I am doing a BER (BIT ERROR RATE) in vhdl and I have to put 2 registers clear on read which counts the number of words in error and the number of bits in error. i want to know how i can code a register clear on read in vhdl? thank you,
vhdl wrote: > I have to put 2 registers clear on read Usually thats a really bad (better: a stupid) idea in real life. Because you will encounter strange behavior when this "automatic reset register" ist used in a processor system. Because then it may be that debugger is told to show some values around this register and it reads this register additionally into a cache for "future use". But by reading the register its reset to 0 --> weird behavior of your software. > i want to know how i can code a register clear on read in vhdl? You must reset the register somewhere near to the line where it cpounts up. The final implementaion depends heavily on what you already have...
Atomic read&clear operations on registers are quite common in signal processing. If you need to count the exact number of events (e.g. uncorrectable bit errors in a Reed Solomon Decoder) over a certain time interval, usually a counting register is used which is cleared upon read. If reading and clearing were non-atomic, there may occur further events in between that are never counted.
This technique can be found in many Devices. (e.g. in the old C64, the MOS6522 and 6526 use this for interrupt bits) Together with a CPU and a simple Bus System, it can be implemented like this:
1 | -- ce : chip enable
|
2 | -- we : write enable
|
3 | -- addr: bus address
|
4 | -- do : bus data output
|
5 | |
6 | if rising_edge(clk) then -- clocking |
7 | |
8 | case addr is |
9 | ..
|
10 | when REG1 => do <= reg1value; |
11 | if ce = '1' and we = '0' then |
12 | reg1value <= (others => '0'); |
13 | end if; |
14 | when REG2 => do <= reg1value; |
15 | reg2value <= (others => '0'); |
16 | if ce = '1' and we = '0' then |
17 | reg2value <= (others => '0'); |
18 | end if; |
19 | ..
|
20 | end case; |
21 | |
22 | end if; -- clocking |
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