 Hello everyone. Myself Prithviraj Pitambay working as an assistant professor in the Department of Electronics Engineering at Valjean Institute of Technology, Solapur. Let us learn today programmable ROM. At the end of this session, you can examine the general structure of programmable read-only memory. You can implement simple Boolean functions using PROM. So let us first understand what is a PROM. So PROM is a conventional programmer logic device and integrated circuit with fixed and array and a programmable R array. So PROMs are used to provide and R that is SOP implementations. So let us learn the structure of PROMs. So we have here input lines which are connected to the fixed and array through some buffers and inverters. The output of this fixed and arrays are connected to programmable R arrays and again the output of R arrays are again made available to the users through these output lines through some inverter and buffer combinations. So if you compare structure of PROM with the remaining PLDs, you will observe that the and array is fixed whereas R arrays are programmable. So this fixed and array is very, very similar to a decoder structure. So let us go further. So first fixed and array structure from n number of inputs and their complements, multi-input AND gate arrays produces maximum turos to n number of AND terms. Again this number is limited but you can see here an example of AND array in PROMs. So here you will see it is a three input function ABC are connected through the inputs of these AND gates and the connections are fixed here. So this black dot represents a fixed connection. So for this particular three input system, you can have maximum eight combinations as two raise to three is the relation. So you can have maximum eight AND terms that is main terms programmable R array. So the AND terms or more specifically mean terms from the AND arrays are fed to the programmable R array and this is used to realize a set of output in SOP form, more specifically in the canonical form. So let us implement some functions. So number of mean terms are fixed as we have saw in the fixed array slide. Here it is implement a simple Boolean function using PROMs. So from this AND array slide, we know that the mean terms are fixed. So a set of Boolean expressions need not to be minimized as AND array is fixed. So we have mean terms output of AND array. So this realization of a set of Boolean functions is totally based on the mean terms and not on the product terms. So here user can program the fuses of R array depending upon the mean terms and the final realization is totally based on SOP expressions. So let us implement some Boolean functions. So here we have f1 is equal to F1 which is function of AB having mean terms 0, 1 and 3. So let us find out the mean terms for this given function. So the mean terms for F1 are A bar, B bar, A bar, B and AB. So F1 is nothing but A bar, B bar plus A bar, B plus AB. Let us implement one more function F2 which is a function of AB. The given mean terms are 0 and 2. Let us write down the mean terms for F2. So F2 is A bar, B bar plus AB bar which is in a canonical form as both the terms are mean terms. So if you observe F1 and F2, both functions are in canonical form. They are made up of only mean terms. So no minimization is required when implementing any Boolean function using PROMs. So this is the final diagram of these two function implementations. So here you will see that for two variable function the PROM used have four AND gates and as we need to implement two functions we require two ORs and the OR array is programmable whereas AND array is fixed. So two inputs, four mean terms or four AND gates. So the function F1 is A bar, B bar plus A bar, B plus AB. So here you will see that the output of the first AND gate is A bar, B bar. The fuse is intact. Output of the second AND gate is A bar, B. Again the fuse is intact. Output of the third AND gate is A, B bar but this term is not part of F1. That is why this fuse is blown. The output of this fourth AND gate is AB. Again this term is a part of F1. So the fuse is intact. So output of the first OR gate is nothing but our function F1. So only three fuses kept intact and one is blown away. Similarly for F2 we have two terms A bar, B bar plus AB bar. So again the output of first AND gate is reused here which is A bar, B bar. So the fuse is kept as it is. Output of second AND fourth AND gates are not part of this equation. That is why those two fuses are blown away. Again output of third AND gate which is AB bar which is a part of F2 function. That is why this fuse is kept as it is. So F2 is made up of these two main terms. That is why these two main terms are kept as it is. The remaining fuses are blown away. So this is how you can implement simple Boolean functions using programmable read-only memories. These are the textbooks which you can refer for the further reading. Thank you.