I'm using the KSZ8081RNA/RND 10BASE-T/100BASE-TX PHY with RMII support.

This is the hardware design checklist of the IC.

On the page 2 of the checklist, there is a power supply connection figure.

What happens if I don't populate the 2.2 uF and 100 nF capacitors on pin VDD_1.2? What is the risk of leaving out those capacitors?

There are more than 10,000 boards that have been done without placing these capacitors, but have passed the functionality tests. The working environment temperature is 65 degrees C. It can get hotter but does not face cold temperatures.

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    \$\begingroup\$ Good question. I don't think anyone can say with certainty what will happen. I have some questions for you though. How much will your company have to pay if the boards stop working later? How much will it cost to fix the boards? If they have already been delivered to customers you are in a real difficult situation. If there is any possible way to fix them before they go to customers, I know I would insist on doing that if I were in your situation. \$\endgroup\$
    – user57037
    Aug 8, 2022 at 5:23
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    \$\begingroup\$ Oh, just to clarify, the board layout has places for the capacitors, right? This is just an assembly snafu, right? \$\endgroup\$
    – user57037
    Aug 8, 2022 at 5:47
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    \$\begingroup\$ Are there any decoupling capacitors for ICs sharing the same power rails very close to the IC in question? And you should also take a sample of boards and leave them running 24/7 under all combinations of operations and monitor them for problems. And are there power and ground planes for the 1.2V right below the IC? \$\endgroup\$
    – DKNguyen
    Aug 8, 2022 at 5:53
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    \$\begingroup\$ Yeah, if possible, make a screenshot of the layout near that pin and add it to the original question. I realize you already selected an answer, though, so if you feel you are finished, so be it. \$\endgroup\$
    – user57037
    Aug 8, 2022 at 6:34
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    \$\begingroup\$ Even without pads, 0402 capacitor could be relatively easily soldered to the IC pins. Having just 2.2µF ceramic capacitor close to the pins is almost as good as having the 2.2µF + 0.1µF combination. \$\endgroup\$
    – jpa
    Aug 8, 2022 at 17:51

5 Answers 5


You can't predict what will happen if there is insufficient decoupling.

You can get any manner of glitches, hiccups, errors, and latch-ups and it may be the same one every time, or a different one every time. They may occur like clock work or just once in a blue moon, and you may be able to replicate them or you may not be able to replicate them.

Basically, if there is any problem in a process involving the IC in any way, you won't be able to identify if the problem is being caused by insufficient decoupling or something else.

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    \$\begingroup\$ @Newbie The concept is marginal behavior. If there is a voltage dip on a power rail, the power rail voltage may become marginally too low. The whole concept of "marginal" is that correct operation cannot be guaranteed. I suppose that any bit anywhere in the digital logic chain could be in an incorrect state due to the power glitch. \$\endgroup\$
    – user57037
    Aug 8, 2022 at 5:41
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    \$\begingroup\$ @mkeith And that power glitch might only occur for a certain sequence(s) of states. So when it does strike, it might strike not just at different parts in the logic chain, but at different times. \$\endgroup\$
    – DKNguyen
    Aug 8, 2022 at 5:42
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    \$\begingroup\$ Once in a blue moon = 1.16699016 × 10^-8 hertz. Sorry, couldn't resist :D \$\endgroup\$
    – Mołot
    Aug 8, 2022 at 13:32
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    \$\begingroup\$ On the other hand the voltage might not become marginally too low. But how do you know? You don't know. Hence what mkeith said about: how much would it cost if it was? \$\endgroup\$ Aug 8, 2022 at 18:59
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    \$\begingroup\$ But how do manufacturers know how much decoupling is required? Are they not just doing testing with certain capacitors across a wide range of operating conditions, samples and voltage regulators? If OP does vigorous testing (maybe even under worse conditions than would be expected in the field) wouldn’t it be good enough? \$\endgroup\$
    – Michael
    Aug 10, 2022 at 6:17

These capacitors smooth the output voltage of the internal linear regulator that powers the logic circuitry. Without these capacitors the voltage will glitch whenever there is a change in current draw (which happens whenever a logic gate changes state). This could cause the device to misoperate in unpredictable ways.

Just because it passed functional tests doesn't mean it won't fail in the field. The tests almost certainly were not comprehensive enough to cover every possibility, since without intimate knowledge of the chip's internal operation you cannot say what situation would cause the greatest current variation or whether the logic would be affected by it. The tests may even have missed actual failures that didn't fall outside its limits, but may cause trouble in the field under different conditions or as the boards age.

This is not just theoretical - anyone who has been in this game for long has stories to tell.

What is the risk of avoiding those caps?

The boards will probably be unreliable in the field. This will upset some of your customers, create ongoing support costs and damage your company's reputation. If the boards are being used in a critical application where failure causes injury or loss of life, those responsible (or who knew about it and didn't fix the error) could be imprisoned.

You know what needs to be done. The boards should all be recalled and reworked or replaced.

  • 12
    \$\begingroup\$ This is an Ethernet PHY chip. Most temporary internal glitches will be indistinguishable from other kinds of packet errors. The Ethernet protocol (and the IP protocols that run on top of it) are fairly robust with respect to packet errors. The primary effect will be increased latency and reduced peak throughput as a result of excessive packet retransmission. \$\endgroup\$
    – Dave Tweed
    Aug 8, 2022 at 14:42
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    \$\begingroup\$ Yep. Functional tests are usually very orderly, and an orderly execution trajectory misses most states. Years ago, I worked at a computer company that was having trouble with field failures: some machines passed comprehensive manufacturing tests but wouldn't run reliably. I wrote a diagnostic that was deliberately as chaotic as possible, with an execution trajectory based on races between asynchronous clocks combined with a bunch of pseudorandom hashing. The engineers hated it because it couldn't say why memory was corrupted or some invariant broke, but it was great for production. \$\endgroup\$
    – John Doty
    Aug 8, 2022 at 16:04

My question is, on the pin VDD_1.2, what if I don't populate the 2.2uF and 100nF capacitors?

The two bypass capacitors have different roles. Every capacitor has internal inductance (often called leakage inductance) that affects the time-domain response of the (CL) capacitor-inductor to changes in the current drawn by an IC. In effect, the bypass device's CL is in parallel with all other devices on the board, through the power cables, and back to the power supply. The difference is that, being closer, this bypass device has less inductance on the PC, so the capacitance will be more effective.

The 2.2uf capacitor has more energy storage, so it can filter our a larger total energy change. It typically has a larger leakage inductance though, so the time-domain response of the CL is worse, and may allow a supply voltage dip that affects device operation.

The 100nf capacitor has less energy storage, but lower leakage inductance. It can more quickly supply or absorb energy.

Depending on how your PC was laid out and constructed, there is also capacitance between the power and ground nets of layers, which also provide a bit of filtering.

To analyze this, you should check the distance from the nearest alternative bypass device. If there is another 100nf capacitor "close" to the IC, you may be OK. If this is the only 2.2uf decoupling on the board, you may have a problem.

If you have a problem, you can expect random state transitions flip-flops, also known as state bits, in your IC. Even if your IC is purely combinatorial, you may get glitches in the outputs or slower propagation delays at seemingly random times. This can affect the setup and hold time requirements of subsequent inputs.

I see that this is an only EtherNet PHY. Most of the chips I've used are PHY/MACs, which have more state that can be corrupted. Even the naked PHY implements 32 registers of 16 bits which contain the device configuration.

There are many bits of state that can be corrupted.

Some of these will make the device inoperable until reprogrammed. Some drivers reprogram these bits frequently, and some program them only rarely, or even only once.

This PHY is probably connected to a MAC, perhaps in an onboard SOC.

Some errors result in packets being rejected, which as you noted, is not a fatal problem with ethernet. It will reduce performance. What is the packet timeout (retransmit time) on your network? With the standard specified values, a dropped packet results in a significant performance reduction, especially for TCP. What are your customer's expectations? Will they notice? If not today, might they later as the customer upgrades various parts of their systems?

Data from the PHY is passed to the MAC through an RMII interface. Does this interface meet the timing specs when the power does not? Probably not. Here the situation may be worse because some corrupted bits may result in corrupted data being passed to the system memory. What happens depends on the ethernet stack, and how careful it is about recomputing (at CPU expense) checksums. If the packet checksum was computed correctly, but the data was corrupted in the buffer, will the system use the corrupted data? What further errors will that corruption cause? Since we "know" the ethernet packet was correct, some drivers and protocol stacks will avoid a further checksum in the interest of efficiency.

Final thoughts:

One of my key rules of engineering: If it takes more than three sentences to describe why something works, it doesn't.

There is much that can go wrong. The process of determining which mitigations you have in place, and adding one you don't, is costly. Add those costs into your cost-benefit analysis when deciding if you need to write a recall, manufacturing deviation authorization, rework instructions, or whatever paperwork you have in your shop. Your name will be associated with the decision. Costs today are quickly forgotten. Recurring problems, loss of company reputation, canceled orders, negative reviews -- these don't have a chance to be forgotten and will stay with you.

Make things right for your customers and your company will grow.

You can even turn this into a positive PR statement, either to customers or at least in your internal communications, to build your quality reputation. Every company makes mistakes. The good ones fix them.


A few days after I wrote this answer, a friend sent me a link to an article about a company in a similar situation -- manufacturing had not built the board with the right set of components, but the board (nearly) worked as required. This company informed its customers, and is building both better internal processes and an external reputation for responsible, quality engineering. Here is the link: Another consequence of supply shortage: mass production mishaps


From listed document:

• Pin 1 is the internally generated 1.2V core power rail. Connect 2.2-µF and 0.1-µF capacitors from Pin 1 to ground using wide traces as appropriate for power distribution. Do not connect an external 1.2V source.

You have no decoupling on what appears to be the core power for the chip. You have 10,000 boards that pass functionality tests. Functionality tests are minimal go/no go tests that prove the board works (as in the pins change state), but actual operation will stress the board beyond these basic tests.

You have no decoupling caps and most importantly, you have no pads for them. An expensive kludge would involve soldering 100nF directly to pins, but 2.2uF would be a stretch.

Take a number of boards. Kludge some of them and test the crap out of them. Log errors and compare errors to boards kludged with capacitors. That will truly answer your question, but deep down you already know the answer.

Odds are that the non-bypassed will have more failures. Now, you won't be completely sure all failures are due to the lack of decoupling capacitors, but both documents listed clearly state the need for decoupling caps on 1.2V core power rail.

The output of the 1.2V regulator will fluctuate with load. Capacitors filter or stabalize the fluctuation. I'd guess a lot of errors would occur when board goes from no-load to full-load.

What happens when users in the field get errors that you have no idea are caused by the lack of capacitors or something else? Your support engineers will have an easy option to blame, which may or may not be the problem. This will not help your client base.


The answer is the same for all components.

If the component is used outside the conditions listed in the data sheet, then the component might not work within the specifications listed in the data sheet. If an important condition is missing, such as bypass caps, the part may not work at all.

If you need a specific answer in what ways the part might malfunction then you need to ask chip manufacturer for that.


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