Unused ICs on Intel network card?

[andru123]

I have bought an Intel i210 network adapter. Looking at the board, I tried to analyze what DC-DC converter it is using.
But, to my surprise, i see that the board is filled with IC components that are not used!
Am i reading it wrong? It does not make sense, why would Intel put unused components on their own reference board?
The central controller - i210-AT - is a gigabit network chip, one of the most reliable and power efficient network chips from Intel of the last 15 years.
It has a built-in Voltage Regulator, and can operate in 2 modes:
- using built-in VR it only needs 3.3V as power source
- with external power supply it needs 3.3, 1.5 and 0.9V
Intel recommends to use the internal VR and just connect 3.3V.
The board draws 3.3V from the PCIe pins. looking at the IC chips on the board I see:
- winbond flash memory module (bottom)
- TI low power voltage detector TPS3803 (https://www.ti.com/product/TPS3803) middle
- TI buck-boost converter TPS6303 (https://www.ti.com/lit/ds/symlink/tps63030.pdf?ts=1691719718933) on the top, marked CEE

The problem is that the converter output does not go anywhere! Neither is the voltage detector.
Why would Intel put non-working parts on their board?
I see that some OEMs that use exact same board layout, remove all these (not needed) elements: https://www.tonitrus.com/en/server/accessories/others/10148012-014-hp-e0x95aa-intel-i210-t1-netzwerkkarte-pci-express-1.000-mbps/
The same card but without any extra ICs. It works fine I guess.

Can you comment please, am I reading it right? Are these DC-DC converters not used? Why would Intel place them on the board then?

The i210 datasheet: https://cdrdv2-public.intel.com/333016/333016%20-%20I210_Datasheet_v_3_7.pdf
I attached a few screenshots from their manual showing the i210 pinout (page 21), Power connectivity design (page 833) and the board layout (page 819). However even there you can see that the DC converter in the top right corner is not connected to anything. What is going on?

[andru123]

Attaching more images

[tychob]

Looking at the boxes with 1 to 6 in the top left of the card, I suspect this is a 6 layer stackup, and that the connections are via an internal layer, not just the top and bottom. It makes sense that they would want a top and bottom ground plane on a board like this, and I suspect from the note on the schematic that there is a large shape on one of the inner layers for transferring power.

[T3sl4co1l]

Are you just assuming because of a capability (internal regulator) that they are using it, and put extra parts on without?  Have you confirmed with the schematic (do they release any documentation for this "reference board" at all?), or at least continuity checks?  If your only inspection is visual, and you aren't aware of multilayer boards, yeah, that would be an oversight.

Tim

[DavidAlfa]

No way, makers would kill to save few cents.
Test connections with the DMM in beeeeeeep mode you'll quickly find out it's for the SoC or something else!
You can clearly see the internal ground (Or power) layers.

[andru123]

Thanks guys!
I did beeep tests (man, these pins are tiny) and I see that:
- pin 39 is connected with 47, and 38 with 32,11, 42, 59
- Vout of DC converter is connected with 3.3V input pins (10,..) of i210 chip
- the PCIe slot incoming 3.3V is connected to Vin of the TI DC-DC converter (via the voltage sensing circuit)

So, i210 is using its internal VR. I assume Intel uses the extra ICs to stabilize the 3.3V from PCI bus, that is the only purpose? Not really needed, modern PSUs provide stable 3,3V, but nice from Intel.

[DavidAlfa]

I guess Intel know one or two things about computers!
ATX psus are pretty noisy, even the best ones, and network cards basically work by (de)modulating very high frecuencies, probably they did so to get the cleanest power possible.

[wraper]

https://www.intel.nl/content/dam/www/public/us/en/documents/schematic/i210-at-82574-nic-reference-dual-design-schematic.pdf

[mariush]

The 2nd contact from the right is 3.3v aux, and there's a bunch of unpopulated pads there...  maybe they were thinking of implementing some higher efficiency stand-by/sleep power for wake on lan or something like that?

Card's from 2016, when group regulated power supplies were more common, maybe that's why they want to stabilize 3.3v ...

[andru123]

@wraper The voltage sensing is on the left? How does it work, the 3,3V is divided by the 2 resistors, to match the sensing circuit range?
Is not it extremely inefficient, as there will be current flowing through the resistors to the ground?

[wraper]

Well, of course there will be some current. But you can calculate it yourself and it will be very small. This is not exact circuit as you have, it's a reference design with different converter IC. However there is a reason written why buck-boost was used and that it may be not necessary. I don't think there is any reason using diodes for power OR-ing in PCI-E card. So IMHO they either just made the same as on reference design or decided that 3.3V PC power may be not accurate enough.

[JustMeHere]

You can see one of the GND or PWR pane in the board. It's the dark green area.  The stiching around the edge suggest there are also internal signal layers. That's probably where most of the GB speed magic happens.  The pair of panes and the stiching makes this a EMI free zone.  That also helps with the card not emitting noise.
 

[andru123]

I have calculated, and with 3.3V / 19.800 Ohm = 166 microAmp. This is quite high, as the circuits self talk about few micro Amps of quiescent currents, or even nano Ampers. I could understand if it was really needed, but here it looks like it is there only to “fit” the range of voltage sensing circuit. Total waste. They should have used another voltage sensor targeted at 3.3V.

[ve7xen]

I guess the reason is that they need to power the chip during sleep mode from 3V3 AUX, however they don't want to rely on this being available (it is optional per PCIe spec), and also it may not provide enough power during full operation to drive the chip (it's limited to around 1W). So they diode OR the two 3V3 rails and then boost it back to 3V3 for the internal regulator. They might use the supervisor IC to turn off the 3V3 AUX when the main 3V3 is available?