Quote:
How does the (say) SRAM OE line go low?
I'm on the bus until tomorrow so I can't draw a schematic but I will try to explain. SRAM OE has two inputs. One of them is from the 74HC244 buffer and the other is from a pin on your FPGA. When the system starts, the FPGA pin is floating and the 74HC244 immediately drives the line high to z-state the SRAM OE. The OE pin of the 74HC244 also has two inputs, a pull
down and an FPGA pin. Because that FPGA pin is also floating on startup, the pull down enables the OE of the 74HC244, which drives the SRAM OE pin high. After your FPGA is done configuring, the FPGA pins driving the SRAM OE and the 74HC244 OE can now be driven by the FPGA. Bringing the 74HC244 OE high z-states the 74HC244 pin driving the SRAM OE since it is now can now be driven by an FPGA pin.
EDIT: Sorry for the poor quality. Try this:
At start up the 74HC244's OE is 0 and the SRAM OE is driven high by 1Y0. After configuration, IO1 drives the 74HC244's OE high, which z-states 1Y0. the SRAM OE is then controlled by IO2. If you are afraid you will make a mistake with the FPGA configuration, you can add a resistor to 1Y0 to prevent contention, but you won't need one otherwise.