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Hi everybody,

I am using Altera DE2 board for the implementation of some VHDL designs. I wonder is it possible to find the way to MEASURE the power consumed by the design? Could I, hypothetically, insert the resistor between the FPGA power supply (VCCINT) and FPGA device and by measuring the current through the resistor to caclulate the power ? http://www.mypicx.com/01312012// This means to unsolder the TC11 capacitor and switch it with the serial connection of the same capacitor and one jumper. Connecting the resistor on the jumper and measuring the voltage on it I will be able to calculate the current my design will consume (see the attachment). Please discuss with me and share your point of view. I would be very grateful to you. Best regards, Bojan http://uploadpic.org/v.php?img=yaM1o2JWkLink Copied

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Dear Dave,

Thank you so much ! Thank you from the bottom of my heart. Your advices were so precious to me. I understand you very well. It is now up to me to switch R53 with the jumper and to find out the allowed drop of the actual 1.2V core supply. Dividing the measured differential voltage on the new R53 resistor with its nominal value I obtain the current value (Idesign). Idesign*1.2V=Ptotal_measured ! Am I right ? :D Cheers mate, Bojan- Mark as New
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My idea is to use some oscilloscope with the differential probe.

1) To apply some 1000 input values to the design, to measure differential voltage and to save it in .txt format. 2) Change the probe ends and repeat the procedure. 3) From the saved measured values (in .txt files) calculate MEAN voltage value (Vr_mean) 4) Idesign=Vr_mean/R 5) Pmeasured_total=Idesign*1.2V 6) :cool:- Mark as New
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