BQ24780S: ACDRV issue

Hey Bill,

Q1 and Q2 are back-to-back input protection FETs.

As far as criteria:
ACDRV to CMSRC will be set to 6V, meaning these FETs cannot be incredibly large such that they will not turn fully on with a 6V gate-source voltage.

Additonally, the FETs cannot be overly small such that the turn-on time is extremely fast; this could create a path from the input directly to the battery as the Break-Before-Make logic of the power path selection is heavily time dependent. You can also increase the gate resistance and gate-source capacitance of each FET in order to slow them down.

Lastly, will want to make sure the RDSon of the FETs is not so large as to create large conduction losses.

As Angelo stated earlier, you may use the Ciss, RDSON, VGS threshold, Rg, and Qg values of the EVM's Q1/Q2 FETs, CSD17308Q3 for comparison when selecting your FETs.


Regards,
Joel H

Hi Bill,

ACDRV is a charge pump output which drives ACFET and RBFET. The current limit is 60 µA, as shown in the table below.

Best regards,

Angelo

Hi Bill,

The purpose of R3 and R4 is to limit the inrush current on the ACDRV and CMSRC pins. A 3 kΩ resistor should also be fine as long as its power rating is high enough.

Please take a look at this related thread which could be helpful to you: https://e2e.ti.com/support/power-management/f/196/t/690509

Best regards,

Angelo

Hi Bill,

The resistor value does affect the Q1/Q2 turn on/off time. A bigger resistor will cause a slower turn on/off time. What I meant with my last reply is that it's acceptable to use a 3 kΩ resistor instead of a 4 kΩ resistor as long as the resistor's power rating is sufficient.

As Joel stated, you can slow down the FETs' turn on/off time by increasing the gate resistance and/or increasing the gate-to-source capacitance.

Best regards,

Angelo

Hi Bill,

The purpose of R3 and R4 is to limit the inrush current on the ACDRV and CMSRC pins, so I wouldn't recommend changing these resistor values too much from the 4.02 kΩ value given in the datasheet.

Assuming a 20 V input, the power dissipated by the resistor would be P = (V^2)/R = 400/R. Therefore, a smaller resistor would need to have a higher power rating. This may mean that you would need to use a larger package size as well.

For example, a 4 kΩ resistor would need to handle 0.1 W, while a 2 kΩ resistor would need to handle 0.2 W. The 2 kΩ resistor would necessitate the use of a larger 1206 package instead.

May I ask why you want to change the component values used on the EVM? What are you trying to achieve?

Best regards,

Angelo

Hi Bill,

You mentioned Q = CV = It previously, which gives some clues about how to adjust the turn on/off time. These are the main approaches you can try:

1) Change the gate-to-source capacitance. A larger Cgs leads to a slower turn on/off.

2) Change the gate resistance. A larger Rg leads to a slower turn on/off.

3) Use a FET with a different gate charge. A higher Qg leads to a slower turn on/off.

However, we generally recommend sticking to the component values used in our EVMs. This is the circuit that we do our testing and validation on, so we can be sure that the device will work correctly with those component values.

Best regards,

Angelo