Hi,

I'm trying to change the input voltage of a Xilinx Spartan-3E FPGA (the default one is 1.2V). The board is a Digilent Nexys 2, and of course a tension regulator is involved, hence I have to consider it.
The schematic is the following:

SW1 is the output of the tension regulator and its value is of 0.8V. VCC1V2 is set using the formula: VCC1V2 = 0.8V * [(R78+R79)/R79].
So, by changing the R78 value I can change the input voltage of the FPGA.

However, a user in the Xilinx Forum suggested another solution (http://forums.xilinx.com/t5/Xilinx-Boards-and-Kits/Nexys-2-board-variable-supply-to-core/m-p/116380/highlight/true#M3707) consisting of: "Lift one side or the other of L2. You can connect a variable output bench supply directly to the VCC1V2 net.".
Since I'm not an Electronic, I would like some additional informations:

• I just lift one side of L2 and connect a different power supply (let's say 0.9V, coming from a dedicated external power supply), without any particular precaution?
• how should I calculate the VCC1V2 value? Using the above 0.9V value, VCC1V2 will still follow the above equation? So, VCC1V2 would be equal to: 0.9V * [(R78+R79)/R79]?

Thanks for any hint,
Pierpaolo

 

Before we get to that, you will be running the FPGA outside its absolute maximum ratings, is there a reason why you want to do this?

There are probably alternatives to what you want to accomplish by doing this.

 

I know that the absolute maximum ratings are of -0.5V and 1.32V, but what I proposed above was just an example Indeed, the recommended operating voltage is between 1.14V and 1.26V. I just want to evaluate a circuit I've done under various operating conditions different than the default one (1.2V). For example, I'd like to "test" my circuit when the FPGA is supplied with 1.14V and 1.26V.
To this aim, and supposing that I'm feeding the FPGA by a mini-USB connector (and no external power supply are used), I should:

• "disable" the Vcc pin of the USB port (it's quite simple, and I've already done it in the past), hence the tension regulator doesn't work;
• lift one side of the L2 inductor;
• connect a supply voltage different than 0.8V (offered by the tension regulator), e.g. 0.84V;
• calculate what the VCC1V2 voltage would be using still the equation introduced above (~1.26V).

Can anybody confirm or deny any of the following steps?

On a side note, I'm not interested of power supplying the IO pins, LEDs etc., but only the FPGA. Therefore, I'm not interested in the VCC2V5 output/input.

Thanks,
Pierpaolo

 

Sorry, 0.8V is within specs, I misread that.

That said, you can do this, but you may run into issues with the regulator not having a closed loop.

I'm not sure why this matters, but sure, go for it.

 

I would change the voltage divider. Then the regulator is a regulator. If you just apply a voltage to the feedback pin it's voltage will change if the input supply varies or if the load changes.