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TL082IDR

HIDE66163
Intellectual 330 points
Other Parts Discussed in Thread: TL082, TL084

Hello,  On the data sheet of TL082 (TL082IDR), Imput bias current is specified 20nA max over full temp range (TA: -40deg C to 85deg C).

In case of using TL082 at 85deg C(Ta), what is the allowable junction temperature(Tj)?

  • 0
    TI__Guru*** 140436 points

    Hello Hide,

    The 20nA spec was based on characterization data where Ta=Tj=85C.
    The input current is junction leakage and is exponentially proportional to the junction temperature.

    The device will function with Tj greater than 85C, but the input current may be higher.

    Regards,
    Ron Michallick

     

  • 0 in reply to Ron Michallick
    Intellectual 330 points

    Hello Ron,

    Thank you for the explanation. 

    Let me confirm about "20nA spec was based on characterization data where Ta=Tj=85C".

    Normally, Tj is higher than Ta due to self-heating of device.  If it keeps input bias current to 20nA, I guess Ta should be lower than 85C, right?

    Regards,

    Hide

  • 0 in reply to HIDE66163
    TI__Guru*** 140436 points

    Hide,

    Yes you are correct. In most applications the TL082 does not dissipate much heat so the difference will be small.
    Typically the input current is much less as seen in figure 17 of the data sheet..

    Regards,
    Ron Michallick

  • 0 in reply to Ron Michallick
    Intellectual 330 points

    Hello Ron,

     

    It seems to me that the difference is not small...

    Power dissipation = 1.4mA * 30V * 4ch = 0.168W  --- (TL084, 4ch)

    Theta Ja = 86C --- (TL084, SOP14)

    Tj - Ta = 86 * 0.168 = 14.448C

     

    If above calculation is correct, Ta shoule be below 71C (85C - 14C) to keep 20nA_Max input bias current, is that right?

     

    Regards,

    Hide

  • 0 in reply to HIDE66163
    TI__Guru*** 140436 points

    Hide,

    Yes, the formulas are correct.

    There are two power formulas for output current.
    If current flows out of TL082, power = ('VCC+' - 'Vout' ) * Iout
    If current flows in to TL082, power = ('Vout' - 'VCC-' ) * Iout

    Regards,
    Ron Michallick

  • 0 in reply to Ron Michallick
    Intellectual 330 points

    Ron,

    Let me confirm if I understand this correctly.

    1) In case of using TL082 where Ta=85C, Tj will be around 100C, right?

    If above my understanding is right, I have a following question;

    2) If I use TL082 where Ta=85C and Tj=100C, which parameter lose coverage specified on the datasheet? (ex. Input bias current?, Input offset current?)

    3) On the above parameter that lose of datasheet (ex. Input bias current?, Input offset current?), can I get estimate value? (estimate value where Ta=85C and Tj=100C)

    Regards,

    Hide

  • 0 in reply to HIDE66163
    TI__Guru*** 140436 points

    Hide,

    1) Right.

    2) The input current specs would seem to be increased by 167%, but the tighter 25C spec over rides 85C spec; so the increase is just 14%.
    2) The open loop gain spec, AVD would drop by 10%

    3) IIB @100C max = 22.8nA;  IIO @100C max = 2.28nA
    3) AVD @100C min = 22.5V/mV