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How many T states are require for LHLD instruction in microprocessor?
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How do you draw timing diagram for 8085 instruction LHLD 5000H?
There is an example of a LHLD 5000H diagram on this website: atelier-drachenhaus.de/timing-diagram-8085. This will provide an idea of how to draw the diagram.
What is 3 byte instruction in microprocessor?
A two-byte instruction gives the specific function instruction in two bytes, or two words. The first specifies the opcode, which tells the microprocessor what operation will occur. The second specifies the operand, or the data that the operation is done on.
What is the different between LXI and LHLD in 8085?
What is the different between LXI and LHLD in 8085?LHLD 2050 Means..copy content of 2050 and 2051 to HL pairif2050 -> 90H2051->5AHLHLD 2050 implies..L -> 90HH -> 5AHLXI means..Load Register Pair with Immediate data.. 16bit dataLXI B ,2050HLoads BC pair with value 2050HB-> 20HC-> 50HIts similar to 2 MVI instructionieMVI H,20HMVI L,50Hwhich eventually implies..HL -> 2050Hthe advantage of LXI over MVI is that LXI is 3 byte and requires 10T states.Two MVI instruction is 4Byte and require 14Tstates...
Addition and subtraction of two 16 bit numbers using 8085 microprocessor?
The 8085 is an 8-bit computer, with only limited capability to do 16 bit arithmetic. In order to add two 16-bit numbers, NUM1 and NUM2, together, and store the result at NUM3, you can use the code... LHLD NUM1 XCHG LHLD NUM2 DAD D SHLD NUM3 If you want to subtract NUM1 from NUM2, you need to take the two's complement first, by inverting it and adding one... LHLD NUM1 MOV A,H CMA MOV H,A MOV A,L CMA MOV L,A INX H ... and then continue with adding NUM2... XCHG LHLD NUM2 DAD D SHLD NUM3
Why do use cisc and risc processor?
in cisc concept instruction set is complex but in case of risc the ins truction are reduced. for example for mov operation many instruction are there as mov,mvi,sta,lda,lhld,shld,stax,ldax etc. but in case of risc only load and store(lda ,sta) are used for all as in case of ARM processors.
How does LHLD transfer insruction work in Intel 8085?
it stands for load HL pair with data from the specified location in the register. eg. LHLD 9200 Aditi Misra 3rd yr,B.tech,CSE Netaji Subhash Engineering College
What is the different between SHLD and LHLD in 8085?
LHLD 2600h is use to store data of 2600h in L register and data of 2601h in H register While SHLD 2600h is use to store data of L in 2600h memory location and data of H in 2601h memory location data of H register
What is a direct address in language?
It means speaking to someone directly. Susan, clean up your room. Bob, hand me a flashlight. I told you already, Mom, I am not going. In these sentences, the direct address is to Susan, Bob, and Mom.
What is the function of immedidate instructions in the Intel 8085?
The immediate instructions in the 8085 contain the operand in the instruction, as the second byte, rather than in a register. For instance, if you want to add 3 to the accumulator, you can use ADI 3. Without the immediate instructions you would need two instructions, LHLD {address-of-a-3}, ADD M, but you might also need to save and restore HL.
Multiplication 16 bit program for 8085?
Lxi b, 0000h lhld 8000h xchg lhld 8002h dcx d l006: lda 8002h add l mov l, a lda 8003h adc h mov h, a jnc l013 l013: inx b dcx d mov a, d ora e jnz l006 shld 8006h mov l, c mov h, b shld 8004h hlt
Write a program to subtract two 16 bit numbers in microprocessor 8085?
LHLD 4000H : Get first 16-bit number in HLXCHG : Save first 16-bit number in DELHLD 4002H : Get second 16-bit number in HLMOV A, E : Get lower byte of the first numberSUB L : Subtract lower byte of the second numberMOV L, A : Store the result in L registerMOV A, D : Get higher byte of the first numberSBB H : Subtract higher byte of second number with borrowMOV H, A : Store l6-bit result in memory locations 4004H and 4005H.SHLD 4004H : Store l6-bit result in memory locations 4004H and 4005H.HLT : Terminate program execution.
Write a program to multiply two 8 bit signed numbers using 8085 microprocessor?
s v i e t k a n u r u .t k MULTIPLICATION OF TWO 8 -- BIT NUMBERS AIM: To perform multiplication of two 8 -- bit numbers using 8051 instruction set. ALGORITHM: 1. Get the data in A -- reg.2. Get the value to be multiplied in B -- reg.3. Multiply the two data4. The higher order of the result is in B -- reg.5. The lower order of the result is in A -- reg.6. Store the results. PROGRAM: ORG 4100CLR CMOV A,#data1MOV B,#data2MUL ABMOV DPTR,#4500MOVX @DPTR,AINC DPTR MOV A,BMOVX @DPTR,AHERE: SJMP HERE
