I am using the STUB4500 USB c PD management chip.

Schematic below shows how I'm implementing it. I guess the 100k to 22k divider is to keep Vgs below 20v when the chip pulls vbus_en_sync to 0. With vbus at 20v, the Vgs of T1 will be about 3.6 volt.

Can I increase R11 to say 44k and make Vgs ~6v in hopes of getting a lower rds on? Will that slow down the turn on time for T1? Is it already too slow?

 

Can I increase R11 to say 44k and make Vgs ~6v in hopes of getting a lower rds on?

Doing that would increase on resistance. It should be in the couple dozen mΩ range. How much lower so you want it to be?

 

Doing that would increase on resistance. It should be in the couple dozen mΩ range. How much lower so you want it to be?

Oh ok yeah I was thinking of it backwards. Because the voltage on the gate would be about 3.6 (with R11 = 22k) then VGS would be about 16.4v, plenty high.

But would a VGS of ~14v really increase the resisitance that much? I'll look at the data sheet. This question is purely academic. I'm fine with it at ~16.4v.

Thank you for the reply

 

would a VGS of ~14v really increase the resisitance that much?

You don't provide sufficient information and neither does the manufacturer. They don't even characterize on resistance at 14V.

 

Oh ok yeah I was thinking of it backwards. Because the voltage on the gate would be about 3.6 (with R11 = 22k) then VGS would be about 16.4v, plenty high.

But would a VGS of ~14v really increase the resisitance that much? I'll look at the data sheet. This question is purely academic. I'm fine with it at ~16.4v.

Thank you for the reply

The RDS(on) max. according to the datasheet is 30mΩ at 10V. At Vgs 14V and 16V it is <= 30mΩ.
So it wouldn't change much.

 

an-1084p1.p65 (infineon.com)

This is the best app note for power mosfet basics that I've found over the years. It has a lot of info in a small form factor. Read it several times.

From the "Gate Charge" Section:
" The advantage of using gate charge is that the designer can easily calculate the amount of current required from the drive circuit to switch the device on in a desired length of time because Q = CV and I = C dv/dt, the Q = Time x current. For example, a device with a gate charge of 20nC can be turned on in 20µsec if 1ma is supplied to the gate or it can turn on in 20nsec if the gate current is increased to 1A. These simple calculations would not have been possible with input capacitance values. "

This will tell you how fast your mosfet will turn on. Is it fast enough? Depends on your load... is it capacitive, resistive, inductive, or a combination?