Oops! You've now smoked two RFX-2400 boards and/or the FPGA ;)
Bzzzt! Ok thanks. Your explanation helped clear up my confusion about the RX and TX nets in the schematic and prevented frying my boards as well. ;)
>(I _believe_ that the high 7-bits are available for your use. It mightbe 8, but I haven't looked at the >schematics in a _long_ time. I'm notkidding when I say _you_ should check the schematics.)
Yes it is only the 7 msbs.
If you can get by with a single RFX-2400, my suggestion is that you
put it on the A-side, and then put a Basic Tx and a Basic Rx on the
B-side. Then you've got a total of 32 uncommitted i/o pins on the
B-side.
This is a good idea. So if I insert one of each Basic Tx and Rx on side B, with the RFX2400 on side A, then the following should work (?)
u = usrp.source_c(0, 64)
u._write_oe(1, 0xffff, 0xffff)
u._write_fpga_reg(FR_DEBUG_EN, bmFR_DEBUG_EN_RX_B | bmFR_DEBUG_EN_TX_B)
and then in usrp_std.v
master_control master_control
(....
//.debug_0(rx_a_a),.debug_1(ddc0_in_i),
.debug_0(rx_debugbus),.debug_1(ddc0_in_i),
.debug_2(rx_a_a),.debug_3(rx_b_a), .... );
Thanks again!
Nikhil