The bit addressable memory in 8051 is compose from 210 bits:
- bit address space: 20H - 2FH bytes RAM = 00H - 7FH bits address;
- SFR registers;
The following addresses are NOT bit addressable, only 1-byte addressable:
- 32 bytes RAM from 00H to 1FH (R0 - R7 registers in all four banks);
- 80 bytes RAM general user from 30H to 7FH.
Because its a microcontroller with an 8 bit data bus width.
It is program status register.It i 8 bit ,bit addressable register.It consists of four maths flag.
The 8051 is a microcontroller, not a microprocessor. To add or subtract, use the ADD or SUBB opcodes.
65536 bytes, because the 8051 family has a 16 bit external address buss.
the 8051 microcontrroller is the name doesnt matter easy or not....but the maiin and imp diff between both is at89s51 represents the 8051 chip no....along with the name of company i.e atmel, so all n all both are same thing ...... 89c51 have RISC architecture and contains less no of opcodes which are easy for programming. so iti is preferred than 8051.
The 16 bytes (128 bits) at internal RAM locations 0x20-0x2F are bit-addressable.
89c51 is an microcontroller of 8051 series it is an 8-bit device packed in 40-pin package.
Because its a microcontroller with an 8 bit data bus width.
It is program status register.It i 8 bit ,bit addressable register.It consists of four maths flag.
Of the 128-byte internal RAM of the 8051, only 16 bytes are bit-addressable. The rest must be accessed in byte format. The bit-addressable RAM locations are 20H to 2FH.
just a note: Clarify your question Bit-addressable registers are registers that its bits can be modified individually. that means if you have register named "ACC" that is bit addressable , you can change its bits (D0 -D7) individually by special instructions SETB and CLR. ex: SETB ACC.3 ; will set bit number 4 (remember bit 0) in the register in 8051 there are many bit-addressable registers such as A (ACC), B, SCON, PCON, TCON, p0,p1,p2,p3 . best wishes, drdigital.
The 8051 is a microcontroller, not a microprocessor. To add or subtract, use the ADD or SUBB opcodes.
An 8051 microcontroller is a popular 8-bit microcontroller that was first introduced by Intel in 1980. It is widely used in embedded systems due to its simple architecture, low cost, and ease of use. The 8051 microcontroller typically includes CPU, RAM, ROM, I/O ports, timers, and serial communication interfaces on a single chip.
There is no such thing as 14-bit byte addressable memory. There IS, however, 16-bit byte addressable memory, which I hope is what you meant. In that case, you simply divide 16 by 4 to get 16/4 = 4 nibbles. If you did in fact mean 14-bit address, then it would be 14/4 = 3.5 nibbles
65536 bytes, because the 8051 family has a 16 bit external address buss.
8 bit
A register is said to be bit addressable if it's individual bits can be set or reset . e.g, let us suppose we have a 8 bit register named D being d7d6d5d4d3d2d1d0 then it is said to be bit addressable if i can set or reset bit dx using some instruction , e.g in 8051 SETB and CLR are used for this purpose , SETB D.1 makes d0 1 , CLR D.4 makesd3 0.