10-10-2011 10:15 AM
I have a question about measuring the current in ML605 board. I have read the Xilinx documents and I was confused between two different options:
1- in the document ug370 in the section "ChipScope Pro Tool and System Monitor" (page 60) it is mentioned that the VAUXP is the external input we should read for measuring the current. For this case is this equation correct:
voltage on auxiliary input 0 = Current (amps) * .005 ??
2- in the document ug534 in the section "ML605 Board Power Monitor" (page 73) it is mentioned that the VAUXP is the external input that we should read to measure the current.
Which one is correct??
Am I right that the first one (VAUXP) is for measuring the current of the FPGA only and the second one (VAUXP) is for measuring the current of the whole board?? If this is correct then is this correct that for measuring the power of the FPGA we need to multiply the first one by VCCINT and for measuring the power of the whole board we need to multiply the second one by VCCAUX or should we multiply it by 12V??
10-10-2011 12:07 PM
The ML605 manual describes how to measure the current. The .005 ohm series resistors can be optioned in, or out using the header jump-jacks. Either use of a voltmeter measuring across the resistors, or use of the system monitor A/D may be used.
Xilinx San Jose
10-13-2011 02:07 AM
There is a little-known feature in chipscope that makes vccint power measurement much easier.
The feature is called VCCINT Power Measurement, and uses the SysMon.
1 - Open the SysMon console in chipscope, by double clicking on the SysMon Console listed under the device name.
2 - Select the external channel from the drop-down list on the menubar -VauxP/N is connected to the current shunt resistor. Note: Jumpers on pins 9-10 and 11-12 are required on J35 to enable the connection to the sysmon.
3 - Right-click on the VauxP/VauxN text displayed to the left of the voltage graph, and select VCCINT Power Measurement.
4 - In the diaglog that pops up, enter the shunt resistance in mOhms, and the number of digits of precision you want displayed
Chipscope will now plot the Vccint power in W - as it automatically multiplies the voltage data from the VCCINT channel, by the current measured through the shunt.
This same feature can be used to enter a custom transfer function for any of the external channels.
Hope this helps,
10-13-2011 10:29 AM
thank you for your thorough answer. It was a great help. I have done the first steps you mentioned before but the last step was nice. This pretty much solves my problem but still a small confusion I have is what does VAUXP measures (ug534 page 73)?? is that the power consumption of the whole board??
10-14-2011 03:21 AM
Yes, VauxP/N is the channel for the board current. VauxP/N is the channel for the board supply (nominally 12v).
In both these cases, there is a transfer function, which must be used in order to compute the actual value.
For the board current, the voltage at the sysmon input is vshunt*50, as there is a 50x amplifier to bring the signal within the sysmons 1V range, which also acts as a level shift. (see page 39 of the ml605 schematics for this circuit).
For the board voltage, the transfer function is via a simple resistor divider of 24:1 (also on page 39). This means 1V out = 25V in, or an attenuation of 25x - so to convert back, simply multiply the sysmon reading by 25.
There is a reference design (EDK based) that has a demo which reads back all of these sensors - you can find it here:
In the demo desgin files, there is C code (SysMon_Demo.c) which directly reads back all of the external sensors and calculates the vccint power, as well as the board supply voltage, current and power.
Note: There were some revs of the ml605 board, where the shunt/resistor divider values were different - so the C code in the reference design may need to be changed to match your board's configuration. These values are defined in the file SysMon_Demo.h.
10-20-2011 11:28 AM
In our work we need very fast power reading from the FPGA and I was thinking of using an oscilloscope for reading the current instead of using chipscope or sysmon (I know we can change the timing of the reading by writing to the registers but still the oscilloscope is fastest way). Based on what I see in Xilinx documents, the AUXP is connected to pin G11 of the virtex 6 and AUXN is connected to F11. Also these pins of the FPGA are connected to GND_107 and HA08_N of the FMC HPC head. Is it possible for me use an oscilloscope and measure the voltage across those pins of FMC conntector and by dividing that by 5mOhm, I get the current?? Also the Vauxn (pin HA08_N) will give me the Vccint ??
08-08-2012 05:50 PM
Also UG534 specifies to connect pin 9 to 11 and pin 10 to 12 using jumpers, not the way it was previously suggested. I don't know what would happen if you did it the wrong way.
Does anybody know how I could access this information internally from the FPGA logic?