Motherboard inductors

Hi, one.

Can't have that.

In general, to store some energy and give it up later. A great many more detailed purposes fall within that rubric.

Au contraire. Without inductance, signals would never get anywhere. Wave propagation would be halted.

Yes, indirectly. Take them out and any number of clocks will stop being generated, as the power forms they help condition or generate collapse.

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--Larry Brasfield
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Above views may belong only to me.

You're right. Those inductors are slowing your motherboard WAY down. The manufacturers put them in there for their low-end boards and if you pay a lot more, they simply take them out and presto, instant speed demon. That way they make a profit because they don't have to build two versions with a lot of different parts.

Jim

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Are you talking about soldered on components, or serpentine traces?

They are part of the switch mode power supply circuitry, used to convert from one supply voltage (e.g. 5V) to another (e.g. 1V).

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John Devereux

Almost invariably as part of voltage step down on board power supplies needed to generate the low voltages needed for modern CPUs and GPUs.

No.

Graham

Just now my 3.3V is running at 3.15V, and my Vcore = 1.52V, for an AMD Athlon64 3200+ processor. This beast's current draw varies dramatically as a function of whatever program is executing, a buck switching regulator located on the PCB must be able to handle this. This switcher is sourced from the +12V supply. The peaks currents are 70A, 120A or whatever, so the switcher is broken into multiple sections in parallel, four is common. There may be another supply for the ram or other chips. That's why you see so many inductors.

--
 Thanks,
    - Win

Out of curisosity, is this breaking-into-sections for form-factor reasons (bigger inductors would stick up to high and block plug-in boards or otherwise stick outside of the ATX form factor), for PCB reasons (for high currents you'd like to have thicker traces but instead you compromise to thinner copper because most traces are logic not power), component price reasons (sweet spot for inductors or semis), or something else?

I'm completely flabbergasted that umpteen-layer ATX motherboards are availble for $30 with all that stuff already on it.

Tim.

Resistance is futile! Capacitance is immoral!

A buck converter is a "switching regulator" that specifically has a lower output voltage than input voltage. There's also boost converters, inverting, and buck-boost, which have self-evident meanings.

Well, when you're talking about e.g., 100A, even with multi-phase converters you still have to pay attention to the capacitors' ripple current specs... :-)

Unbelievable garbage response...

[...]

Especially for writing ascii text messages to send to a newsgroup:)

Mike Monett

There are two reasons for dividing the load among multiple stages. It is much easier to design a 30 Amp switcher than a 120 Amp one and the efficiency is likely to be much higher. The main reason is that in a buck converter current is only drawn for a small proportion of the time. When converting from 12 V to 3 V the switching device conducts for 25% of the time. This means you need capacitors with enormous ripple current ratings. By running four stages in synchronism but each phased 90 degrees after the previous one, the input and output capacitors see almost constant current.

You're probably right about the thickness of the traces. 120 Amps is a lot for a pcb.

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Graham

While I have had a fair bit of experience with sitch mode, I haven't seen on of these mother boards. Two questions: Is what you are calling a a "buck converter" what I have always called a switching regulator? It sounds like it - i.e in your example, switch on 25% connecting 12V to LC filter?

This means you need capacitors with enormous ripple current

I assume, then, that four switches feed four inductors from one input capacitor and one output capacitor. Neat idea! The caps should see only minor switching transients.

Ted

Makes perfect sense. My experience with buck regulators is at the much-lower-current end (a few amps, not hundreds!)Thanks, Graham!

Tim.

In message , Joel Kolstad writes

Yes you do, but the capacitors have a much easier life. With four stages each running at 30 Amps the current through the first inductor might go from 25 to 35 Amps during the time it is being energised. Then the second inductor does the same and so on. The current drawn from the supply is a sawtooth waveform of 30 Amps d.c. plus or minus 5 Amps. The input cap' only sees the +/- 5 Amp bit.

A single stage with the same percentage ripple would draw zero current before the inductor is energised. This would jump to 100 Amps when the switching device turns on and then rise to 140 Amps before returning rapidly to zero for the remaining 75% of the cycle. Now THAT is what I call ripple current.

The output cap' also has a much easier time of things.

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Graham

To add a bit of concrete information to this discussion, here are some details about the switching supply in my last computer's mobo, an ASUS P4C800-E, with socket 478 for a 3.2GHz Intel P4 cpu. I've posted a photo on a.b.s.e., asus_p4c800.jpg Note the three big inductor coils. The ASUS P4 cpu motherboard uses a three-phase design similar to the one described in this Intel document,

See also,

The switcher starts with an inductor located at the ATX12V jack to reduce switcher noise radiated from the input power cable. This feeds a 1/4 to 1/2" copper 12V bus with four tall bypass electrolytics, 1200uF 15V, which provide the pulses of current. The large copper area is in part to provide cooling for the FETs.

The output of the switcher circuitry is a 1"-wide 3"-long copper bus, with a bank of 8 Nichicon 1500uF 6.3V HD-type electrolytics to ground,

According to the datasheet, these caps have an esr of 0.023 ohms at 100kHz, so eight would provide a 2.9 milli-ohm stiffness. If the load changes suddenly (too fast for the switcher to respond), say by 50A, these caps limit the voltage step to 150mV. This is a *very* critical issue, because if the voltage drops too much the processor will crash.

The P4 cpu runs at 1.52V, with currents in the 75A region, IIRC. The three-phase switcher on the ASUS mobo runs at 1MHz, using an Analog Devices ADP3180 controller IC. The '3180 clock is set to run at 3MHz with Rt = 56k. A new phase starts every 330ns, and each phase has a 1us period, so the three phases overlap.

At 75A cpu load, each inductor carries 25A of roughly-continuous current, and its input is switched between 12V and ground with two TO-252 D-PAK power MOSFETs, at a roughly 15% duty cycle. It's interesting that the ASUS designers, who are masters at optimizing performance vs cost, didn't use the same type of MOSFET for the 12V and ground switches. The 12V side is a Silicon Standard Corp ssm60T03H,

which is a FET with commonplace performance (30V 12-milliohms 45A), from a fabless semiconductor house - low cost, no doubt.

By contrast, the ground-side MOSFET is an ST std90nh02L, a 20V 5.2-milliohm 60A part that's very impressive compared to the 12V part.

This beast commands a higher price, and has been highly refined; its datasheet went through four versions in a short 12-month period.

Fairchild's fdd6670A or ISL9n308 are would-be contenders for the coveted low-side FET slot. They have higher Ron, but lower Ciss, allowing for reduced switching loss, which we haven't discussed.

.. +12 copper bus .. ---+-------------------------+------------------ .. | | two other phases .. +-|>|-+-||-, | high-side ,--#####-- .. _|_____|_ | |--' ssm60T03H | .. | |--|-- 2.2 ---|| +--#####-- .. | | | |--s inductor | .. --| ADP |--+--------------+----#####----+--- CPU cap bank .. | 3418 | |--' 1.52V Pentium P4 .. | |-------------|| low-side .. |_________| |--s std90nh02L .. | | (two pad sets) .. '-------------------+ .. | .. mobo gnd plane

Another interesting tidbit, the ASUS mobo has blank spots for an extra FET in parallel with each of the three ground-side FETs.

Let's do a simple analysis of the two FETs as they do their job. First, the '60T03 high-side 12V part, with its 12-milliohm Rds(on), will drop 300mV at 25A. That may seem a lot, but the 12V switch output is still 11.7V, so it's not so bad. Second, consider the '90nh02 on the ground side: its 5.2 milliohms drops only 130mV, but compared to the 1.52V output it's significant (8.6%). Now we begin to see why the ASUS designers carefully selected this part. Oh, one more thing, during the low part of the cycle the FET drain voltage is negative 130mV, not +130mV. Think about why this is so. And remember, these are N-channel FETs (just a little subtlety).

The voltage across the inductor is 11.7 - 1.5 = 10.2V during the short high part of the cycle, and -1.52 -0.13 = -1.65 during the longer low part, ignoring the resistive drop in the inductor. Since inductor dI = V/L dt, and at equilibrium +dI = -dI for each cycle, we know the +t and -t times are inversely proportional to inductor voltage. Now we can calculate the high-side duty cycle, 1.65/(10.2+1.65) = 13.9%, or about 15% of the time at 12V, and 85% of the time at ground. This means the inductor current rises fast for 150ns, then drops slowly for 850ns back to where it started. Also, each 850ns portion overlaps several 330ns phase-intervals.

The relative power dissipation of the two FETs is 12*15% = 1.8 for the high one, compared to 5.2*85% = 4.4 for the low one. That's 2.4x higher for the low one, even tho it's a much lower Ron part. Ahah! Now we see another reason the ASUS engineers were so picky about the ground-side clamp FETs, and why they designed the PCB to allow for mounting two paralleled low-side MOSFETs for each phase.

Each pair of FETs is driven its own Analog ADP3418 mosfet driver,

This is a single dricer, so three are required. The designers may have been tempted to use cheaper dual driver ICs, but the high FET gate capacitance (2850pF for the '90nh02) and the 1MHz driving frequency places considerable thermal-dissipation stress on the driver. This issue also limits the size of the low-side FET, and hence its Rds(on) performance.

The 12V FET's gate has a 2.2-ohm series resistor, but the bottom one is driven directly. Each driver IC is located within 1" of its FETs, but the controller IC is located 4" away from the most distant IC.

OK, gang, that's the mobo cpu-switcher teardown story, pictures at 11.

--
 Thanks,
    - Win

A followup zoomed photo is on a.b.s.e., "Re: ASUS P4 mobo photo, for s.e.d. "Re: Motherboard inductors" - asus_p4c800.jpg"

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 Thanks,
    - Win

The inductors are wound with three paralleled #14 wires, for a calculated Rdc = 0.8 milliohms (and measured, below 1 milli-ohm). We know the inductor's ac resistance may be higher due to skin and proximity effects, so this issue bears more detailed examination.

--
 Thanks,
    - Win

Thanks very much for the description of this power supply Win! I found it very interesting. Posts like this one make s.e.d really worth the reading time :)

Regards, Alan

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