LC66354C 데이터 시트보기 (PDF) - SANYO -> Panasonic

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LC66354C

Four-Bit Single-Chip Microcontrollers with 4, 6, and 8 KB of On-Chip ROM

SANYO -> Panasonic 

LC66354C, 66356C, 66358C

Mnemonic

Instruction code

D7 D6 D5 D4 D3 D2 D1 D0

[Accumulator manipulation instructions]

CLA

Clear AC

1000 0000 1

DAA

Decimal adjust AC

in addition

1100 1111

0010 0110 2

DAS

Decimal adjust AC

in subtraction

1100 1111 2

0010 1010

CLC

STC

Clear CF

Set CF

0001 1110 1

0001 1111 1

CMA Complement AC

0001 1000 1

IA

Increment AC

DA

Decrement AC

0001 0100 1

0010 0100 1

RAR

Rotate AC right

through CF

0001 0000 1

RAL

Rotate AC left

through CF

0000 0001 1

TAE

Transfer AC to E

0100 0101 1

TEA

Transfer E to AC

0100 0110 1

XAE

Exchange AC with E 0 1 0 0 0 1 0 0 1

[Memory manipulation instructions]

IM

Increment M

0001 0010 1

DM

Decrement M

0010 0010 1

IMDR i8 Increment M direct

1100 0111

I7 I6 I5 I4 I3 I2 I1 I0 2

1100 0011

DMDR i8 Decrement M direct I7 I6 I5 I4 I3 I2 I1 I0 2

SMB t2 Set M data bit

0 0 0 0 1 1 t1 t0 1

RMB t2 Reset M data bit

0 0 1 0 1 1 t1 t0 1

[Arithmetic, logic and comparison instructions]

AD

Add M to AC

0000 0110 1

ADDR i8 Add M direct to AC

1100 1001

I7 I6 I5 I4 I3 I2 I1 I0 2

ADC

Add M to AC with CF 0 0 0 0 0 0 1 0 1

ADI i4

Add immediate data 1 1 0 0 1 1 1 1

to AC

0 0 1 0 I3 I2 I1 I0 2

SUBC

Subtract AC from M

with CF

0001 0111 1

ANDA

And M with AC then 0 0 0 0 0 1 1 1 1

store AC

ORA

Or M with AC then

store AC

0000 0101 1

Operation

Description

Affected

status

bits

Note

AC ← 0

1 (Equivalent to LAI 0.) Clear AC to 0.

ZF

AC ← (AC) + 6

2 (Equivalent to ADI 6.) Add six to AC.

ZF

AC ← (AC) + 10

2 (Equivalent to

Add 10 to AC.

ZF

ADI 0AH.)

1 CF ← 0

Clear CF to 0.

CF

1 CF ← 1

Set CF to 1.

CF

1 AC ← (AC)

Take the one’s complement

of AC.

ZF

1 AC ← (AC) + 1

Increment AC.

ZF, CF

1 AC ← (AC) – 1

Decrement AC.

ZF, CF

AC3 ← (CF),

1 ACn ← (ACn + 1),

CF ← (AC0)

AC0 ← (CF),

1 ACn + 1 ← (ACn),

CF ← (AC3)

1 E ← (AC)

Shift AC (including CF) right. CF

Shift AC (including CF) left. CF, ZF

Transfer the contents of AC to E.

1 AC ← (E)

Transfer the contents of E to AC. ZF

1 (AC) ↔ (E)

Exchange the contents of

AC and E.

Has a vertical

skip function.

1

M (HL) ←

[M (HL)] + 1

1

M (HL) ←

[M (HL)] – 1

Increment M (HL).

Decrement M (HL).

2 M (i8) ← [M (i8)] + 1 Increment M (i8).

ZF, CF

ZF, CF

ZF, CF

2 M (i8) ← [M (i8)] – 1 Decrement M (i8).

ZF, CF

1 [M (HL), t2] ← 1

1 [M (HL), t2] ← 0

Set the bit in M (HL) specified

by t0 and t1 to 1.

Clear the bit in M (HL)

specified by t0 and t1 to 0.

ZF

1

AC ← (AC) +

[M (HL)]

Add the contents of AC and

M (HL) as two’s complement

values and store the result

in AC.

ZF, CF

Add the contents of AC and

M (i8) as two’s complement

2 AC ← (AC) + [M (i8)] values and store the result

in AC.

ZF, CF

AC ← (AC) +

1 [M (HL)] + (CF)

Add the contents of AC,

M (HL) and C as two’s

complement values and

store the result in AC.

ZF, CF

AC ← (AC) +

2 I3, I2, I1, I0

Add the contents of AC and

the immediate data as two’s

complement values and store ZF

the result in AC.

AC ← [M (HL)] –

1 (AC) – (CF)

Subtract the contents of AC

and CF from M (HL) as two’s

complement values and store

the result in AC.

ZF, CF

1 AC ← (AC)

[M (HL)]

Take the logical and of AC

and M (HL) and store the

ZF

result in AC.

1 AC ← (AC)

[M (HL)]

Take the logical or of AC and

M (HL) and store the result ZF

in AC.

CF will be zero if

there was a

borrow and one

otherwise.

Continued on next page.

No. 5484-14/21

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