I would aslo use some larger (possibly tantalum) capacitors for C1 and C2, especially fo C1. I know these are what's suggested as minimum values in the TI LDO datasheet, but the circuit is going to be connected to a long USB cable which makes a charge capacitor more important. I guess there's room for a larger C1 since you have made it leaded, but might be a good idea to chage it to round polarized capacitor so you can see what's + and what's - if you choose to use a tantal capacitor here.
Here's what TI writes about C1:
http://focus.ti.com/lit/ds/symlink/tps71533.pdf
And the device must be said to possibly be more than just several inches away from the
power source which is the PC through a long USB cable.
Your have to move your USB connector closer to the edge of the PCB, otherwise there's no romme to put a plug in the connector.
You could use a leaded crystal and the place the two crystal capacitors directly below the crystal on the blue layer. The footprint of a leaded HC49 crystal is smaller than the footprint for an SMD HC49 crystal anyway becuse you don't need large pads to go outsie the crystal house. Doing this would make the traces for the high frequency crystal shorter and would take away board space on the left side. If you are not mounting the PCB by
machine, then it doesn't matter if you have placed all components on the same side or not. Then you could take away some of the PCB on the left side which would also move the USB connector closer to the edge without moving it further away from AVR. You could also move J2 closer to the AVR.
You should remove D2 from your your circuit again, as it is a redundant diode and serves no purpose (see AVR042 figure 3-1).
The reset pin already has a diode to GND internally. It's only the diode to VCC that's missing to allow high voltage programming.
Everything inside the "Microcontroller" box is inside the AVR. Only C, R and the diode to VCC is not inside the AVR.
You should also move D1 further away from the screw terminal if you want to be able to solder D1. By removing D2 you will have room to turn D1 45 degrees which will solve this problem.
You can also safely remove either C5 or C6 and route the two digital VCC pins pin 4 and pin 6 together under the IC. One cap here is plenty, no reason to use two caps here. The leaded part doesn't even have two VCC pins, it's just because the SMD package has 32 pins instead of 28 pins, so they have to use the remaning pins for something. They might even be connected directly to each other internally. You can try to test it with an ohmmeter.
After looking at your PCB layout I don't see any purpose of the 0 ohm through hole resistor R6. Instead just place a single via where R6 is placed to move the red trace from the AVR to the blue trace of the ISP connector. No reason to add two vias (holes) and a resistor when a single via is plenty.
You have also routed pin 2 of the ISP connector to the wrong place, this should have been routed to VCC (3.3V) instead of the USB supply voltage as you have done. You don't have to worry about damaging the LDO with reverse current while you power VCC (3.3V) directly from the ISP with the USB connector disconneted.
Here's what the LDO datasheet says about this:
http://focus.ti.com/lit/ds/symlink/tps71533.pdf
That said, have you considered turning the ISP connetor 180 degrees? Looks like this will offer better rounting.
Looking at your PCB layout and 3D rendering I can see the SMD caps and resitors are two different sizes. Maybe you should change them to all be either 0603 or 0805.
Have you considered designing the PCB to fit a small standard plastic case?
Looks like the schematic is from CadSoft Eagle?
Can Eagle also make 3D renderigns now, didn't know that? Looks like the 3D renderings from Altium Designer.
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