 Hello everyone. This is Vishwana Chauhan, Assistant Professor, Department of Computer Science and Engineering, Walshan Institute of Technology, Solar Port. Now I am here to explain the error correction in semiconductor memory. At the end of this session, students will be able to identify the error correction which is subjected to semiconductor memory. Let us see the different types of semiconductor memory like RAM, ROM, P-ROM, EP-ROM, Electrically, EP-ROM, Flash memory. Now the category based on read-write. RAM, it allows for read-write memory and ROM, read-only, and other these read mostly memory. And the next factor which is regarding erasing. So RAM can be erased electrically, byte-level, ROM and P-ROM not possible. And for other remaining three like UV light, chip-level, electrically, byte-level and electrically, block-level respectively. Regarding writing mechanism, RAM electrically, ROM, mask and remaining electrically. With respect to volatile, RAM is volatile and other are non-volatile. Let us see the different errors. There are mainly two types of errors. One is hardware failure and second one is software failure. Let us consider error correction code function. Now initially data in which is M is fed input to function F which leads to K, M and K are stored in memory. M is picked which is passed to function F, K is generated. So comparison between these two are carried out by comparator. If it mismatches then there is a error through this signal it is detected. If not the data is passed through corrector and that will be the actual data out. The comparison is three results. The first one is there is no error detected. Second one is an error is detected but it is possible to correct that error. Third case an error is detected but it is not possible to correct it. Let us see the hamming code and its syndrome to find the error. So this diagram talks about hamming error correcting code. So there are three circles. Let us consider the three digits like 1, 1, 1. These are the information and this is 0. So we will see another diagram which is focusing on hamming error correcting code. See this in this circle first circle one is inserted because already this circle covers three ones. Adding one which makes equal even number of ones. So that is why this one is added. If you consider second circle now second circle covers two ones and this is zero. So now better to put zero so that it becomes even number of ones. Similarly for third circle already two number of ones are there and hence insert zero. So by putting one zero zero in the three respective circle which makes even number of ones in their respective circles. So because to detect the error let us consider this bit which is zero if you check the previous it is one. So this is the actual data which is one supposed to be transmit. But because of due to some problem error is occurred which makes one to zero. So under this case now even parity concept is applied. So if you consider the first circle there are three number of ones and this is zero which is not having even number of ones. Means there is a error. Similarly for second circle it remains as it is because the zero is not a part of second circle. And coming to third circle initially it was one and this one which makes two number of ones for third circle. Because of error now it makes only one. So that indicates that for circle one and three it is an even. So that indicates that this part is under error. Next data bits and check bits. Data bits and check bits. Let us consider four check bits which leads eight number of data bits. There is one formula. Let us consider m that indicates data bits size of data bits eight. And initially let us consider k which is equal to three. If you substitute these values in this equation two raised to k minus one which is greater than or is equal to m plus k. So if we substitute m is equal to eight and k is equal to three in this equation so it leads less than eight plus three. Now it is better to take next value which is k is equal to four. If we put four in this equation it satisfies which is greater than eight plus four. So according to this formula it is greater and hence it requires four bits of check bits for eight bit data. Similarly according to this formula for sixteen it is five and for thirty two it is six. Sixty four seven one twenty eight eight two fifty six nine like this. Now let us consider next which is layout of data bits and check bits. Total m eight plus k four total twelve one to twelve bit position. So this is represented in binary and third row indicates data bit. Data bit d one d two d three d four d five d six d seven d eight and check bit. Four check bits c one c two c four and c eight. So like this total twelve bits. Now the check bit c one is calculated like this. D one xor with d two xor with d four or xor with d five xor with d seven. This is because now c one that indicates that the right most bit position is one. So out of these twelve values the right most bit position which is one is taken for xor operation. So that's what it is. D one next it is d two d four d five d seven like this. Similarly for other c two c four and c eight. Think about this question and pause the video and answer for this question. What are the results generated by comparison? The answer is there is no error and error is detected and it is possible to correct the error. Third case and error is detected but it is not possible to correct it. So we are already familiar with c one c two c four and c eight. According to this formula we are putting the respective values. Let us consider if eight bit is like this. Then the c one c two c four c eight looks like this. And hence the final value which is zero one one one for the given bit. Assuming that if there is a error if it occurs in this position then what happens? We will see if it is under error bit three then it becomes one instead of zero. Then the c one c two c four and c eight values are like this zero zero zero one. That is zero zero zero one. So the previous value zero one one one XOR with zero zero zero one and finally it is zero one one zero. That indicates six means six position is nothing but d three position. These are the references. Thank you.