Intel 8085

The two men that designed the MITS Altair 8800 computer kit in 1975 were Ed Roberts and Forrest M. Mims III.

The Altair 8800 used the Intel 8080 as it CPU. Roberts and Forrest did not invent the 8080 - Intel did - they simply used the 8080 in their product.

disrupted

To ensure that any chemicals, hazardous substances and dangerous goods used as part of the companies operations are identified and controlled in a manner to minimise the risk of adverse health effects to employees, customers, contractors and the public.

There are four T states during op-code fetch in the 8085. T1 is the ALE state, where the address is emitted and the RD- line goes true; TWAIT is zero or more wait states based on READY; T2 is the middle of the fetch, and the internal strobing of the op-code; T3 is the completion of the fetch and stabilization of the bus signals; and T4 is the internal processing cycle.

The DAA (Decimal Adjust Accumulator) instruction in the 8085 conditionally adds 06H to A in order to normalize the low order nibble to a BCD value betwen 0H and 9H. It then conditionally adds 60H to A in order to similarly normalize the high order nibble.

This is done as part of supporting multi-precision decimal arithmetic.

The direct memory access or DMA mode of data transfer is the fastest amongst all the modes of data transfer.In this mode ,the device may transfer data directly to/from memory without any interference from the CPU.The device requests the CPU through a DMA controller to hold its data,address and control bus so that the device may transfer data directly to/from memory.

The difference between the 8086 and the 8088 is that the 8086 has a 16 bit data bus and that the 8088 has an 8 bit data bus.

Both processors are the same 16 bit processor, and both have a 20 bit address bus. The 8086 is twice as fast as the 8088 in terms of data transfer rate on the bus for the same bus clock speed.

because its an 8bit micro processor similar to those used in micro controller .

capable of performing 16 bits double register calculations .

something easy to start with

16 bit addition
:
mov al ,bcd1
add al,bcd2
daa

mov result,al

mov al, bcd1+1
adc al,bcd2+1
daa
mov result+1,al

0FH is a literal constant for the integer value 15. In the 8086/8088, as an opcode, it is the lock prefix, which means that no other CPU or bus master can take the bus during the execution of the instruction.

Jump on Carry is simply a "JUMP" instruction which will transfer the control to some specific location if carry flag is set.

For Example:

JC 2004H

This instruction will take the control to address location 2004H if carry flag is set.

if we have 10 address bits then we can have a memory module of

2^10 = 1024 b = 1kb

so for 16kb we need

16*2^10 = 2^14=16kb

here we need 2 chips as 1 chip only provid us 2^12 memory.

Address and data bus are multiplexed to reduce complexity.

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The LDAX/STAX instuctions in the 8085 load and store the accumulator. The target address used is contained in either the BC register pair (LDAX D), or the DE register pair (LDAX D).

The carry flag indicates that an arithmetic operation between two numbers resulted in an unsigned carry or borrow. The overflow flag indicates that an arithmetic operation between two numbers resulted in an unexpected change in sign, i.e. an overflow.

Carry is also like overflow. The difference is in interpretation. Do you consider the numbers signed or unsigned? Its all in interpretation because the hardware logic needed to add two numbers works the same, no matter if they are considered signed or unsigned.

Both flags are set or cleared together, so you can use whichever you wish.

780

bcz multiplication and division done in 16bit operation so we must ve to use pair of reg so we ve to use ax reg ....... thats why final value store in a and b pair of reg....

In order to determine the instructions per second in an 8085 microprocessor, you need to know how long each instruction takes to execute. Some are as short as 4 T cycles. Some are as long as 18 T cycles. This is dependent on how the program is written. Add up the T cycles for each instruction. Divide the clock frequency in hertz by the number of T cycles, and you get instructions per second. Note that clock frequency is one half of the crystal frequency. Note also that you must include Twait cycles in your calculation.

Because of it can process 32 bit data at a time i.e. simultaneously

summary Because of word length is 32 bit i.e perform operation on 32 bit data at a time(so it is depend on data bus hence data bus is 32 bit)

Another reason is because of It has 32 Bit ALU in which 32 bit operation can be performed simultaneously.

for eg: JNZ 8050 so when the condition is true it would take

1. opcode fetch- 4T

2. Memory read- 3T

3. memory read- 3T

total 10 T states and when the condition is false it takes

1. opcode fetch-4T

2. 3T states for checking flags and incrementing PC.

INR affect the carry flag.

DMA (Direct Memory Access) does not actually interrupt the CPU - it requests control of the bus, so that it can perform the transfer itself. It becomes a bus-master. This is done using the HOLD and HLDA (Hold Acknowledge) control pins.

This is not the same thing as an interrupt, which is where an external device requests the CPU's attention, and the CPU goes off and performs some code to service that request. In the case of DMA, the CPU actually freezes for the few cycles that the DMA controller requires, which is much, much more efficient than using an interrupt service routine.

No. Some programmed computers, such as the DEC PDP-11, are minicomputers. Some, such as the IBM Z/800, are mainframes. The distinction is somewhat vague, however, as many large computers are implemented as multiple microprocessors, or multiple bit sliced RISC machines.