High side switches reverse battery protection

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Hi everybody,

I see that the Infineon smart high side switches provide battery reverse protection, by connecting a 150 ohm resistor in GND as a current limiter.
Just as a reference see the note at the bottom of the first page in the ITS711 datasheet.

http://goo.gl/UYabFM

Nothing is stated about the power dissipation of this resistor, but, from the diagram at page 9 of the same datasheet, it appears to be

(Vbb - Vtvs) ^2 / 150

This is 5.68W, with 30V reverse battery voltage and an estimated 0.8V drop on the internal TVS diode, and then quite a large value.


Is my estimation correct?

Many thanks in advance for any comment.


Marco
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8 Replies
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Hi Marco,

I think the internal zener diode is in blocking direction, how do you come up with 0.8 V drop voltage?

Luxun
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Hi Luxun,

the internal zener diode is conducting because it is forward biased, i.e. with plus on anode and minus on cathode and then, in this situation, it acts as a normal diode.

Please see attachment.



Marco
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It would be nice to have some more comments ... 🙂


Marco
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Hi,

I think you can use a schottky diode instead of 150 ohm resistor for this purpose.
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Hi b123,

first of all, thanks for your comment.

Your porposal seems to me very appropriate, but what makes me puzzled is why Infineon suggests the use of a resistor that ends up to be unreasonabling huge.
What do you think?

Marco
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Hi Damon,

I agree with you, but why Infineon reccomends to use a 150 ohm resistor in a lot of datasheets?


Marco
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User9050
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First solution authored
1986.attach

My guess is that this zener diode in the ITS711 datasheet (here in the picture VZ2) should explain the overvoltage protection. Regarding the reverse power protection there are only the path of the output with the load, where the load has to limit the current, and the path of GND to the Status pin ST, where RGND has to limit the current. Do you have a sample on hand to check this and measure the current into the pin GND with R=150Ohms connected?
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Hi Schulz,

tanks for your comment 🙂


Currently I don't have a chip to test but, from the schematic that you posted, we can see that, in the event of battery reverse, the main path that causes power dissipation on the board, is through VZ2 and Rgnd.
In this situation VZ2 is conducting because it is forwarded biased, and then it acts as a normal diode with a drop voltage that we can estimate of about 0.7 1 V.

So Rgnd is the only current limiting element, and it has to dissipate several watts.


Marco
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