Circuit modification for 6EDL7141 driver

Tip / Sign in to post questions, reply, level up, and achieve exciting badges. Know more

cross mob
Ksmruti
Level 1
Level 1
First like received First like given First reply posted

Hi,
I am using the 6EDL7141 Gate Driver IC for one of the battery-operated Gardening tool applications where the battery is removable. Battery voltage is in the range of 24V. I want the gate driver to be activated once I plug in the battery in the system. What circuit modifications need to be done on CE pin in that case?

0 Likes
1 Solution
SuhasBobade
Moderator
Moderator
Moderator
100 sign-ins 50 replies posted 25 solutions authored

Hi Smruti,

Thanks for posting!

Gate Driver starts up if you apply the voltage on the Chip Enable pin of 6EDL7141 IC. 

As per recommended operating conditions mentioned in the datasheet, maximum voltage up to 6V can be applied on the pin.

You can design a voltage divider circuit to step down 24V - 5 V and output of the divider can be given to CE pin. Input to the divider can be tapped from the battery terminal itself. So that as soon as you connect the battery to the circuit, divider circuit will power up and it will fed voltage to CE pin. Once CE pin gets voltage, gate driver will activate.

Hope this clarifies your doubt.

Still you have any question, feel free to contact us on the forum.

Thanking you. 

 

View solution in original post

0 Likes
2 Replies
SuhasBobade
Moderator
Moderator
Moderator
100 sign-ins 50 replies posted 25 solutions authored

Hi Smruti,

Thanks for posting!

Gate Driver starts up if you apply the voltage on the Chip Enable pin of 6EDL7141 IC. 

As per recommended operating conditions mentioned in the datasheet, maximum voltage up to 6V can be applied on the pin.

You can design a voltage divider circuit to step down 24V - 5 V and output of the divider can be given to CE pin. Input to the divider can be tapped from the battery terminal itself. So that as soon as you connect the battery to the circuit, divider circuit will power up and it will fed voltage to CE pin. Once CE pin gets voltage, gate driver will activate.

Hope this clarifies your doubt.

Still you have any question, feel free to contact us on the forum.

Thanking you. 

 

0 Likes
Ksmruti
Level 1
Level 1
First like received First like given First reply posted

Hi,

Understood.

Thanks for quick reply.

0 Likes