Hi everyone!
I am designing a buck converter that would charge a battery and planning to use an N type MOSFET on high side and FAN7085_GF085 driver to go with it.

Datasheet:
https://www.fairchildsemi.com/datasheets/FA/FAN7085_GF085.pdf

Application notes:
https://www.fairchildsemi.com/application-notes/AN/AN-9052.pdf
https://www.fairchildsemi.com/application-notes/AN/AN-4171.pdf

If the converter happens to go to discontinuous mode however, the diode stops conducting -> voltage at MOSFET source becomes something >0, and the internal recharge path of FAN7085 activates (as much as I understand it according to the application note 9052, page 2). Doesn't this connect the battery through an inductor to VS pin of the driver circuit (which is connected to the ground at that moment)? What if I put a diode in series with an inductor (like I did in this capture of the schematic)?
Thank you in advance.

 

Hi everyone!
I am designing a buck converter that would charge a battery and planning to use an N type MOSFET on high side and FAN7085_GF085 driver to go with it.

Datasheet:
https://www.fairchildsemi.com/datasheets/FA/FAN7085_GF085.pdf

Application notes:
https://www.fairchildsemi.com/application-notes/AN/AN-9052.pdf
https://www.fairchildsemi.com/application-notes/AN/AN-4171.pdf

If the converter happens to go to discontinuous mode however, the diode stops conducting -> voltage at MOSFET source becomes something >0, and the internal recharge path of FAN7085 activates (as much as I understand it according to the application note 9052, page 2). Doesn't this connect the battery through an inductor to VS pin of the driver circuit (which is connected to the ground at that moment)? What if I put a diode in series with an inductor (like I did in this capture of the schematic)?
Thank you in advance.
View attachment 126161

I don't think Vs is tied to ground. It is floating at the source voltage isn't it?

 

Hi everyone!
I am designing a buck converter that would charge a battery and planning to use an N type MOSFET on high side and FAN7085_GF085 driver to go with it.

Datasheet:
https://www.fairchildsemi.com/datasheets/FA/FAN7085_GF085.pdf

Application notes:
https://www.fairchildsemi.com/application-notes/AN/AN-9052.pdf
https://www.fairchildsemi.com/application-notes/AN/AN-4171.pdf

If the converter happens to go to discontinuous mode however, the diode stops conducting -> voltage at MOSFET source becomes something >0, and the internal recharge path of FAN7085 activates (as much as I understand it according to the application note 9052, page 2). Doesn't this connect the battery through an inductor to VS pin of the driver circuit (which is connected to the ground at that moment)? What if I put a diode in series with an inductor (like I did in this capture of the schematic)?
Thank you in advance.
View attachment 126161

Oh, I think I see what bothers you. The initial charge switch inside the FAN. I think the resistor is very large. So not really a short.

 

Hi, thank you for the reply.
Direct quote from the application note:
"Although the recharge circuit shown in the functional block diagram is a switch in series with a resistor, it is physically implemented as a MOSFET. The switch represents a gate controlled MOSFET and the resistor represents the RDS(on) of that MOSFET."

So my question would essentially be: What causes the internal switch to activate? A detection of voltage > 0 at VS or something else? (as the application note only states this: "In the event that both of the switch (S1 and S2) are turned off, the internal recharge switch of the FAN7085 will provide the path to charge the bootstrap capacitor.")

Here is an internal block diagram:

 

I dont know enough about bootstrap drivers to say when that reset path might close, but a diode seems like a good plan anyway if only to protect against an input short... Do that with no diode and the battery will discharge via the inductor and the FET body diode.

Two other comments...
You are going to need some capacitance from the output of the inductor to ground, quite a bit I suspect.
You will want to measure battery current after this so it needs to be before the ACS...
It looks as if the high side driver does not supply the bootstrapped voltage IE VB will need to be at VDD + VGS continually, hence the referance to the charge pump in the datasheet.

I will freely admit I m ay be wrong but I recomed you check.
Al

 

Hi, thank you for the reply.
Direct quote from the application note:
"Although the recharge circuit shown in the functional block diagram is a switch in series with a resistor, it is physically implemented as a MOSFET. The switch represents a gate controlled MOSFET and the resistor represents the RDS(on) of that MOSFET."

So my question would essentially be: What causes the internal switch to activate? A detection of voltage > 0 at VS or something else? (as the application note only states this: "In the event that both of the switch (S1 and S2) are turned off, the internal recharge switch of the FAN7085 will provide the path to charge the bootstrap capacitor.")

Here is an internal block diagram:
View attachment 126164

I think...... The only purpose of the switch is to provide a path at turn on. If you don't have anything to pull Vs low to start with (like a passive load or a low side fet) it will turn on to provide a path to charge the boost cap. The rest of the time I think it just rests.
Mind you, I'm not sure about this.