Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
-
Economics & Finance
8085 Articles
Page 7 of 36
Representation of fractions
To represent fractions may be necessary quite often inside the computer. For example, it may be needed to represent inside a computer a value like +0.610 or -0.610. To represent signed fractions, it is necessary to assume the binary point just after the MSB in the bit sequence. Such numbers where the binary point is assumed to be at a fixed position in the bit sequence are called fixed-point numbers.Unsigned fractions will have the assumed binary point at the extreme left. SM, 1's complement, and 2's complement fractions will have this imaginary binary point just to the right of the ...
Read More2's complement fractions
As an example, the value of 1 001, if the interpretation is that it is a 2's complement fraction will be as follows -It is 1.001 assuming the binary point after the MS bit. As the MS bit is 1, it is a negative number. Then the remaining bits do not specify the magnitude directly. The 2's complement of 1 001 is 0110+ 1 = 0 111. This is a positive fraction with the value 1 × 2−1 +1 × 2−2 +1 × 2−3 = 0.5 + 0.25 + 0.125 = 0.875 decimal. Thus, -0.875 is the value of 1001.If ...
Read MoreInstruction set of 8051
The instructions of 8051 Microcontroller can be classified into five different groups. These groups are like belowData Transfer GroupArithmetic GroupLogical GroupProgram Branch GroupBit Processing GroupThis Bit-Processing group is also known as Boolean Variable Manipulation.Like 8085, some instruction has two operands. The first operand is the Destination, and the second operator is Source.In the following examples, you will get some notations. The notations are like −Rn = Any register from R0to R7 Ri = Either R0 or R1 d8 = Any 8-bit immediate data (00H to FFH) d16 = 16-bit immediate data a8 = 8-bit address bit = 8-bit address of ...
Read MoreSigned floating point numbers
Were present real numbers in our daily life is not convenient for representing very small numbers, like +0.00000012347650. This same number can be more conveniently represented in scientific notation as +1.23476× 10−07. But this actually stands for +0.000000123476. So there is an error of 0.00000000000005, which forms a very small percentage error.Floating-point representation is similar in concept to scientific notation. Logically, a floating-point number consists of:A signed (meaning positive or negative) digit string of a given length in a given base(or radix).This digit string is referred to as the significand, mantissa, or coefficient.A signed integer exponent which modifies the magnitude of ...
Read MoreData transfer group in 8051
In 8051 Microcontroller there is 28 different instructions under the Data Transfer Group. In total there are 79 opcodes. The flags are not affected by using the data transfer instructions, but the P (Parity) flag may change if the value of A register is changed using Data Transfer Instruction. Similarly, when a data is transferred to the PSW register, the flags will change.In the following table, we will see the Mnemonics, Lengths, Execution Time in terms of the machine cycle, Number of Opcodes etc.MnemonicsByte CountExecution TimeOpcode CountMOV A, Rn118MOV A, a8211MOV A, @Ri112MOV A, #d8211MOV Rn, A118MOV Rn, a8228MOV Rn, ...
Read MoreArithmetic group in 8051
In 8051 Microcontroller there are 24 different instructions under the Arithmetic Group. In total there are 64 opcodes. The Carry Flag (CY), Auxiliary Carry (AC)and Overflow flag (OV) are affected based on the result of ADD, ADDC, SUBB etc. instructions. The multiple and divide instructions clear the Carry flag, and also does not affect the AC flag. After execution of multiplication, the OV flag will be 1 when the result is greater than FFH. Otherwise, it is 0. Similarly, after division OV flag is 1 when the content of B is 00H before division, otherwise it is 0. The DA ...
Read MoreUnsigned binary integers
Unsigned binary integers are numbers without any ‘+’or ‘-’ sign. Here all bits representing the number will represent the magnitude part of the number only. No bits will remain reserved for sign bit representation. An unsigned binary integer is a fixed-point system with no fractional digits.Some real life Examples are −Number of tables in a class, The number of a member of a family.Obviously, they are unsigned integers like 10 and 5. These numbers have to be represented in a computer using only binary notation or using bits.Numbers are represented in a computer using a fixed size, like 4, 8, ...
Read MoreSign Magnitude notation
The sign-magnitude binary format is the simplest conceptual format. In this method of representing signed numbers, the most significant digit (MSD) takes on extra meaning.If the MSD is a 0, we can evaluate the number just as we would any normal unsigned integer. And also we shall treat the number as a positive one.If the MSD is a 1, this indicates that the number is negative.The other bits indicate the magnitude (absolute value) of the number. Some of the signed decimal numbers and their equivalent in SM notation follows assuming a word size of 4 bits.Signed decimalsign-magnitude +6 0110 ...
Read MoreShift a multi-byte BCD number to the right in 8051
Here we will see a problem to shift some multi-byte BCD number to the right. The BCD numbers are shifted by two digits (8-bit). Let us consider a four-byte BCD number (45 86 02 78) is stored at location 20H, 21H, 22H, 23H. The address 10H holds the number of bytes of the whole BCD number. So after executing this code, the contents will be shifted to the right and 20H will hold 00H.AddressValue...20H4521H8622H0223H78...Program CLRA;Clear the Register A MOVR2, 10H;TakeByte Count INCR2;Increase R2 for loop MOVR1, ...
Read MoreBinary to BCD conversion in 8051
In this problem, we will see how to convert an 8-bit binary number to its BCD equivalent. The binary number is stored at location 20H. After converting, the results will be stored at 30H and 31H. The 30H will hold the MS portion, and 31H will hold the LS portion. So let us assume the data is D5H. The program converts the binary value of D5H to BCD value 213D.AddressValue...20HD521H...ProgramMOVR1, #20H;Takethe address 20H into R1 MOVA, @R1;Takethe data into Acc MOVB, #0AH;LoadB with AH = 10D DIVAB ;DivideA with B MOVR5, B;Storethe remainder MOVB, #0AH;LoadB with AH = 10D DIVAB ;DivideA ...
Read More