LM5008: Shutdown design issue

Akilan G14

Part Number: LM5008

Hello sir,

We are using LM5008 in BMS and the sleep current is very important for us.

Because of that, we wanted to shutdown the LM5008 when the Battery is under voltage (38V) which is higher than ICs internal Under Voltage lock (5.3V).

But If I pull down RT/SD to ground according to datasheet, then the current through the RT will be 210 uA in my case. Again which is higher than ICs Quiescent current 110 uA.

We need to have shutdown current less than Quiescent current which is around 10 uA. Please let us know how to achieve this.

Thanks,
Akilan G

over 7 years ago

0 Frank De Stasi over 7 years ago

TI__Guru 73235 points

I do not think you can achieve such a low shutdown current of 10uA with this device. If you just
need to save the current in the RT resistor, then you could switch it out when in shutdown.
If you need very small shutdown current you could use a series pass switch in front of the regulator to
disconnect it from the input.

If you only need an input up to 60V, you might try another device such as the LM4600x.
You could also try a search on the TI website.

0 Frank De Stasi over 7 years ago in reply to Frank De Stasi

TI__Guru 73235 points

I am closing this post

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LM5008: Shutdown design issue