hello everyone in this session we'll discuss computer registers to design any of the basic computer we require a certain minimum number of registers so to design our basic computer today we'll discuss the processor registers which are required for our system so the first register in our list is PC program counter program counter is of size 12 bits which holds the address of the next instruction which is to be executed next we are having address register a R which is also of size 12 bits so address register will holds the address of the memory that means this will point to the memory next we are having is IR which is of size 16 bits eye R stands for instruction register and this in Sturgis tur will holds the instruction code then we are having TR TR is of size 16 bits which stands for temporal register which holds the temporary data generated during certain micro operations carried out also it K it is used during the call and return of some subroutines then we are having data register which is also size 16 bits now this D our data register will holds the memory operand that means the data which is fetched from the memory will be getting stored into this memory will be getting stored into data register then we are having AC accumulator which is of size 16 bits it is also termed as processor register that means when we perform any of the micro operations like arithmetic shaved logic the in result which is getting generated will be stored into this accumulator and also the date processing will be carried out on to this register only so this is one of the important register for our basic computer then we are having is out R stands for output register which is of size 8 bits now this will holds the output character that means the character which we want to send on to our output device would be stored into this particular out our register next we are having is InP our register which is of size 8 bits stands for input register now this input resistor will hold input character that means the data which is given input from input device would be stored in this I n PR and then it would be transferred to the corresponding register or memory the next we are having is the important unit without which our computer cannot work that is memory in our design of basic computer the memory size what we are considering is 4 0 9 6 words and each word is of 16 bits so we can say that the memory size is 4 0 9 6 cross 16 means 4 0 9 6 cross 16 bits can be stored into this memory these are the registers and memory required for our basic computer but now let us see the arrangement of this how they are inter mutually connected with each other so for that we need to understand the common bus system of a basic computer so the common bus system of this basic computer is something like the arrangements of memory and registers we had seen all this registers and memory now these registers and memory are connected to this bus by some arrangement let's see one by one but this before moving to that arrangement all this memory and registers are given certain signals so starting with memory memory will have two signals read and write because in memory we can perform only two operations either we can read from memory or we can write to the memory so read or write then we are having a our pc d r n AC all this for register will have 3 signals that is load increment and clear so load means it will allow the data to store into this register means it will load the registers increment INR this INR signal will increment the content of the corresponding register that means if the signal is high on INR the content of any of this register would be incremented by one and the last is clear so this clear signal will clear the data of that particular register that means it will make all the bits to be zero so it will reset the register then we are having IR register now IR register is having only one signal that is load because IR need not require any kind of increment or clear operation so it is having only one signal that is load TR will again have all three signals an outer will have one signal load because it will load the data from accumulator during its transferring of data to output device so outer will have only one signal that is load now as you can see I NPR do not have any kind of signal like load or increment or clear because in a NPR register input register is always connected to the input device because the data coming from input device would be stored in this InP R and it is asynchronous it does not depends on any of the signal now all these registers and memory except I on PR are connected to some clock because without the synchronization they cannot work so the clock is provided to this particular registers now if we move further these registers and memory will output the data to bus so they are connected to the bus and again these outputs will be given to bus on some signals because our bus is multiplexed which will be coming into the later part but the one register output of this error register is given to the memory to point to the memory so this is given as an input part to the memory by means of a R then the input part to this all registers from bus that means memory and address register PCD year then ir t our outer all are having some inputs except to register AC and i NPR they cannot get input from the bus so how they would be loaded from where the data will come let us see one by one this is one more unit named as adder and logic unit so ALU a computer cannot work siva without ALU it is important part so arithmetic and logic unit so adder and logic that is one part added over here so this adder and logic unit will help three inputs the first input would be coming from data register as you can see when the data register would be transferring to the data to bus it will also transfer the data to adder in logic unit now this adder and logic unit will also get the second input from the accumulator itself as you can see the output of accumulator is given to bus as well as it is given to the adder logic unit and the third input to this adder logic unit would be coming from I NPR so the output of iron PR is given to this a deranged logic unit now this adder in logic unit will be giving its output to accumulator so this is the only way of transferring the data from anywhere to this accumulator this three are the only ways from by which we can load the accumulator then we are having one more thing that is e e is a flip-flop bit extra bit Easton's for extra bit because during certain micro operations like arithmetic micro operations addition subtraction some carry bits or borrow bits are generated so they would be stored in this e this e bit will be having that extra bits and it is also used for overflow flags so this e would be getting the output from a deranged logic unit so these is the general arrangement of our registers and memory now as I told you earlier that our bus is multiplexed so there are three multiplexing signals select lines given to this bus that is s 2 s 1 and s 0 so this s 2 and s 1 and s 0 based on these three select lines any of the output lines from here would be selected if I provide the Select lines to be 0 1 1 at that time the corresponding line 3 would be enabled that is the output from data register would be transferred to the bus and that data would be moving on to the bus so bus is loaded with the content of tier similarly if I provide all the 3 values for s 2 s 1 and s 0 to be 1 1 1 lines in decimal value corresponding to it is 7 so memory data would be transferred to bus similarly when I provide one address register content would be on the bus 2 then the content of PC would be there on bus for AC 5 instruction register 6 temporary register so based on this multiplexing values the corresponding output of the register would be transferred to the bus so as you can see this is the general arrangement of common bus let us take one example we want to transfer the content from memory from memory to accumulator as you can see there is no direct way available that means if we put the data on bus and it gets loaded to the accumulator no you can see there is no input directly from bus to accumulator so what we can do so what we are having is the way the first of all what we'll do is we'll be loading the bus with the content of memory by providing the Select lines to be 1 1 1 so on giving the Select line inputs the content of memory would be transferred to the bus so now the bus will have the data from the memory and it is moving on to the bus the question is now we need to load it to accumulator so an intermediate or estep what we'll perform is we will load that data into D our first data or distr so the bursts data would be transferred to dr by enabling the load signal and the data would be loaded into da so one intermediate step is performed now changing the select lines to be 0 1 1 the content of data register would be transferred to the bus and now file transferring the data to burst as you can see I don't my main point is not that while transferring the data to the bus you can see that the data register is also connected to a drain logic unit so otha data from dr has been transferred to adder in logic unit and this adder and logic unit will transfer this data to accumulator on enabling the load signal of AC so this is how the data is transferred to memory from memory to accumulator so this is what is all about common bus system of a basic computer