First PCB design, PWM Motor Controller feedback

Question

EasyEDA Project

I made my first PCB design and I would like to hear a quick feedback to it. It is a motor controller board, with two MOSFETs parallel, driven by a MOS driver IC and the PWM signal comes from a Attiny85 IC. The USV signal will measure the battery voltage and stops operating the device if below a certain voltage. USV is a simple voltage divider and will drain the battery, I am okay with that design flaw. AIN will measure a analog signal and adjust the PWM regarding to that. There is also a ISP connector to program the Attiny.

The DC-DC Converter should be a standard application. Everything regarding BAT DP is experimental, I don't know if this will work, I can't even explain how it will work since I have no insight of the battery circut. I found a old reference design and copied it, hopefully it's okay.

The design should operate with 180W and I will test it up until ~600W and see what happens. VCC will be from 14-20V.

If someone feels like giving me tips or finds mayor errors, I would like to hear that.

Schematic:

PCB design:

Datasheets:

• Q1, Q2
• Q3
• Q4, Q5
• U1
• U2
• U3
• D4

Answer 1

What's the purpose of the capacitor on the output of gate driver IC, and where did you found that 100 ohm gate resistors are good choice.

Next, why the gate driver is so far away from MOSFETs? Also, gate traces may be too thick, the inductance is increased, however you didn't mention the PWM frequency, if enough low this won't be an issue.

I don't see that MOSFETs have both drain and source connected on PCB layout.

EDIT:

The absolute maimum rating for U3 is 20V, while datasheet says that the working voltage is 4.5 to 18V, therefore you can't use 20V as Vcc.

The cap on the U3 output mentioned on page 10 of its datasheet is used for a test, it simulates the gate capacatance of the MOSFEET, but it is not intended that you mount a capacitor on the output in real project.

The gate resistor is usually very low from 3 to 10 ohm, surely not 100.

Answer 2

Q2 has the same gate drive Voltage as the drain Voltage (both are Vcc), so it's being used like an emitter follower (common collector) NPN transistor. With a typical MOSFET, this means it will drop about 7 Volts, which is way too high. When that burned out the lower MOSFET burned out too for obvious reasons. There is a simple circuit that uses diode + zener + capacitor + 2 resistors + NPN transistor to boost the gate drive Voltage. There is also the IR2110 and similar ICs to do it for you. D1 is the diode for the buck converter, and it will blow if output current is high and the output Voltage is low compared to the input to the buck converter. In this condition the buck converter will turn on with a small duty cycle and rest of the time all the current goes through the diode.