Virtex® Family FPGAs

[mcgett]

No jumpers exist on the supply rails.

[stuartdu]

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?

[bassman59]

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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?

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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.

[stuartdu]

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

[bassman59]

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stuartdu wrote:

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

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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.

[stuartdu]

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?

[mcgett]

> 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.

[stuartdu]

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?

[rcingham]

> 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.

[bassman59]

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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.

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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 :)