Xinje XD5 - Page 150

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< When is word device > n≤4
D0DECO D1 K3
X0
n
S· D·
0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
7 6 5 4
2
1 0
D0
D1
bit15
bit15 bit0
bit0
4
2
1
3
11
10 9
8
13
12
15
14
Ignore high 8-bit, all to
be 0
The low n-bit (n ≤ 4) of the source address is decoded to the target address. When n ≤ 3, the
high 8-bit of the target turns to 0.
If n = 0, the instruction will not be executed. If n is out of 0 ~ 4, the instruction will not be
executed.
N = 3, so the decoding object in D0 is bit2-bit0, and the maximum value it represents is 4 + 2
+ 1 = 7.
N = 3, so in D1, 2
3
= 8 bits are needed to represent the decoding result, that is, bit7 ~ bit0.
When bit2 and bit1 are both 1 and bit0 are 0, the value is 4+2=6, so bit6 in D1 is ON.
< is word soft component > n≤4
D0DECO D1 K4
X0
n
S· D·
0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
7 6 5 4
2
1 0
D0
D1
bit15
bit15 bit0
bit0
4
2
18
3
11 10 9 813
12
15
14
The low n-bit (n ≤4) of the source address is decoded to the target address. When n ≤ 3, the
high 8-bit of the target turns to 0.
If n = 0, the instruction will not be executed. If n is out of 0 ~ 4, the instruction will not be
executed.
N = 4, so the object of decoding in D0 is bit3 ~ bit0, which represents the maximum value of
8 + 4 + 2 + 1 = 15.
N = 4, so in D1, 2
4
= 16 bits are needed to represent the decoding result, that is, bit15 ~ bit0.
When bit3, bit1 and bit0 are all 1 and bit2 is 0, the numerical value is 8+2+1=11, so bit11 in
D1 is ON.
D·
D·