TEC-2 micro-programming and testing

Topic Requirements

The memory unit data is subtracted from the contents of the memory unit represented by the absolute address addr, and the result is saved to the memory unit specified by the DR content.

Instruction format

D4	DR	X
ADDR
DATA

Command function

　　　　data-[ADDR]→[DR]

Design analysis

According to the instruction function and instruction format, the first reading address addr unit content is temporarily placed in the Q register. By putting Pc→ar, the content in MEM is the data of the memory unit. At this time, Mem-q→q, and will Dr→ar, then address register AR storage is exactly the DR, so as long as the Q register content to write memory can complete the command function. In addition, the micro-program must achieve two PC+1→PC operation in order to ensure the correct direction of the PC.

Micro Program

1  -: Pc→ar, pc+1→pc;0000 0E00 a0b5 54022 101: Mem→ar;0000 0E00 10f0 00023 102: Mem→q;0000 0E00 00f0 00004 103: Pc→ar, pc+1→pc;0000 0E00 a0b5 54025 104: Mem-q→q;0000 0E00 02E0 00006  the: Dr→ar;0000 0E00 90b0 000A7 106: Q→mem, cc#=0      ;0029 0300 1020 0010

Microcode detailed pc→ar, pc+1→pc 0000 0E00 a0b5 5402

ci3-0

/mio req/we

mi8-0

A

B

Sci

DC1

DC2

No action

f→b Y=a

r+s

r=0 S=b

pc

pc

cin value 1

no tube

address register ar

 

/we set to 1 send y (PC) to the internal bus so that AR receives

F=0+PC+CIN→PC Y=A=PC The contents of Y are is sent to the internal data bus

 

 

pc default is R5

pc default is R5

pc+1→pc

 

Pc→ar (y→ar)

1110

101

0 1 0

0 0 0

0 1 1

0101

0101

01

000

010

Note: The following a, B are 0000, do not operate

Mem→ar 0000 0E00 10f0 0002

ci3-0	/mio req/we			mi8-0			Sci	DC1	DC2
Sequential execution	Memory Read			No F	R+s	R=d S=0	CIN value 0	No tube.	Address Register AR
	MEM's data Send to internal bus			Y=f=d+0 Send to internal bus	D+0	The D-Input is Data sent from the internal bus			Mem→ar (Y=d→ar)
1110	001			0 0 1	0 0 0	1 1 1	00	000	010

mem→q 0000 0E00 00f0 0000

ci3-0	/mio req/we			mi8-0			Sci	DC1	DC2
Sequential execution	Memory Read			F→q Y=f	R+s	R=d S=0	CIN value 0	No tube.	Not used
	MEM's data Send to internal bus			F=d+0→q	D+0	The D-Input is Data sent from the internal bus
1110	001			0 0 1	0 0 0	1 1 1	00	000	000

Pc→ar, pc+1→pc 0000 0E00 a0b5 5402 (slightly) mem-q→q 0000 0E00 02E0 0000

ci3-0	/mio req/we			mi8-0			Sci	DC1	DC2
Sequential execution	Memory Read			F→q Y=f	R-s	R=d S=q	CIN value 0	No tube.	Not used
	MEM's data Send to internal bus			F=d-q→q	D-q	The D-Input is Data sent from the internal bus
1110	001			0 0 0	0 1 0	1 1 0	00	000	000

Dr→ar 0000 0E00 90b0 000A

ci3-0	/mio req/we			mi8-0			Sci	Sa	SB	DC1	DC2
	No action	None y=f	r+s	r=0 S =b	cin Value 0	A=a	B=dr	no tube	Address register AR
 	/we set to 1 send Y to the internal bus so that AR receives			y= F=0+b To internal bus		sa=0 The value of the  a port is from the field of a port sa=1 The value of the  a port from the field of the SR	sb=0  b port values from B-Port fields sb=1  b Port values from the Dr field	 	dr→ar (y→ar)
1110	101			0 0 1	0 0 0	1 1 0	00	0	1	000	010

Q→mem, Cc#=0, 3#, a4h 0029 0300 1020 0010

  send data to internal bus IB

so the mem can get Y (Q)

TD style= "Text-align:center;" >

Next Address mar9-0	ci3-0	/mio req/we			mi8-0			Sci	DC1	DC2
a4h a4h The function of the micro-instruction is based on whether there is an interrupt request, The determines whether to enter the interrupt processing process or to execute the next instruction sequentially. This is an action that should be performed after each machine instruction is completed.		Memory Write	None y=f	r+s	r=0 s=q	cin value 0	output of the operator	not using
final conditional transfer to a4h execution of subsequent handlers if interrupted: →adh∽b6h→a5h without interruption: →a5h					y=f=q	0+q		 	 
10100010 (forward 0)	0011	000			0 0 1	0 0 0	010	00	20s	000

Episode

On how to get the microcode, first we have to use the micro-instruction parser . But when I poked the little penguin, something went wrong ... Such

This is the reason why Tabctl32.otx is not in the correct directory and has not yet been loaded to find a solution on your own. Say so or give the link. "

Test tools:

You can open the TEC-2 simulator, and then open the monitoring program, or you can open the monitor~ in the folder directly

Steps

1. First, enter the microcode into a memory unit that starts with 0900H. Use the e command to input microcode, enter after the transmission of the program. Separate each value with a space and press ENTER when finished. Such as:

>e0900

2. Load microcode and enter microcode with a command. Such as:

>a0800
0800：MOVR1, the       ;microcode in-memory first address 0802：MOVR2,7         ;Micro-instruction bar number 0804：MOVR3, -       ;microcode in the micro-control memory of the first address 0806: LDMC;load microcode instructions to load microcode instructions into micro-control memory 0807：RET 0808：

3. Run the program with the G command to load the microcode. Such as:

>g0800

4. Use the a command to enter a new program for testing. Such as:

>a08200820：MOVR0,0011        ;0011→r00822：MOV[0890],r0;R0 content → Memory 0890 address0824：MOVR1,0891        ;0891→R10826：NOP                 ;NOP is an empty operation instruction, reserved unit0827：NOP0828：NOP0829：RET

5. Call new instructions at the beginning of NOP to test. Such as:

0890 0111            ; instruction Format: D4drx ADDR DATA (D410 1 is the R1 register, where the results are sent. In 4, we used MOV R1, 0891 command, so you can see the contents of R1 in 0891.

This invokes the operation code D4 of the new instruction, with the operand addr=0890 data=1111. And then:

>g0820                     ; executes a micro program with a first address of 0820

6. View the results and use the D or r command to view the memory or register status after the program is run.

>d0891

7. Run:

Reference

1. Http://www.cnblogs.com/joyeecheung/p/3687773.html
2. Http://www.cnblogs.com/chenshiyu/p/4457211.html

TEC-2 micro-programming and testing