Yellow is the critical current flow out, not sure where it goes from the output cap, C3.
Pink is the ground return, kinda sorta, theres a few critical loops in a circuit like this, some follow the path of least resistance, some follow the path of least impedance (follow their supply trace).
Keep in mind your goal is to minimize the size of this loop and minimize the resistance between yellow/pink at high frequency, (minimize impedance to ground).
Lets just quote the datasheet since it covers all of this:
"When planning layout there are a few things to consider when
trying to achieve a clean, regulated output. The most impor-
tant consideration when completing the layout is the close
coupling of the GND connections of the CIN capacitor and the
catch diode D1. These ground ends should be close to one
another and be connected to the GND plane with at least two
through-holes. Place these components as close to the IC as
possible. "
C1 and D2 in this case. C1 has a really, really, long path to ground for a power supply decoupling cap. Its also on the other side of the circuit from D2's ground. When your talking about high frequency PCB design when it says "tightly coupled" that doesn't mean just both tied to a ground plane. It means both ground pins are right next to each other, with a surface polygon pour connecting them and multiple via's to a ground plane right next to the pads.
Incidentally aren't your diode silkscreen's backwards?
"Next in importance is the location of the GND con-
nection of the COUT capacitor, which should be near the GND
connections of CIN and D1."
C3 is the output cap and its ground is roughly as far away from the other 2 as you can get.
"The FB pin is a high impedance node and care should be
taken to make the FB trace short to avoid noise pickup and
inaccurate regulation. The feedback resistors should be
placed as close as possible to the IC, with the GND of R2
placed as close as possible to the GND of the IC. The VOUT
trace to R1 should be routed away from the inductor and any
other traces that are switching."
Think your ok with this one, your probably better off running the trace from C3 further away from the inductor but its probably ok.
"High AC currents flow through the VIN, SW and VOUT traces,
so they should be as short and wide as possible. However,
making the traces wide increases radiated noise, so the de-
signer must make this trade-off. Radiated noise can be de-
creased by choosing a shielded inductor."
If were you i'd just use polygon pours for most of these connections, just make sure you have appropriate filtering in place and a shielded inductor.
"The remaining components should also be placed as close
as possible to the IC. Please see Application Note AN-1229
for further considerations and the LM2734 demo board as an
example of a four-layer layout."
If using 4 layer why not reference that app note? pretty much covers using a big ol' ground pour for the critical ground return and a pour for the SW output.
First, it's absolutely expected that the output current and input current (averaged over a switching cycle) are not equal in a switching converter. If the currents were equal, the efficiency couldn't be any better than a linear regulator's.
Now, let's look at a simple buck regulator:
When the switch (Q1) is closed the input current does indeed go into the load. But part of it also goes to recharging Cout, whose voltage has drooped during the "off" part of the cycle.
When the switch is open, the load still receives current, but it's supplied by D1 and Cout.
So there's no concern that by not drawing input current during part of the cycle you might fail to provide power to the load. It's just part of how a buck converter works.
Would a large capacitor on the input of my step-down converter do the trick?
A large capacitor (Cin in the schematic) won't change the fact that when the switch is open, no input current is drawn.
What it will do is, when the switch is closed, allow much of the input current to come from Cin instead of from the upstream voltage source. This current will be flowing in a relatively small loop, and so not produce so much EMI issues as if it had to flow from the upstream source, however far away that might be.
It also means that whatever inductance there is in the lines from the upstream source to your circuit won't cause Vin to droop during the "on" part of the switching cycle and interfere with the converter's operation.
Edit
I realized you're worried about the peak current drawn during the "on" part of the cycle being higher than the PoE can supply.
Yes, a larger Cin will help with that, by smoothing out the current drawn over the switching cycle. But basically any capacitance on the load side of the PoE will also help.
The choice of operating frequency and L1 value will also affect the peak current draw on the input.
Best Answer
I fixed the problem by using a higher value inductor(10uH) as suggested by the datasheet and the regulator does not heat up at all under light and no load conditions. I have tested it at half the maximum expected load (250ma) and I am getting 20mv pp ripple. I'll do more testing at the max load, but the results are very good so far.