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Visitor
Visitor
9,078 Views
Registered: ‎10-28-2008

Power (current) measurement of Virtex 5 on ML505 (XUPV5-LX110T)

Hi All,

 

I want to measure the power consumption of my Virtex 5 FPGA on ML505 board (the board is actually XUPV5-LX110T) and by searching in this forum it seems the only way is to measure the current (voltage) of a specific resistor on the board. I have seen a post in this forum about ML405 board which they have mentioned the resistors that should be considered are R182 and R181 .

Can someone tell me the resistors I need to consider on my ML505 board for current (voltage) measurement??

 

Thanks,

Hessam

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Xilinx Employee
Xilinx Employee
9,066 Views
Registered: ‎01-03-2008

While the ML405 has current sense resistors and the ML605 has current resistors and on board instrumentation to read thec current, the ML505 (and by extension the ML506 and ML507) does not.

 

Sorry.

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Newbie
Newbie
9,053 Views
Registered: ‎12-23-2010

I am actually trying to measure peak or average power when running an application on the ML505 board. is there any other way to measure this? any work around? 

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Xilinx Employee
Xilinx Employee
9,051 Views
Registered: ‎01-03-2008

The only workaround would be to remove the supplies from the ML505 and source them externally. 

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Newbie
Newbie
9,030 Views
Registered: ‎12-23-2010

thanks for the reply, which supllies do you mean? if you mean the external power suply, then measuring the power would not be very inacurate? cause the power read would be for the entire board, including the lcd, flash memories, ports and the FPGA. it would be almost impossible to distinguish the fpga power, correct me plz if I am wrong.

 

thanks,

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Instructor
Instructor
9,028 Views
Registered: ‎07-21-2009

which supplies do you mean?

It depends on which supplies you want to measure.

 

It sounds like you fully understand the limitations of measuring an entire power supply, rather than specific loads for each power supply.  Apparently the ML505 board was not designed with specific, isolated power load measurements in mind.  In fact, it's downright ugly.

 

  • The VCCAUX (2.5V supply) regulator is a leadless surface mount package, it's not easy to remove without potentially lifting pads on the circuit board.
  • The VCCINT (1.0V) supply is a switcher module with pins. (note: I don't have a photo of the ML505 board, so I can't tell if the switcher modules are soldered in or socketed).
  • The VCCO supplies are generally impossible to separate from the rest of the board circuitry, with the exception of the VCCO_EXP supply which feeds 2 IO banks via a jumper header.

There might be other development boards available which will provide you with better access to the measurements you seek, or you can design your own (perhaps based on the ML505 design, with your changes added in).

 

Are you primarily concerned with cooling (package thermal dissipation) or power supply current?  You should consider a "paper" design in ISE, and then let the ISE Power Analyser guide you.  Sure, the ISE Power Analyser is only an approximation... but then again any measurement you take is also a one-off sample case rather than a rigorous worst-case analysis.

 

- Bob Elkind

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Xilinx Employee
Xilinx Employee
8,995 Views
Registered: ‎01-03-2008

> thanks for the reply, which supllies do you mean?

 

I meant removing the DC-DC power regulators on the board for VCCINT and VCCAUX and then wire in an external supply that has current measurement.  This won't be an easy task.

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Newbie
Newbie
8,975 Views
Registered: ‎12-23-2010

well, I doubt if I can make a good external voltage regulator as the on-board voltage regulator. 

from the answers I got I believe if not impossible it seems very difficult to have somewhat an accurate power measuremt. I looked at the xpower analyzer, and as it appears i need to generate a VCD for my application and then use xpower analyzer to measure the peak power when the applicaiton is running. generating the VCD file for a C code and then dump it to the fpga board is then another quesiton I have, can you help me on this.

I am using benchmakrs from http://www.eecs.umich.edu/mibench/source.html which are embedded program for mobile applications. so there are C codes (executable + input) and my question is how to generate these VCD file and then how to dump it to the FPGA board? any idea or thought on this?

 

thanks,

 

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Instructor
Instructor
8,973 Views
Registered: ‎07-21-2009

well, I doubt if I can make a good external voltage regulator as the on-board voltage regulator.

Perhaps not, but it's not difficult to solder in something which is perfectly usable.  If you are uncertain how to do this, it's best that you don't risk mucking up the board.

... use xpower analyzer to measure the peak power when the applicaiton is running

Xpower Analyzer needs a placed and routed design for a start.  It doesn't use a simulator or emulator, so there's nothing "running".  You can improve and tweak the results by providing estimated operational profiles (clock frequencies, toggle rates, etc.).

 

See the XPA summary page.  And this page.  Or better yet, put together a quick and simple design, and use the design to acquaint yourself with XPower Analyzer.

 

- Bob Elkind

SIGNATURE:
README for newbies is here: http://forums.xilinx.com/t5/New-Users-Forum/README-first-Help-for-new-users/td-p/219369

Summary:
1. Read the manual or user guide. Have you read the manual? Can you find the manual?
2. Search the forums (and search the web) for similar topics.
3. Do not post the same question on multiple forums.
4. Do not post a new topic or question on someone else's thread, start a new thread!
5. Students: Copying code is not the same as learning to design.
6 "It does not work" is not a question which can be answered. Provide useful details (with webpage, datasheet links, please).
7. You are not charged extra fees for comments in your code.
8. I am not paid for forum posts. If I write a good post, then I have been good for nothing.
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Visitor
Visitor
8,934 Views
Registered: ‎10-28-2008

Hi,

 

Is there any jumper on the ML505 which bridges the Vcc for the FPGA?? If there is such a jumper we can open it and connect a very small resistor in between, so the voltage drop can be very small and we can measure the voltage over that resistor.

 

Thanks,

Hessam

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Xilinx Employee
Xilinx Employee
8,587 Views
Registered: ‎01-03-2008

No jumpers exist on the supply rails.
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Adventurer
Adventurer
8,382 Views
Registered: ‎10-13-2011

What about cutting the pin and putting a small resistor between the output voltage of the 1V0 Regulator and VCCint? I think we could use this to measure current and also measure the voltage across it. Can anybody think of a reason why this woudn't be accurate?

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Historian
Historian
8,376 Views
Registered: ‎02-25-2008


@stuartdu wrote:

What about cutting the pin and putting a small resistor between the output voltage of the 1V0 Regulator and VCCint? I think we could use this to measure current and also measure the voltage across it. Can anybody think of a reason why this woudn't be accurate?


That's the usual method of measuring current, although of course you probably want to have a high-gain instrumentation amplifier measuring the drop across the resistor.

----------------------------Yes, I do this for a living.
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Adventurer
Adventurer
8,372 Views
Registered: ‎10-13-2011

What do I need to consider when choosing a resistor? Size, heat allowance, accuracy?

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Historian
Historian
8,369 Views
Registered: ‎02-25-2008


@stuartdu wrote:

What do I need to consider when choosing a resistor? Size, heat allowance, accuracy?


You have to balance voltage drop across the resistor against accuracy of your measurement. If you're trying to measure the current in a 1V supply, take care to ensure that the drop across the resistor doesn't reduce the supply below whatever its minimum might be. (RTFDS)

 

The higher the resistance, the larger the drop and the less gain you have to apply to it to be able to measure it easily. At the same time, the higher the drop across the resistor, the lower your rail.  

 

Power, of course, goes as I^2 * R, so the higher the resistance, the greater the power dissipation in the resistor.

----------------------------Yes, I do this for a living.
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Adventurer
Adventurer
8,357 Views
Registered: ‎10-13-2011

The Xilinx datasheet states that the current going into VCCint is 16A maximum and I found that the Quiescent current is 1.154A for my FPGA. I assume I would need to choose a resistor that only drops less than 50 mV at a current of 16A (3 mOhms or less). Or is there a way to define a smaller current range?

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Xilinx Employee
Xilinx Employee
8,353 Views
Registered: ‎01-03-2008

> The Xilinx datasheet states that the current going into VCCint is 16A maximum

I think that you mean that the ML505 schematics shows a maximum of 16A from the VCCINT power supply.

 

The output of the power regulator is a pin that directly attaches to the VCCINT power plane,  so it will be nearly impossible to add a current sense resistor.   If you must make this current measurement then you will need to remove the power regulator entirely and solder in an external power supply.

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Adventurer
Adventurer
8,351 Views
Registered: ‎10-13-2011

Yes I meant the schematics for the ML50x.

 

I am planning on cutting the pin and inserting a resistor between the output voltage of the 1V0 Regulator and VCCint. Why would I need to remove the entire regulator?

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Teacher
Teacher
8,345 Views
Registered: ‎09-09-2010

> I am planning on cutting the pin and inserting a resistor between the output voltage of the 1V0 Regulator and VCCint.

If that works out badly, an external supply with a current reading display would be a good thing to try.

------------------------------------------
"If it don't work in simulation, it won't work on the board."
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Historian
Historian
8,342 Views
Registered: ‎02-25-2008


@mcgett wrote:

 

The output of the power regulator is a pin that directly attaches to the VCCINT power plane,  so it will be nearly impossible to add a current sense resistor.   If you must make this current measurement then you will need to remove the power regulator entirely and solder in an external power supply.



Ed, where's your sense of adventure?!

 

Lift the regulator pin and squeeze an SMT resistor of appropriate value between the pin and the pad :)

----------------------------Yes, I do this for a living.
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Xilinx Employee
Xilinx Employee
8,167 Views
Registered: ‎01-03-2008

> Lift the regulator pin and squeeze an SMT resistor of appropriate value between the pin and the pad :)

 

The pin can't be lifted as it is a through hole pin and access to cut it as the OP wants to do is mostly blocked by other components.  The pin in question has been highlight in RED here:

u37_5.JPG

 

If the pin was cut then a 3mOhm sense resistor (an SMT 0402 wouldn't be precise enough or have the necessary wattage rating) wouldn't fit between the power module and the board.  Pulling the whole regulator off of the board is likely the easiest and quickest approach with the least chance of damage, but the OP is free to do as they please with their board.

 

> Ed, where's your sense of adventure?!

Sorry, I'm a bit of a kill joy. I said "nearly impossible" and maybe I should have said "extremely hard".

 

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Adventurer
Adventurer
8,161 Views
Registered: ‎10-13-2011

If I pull the whole regulator off and source the 1V0 VCCint with an external source, is there anything else I need to worry about? Would I need a source that can provide up to 16A? I know the regulator has a lot of other pins (track, SYNC, TT, +/- Sense, Inhibit). Would I still need to worry about these?

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Historian
Historian
8,152 Views
Registered: ‎02-25-2008


@mcgett wrote:

> Lift the regulator pin and squeeze an SMT resistor of appropriate value between the pin and the pad :)

 

The pin can't be lifted as it is a through hole pin and access to cut it as the OP wants to do is mostly blocked by other components.  The pin in question has been highlight in RED here:

u37_5.JPG

 

If the pin was cut then a 3mOhm sense resistor (an SMT 0402 wouldn't be precise enough or have the necessary wattage rating) wouldn't fit between the power module and the board.  Pulling the whole regulator off of the board is likely the easiest and quickest approach with the least chance of damage, but the OP is free to do as they please with their board.

 

> Ed, where's your sense of adventure?!

Sorry, I'm a bit of a kill joy. I said "nearly impossible" and maybe I should have said "extremely hard".

 


Hah, I didn't check to see how the regulator was installed.

"Extremely hard."

If it was easy, everyone would be doing it.

 

How about drilling out the pin? YEAH!

 

Every engineer has tales of heroic rework.

----------------------------Yes, I do this for a living.
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Xilinx Employee
Xilinx Employee
8,150 Views
Registered: ‎01-03-2008

> Would I need a source that can provide up to 16A?

You would need to have a source that can provide all of the current that your design needs.  How much that is only you can know.

 

> I know the regulator has a lot of other pins (track, SYNC, TT, +/- Sense, Inhibit). Would I still need to worry about these?

Only GND and VCC are needed.  If you have remote sense power supply then connecting the sense pins to +/- Sense would make sense. :-)

 

@bassman:

> How about drilling out the pin? YEAH!

The pin is fairly large (larger than a 100mil header), so I would be worried about drill wander and destroying the module.   You can't tell from the image, but the pin is larger under the module than the hole it is in, so it isn't possible to just desolder pin and pull it out. 

 

Now, if you removed the whole module from the board then it would be easy to desolder and remove the pin and the resolder the module into the board, but then you are back to figuring out how to squeeze a sense resistor into the board.  Inserting an amp meter between the module and the board would be a reasonable option in this case, but I still like the external power supply.

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Adventurer
Adventurer
8,141 Views
Registered: ‎10-13-2011

I don't think using an Ammeter would be a good idea, because typically the burden voltage is too high. The VCCint has to stay between 0.95 and 1.05 V.

 

I was just going to solder leads onto each end of the pin (after I had cut it) and insert a shunt resistor to measure the current across. I wasn't going to squeeze a tiny resistor in there because that sounds too hard. Hopefully the leads don't have too much resistance.

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Observer
Observer
8,126 Views
Registered: ‎06-14-2010

Hi,

I have experimented similar current measurement using clamp meter. I have removed the power module and cut the out put pin and soldered a thick wire with 1.5 inch length to the power module so that the clamp meter goes in the wire loop and soldered back the module on to the PCB. With clamp meter i am able to mesure the current on the VCCint.

 

regards

 

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Adventurer
Adventurer
8,115 Views
Registered: ‎10-13-2011

Was it difficult to solder the module back onto the PCB? Isn't VCCint a DC current and aren't clamp meter's not very good at measuring DC signals?

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Observer
Observer
8,111 Views
Registered: ‎06-14-2010

Clamp meters are accurate in measuring the DC currents also. Agilent clamp meter( agilent U1210 series) is used for measuring the current.

We have not faced any problem in soldering back the module.

 

 

 

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Historian
Historian
8,106 Views
Registered: ‎02-25-2008


@stuartdu wrote:

 Isn't VCCint a DC current?



It's connected to a DC supply, but the current draw is absolutely dynamic, and it's instructive to watch it on a 'scope display.

----------------------------Yes, I do this for a living.
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