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TL431LI-Q1: What is the off-state current of TL431LI-Q1 at 125°C junction temperature?

Part Number: TL431LI-Q1

I would like to ask the maximum off-state current of TL431LI-Q1 at 125°C.

There are contradictory information in the datasheet. I use the following datasheet:

https://www.ti.com/lit/ds/symlink/tl431li-q1.pdf

There is a table item in 7.5 (Off-state cathode current )

which says that the off-state cathode current is in the range 0.1...1 uA, where 0.1 uA is the typical value.

However, according to the following figure (figure 4 in the above datasheet), the off-state current at 25°C is approximately 0.001 uA (or probably less, it is around zero).  

The two typical values (0.1 uA in the table 7.5 and the value 0.001 uA in Fig 4) are very far from each other. Figure 4 shows that the off-state current is 0.008 uA at 125°C but because the current at 25°C is very far from the typical value, I don't know how to extrapolate for a reliable but realistic worst-case upper bound.

Do you have some measurement or other reliable estimation for 125°C (or similar temperature range) for the off-state current?  

  • Hello,

    The plots are taken from typical values of a random sample of parts and do not show device variations. We only guarantee the specs in the electrical characteristics tables which take into account all device variations. Please use the 0.1µA typ, 1µA max values for your design.

    Thanks,

    Walter

  • Dear Walter,

    Thank you for your answer.

    My problem is that the "Electrical Characteristics" explicitly defines the values for 25°C. However, the temperature dependency of semiconductors could be seriously high. I must take into account the temperature dependency in my analysis, because our product should work at 125°C. 

    Do you have the graph in figure 4 with higher resolution, where I can see the Ioff current at 25°C? So that I could calculate the temperature dependency more exactly, and I could extrapolate the maximum Ioff current from 25°C to 125°C.  

    My other idea is that I can see from fig. 4 that the Ioff current increases from 0.008uA@125°C to 0.048uA @150°C, so Ioff increases by factor 6 per 25°C. Hence:

    Ioff(125°C)/Ioff(25°C)=6^((125°C-25°C)/25°C)=6^4=1296.

    But if I multiply the maximum current at 25°C by this factor, I get: 

    Ioff,max(25°C)*1296=1uA*1296=1.3mA which seems to be pessimistic.

    So if you could share the graph in Figure 4 with higher resolution at 25°C may be more informative.

    Thank for your answer in advance.

  • Hello,

    I will look if we have a different graph or data.

    Please allow me a few days to gather the information.

    Thanks,

    Walter

  • Hello,

    Looking at maximum values for tested data across temperature, at 125C Ioff can be close to 0.9uA max and at 25C 0.02uA max. The curve shape will follow the one in figure 4. 

    I hope this data helps however we only guarantee the values in the electrical characteristics table. 

    Best,

    Walter

  • Dear Walter,

    Thank you, it is very close to the solution. My problem is that if we have a limited number of measurement, we need some postprocessing to get a reliable result. I should extend your result (0.9uA). To do this, I need some additional data.

    My first question is:

      *1) How many measurement results do you have? (at least approximately)

    My second question depends on how many information can you share with me. I write my question in decreasing order of priority. It would be great if you could answer at least one of the following questions:

     *2/a) Can you share these statistical data with me somehow? (e.g., in excel data, a set of graphs...)

     *2/b) If you can't share the dataset, do you have the mean and variance of leakage current at 125°C? 

     *2/c) If you can't share the dataset, can you share the minimum and maximum of leakage current at 125°C? 

     Thank you.

    György Orosz

  • Hello György,

    Unfortunately I cannot share more data or the number of samples but the maximum result or 0.9uA at 125C includes multiple deviations away from the mean.

    Thanks,

    Walter 

  • Dear Walter,

    Thank you for the answer.

    Best regards,

    György Orosz