Why use small resistor for Vcco voltage regulator

In Xilinx application notes XAPP457 (

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the resistors used to adjust output voltage is only 22.6 and 38.3, shown in Figure 1 in the pdf file.

According to LT1763 data sheet(

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, shown in Figure 2. Adjustable Operation, the R1 should not greater than 250K.

My question is why in Xilinx application notes XAPP457 using such a small resistors? Is it better to use 22.6K and 38.3K resistors when the power consumed by resistors and the current bypass are much small?

Reply to
jasonL
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,

How much current passes thru 22.6K ? How much current passes thru 250K ? What is the ADJ pin current ?

Now, why does that difference bother you ?

One other reason to lower the dividers, is to make the node less susceptable to noise pickup.

If someone touches that PCB node, whilst it is on, do you want the FPGA to pop its lid ?

-jg

Reply to
Jim Granville

Hi Jason, It's because the app. note probably has a mistake. The values given will work, I guess, as they are less than 250k, but I would use 22.6k and 38.3k. You should email Eric, the XAPP author, and see what he says. You can find his email address in the C.A.F archive somewhere. Cheers, Syms.

Reply to
Symon

Sometimes low values are good to provide a minimum output load but these are so low that they would need to be probablly 1206 sized resistors, or bigger, to take the power dissapated.

Higher values also react more with parasitic capacitance to form a filter that might slow down response of the regulator and sometimes this is a reason to use lower values.

John Adair Enterpo> In Xilinx application notes XAPP457

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Reply to
John Adair

From the appnote:

"Careful inspection of the original design reveals that the impedance of the feedback network is atypically low. Even with zero load, approximately 50 mA is lost through the feedback network. At first, this loss seems undesirable, but it serves an important purpose." >

-- Bas

Reply to
Bas Laarhoven

So now we know! I knew Eric wouldf have his reasons. Ta, Syms.

Reply to
Symon

50mA does not stack up, but checking I see the OP has made an error.

The values quoted in XAPP457 are NOT K, but 22.6 and 38.3 OHMS!.

I'd call that an atypical app-note design choice:

  • It commits to 150mW of heating, in what are voltage divider components.
  • It means that later change of the 50mA level, requires change of TWO components that also determine the Vcc.
  • If someone removes the ADJ regulator, and drops in a 3.3V one, they may completely miss the LOAD element of that divider.

Such casual addition of 50mA of base load, reflects on some of todays FPGAs current consumption.....

-jg

Reply to
Jim Granville

Hello,

The resistor selection is intentional. In this application, it is desirable to shunt 50 mA directly to ground. You can do that by adding a third resistor, or by sizing the voltage divider to accomplish the same effect. This "wasted" current prevents the regulator output current from dropping to zero when there is a lot of overshoot. If you are interested in reading more about it, the application note XAPP457 does contain a discussion on this topic.

Also, if you can somehow guarantee other components powered by VCCO will always be drawing some minimum current, you can subtract that from the 50 mA and size the voltage divider to be less leaky...

Eric

Reply to
Eric Crabill

Eric Crabill schrieb:

"Careful inspection of the original design reveals that the impedance of the feedback network is atypically low. Even with zero load, approximately 50 mA is lost through the feedback network. At first, this loss seems undesirable, but it serves an important purpose. The I/O protection diodes in Spartan-3 Generation FPGAs form a charge pump that can inject excess bus switching energy into the VCCO and GND rails. This effect is pronounced when the device is not participating in a bus transaction, but switching activity from other devices generates overshoot/undershoot on the bus signals. This effect poses a problem for a regulated VCCO rail. Most regulators do not tolerate reverse current and might lose voltage regulation under such circumstances. One way to solve this problem, at the expense of slightly higher power consumption, is to shunt current to the GND rail as shown in Figure 2."

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For quick access. Sounds logic.

Regards Falk

Reply to
Falk Brunner

Hi Eric,

Yes, Bas pasted that. My point was that saving one resistor in a App note is perhaps the wrong emphasis, and you would be better 'protecting the novices from themselves' ?.

If that 50mA really is important, the biggest risk seems to be someone cutting out the ADJ block, and using a Fixed 3,3V reg [thus saving two more pesky resistors ;) ] - but not realising they lost the 50mA load at the same time.

-jg

Reply to
Jim Granville

Thank you all, I get the point to have small resistors. If a little higher cost is allowed, are there any other solutions consuming less power to better handle overshooting?

Reply to
jasonL

Sure

- find a regulator that is able to also sink current. The DDR termination regulators are designed to do just that.

You could also measure your own overshoot energy in each design, and target that.

-jg

Reply to
Jim Granville

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