This project is designed to show the internal workings of computer.
Topics that will be covered, include:
- Machine Language
- The Central Processing Unit (CPU)
- How a CPU performs its operations
- Instruction Fetch-Execute sequence
- How they are implemented
- The Control Unit
- Random Access Memory (RAM)
- Data Storage
Due to the nature of this project, this manual is subject to change over time to adapt to the progress of the students.
In this project you will be building a simple computer with a central processing unit with support for a few instructions and some memory. The instructions that will be implemented are simple operations such moving data between registers and performing simple addition operations.
By the end of the project, your computer should be able to demonstrate the operation of the following program:
MOV A, B
INC B
MOV B, A
INC B
In this project, we will be using a standard 5 volt direct current power supply. We will be connecting the positive terminal to our supply rail called VCC
and this is by convention red colored. The negative terminal will be connected to our reference rail called ground or GND
and is by convention black but may appear at times as blue.
In this lab, we will use VCC
, 5V
and +5V
In this project, we will be using a prototyping project board more commonly known as a breadboard. It contains small holes which share an electric connection with holes in the same row or column. There are two areas on the breadboard: the prototyping area and the supply rails.
The prototyping area located in the middle is where you will place the components in this project. The prototyping area has a grid of holes, identified by a letter and number written on the top and side. These holes are electrically connected in a line of five (i.e. pins 1A through 1E are all connected but not to 1F or 2A). There is a gap in the middle which will be where you put the center of your DIP (Dual Inline Package) components so that you can make connections to their pins.
The supply rails are arranged in two rows connected in a long line with a red or blue stripe next to them. They are electrically connected as indicated by the line next to them (i.e. a break in the line means a break in the connection). These are used for your power supply connection and you should attach them to VCC
or GND
for red and blue respectively.
We will be using 22 gauge solid core copper wire for this project.
The wires should be properly cut and the plastic insulation must be stripped from the inner conductor before use. The amount of insulation stripped should nominally expose 6mm (1/4 inches) of wire.
We will be using standard 300mil plastic Dual Inline Package (DIP) Integrated circuit chips for this project.
We will be using 330 &Ohm; resistors as a means to limit current through LEDs and to serve as pull-ups on switches. The clock circuit will make use of two 22 K&Ohm; resistors as part of a resistor-capacitor (RC) tank circuit.
We will be using a capacitor in the clock circuit as part of the RC tank circuit.
We will be using a strip of Light Emiting Diodes (LEDs) as a means of visualizing binary data values. These must always be using in series with a resistor so that they will not burn out.
We will be using a strip of switches to toggle between data values.
Inside the computer, we need a way of representing information, however, the only means we have is through voltage levels. The information we can store on a register is called a bit and is the absence or presence of a voltage potential relative to a common reference. The convention we have adopted is that the presence of a voltage represents a logical 1
and the absence of a voltage represents a logical 0
Decimal | Binary | Decimal | Binary | Decimal | Binary | Decimal | Binary |
---|---|---|---|---|---|---|---|
0 | 0000 |
1 | 0001 |
2 | 0010 |
3 | 0011 |
4 | 0100 |
5 | 0101 |
6 | 0110 |
7 | 0111 |
8 | 1000 |
9 | 1001 |
10 | 1010 |
11 | 1011 |
12 | 1100 |
13 | 1101 |
14 | 1110 |
15 | 1111 |
All logic on an IC is built from logic gates. Logic gates are a combination of transistors to form logical boolean operations. The simplest gates to produce are the NOT gate, NAND gate and NOR gate. Using these gates, more familiar gates such as AND and OR can be built.
In this lab, we will introduce ourselves to the equipment in the lab. We will also look at what is required to represent data electrically and how to get output from it. We will set up a simple circuit to produce data values with a switch, visualize the value with an LED and then use the value produced as an input to a logic gate and then visualize the output of a logic gate.
You will have been issued a lab kit containing:
- A project board
- A power supply
- A wire cutting and stripping tool
- A chip removing tool
The project kits issued this year have the project board integrated with the power supply, an additional breadboard is also included in case you run out of space on your project board later on in the lab.
This is a lab which deals with electricity, your lab demonstrator should have given you a quick talk about the safety of dealing with electricity. While the voltage and current levels are not high enough to cause serious injury to anyone caution must still be taken with this equipment. The equipment may also become damaged if wired improperly. If at any point you are not sure about the setup of your project board, call over your lab instructor.
If something is not working or components become hot, turn off the power. Look for the issue or call over your lab instructor if needed.
IF YOU SMELL SMOKE OR BURNING, CUT THE POWER ON YOUR POWER SUPPLY IMMEDIATELY AND CALL OVER YOUR LAB INSTRUCTOR!
- Take out your project board from the kit that was issued to you, verify that it works.
- Hook up all supply rails to +5V and Ground on the power supply.
- Hook up a switch pack and LED pack.
- Use the switch to control the LED.
- Hook up a logic gate and use switches for its input and an LED for its output.
In this lab, we will look at the circuits involved in forming logical equations. The circuit we will be using is called the half-adder which is used for adding two bits together to get a result.
In using two bits, we need two inputs, one to represent each bit. Because we are using the binary number system, we can potentially end up with two digits in the result since our addition overflows or carries into another digit (in the case of 1+1). This means we will need two outputs to represent the result.
We will be representing our inputs as A and B and our outputs as C and S. S will be the sum of the two digits and C will be the value that carries over.
A
+ B
---
CS
In order to construct our circuit, we will need to know the equation that represents each bit in the result. To do this, we can construct a truth table which can help us in determining the equations.
A | B | • | C | S |
---|---|---|---|---|
0 | 0 | = | 0 | 0 |
0 | 1 | = | 0 | 1 |
1 | 0 | = | 0 | 1 |
1 | 1 | = | 1 | 0 |
The derived equations are constructed from the primitive Boolean algebra operations AND OR and NOT. These operations have corresponding logical gates which are available as DIP chips in the lab.
- The NOT gate is part number 7404
- The AND gate is part number 7408
- The OR gate is part number 7432
You will need to construct the truth table above as a configuration of logic gates such that the inputs A and B are provided by switches and the outputs C and S are visible on LEDs.
- To do this, you will need to construct the Boolean algebra equation for C and S using only the variables A and B.
- Using these equations you will modify them to use only the AND, OR and NOT operators.
- This equation will allow you to draw the gate diagram with the gates availible to you.
- Use this gate diagram and the IC pinouts provided at the end of the manual to construct the circuit.
Reminder:
- The part numbers on the chip may vary from chip to chip so ignore the letters.
- SN74HCT04 would be referred to simply as 7404.
In the last lab, we looked at how we can construct a circuit for any given truth table. In this lab, we will look at how we can use this and some new parts to build other components of a computer.
A computer does not always require that the input for its operations be present at all times. It uses memory to hold inputs for when it needs them and can store results to memory for later. Latches provide the facility to do this and are build by a cross-coupled pair of NAND gates or NOR gates. Cross-coupling is when the output of one gate is used as the input for the other and vice-versa. This creates a feedback in the circuit which is used to store a one bit value. The other inputs allow the feedback to be changed thus changing the stored bit value. This is what is known as an R-S latch. It has two inputs (R and S) and two outputs (Q and ~Q).
R | S | Latch Behavior |
---|---|---|
0 | 0 | Retains value |
1 | 0 | Stores a 0 |
0 | 1 | Stores a 1 |
1 | 1 | Undefined |
The R-S latch does not provide the ideal means for sampling a value to store. For this, we add some logic in front of the RS latch to create a D latch. With the D latch we have inputs D and E which feed R and S so that when E is 0, we retain a value and when E is 1, we store the value on D.
D | E | • | R | S |
---|---|---|---|---|
X | 0 | = | 0 | 0 |
0 | 1 | = | 1 | 0 |
1 | 1 | = | 0 | 1 |
X in this case is used to denote the "Don't Care" input where the value can be either 0 or 1 for this particular output as when E is 0 we are retaining a value and D is not used under this case.
In this part of the lab, you will construct a D type latch and test its ability to store a value.
Latches are level-triggered which in the case of D latches, means that when the level of E is high (E = 1) it stores the value from D which may change change while E is high. This is not an ideal behaviour because we would like to sample a particular value at a particular time. This behavior is called edge-triggered and is the moment when a value changes in a particular direction. The positive edge or rising edge is when a value goes from 0 to 1 and the negative edge or falling edge is when a value goes from 1 to 0.
A flip-flop is two chained D-type latches such that the D input for the second comes from the Q output of the first. E is inverted for the second one. This means that at all times, E is keeping the value locked for exactly one of the latches in the flip-flop.
- An explanation of how data is represented electrically
- How logic gates relate to Boolean Algebra
- The equations that you have derived for the half-adder
- A gate diagram for those equations
- A truth table for a full-adder circuit
- A gate diagram for and an explanation of a D latch
Lab reports are due two weeks after they are assigned. Lab report 1 is due the week of February 10.
There will be a 10% deduction on the lab grade if you fail to include your name or group number on your report. There will be a 40% deduction in lab grade if you fail to complete the labs before the lab report deadline.
-
Always use a color convention!!! ( VCC = RED, Ground = BLACK)
TIP : it will help you to use the same color when you want to propagate your clock signal throughout your breadboard
-
Please use shorter wires when you want to connect something. Your breadboard will look much cleaner and will cause you less problems if you even need to do some debugging later on.
-
Always try to wire your circuit as neatly as possible.
-
Make sure that you are using the correct chip and that its pins aren't bended.
-
Check if your ground and power pins of all your logic chips are connected correctly.
-
During the construction of your circuit, you may have incorrectly connected a cable. To find this error, you will need to create a "debugging cable".
- Connect 1 long red cable to a LED.
Since red is used for vcc in your color convention, no one will be mislead in error that it is a required cable in your circuit.
- When you will connect this cable to a random pin on your breadboard, it will show you its data.
- E.G.
You want to make sure that you have spread your clock signal correctly on your breadboard.
- Connect your 'debug cable' to a pin that is supposed to be receiving a clock signal. If your debug LED is not showing you the clock signal, then you have found your problem, however, if it is showing the correct pattern, then you can proceed to the next pin to be tested.
- This process can be used to verify if a chip is receiving the correct input or producing the good output.
Faulty chip can be found using this method
- You may also add more LED on you breadboard to show intermediate result. It could help you isolate problem later on.
At this phase, we will start building the lab project for the rest of the course. You may clear off everything on your project board up until now as it will not be needed. Pay close attention to wiring as from now on we will be keeping everything on the project until the end of the project and poor wiring will make locating mistakes in the lab much more difficult.
In this lab, you will be building system clock for your computer. The system clock is an alternating signal changing from 0 to 1 and back at a regular interval. This signal is used to provide a changing input which can be used to make the computer work autonomously. Without this signal, the computer would need the user to make a change in inputs for every phase of computation similarly to how in the preliminary labs, our outputs only ever changed when an input was changed.
The primary function of a CPU is to perform a set of operations called instructions. These instructions have an almost one-to-one relationship with lines of code in assembler language (which will be seen in the tutorial for this class) and perform operations such as copying a value from one register to another or more complex operations such as arithmetic and memory access. These instructions may need to occur in several phases so the system clock needs to be expanded into several clock signals where we will have one for each phase. In this lab, we will be constructing a fixed four phase clock generator (meaning that each instruction has four phases, even if some do not use all of them and that number is fixed).
We will introduce two new logic chips for this purpose.
-
The 555 chip which is a chip used to control timing. We will wire it up in asyncronous mode so that it generates the alternating signal from one to zero. This setup uses an RC tank circuit to control the timing of the signal meaning the values of the resistors and capacitor control how fast the value changes.
-
The 74164 chip which is a serial-in-parallel-out (SIPO) shift register. A SIPO shift register is a chip with several flip-flops chained up internally so that all the outputs of each flip-flop are accessible on their own pin and where the inputs are provided by the preceding flip-flop's output. The first input is made accessible on a pin which we use to shift values in.
We will be using the clock signal to generate a clock for the shift register so that it can shift values on its own. The value that we shift in will be based on the values that are already in the register and will be such that there is only one zero in the register (i.e. shifting in 1 unless the register contains all 1s).
In this lab, you will be building the CPU buses for your computer. A CPU bus is a common system such that registers can move values between each other. In this lab we will build two buses, one where all the registers outputs are attached to one of the ALU inputs, and one where the register inputs are attached to the ALU output. Since the ALU has two inputs, the remaining input will be attached to a controllable 0 or 1.
A Register is a Flop-Flop with a Tri-State buffer attached to make the output controllable. This gives a device which can be attached to a bus for both inputs and outputs. The register will read off the bus attached at its input just as a flip flop would when its enable line is triggered, this works because one output can feed several inputs. The Tri-State buffer allows the output to be controlled by electrically disconnecting it so that we can allow only one value to be output on the bus at a time. Without it, the bus would be attached to more than one output leading to conflicting signals on the bus.
...
- An explanation of why a clock signal is needed in a computer
- The relation between the clock signal and instruction phase signals and why we need instruction phases
- The clock signal is the signal that alternates between zero and one
- The phase signals are the four signals that we brought out to the LEDs
- How we derived the phase signals (hint: shift register and feedback)
- hint: shift register and feedback
- An explanation of the general architecture of the computer
- Describe the functions of an ALU and compare it to what we have built in the lab
- Describe how one would perform the register transfer language operation A -> B
Lab reports are due two weeks after they are assigned. Lab report 2 is due the week of March 10.
There will be a 10% deduction on the lab grade if you fail to include your name or group number on your report. There will be a 40% deduction in lab grade if you fail to complete the labs before the lab report deadline.
In this lab, you will be building the RAM for your computer, it will be comprised of 4 registers selected through a decoder. This will give us a 4x4 memory unit with 4 data lines and 2 address lines.
In a computer, we only have a small number of registers to store data. Typically, we need to store much more information than the registers provide us. For this reason, we use main memory. Main memory is a large array of memory units (typically flip flops) that we can access by choosing which unit we want access to through a register and performing operations to that unit as we normally would to a register. This means we need two special purpose registers in our CPU. We have a Memory Address Register (MAR) which picks which memory unit we want to access using its address (similar to an index for an array) and a Memory Data Register (MDR) which exposes the contents of a particular memory address to the CPU. It is important to note that the MAR is a write only register since we only ever specify memory locations and the MDR is not actually a register but a means of exposing memory as a register.
We are using four register as our memory units in this lab. These registers are are selected though a decoder. The decoder gets an address from the MAR and uses it to select which register should be active.
In this lab, we will also be including our Instruction Register (IR) which will be used in the next lab.
...
- An explanation of how memory works
- The difference between registers and memory
- Did we build static memory or dynamic memory, what are the differences between them?
- How many address lines did we have and how many addresses did that give us?
- Assume that we are still using a 4 bit word size. How many address lines and data lines would we need if we were to have had 8 words of memory? and for 16 words of memory?
Lab reports are due two weeks after they are assigned. Lab report 3 is due the week of March 17.
There will be a 10% deduction on the lab grade if you fail to include your name or group number on your report. There will be a 40% deduction in lab grade if you fail to complete the labs before the lab report deadline.
If you are unable to complete the lab before the deadline for technical reasons, there will be no grade penalty.
Please inform your lab instructor before the deadline if this is the case
In this lab, we will be building the control unit for your computer. The control unit is what will control how the cpu handles machine language instruction.
add1
= ¬( phase1
∧ ( phase3
∨ ir1
) )
pc out
= phase1
pc in
= phase1
a out
= phase3
∨ ¬ir4
a in
= phase3
∨ ir2
b out
= phase3
∨ ir4
b in
= phase3
∨ ir3
led latch
= ? in
∨ clock
mar latch
= phase1
∨ clock
Note: You will need to remove any existing signals that were being fed to the chips that the control unit is not providing. This is to ensure that each input only gets one value.
After implementing the control unit, the four LEDs that were previously counting might not be, this is normal. Attach ?_in
to pc_in
and they should continue counting.
- Lab 4: Clock Signal Generator
phase1
is the first clock signal brought out to the LEDsphase2
is the second clock signal brought out to the LEDsphase3
is the third clock signal brought out to the LEDs
- Lab 5: Registers
add1
is pin 6 on the 74283pc in
is pin 9 on the 74173 representing the PC registerpc out
is pin 1 on the 74173 representing the PC registera in
is pin 9 on the 74173 representing the A registera out
is pin 1 on the 74173 representing the A registerb in
is pin 9 on the 74173 representing the B registerb out
is pin 1 on the 74173 representing the B registerled latch
is pin 9 on the 74175
- Lab 6: Memory
mar latch
is pin 9 on the 74175 representing the MARir1
comes from the bit 1 in the instruction registerir2
comes from the bit 2 in the instruction registerir3
comes from the bit 3 in the instruction registerir4
comes from the bit 4 in the instruction register
ir1
:+1
ir2
:Ain
ir3
:Bin
ir4
:Bout
/¬Aout
In this lab, we will store a program in memory and have your computer run it.
In the previous lab, we set up the control unit to have the computer control how data was moving through the bus. The control unit used a machine language instruction along with a the clock signal phase to control which registers were storing values and which output their value to control the data flow from the registers.
Recall that in the last lab, we assigned names to the bits of the instruction register (ir3
became Bin
and so on). These names were given since they signify a particular signal is to be active when that bit is active.
If we create a table of all the combinations for the instruction register, we can see which signals should be active for that instruction register value. With that knowledge we can see which registers are moving values and then derive register transfer notation for each of our instruction and subsequently name them accordingly.
We only want to look at the signals active during phase3 since phase1 is always adding one to PC and reading the instruction.
Value † | Active signals | Transfer Notation | Instruction Name |
---|---|---|---|
0000 | add1 , a in , b in , b out |
B + 1 -> A, B | * |
0001 | a in , b in , b out |
B -> A, B | * |
0010 | add1 , b in , b out |
B + 1 -> B | Inc B |
0011 | b in , b out |
B -> B | NOP |
0100 | add1 , a in , b out |
B + 1 -> A | * |
0101 | a in , b out |
B -> A | Mov A, B |
0110 | add1 , b out |
B + 1 -> ∅ | * |
0111 | b out |
B -> ∅ | * |
1000 | add1 , a in , b in , a out |
A + 1 -> A, B | * |
1001 | a in , b in , a out |
A -> A, B | * |
1010 | add1 , b in , a out |
A + 1 -> B | * |
1011 | b in , a out |
A -> B | Mov B, A |
1100 | add1 , a in , a out |
A + 1 -> A | Inc A |
1101 | a in , a out |
A -> A | * |
1110 | add1 , a out |
A + 1 -> ∅ | * |
1111 | a out |
A -> ∅ | * |
† Value is in the order: ir4
ir3
ir2
ir1
( * ) Undocumented Instruction: Instruction not intended for use
We need to verify that all our defined instructions (Inc B
, Mov A, B
, Mov B, A
) work. To do this we program that instruction into memory and check if it behaves as it should. In the case of Inc A
which cause A to increase by one, we would need to attach ?_in
to a_in
so that the LED latches the value of A so that we can see if A is in fact counting up by one.
Once all the instructions work, program a simple program into memory. The instruction register will need to latch instruction from memory based on the address stored in the program counter. This means that the MAR cannot be attached to the switches used for programming. Instead, the MAR should be attached to the in-bus. To do this, move the wires on pins 4 and 5 on the MAR flip-flop so that they connect to pins 4 and 5 on the LED flip-flop. Pins 4 and 5 on the LED flip-flop connect to the lower order bits on the in-bus.
Program into memory the following program:
MOV A, B
INC B
MOV B, A
INC B
? in
in lab 7 should be attached to b in
so that the lab instructor can see the value stored in register B and verify the proper running of the program.
SUBJECT TO CHANGE
- The architecture of the computer
- A description of what the control unit does
- The functionality of the control unit implemented in the lab
- How a program was loaded into memory
- How a program in memory was run
Lab report deadline will be discussed during the lab session.
There will be a 10% deduction on the lab grade if you fail to include both your name or group number on your report. There will be a 40% deduction in lab grade if you fail to complete the labs before the lab report deadline.
The part numbers on the chips may vary from chip to chip. For this project, the parts have been picked so that you may ignore the letters in the part number. This means the parts SN74HCT04 and 74LS04 would be referred to simply as 7404 and perform the same task.
Chips are numbered counter-clockwise starting at the notch on the chip (on the top of the diagram).
The 555 Timer Chip is a chip used to generate various output waveforms. For our purpose, we will be using it to generate our clock signal in the Clock Signal Generator lab. In depth knowledge of this chip is not needed for this course. However, more information can be found on the 555 Timer IC Wikipedia page.
.-._.-.
GND [| |] VCC
TRIG [| 555 |] DIS
OUT [| |] THR
~RES [| |] CTRL
'-----'
VCC
: +5V Supply VoltageGND
: GroundOUT
: Output waveformTRIG
,RES
,CTRL
,THR
,DIS
: These pins determine how the chip should generate its output waveform.
The 7404 is a chip that contains 6 NOT gates, also known as inverters.
Yn
= ~An
.-._.-.
A4 [| |] VCC
Y4 [| 74 |] A1
A5 [| 04 |] Y1
Y5 [| |] A2
A6 [| |] Y2
Y6 [| |] A3
GND [| |] Y3
'-----'
VCC
: +5V Supply VoltageGND
: GroundA#
: Input for correspondingly numbered gateY#
: Output for correspondingly numbered gate
The 7408 is a chip that contains 4 AND gates.
Yn
= An
∧ Bn
.-._.-.
A3 [| |] VCC
B3 [| 74 |] A1
Y3 [| 08 |] B1
A4 [| |] Y1
B4 [| |] A2
Y4 [| |] B2
GND [| |] Y2
'-----'
VCC
: +5V Supply VoltageGND
: GroundA#
: First Input for correspondingly numbered gateB#
: Second Input for correspondingly numbered gateY#
: Output for correspondingly numbered gate
The 7432 is a chip that contains 4 OR gates.
Yn
= An
∨ Bn
.-._.-.
A3 [| |] VCC
B3 [| 74 |] A1
Y3 [| 32 |] B1
A4 [| |] Y1
B4 [| |] A2
Y4 [| |] B2
GND [| |] Y2
'-----'
VCC
: +5V Supply VoltageGND
: GroundA#
: First Input for correspondingly numbered gateB#
: Second Input for correspondingly numbered gateY#
: Output for correspondingly numbered gate
.-._.-.
0 [| |] VCC
1 [| 74 |] A1
2 [| 42 |] A2
3 [| |] A3
4 [| |] A4
5 [| |] 9
6 [| |] 8
GND [| |] 7
'-----'
VCC
: +5V Supply VoltageGND
: GroundA#
: Input bitn
of a binary number#
: Output pin would be 0 ifA1
-A4
match#
. Otherwise, it would be 1.
A4 | A3 | A2 | A1 | • | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 0 | = | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 0 | 0 | 1 | = | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | = | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | = | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | = | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | = | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | = | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | = | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 0 | = | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | = | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 |
.-._.-.
In1 [| |] VCC
In2 [| 74 |] Q8
Q1 [| 164 |] Q7
Q2 [| |] Q6
Q3 [| |] Q5
Q4 [| |] ~RES
GND [| |] CLK
'-----'
VCC
: +5V Supply VoltageGND
: GroundIn1
,In2
: Input bit to shift in on the next rising edge (transition from 0 to 1) ofCLK
Q#
: Output pin containing the current stored valueRES
: Reset all values to 0
On shift, the value in In1
∧ In2
will be stored in Q1
, Q2
will contain the previous value of Q1
, Q3
will contain the previous value of Q2
and so on until Q8
.
.-._.-.
~OE1 [| |] VCC
~OE2 [| 74 |] RES
Q1 [| 173 |] D1
Q2 [| |] D2
Q3 [| |] D3
Q4 [| |] D4
CLK [| |] ~In1
GND [| |] ~In2
'-----'
VCC
: +5V Supply VoltageGND
: GroundD#
: Inputs to Latch onCLK
rising edgeQ#
: Outputs of register onOE1
andOE2
In1
,In2
: Enable this to store values fromD#
on the next rising edge ofCLK
OE1
,OE2
: Enable this to output values fromQ#
CLK
: Clock signalRES
: Reset all values to 0
.-._.-.
~RES [| |] VCC
Q2 [| 74 |] Q3
~Q2 [| 175 |] ~Q3
D2 [| |] D3
D1 [| |] D4
~Q1 [| |] ~Q4
Q1 [| |] Q4
GND [| |] CLK
'-----'
VCC
: +5V Supply VoltageGND
: GroundD#
: Inputs to Latch onCLK
rising edgeQ#
: Outputs of Flop-FlopCLK
: Clock signal to latch on rising edgeRES
: Reset all values to 0
.-._.-.
S2 [| |] VCC
B2 [| 74 |] B3
A2 [| 283 |] A3
S1 [| |] S3
A1 [| |] A4
B1 [| |] B4
Cin [| |] S4
GND [| |] Cout
'-----'
.-._.-.
[| |]
[| |]
[| |]
[| |]
'-----'