Find the Square Root of a given number-Free 8085 Microprocessor projects

Source Program:

LDA 4200H : Get the given data(Y) in A register

MOV B,A : Save the data in B register

MVI C,02H : Call the divisor(02H) in C register

CALL DIV : Call division subroutine to get initial value(X) in D-reg

REP: MOV E,D : Save the initial value in E-reg

MOV A,B : Get the dividend(Y) in A-reg

MOV C,D : Get the divisor(X) in C-reg

CALL DIV : Call division subroutine to get initial value(Y/X) in D-reg

MOV A, D : Move Y/X in A-reg

ADD E : Get the((Y/X) + X) in A-reg

MVI C, 02H : Get the divisor(02H) in C-reg

CALL DIV : Call division subroutine to get ((Y/X) + X)/2 in D-reg.This is XNEW

MOV A, E : Get Xin A-reg

CMP D : Compare X and XNEW

JNZ REP : If XNEW is not equal to X, then repeat

STA 4201H : Save the square root in memory

HLT : Terminate program execution

Subroutine:

DIV: MVI D, 00H : Clear D-reg for Quotient

NEXT:SUB C : Subtact the divisor from dividend

INR D : Increment the quotient

CMP C : Repeat subtraction until the

JNC NEXT : divisor is less than dividend

RET : Return to main program

Note: The square root can be taken y an iterative technique. First, an initial value is assumed. Here, the initial value of square root is taken as half the value of given number. Te new value of square root is computed by using an expression XNEW = (X + Y/X)/2 where, X is the initial value of square root and Y is the given number. Then, XNEW is compared wit initial value. If they are not equal then the above process is repeated until X is equal to XNEW after taking XNEW as initial value. (i.e., X ←XNEW)