Thus, any one zener must be able to crowbar the voltage to its zener voltage.

    The power dissipation of the zener must be able to withstand the maximum sustained current possible in the circuit. In this case, we have a 250mA fuse.  The rule is 1.7*(Rated fuse current)*(maximum zener voltage) = Power dissipation of any single zener.

    If * (req_UL_multiplier) * Vzmax

    0.250 A * (1.7) * (7.5 V * (1 + 6%)) = 3.38 Watts

    This requires us to used 5 Watt zeners.

    Yes, would *could* use a zener as high as 11 Volts and get away with 5 Watt Zeners.

    0.250 * 1.7 * 11 * 1.06 = 4.96

    BUT, getting through U.L. testing unscathed means you do not skirt the edges of what you can do. We want to keep the Vz as low as possible.  

    I did try 8.2 V Zeners with the same deadly results to the regulator.

  • 0 in reply to Ralph Gable
    TI__Mastermind 40080 points

    Thanks for the explanation, did you measure the SW waveform before adding the zener? How about 9.1V or 10V rating zener diode?

  • 0 in reply to Helen chen
    Prodigy 115 points

    I *think* it has to do with the *capacitance* of the Zener as opposed to the *voltage* of the Zener.  I tried a few different 8.2 V Zeners and discovered one manufacture's to work.  Unfortunately, the one that works is not a 5 Watt Zener :-(

    Got *unofficial* word from U.L. that we *may* be able to move the zener to the load side of the output diode **IF** we used three output diodes in parallel (that way, two can fail and there is still a path for an overvoltage to get to the Zeners ... also three of them).  We have to pay them almost $3000 for them to *officially* review it and provide an *official* OK to do that. :-(  This is for ONE PAGE of the schematic! Ugh!

  • 0 in reply to Ralph Gable
    TI__Mastermind 40080 points

    $3000/one page is really expensive!!

    What is the Zeners installed on the output side of the inductor kills performance mean? Did the TPS61040 broken(fail) during this application. The zener diode's capacitance will make the total loss of the TPS61040 bigger.

    BR,

    Helen