Question

Use the 74LS138 , to design an address decoder that will place 8K by 16bit RAM at a starting address of C000h and 8K RAM at a starting address of 6000h.

Answer 1

A total of 13 bits is required to address any memory location in a 8K memory as 2¹³ = 8192 or 8K locations.

Hence starting address of a 8K memory will be 13 times 0 i.e. 0000000000000 and

End Address of a 8K memory will be 13 times 1 i.e. 1111111111111

The Address bus width given in the question is in hex form and its size is 4 digits which is equivalent to 16 bits.

For 8K memory, 13 (A0 - A12) bits are used and hence MSB 3 bits (A13-A15) remaining is used for absolute addressing.

Memory	A15	A14	A13	A12	A11	A10	A9	A8	A7	A6	A5	A4	A3	A2	A1	A0	Address
RAM1	1	1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	Starting Address C000h
	1	1	0	1	1	1	1	1	1	1	1	1	1	1	1	1	End Address DFFFh
RAM2	0	1	1	0	0	0	0	0	0	0	0	0	0	0	0	0	Starting Address 6000h
	0	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1	End Address 7FFFh

74LS138 is a 3:8 decoder. Hence A13, A14 and A15 is input to the decoder as shown below:

For RAM1 , A15A14A13 = 110 Hence Y6 of decoder is connected to chip Select line.

For RAM2,  A15A14A13 = 011 Hence Y3 of decoder is connected to chip Select line.