Zedboard cannot be powered on

[Leo_W]

Hi, digilent,
This zedboard could not be powered on. After testing, I determined that the IC27 and IC28 were damaged.

It can be started normally after replacement. However, the power supply output of IC28 was abnormal: 1V5 and 1V0 were marked on the drawing, but the measured results were 1V0 and 1V5, which were just reversed.

I replaced it with a new IC28 and it's still upside down. Is this the new zedboard design?

I have a zedboard that I bought a long time ago and the voltage matches the drawings. I suspect this may be the cause of the damage to IC27 and IC28.

Thanks!

[JColvin]

Hi @Leo_W,

I'm looking at the latest version of the schematic for the Zedboard , https://digilent.com/reference/_media/reference/programmable-logic/zedboard/zed_sch_rev_f1-public.pdf, and on the first page under Rev E changes it lists that "Interchanged VCC1V0 and VCC1V5 on power supply channels to allow for larger current on VCC1V0".

This appears to be reflected on page 15 of the schematic where on Rev D the VCC1V5 is associated with pins 3 and 6 on IC28 (and VCC1V0 is associated with pin 18), but this arrangement is flipped on Rev E and Rev F, with the primary change of note being the swap in the resistor divider used in the feedback line (R252 and R256; Rev D uses a 0.1% 6.7k Ohm as R252 and a 0.1% 15k Ohm resistor as R256. The location of these two resistors are swapped in Rev E and Rev F).
Rev F also looks like it had a further revision by changing the 0.1% 15k Ohm resistor on R252 to instead be a 0.1% 14.3k Ohm resistor to "Increase VCC1V0 nominal voltage by 19 mV to allow for power distribution plane drop in high-current applications".

My understanding is that all of the power routing based on this switch was accounted for (or otherwise the DDR3 memory would likely not work if it was instead supplied at 1.0 V rather than the nominal 1.5 V), meaning that I would like to believe that this swap in power supply routing is not the cause of failure in this situation.

Let me know if you have any questions.

Thanks,
JColvin