BTG7050-2EPL reverse battery protection

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ElecCheng
Level 2
Level 2
10 sign-ins 5 questions asked 10 replies posted

Hi Expert,

I see in BTG7050-2EPL d/s that it has reverse battery circuit internally. So is it necessary to add external Res//Diode network? Thanks.

ElecCheng_0-1675405738156.png

BR, CHENG

 

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1 Solution
Anshika_G
Moderator
Moderator
Moderator
First question asked 50 likes received 250 replies posted

Hello,

According to the explanation, placing a diode in the GND path of the device will not help in preventing the current flow through the DMOS body diode.

The reverse battery protection needs to be added to the positive battery terminal. The current flow in reverse will be from the capacitor, through the DMOS body diode and to the positive battery terminal (now negative when the terminals are reversed).

Best Regards,
Anshika

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Anshika_G
Moderator
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First question asked 50 likes received 250 replies posted

Hello Cheng,

From datasheet :
"In reverse polarity condition (also known as reverse battery), power dissipation is caused by the intrinsic body diode of the DMOS channel. The reverse current through the output stages must be limited by the connected loads." 

The external reverse polarity protection would be needed if you are using big capacitors loads.
Let me explain with you an example for a load of 1mF capacitor:

During reverse polarity condition two things happen:

  • The VS to GND voltage (as well as the digital inputs) is negative [DS chapter 7.4.1]
  • The VS to OUT voltage is also negative [DS chapter 6.3.1]

During the reverse polarity condition, the DMOS bulk diode is forward biased in inverse condition (if the output is not turned ON), as we have to discharge the load capacitor due to a low or even negative battery voltage.

What is important here, is to calculate the total energy of the IC in case of a charged 1 mF load capacitor at reverse battery condition (inverse output current).

For example, let's consider that the capacitor is charged at 16 V, and we apply -16 V between VS and the output load GND (which is the load capacitor GND). The IC has to handle an energy equal to: ½ * C * V2 = ½ * 1mF * (16V – (-16V))2 = 512 mJ

This is too much for the IC as the maximum allowed energy dissipation is for single pulse is EAS = 12 mJ [PRQ-615] and for repetitive pulse EAR = 2.5 mJ [PRQ-616].

Therefore, a reverse polarity diode is required.

Best Regards,
Anshika



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Hi Anshika.

Thanks for your reply!

When you mention "a reverse polarity diode is required", is it applied at the battery terminal or GND terminal? 

My question is, a diode + resistor network  is needed at chip GND terminal or not? Thanks.

BR,

CHENG

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Anshika_G
Moderator
Moderator
Moderator
First question asked 50 likes received 250 replies posted

Hello,

According to the explanation, placing a diode in the GND path of the device will not help in preventing the current flow through the DMOS body diode.

The reverse battery protection needs to be added to the positive battery terminal. The current flow in reverse will be from the capacitor, through the DMOS body diode and to the positive battery terminal (now negative when the terminals are reversed).

Best Regards,
Anshika

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