BQ2970: CHG and DSG FET Connection

Hi Allen,
The principles of operation of a fet requires the connection to be that way otherwise the fets wouldn't be able to function.

thanks
Onyx

Hi Onyx,

Thanks for your reply. May I know why the FETs can't be connect as common source? Under what specific condition the FETs  will n't be able to function? 

Both of the FET drivers(Cout and Dout) reference VSS as ground rather than V-? Aren't they?

Please see the schematic below to understand what I mean. 

Alen

Hi Alen,

The diagram helped explain what you are trying to do. Looking at the circuit, it appears it would work and i can't think of any reason it wouldn't. However given this wasn't tested and characterized i wouldn't recommend using it except you test it out properly.

Is there any reason why the normal configuration wouldn't work for you?

thanks

Onyx 

Hi Onyx,

Thanks for your reply.

I just curious why we connect the FETs as a pair of common drain BBFETs instead of common source BBFETs. It seems there is no specific reason.

Do you know where do the FET drivers(Cout and Dout) reference? VSS or V-?

Alen

Hi Alen, if it is common source, source voltage is floating before FET is on in N-FET(VGS > VT). So, it may not turn on at start-up condition since source voltage is not biased.

Hi Jasper

Thanks for your reply.
Yes, if it is common source, source voltage is floating before FET is on.
but event if the source voltage is floating the voltage, the source voltage should be lower than Vss + 0.7V(because the body diode). So I think the driver is able to turn on the FETs.
How do you think about it?

Alen

Hi Alan,
Diode is turned on when Source(Anode) is Drain(Cathode)+0.7(Vdiode). If Diode is turne on, you're right but we can't assume the diode is turned on since source voltage is floating. It may turn on FET even we use common source but can't say it is always turned on.

Hi Jasper,


In common source condition, even the source is floating, the source voltage should be within 0.7V ~ - Vth. (assume Vss as ground, If the voltage is lower than -Vth the FET will slight on cause source voltage increase to -Vth or higher ). So Vcout is high enough to turn on the FET.
For example, Vsource is -0.5V(body diode is off), Vcout from 0V to 4V. So the VGS at the beginning is 4.5V. The FET is fully on.

Alen

Hi Alen,
There is a functional diagram on page 17 of the data sheet. Cout and Dout are referenced to Vss.
thanks
Onyx

Noted, thanks!