Computer Hardware

That really depends on several factors, one of which is the language you are going to use. There are literally hundreds of programming languages, for different purposes; each has different requirements.

For example, Java is a programming language that works on different platforms - that is, it works on a variety of types of hardware, and operating systems; the most basic thing you need is the (free) JDK (Java Development Kit), which includes the Java compiler and the Java Runtime; and any basic text editor such as NotePad.

Helllo you Infocus friend. I had the same kind of problem with my Infocus L600.

You should change five capacitors on the power card:

* 4 x 100 microfarad 16volt capacitor

* 1 x 4.7 microfarad capacitor Today I managed to this operation and the Infocus started again.

It takes about 1 hour to do this operation and capacitors cost 3 euros ......

A stack is a concept in computer science that works like this:

You push value A into the stack, and it takes position 1. Then you pop the stack, and get A. If you were to push A, then B, A would be pushed into position 2, and B would be in position 1. When you pop, however, you can only get the value in position 1 (B) and A would pop to position 1. This is useful for data storage, but, expanding, you can do simplified arithmetic if you store data in positions 1 and 2, and define that the ADD command sums positions 1 and 2, erases 2, and stores the sum in 1. And so on.

A software stack is a stack that is implemented in the software of a computer: A large series of commands is given to the processor that make it store data in a "stack" somewhere in memory, and are usually sequential.

Pros: Simple to implement.

Cons: Slow and resource consuming.

A hardware stack is a series of memory units, often built into the processor, that quite literally are a stack. The only way to store data in them is to push, and the only way to access it is to pop from the first position (top of the stack).

Pros: Very fast, and is useful to the processor, even when you aren't actively using it.

Cons: Has to be engineered, and is often expensive to make.

hardware,software and 3rd is data. explain in detail then this question will be completed.

8085 Instruction Set Page 1

8085 INSTRUCTION SET

INSTRUCTION DETAILS

DATA TRANSFER INSTRUCTIONS

Opcode Operand Description

Copy from source to destination

MOV Rd, Rs This instruction copies the contents of the source

M, Rs register into the destination register; the contents of

Rd, M the source register are not altered. If one of the operands is a

memory location, its location is specified by the contents of

the HL registers.

Example: MOV B, C or MOV B, M

Move immediate 8-bit

MVI Rd, data The 8-bit data is stored in the destination register or

M, data memory. If the operand is a memory location, its location is

specified by the contents of the HL registers.

Example: MVI B, 57H or MVI M, 57H

Load accumulator

LDA 16-bit address The contents of a memory location, specified by a

16-bit address in the operand, are copied to the accumulator.

The contents of the source are not altered.

Example: LDA 2034H

Load accumulator indirect

LDAX B/D Reg. pair The contents of the designated register pair point to a memory

location. This instruction copies the contents of that memory

location into the accumulator. The contents of either the

register pair or the memory location are not altered.

Example: LDAX B

Load register pair immediate

LXI Reg. pair, 16-bit data The instruction loads 16-bit data in the register pair

designated in the operand.

Example: LXI H, 2034H or LXI H, XYZ

Load H and L registers direct

LHLD 16-bit address The instruction copies the contents of the memory location

pointed out by the 16-bit address into register L and copies

the contents of the next memory location into register H. The

contents of source memory locations are not altered.

Example: LHLD 2040H

8085 Instruction Set Page 2

Store accumulator direct

STA 16-bit address The contents of the accumulator are copied into the memory

location specified by the operand. This is a 3-byte instruction,

the second byte specifies the low-order address and the third

byte specifies the high-order address.

Example: STA 4350H

Store accumulator indirect

STAX Reg. pair The contents of the accumulator are copied into the memory

location specified by the contents of the operand (register

pair). The contents of the accumulator are not altered.

Example: STAX B

Store H and L registers direct

SHLD 16-bit address The contents of register L are stored into the memory location

specified by the 16-bit address in the operand and the contents

of H register are stored into the next memory location by

incrementing the operand. The contents of registers HL are

not altered. This is a 3-byte instruction, the second byte

specifies the low-order address and the third byte specifies the

high-order address.

Example: SHLD 2470H

Exchange H and L with D and E

XCHG none The contents of register H are exchanged with the contents of

register D, and the contents of register L are exchanged with

the contents of register E.

Example: XCHG

Copy H and L registers to the stack pointer

SPHL none The instruction loads the contents of the H and L registers into

the stack pointer register, the contents of the H register

provide the high-order address and the contents of the L

register provide the low-order address. The contents of the H

and L registers are not altered.

Example: SPHL

Exchange H and L with top of stack

XTHL none The contents of the L register are exchanged with the stack

location pointed out by the contents of the stack pointer

register. The contents of the H register are exchanged with

the next stack location (SP+1); however, the contents of the

stack pointer register are not altered.

Example: XTHL

8085 Instruction Set Page 3

Push register pair onto stack

PUSH Reg. pair The contents of the register pair designated in the operand are

copied onto the stack in the following sequence. The stack

pointer register is decremented and the contents of the highorder

register (B, D, H, A) are copied into that location. The

stack pointer register is decremented again and the contents of

the low-order register (C, E, L, flags) are copied to that

location.

Example: PUSH B or PUSH A

Pop off stack to register pair

POP Reg. pair The contents of the memory location pointed out by the stack

pointer register are copied to the low-order register (C, E, L,

status flags) of the operand. The stack pointer is incremented

by 1 and the contents of that memory location are copied to

the high-order register (B, D, H, A) of the operand. The stack

pointer register is again incremented by 1.

Example: POP H or POP A

Output data from accumulator to a port with 8-bit address

OUT 8-bit port address The contents of the accumulator are copied into the I/O port

specified by the operand.

Example: OUT F8H

Input data to accumulator from a port with 8-bit address

IN 8-bit port address The contents of the input port designated in the operand are

read and loaded into the accumulator.

Example: IN 8CH

8085 Instruction Set Page 4

ARITHMETIC INSTRUCTIONS

Opcode Operand Description

Add register or memory to accumulator

ADD R The contents of the operand (register or memory) are

M added to the contents of the accumulator and the result is

stored in the accumulator. If the operand is a memory

location, its location is specified by the contents of the HL

registers. All flags are modified to reflect the result of the

addition.

Example: ADD B or ADD M

Add register to accumulator with carry

ADC R The contents of the operand (register or memory) and

M the Carry flag are added to the contents of the accumulator

and the result is stored in the accumulator. If the operand is a

memory location, its location is specified by the contents of

the HL registers. All flags are modified to reflect the result of

the addition.

Example: ADC B or ADC M

Add immediate to accumulator

ADI 8-bit data The 8-bit data (operand) is added to the contents of the

accumulator and the result is stored in the accumulator. All

flags are modified to reflect the result of the addition.

Example: ADI 45H

Add immediate to accumulator with carry

ACI 8-bit data The 8-bit data (operand) and the Carry flag are added to the

contents of the accumulator and the result is stored in the

accumulator. All flags are modified to reflect the result of the

addition.

Example: ACI 45H

Add register pair to H and L registers

DAD Reg. pair The 16-bit contents of the specified register pair are added to

the contents of the HL register and the sum is stored in the

HL register. The contents of the source register pair are not

altered. If the result is larger than 16 bits, the CY flag is set.

No other flags are affected.

Example: DAD H

8085 Instruction Set Page 5

Subtract register or memory from accumulator

SUB R The contents of the operand (register or memory ) are

M subtracted from the contents of the accumulator, and the result

is stored in the accumulator. If the operand is a memory

location, its location is specified by the contents of the HL

registers. All flags are modified to reflect the result of the

subtraction.

Example: SUB B or SUB M

Subtract source and borrow from accumulator

SBB R The contents of the operand (register or memory ) and

M the Borrow flag are subtracted from the contents of the

accumulator and the result is placed in the accumulator. If

the operand is a memory location, its location is specified by

the contents of the HL registers. All flags are modified to

reflect the result of the subtraction.

Example: SBB B or SBB M

Subtract immediate from accumulator

SUI 8-bit data The 8-bit data (operand) is subtracted from the contents of the

accumulator and the result is stored in the accumulator. All

flags are modified to reflect the result of the subtraction.

Example: SUI 45H

Subtract immediate from accumulator with borrow

SBI 8-bit data The 8-bit data (operand) and the Borrow flag are subtracted

from the contents of the accumulator and the result is stored

in the accumulator. All flags are modified to reflect the result

of the subtracion.

Example: SBI 45H

Increment register or memory by 1

INR R The contents of the designated register or memory) are

M incremented by 1 and the result is stored in the same place. If

the operand is a memory location, its location is specified by

the contents of the HL registers.

Example: INR B or INR M

Increment register pair by 1

INX R The contents of the designated register pair are incremented

by 1 and the result is stored in the same place.

Example: INX H

8085 Instruction Set Page 6

Decrement register or memory by 1

DCR R The contents of the designated register or memory are

M decremented by 1 and the result is stored in the same place. If

the operand is a memory location, its location is specified by

the contents of the HL registers.

Example: DCR B or DCR M

Decrement register pair by 1

DCX R The contents of the designated register pair are decremented

by 1 and the result is stored in the same place.

Example: DCX H

Decimal adjust accumulator

DAA none The contents of the accumulator are changed from a binary

value to two 4-bit binary coded decimal (BCD) digits. This is

the only instruction that uses the auxiliary flag to perform the

binary to BCD conversion, and the conversion procedure is

described below. S, Z, AC, P, CY flags are altered to reflect

the results of the operation.

If the value of the low-order 4-bits in the accumulator is

greater than 9 or if AC flag is set, the instruction adds 6 to the

low-order four bits.

If the value of the high-order 4-bits in the accumulator is

greater than 9 or if the Carry flag is set, the instruction adds 6

to the high-order four bits.

Example: DAA

8085 Instruction Set Page 7

BRANCHING INSTRUCTIONS

Opcode Operand Description

Jump unconditionally

JMP 16-bit address The program sequence is transferred to the memory location

specified by the 16-bit address given in the operand.

Example: JMP 2034H or JMP XYZ

Jump conditionally

Operand: 16-bit address

The program sequence is transferred to the memory location

specified by the 16-bit address given in the operand based on

the specified flag of the PSW as described below.

Example: JZ 2034H or JZ XYZ

Opcode Description Flag Status

JC Jump on Carry CY = 1

JNC Jump on no Carry CY = 0

JP Jump on positive S = 0

JM Jump on minus S = 1

JZ Jump on zero Z = 1

JNZ Jump on no zero Z = 0

JPE Jump on parity even P = 1

JPO Jump on parity odd P = 0

8085 Instruction Set Page 8

Unconditional subroutine call

CALL 16-bit address The program sequence is transferred to the memory location

specified by the 16-bit address given in the operand. Before

the transfer, the address of the next instruction after CALL

(the contents of the program counter) is pushed onto the stack.

Example: CALL 2034H or CALL XYZ

Call conditionally

Operand: 16-bit address

The program sequence is transferred to the memory location

specified by the 16-bit address given in the operand based on

the specified flag of the PSW as described below. Before the

transfer, the address of the next instruction after the call (the

contents of the program counter) is pushed onto the stack.

Example: CZ 2034H or CZ XYZ

Opcode Description Flag Status

CC Call on Carry CY = 1

CNC Call on no Carry CY = 0

CP Call on positive S = 0

CM Call on minus S = 1

CZ Call on zero Z = 1

CNZ Call on no zero Z = 0

CPE Call on parity even P = 1

CPO Call on parity odd P = 0

8085 Instruction Set Page 9

Return from subroutine unconditionally

RET none The program sequence is transferred from the subroutine to

the calling program. The two bytes from the top of the stack

are copied into the program counter, and program execution

begins at the new address.

Example: RET

Return from subroutine conditionally

Operand: none

The program sequence is transferred from the subroutine to

the calling program based on the specified flag of the PSW as

described below. The two bytes from the top of the stack are

copied into the program counter, and program execution

begins at the new address.

Example: RZ

Opcode Description Flag Status

RC Return on Carry CY = 1

RNC Return on no Carry CY = 0

RP Return on positive S = 0

RM Return on minus S = 1

RZ Return on zero Z = 1

RNZ Return on no zero Z = 0

RPE Return on parity even P = 1

RPO Return on parity odd P = 0

8085 Instruction Set Page 10

Load program counter with HL contents

PCHL none The contents of registers H and L are copied into the program

counter. The contents of H are placed as the high-order byte

and the contents of L as the low-order byte.

Example: PCHL

Restart

RST 0-7 The RST instruction is equivalent to a 1-byte call instruction

to one of eight memory locations depending upon the number.

The instructions are generally used in conjunction with

interrupts and inserted using external hardware. However

these can be used as software instructions in a program to

transfer program execution to one of the eight locations. The

addresses are:

Instruction Restart Address

RST 0 0000H

RST 1 0008H

RST 2 0010H

RST 3 0018H

RST 4 0020H

RST 5 0028H

RST 6 0030H

RST 7 0038H

The 8085 has four additional interrupts and these interrupts

generate RST instructions internally and thus do not require

any external hardware. These instructions and their Restart

addresses are:

Interrupt Restart Address

TRAP 0024H

RST 5.5 002CH

RST 6.5 0034H

RST 7.5 003CH

8085 Instruction Set Page 11

LOGICAL INSTRUCTIONS

Opcode Operand Description

Compare register or memory with accumulator

CMP R The contents of the operand (register or memory) are

M compared with the contents of the accumulator. Both

contents are preserved . The result of the comparison is

shown by setting the flags of the PSW as follows:

if (A) < (reg/mem): carry flag is set

if (A) = (reg/mem): zero flag is set

if (A) > (reg/mem): carry and zero flags are reset

Example: CMP B or CMP M

Compare immediate with accumulator

CPI 8-bit data The second byte (8-bit data) is compared with the contents of

the accumulator. The values being compared remain

unchanged. The result of the comparison is shown by setting

the flags of the PSW as follows:

if (A) < data: carry flag is set

if (A) = data: zero flag is set

if (A) > data: carry and zero flags are reset

Example: CPI 89H

Logical AND register or memory with accumulator

ANA R The contents of the accumulator are logically ANDed with

M the contents of the operand (register or memory), and the

result is placed in the accumulator. If the operand is a

memory location, its address is specified by the contents of

HL registers. S, Z, P are modified to reflect the result of the

operation. CY is reset. AC is set.

Example: ANA B or ANA M

Logical AND immediate with accumulator

ANI 8-bit data The contents of the accumulator are logically ANDed with the

8-bit data (operand) and the result is placed in the

accumulator. S, Z, P are modified to reflect the result of the

operation. CY is reset. AC is set.

Example: ANI 86H

8085 Instruction Set Page 12

Exclusive OR register or memory with accumulator

XRA R The contents of the accumulator are Exclusive ORed with

M the contents of the operand (register or memory), and the

result is placed in the accumulator. If the operand is a

memory location, its address is specified by the contents of

HL registers. S, Z, P are modified to reflect the result of the

operation. CY and AC are reset.

Example: XRA B or XRA M

Exclusive OR immediate with accumulator

XRI 8-bit data The contents of the accumulator are Exclusive ORed with the

8-bit data (operand) and the result is placed in the

accumulator. S, Z, P are modified to reflect the result of the

operation. CY and AC are reset.

Example: XRI 86H

Logical OR register or memory with accumulaotr

ORA R The contents of the accumulator are logically ORed with

M the contents of the operand (register or memory), and the

result is placed in the accumulator. If the operand is a

memory location, its address is specified by the contents of

HL registers. S, Z, P are modified to reflect the result of the

operation. CY and AC are reset.

Example: ORA B or ORA M

Logical OR immediate with accumulator

ORI 8-bit data The contents of the accumulator are logically ORed with the

8-bit data (operand) and the result is placed in the

accumulator. S, Z, P are modified to reflect the result of the

operation. CY and AC are reset.

Example: ORI 86H

Rotate accumulator left

RLC none Each binary bit of the accumulator is rotated left by one

position. Bit D7 is placed in the position of D0 as well as in

the Carry flag. CY is modified according to bit D7. S, Z, P,

AC are not affected.

Example: RLC

Rotate accumulator right

RRC none Each binary bit of the accumulator is rotated right by one

position. Bit D0 is placed in the position of D7 as well as in

the Carry flag. CY is modified according to bit D0. S, Z, P,

AC are not affected.

Example: RRC

8085 Instruction Set Page 13

Rotate accumulator left through carry

RAL none Each binary bit of the accumulator is rotated left by one

position through the Carry flag. Bit D7 is placed in the Carry

flag, and the Carry flag is placed in the least significant

position D0. CY is modified according to bit D7. S, Z, P, AC

are not affected.

Example: RAL

Rotate accumulator right through carry

RAR none Each binary bit of the accumulator is rotated right by one

position through the Carry flag. Bit D0 is placed in the Carry

flag, and the Carry flag is placed in the most significant

position D7. CY is modified according to bit D0. S, Z, P, AC

are not affected.

Example: RAR

Complement accumulator

CMA none The contents of the accumulator are complemented. No flags

are affected.

Example: CMA

Complement carry

CMC none The Carry flag is complemented. No other flags are affected.

Example: CMC

Set Carry

STC none The Carry flag is set to 1. No other flags are affected.

Example: STC

8085 Instruction Set Page 14

CONTROL INSTRUCTIONS

Opcode Operand Description

No operation

NOP none No operation is performed. The instruction is fetched and

decoded. However no operation is executed.

Example: NOP

Halt and enter wait state

HLT none The CPU finishes executing the current instruction and halts

any further execution. An interrupt or reset is necessary to

exit from the halt state.

Example: HLT

Disable interrupts

DI none The interrupt enable flip-flop is reset and all the interrupts

except the TRAP are disabled. No flags are affected.

Example: DI

Enable interrupts

EI none The interrupt enable flip-flop is set and all interrupts are

enabled. No flags are affected. After a system reset or the

acknowledgement of an interrupt, the interrupt enable flipflop

is reset, thus disabling the interrupts. This instruction is

necessary to reenable the interrupts (except TRAP).

Example: EI

8085 Instruction Set Page 15

Read interrupt mask

RIM none This is a multipurpose instruction used to read the status of

interrupts 7.5, 6.5, 5.5 and read serial data input bit. The

instruction loads eight bits in the accumulator with the

following interpretations.

Example: RIM

Set interrupt mask

SIM none This is a multipurpose instruction and used to implement the

8085 interrupts 7.5, 6.5, 5.5, and serial data output. The

instruction interprets the accumulator contents as follows.

Example: SIM

BIOS = Basic Input Output System

CMOS = Copper-Metal Oxide Semiconductor

BIOS is the interface which is built-in to a computer system's firmware used to configure the computer system hardware in very basic ways. BIOS can be used to enable or disable hardware integrated into the motherboard such as IDE controllers, USB hosts, audio controllers, video, and more.

It is often accessed by pressing F2 during startup, but this can change from manufacturer to manufacturer.

CMOS is a type of circuitry often powered by a battery which allows the information in BIOS to be stored. If the CMOS battery is drained or removed all the system configuration data in BIOS will be reset to factory defaults every time the computer is shut off. These small batteries are also commonly used in calculators, hearing aides, and wristwatches.

Due to the close relationship between the BIOS and the CMOS in the computer system, the two phrases are often incorrectly used as synonyms for each other. Some users will refer to opening the BIOS menu as "going into CMOS" or replacing the CMOS battery as "fixing the BIOS battery." However, none of this is seen as grammatically incorrect in common day-to-day conversation.

when your PC has been inoperation for a while the components inside start to heat up this heat can cause damage to those components a heat sink turns the heat into air and vents it out of the PC making sure your compnents eg processors are still in working order

Transistor transistor logic is one type of many different types of bipolar transistor based digital logic circuitry. It is very efficiently implemented in integrated circuit chips, needs only one power supply voltage, and operates at reasonably high speeds.

Transistor transistor logic was first developed in the middle 1960s as a modification of the diode transistor logic, then in use in some digital logic integrated circuit chips but dating back to the earliest discrete bipolar transistor logic developed in the late 1950s and derived from vacuum tube point contact diode logic used in many early first generation computers.

Transistor transistor logic integrated circuits dominated the computer and electronic digital controller market from the late 1960s until the middle 1980s, when metal oxide semiconductor field effect transistor based microprocessors and microcontrollers began to replace it. By the early 1990s transistor transistor logic and other bipolar transistor based digital logic integrated circuits had been replaced with equivalent complementary metal oxide semiconductor field effect transistor integrated circuits that were both faster and consumed less power (thus running much cooler) or with programmable logic devices of various types.

In general transistor transistor logic is now considered obsolete.

There are no 'watts' in a laptop.

A laptop, or any other electrical device uses electricity, and Watts measure the rate at which energy is used. Asking how many watts a laptop uses is like asking how many watts a light bulb uses. There are light bulbs that use a small fraction of a watt to those that use more than 1000 watts. As Watts measure energy consumption, it does not necessarily mean electricity: A person having a mass of 100 kilograms who climbs a 3-meter-high ladder in 5 seconds is doing work at a rate of about 600 watts.

It depends on the laptop as to how many watts it uses.

In electrical terms, one watt is the rate at which work is done when one ampere of current flows through an electrical potential difference of one volt.

When W=Watts; V=Volts; I=Amps; R=resistance in Ohms, then W = VI or W = I2R or E2/R

I have a netbook that uses a 19 volt, 1.5amp power supply. As Watts = Volts x Amps, it would appear that it uses 19 x 1.5 or 28½ watts. But that is only the power supply rating, The maximum the netbook uses is probably closer to half of that. The power supply is most likely very overrated. It operates nicely with the power supply when the battery is exhausted and being charged at the same time as being used.

Verizon FiOS & Verizon Broadband Services offer blazing speed & crystal clarity of the 100% fiber optic Fios network & proven reliability for their TV, Internet & phone products that will exceed your expectations today & keep you connected down the road.

cutt.ly/3hVTqFz

An analog computer is any computer that represents its data in the form of continuously variable signals instead of the discontinuous encoded symbols used by digital computers.

Interleaving is an advanced technique used by high-end motherboards/chipsets to improve memory performance. Memory interleaving increases bandwidth by allowing simultaneous access to more than one chunk of memory. This improves performance because the processor can transfer more information to/from memory in the same amount of time, and helps alleviate the processor-memory bottleneck that is a major limiting factor in overall performance.

The motherboard aka mainboard

A multi-core processor is one which combines what are essentially multiple CPUs into a single chip. As far as the operating system and other software are concerned, it is the same as a dual or quad-cpu computer system, even though the cpus are physically one single unit.

A line is basically a combination of two joined rays pointing in opposite directions. A line has no endpoints and a line segment has two.

<--------------> line

---------------> or <------------ ray

.________. line segment (It's supposed to be a line with dots at the end to signify endpoint)

If by program you mean OS, your available options depends on your OS. If you have a PC (aka Windows), you can install Ubuntu, Kubuntu, and other Linux OSs to run alongside Windows. If you have a Mac, you can install Windows and Linux. Virtual Machine software allows you to do this without rebooting your computer to change the OS. There are many different applications that can let you do this. Some you have to pay for, like Parallels Desktop or VMWare Fusion. There are also open source (free) ones, like Oracle VirtualBox.

A couple scenarios:

Install Windows on a Mac: You can use Boot Camp, a preinstalled application that walks you through the installation process. You can also use any of the VM apps listed above.

Install Ubuntu on a PC: You can use the Wubi installer provided with all versions of Ubuntu when you download them in the related link, or you can use a VM app.

Install Ubuntu on a Mac: You cannot use the Wubi on a Mac, you will need to create a bootable USB drive (steps in the related link). You can also run Ubuntu as a VM.

A piece of intrinsic (pure) silicon at room temperature has, at any instant, a number of conduction-band (free) electrons bthat are unattached to any atom and are essentially drifting randomly throughout the material. Also, and equal number of holes are created in the valence band when these electrons jump into the conductance band.

1111111111111111 (216 - 1 = 65535)

It doesn't matter what is the speed is on the USB it depends on the computer's speed on its hardrive, internet, or both the speed is not based on the 2.0 USB, is the computer either its old or new or in between.

A microprocessor is a multipurpose, programmable device that uploads digital data, processes it according to memory-stored instruction, and provides output results. They operate on numbers and symbols represented in binary code.

17

ohmmeter has it's own power supply

That varies dramatically depending on the revision of the USB used on the port.