Question 1 |

An 8 Kbyte ROM with an active low Chip Select input (\overline{CS}) is to be used in an 8085 microprocessor based system. The ROM should occupy the address range 1000H to 2FFFH. The address lines are designated as A_{15} \; to \; A_{0}, \; where \; A_{15} is the most significant address bit.
Which one of the following logic expressions will generate the correct \overline{CS} signal for this ROM?

A_{15} + A_{14} + (A_{13}\cdot A_{12}+ \bar{A_{13}}\cdot \bar{A_{12}}) | |

A_{15} \cdot A_{14} \cdot (A_{13}+ A_{12}) | |

\bar{A_{15}} \cdot \bar{A_{14}} \cdot (A_{13}\cdot \bar{A_{12}}+ \bar{A_{13}}\cdot A_{12}) | |

\bar{A_{15}} + \bar{ A_{14}} +A_{13}\cdot A_{12} |

Question 1 Explanation:

\begin{aligned} & \text {8kB ROM is given }\\ &\therefore 2^{n}=8 k B=2^{3}\left(2^{10}\right)=2^{13} \end{aligned}

\therefore 13 Address lines are required for memory chip.

But the address range given as 1000H-2FFF H

In order to get \overline{C S} as low, the condition is

A_{15}=A_{14}=0 and A_{13}=0 / 1, A_{12}=1 / 0

The circuit to generate an active low chip select signal can be given as shown below:

The logical expression can be written as,

\begin{aligned} F &=A_{15}+A_{14}+\left(A_{13} \odot A_{12}\right) \\ &=A_{15}+A_{14}+\left(A_{13} \cdot A_{12}+\bar{A}_{13} \cdot \bar{A}_{12}\right) \end{aligned}

\therefore 13 Address lines are required for memory chip.

But the address range given as 1000H-2FFF H

In order to get \overline{C S} as low, the condition is

A_{15}=A_{14}=0 and A_{13}=0 / 1, A_{12}=1 / 0

The circuit to generate an active low chip select signal can be given as shown below:

The logical expression can be written as,

\begin{aligned} F &=A_{15}+A_{14}+\left(A_{13} \odot A_{12}\right) \\ &=A_{15}+A_{14}+\left(A_{13} \cdot A_{12}+\bar{A}_{13} \cdot \bar{A}_{12}\right) \end{aligned}

Question 2 |

A 16 Kb (=16,384 bit) memory array is designed as a square with an aspect ratio of one (number of rows is equal to the number of columns). The minimum number of address lines needed for the row decoder is _______.

6 | |

7 | |

8 | |

9 |

Question 2 Explanation:

Memory size =16 \mathrm{kB}=214 \mathrm{bits }

Number of address lines = Number of data lines

\begin{aligned} 2^{n} \cdot 2^{n} &=2^{14} \\ \quad n &=7 \end{aligned}

Number of address lines = Number of data lines

\begin{aligned} 2^{n} \cdot 2^{n} &=2^{14} \\ \quad n &=7 \end{aligned}

Question 3 |

For the 8085 microprocessor, the interfacing circuit to input 8-bit digital data
(DI_{0}-DI_{7})h from an external device is shown in the figure. The instruction for
correct data transfer is

MVI A, F8H | |

IN F8H | |

OUT F8H | |

LDA F8F8H |

Question 3 Explanation:

To transfer, the data should be present in accumulator and proper instruction should be executed.

For the proper working, AND gates 1 and 2, should produce output as 1.

Since \bar{DS}_{1}, is connected to output 0 of decoder

so input selection should be 000, therefore address lines have inputs as i.e. accumulator should be fed with the contents of address F8F8 H i.e. LOA F8F8 H.

\begin{aligned} A_{15} & A_{14}& A_{13}& A_{12}& A_{11}& A_{10}& A_{9}& A_{8} &A_{7} & A_{6}& A_{5}& A_{4} & A_{3} & A_{2}& A_{1}& A_{0} \\ 1&1&1&1&1&0&0&0&1&1&1&1&1&0&0&0 \end{aligned}

If IO/\bar{M} =0; to activate \bar{G}_{2A}I/O is treated as M/M mapped I/O hence I/O has 16-bit address.

For the proper working, AND gates 1 and 2, should produce output as 1.

Since \bar{DS}_{1}, is connected to output 0 of decoder

so input selection should be 000, therefore address lines have inputs as i.e. accumulator should be fed with the contents of address F8F8 H i.e. LOA F8F8 H.

\begin{aligned} A_{15} & A_{14}& A_{13}& A_{12}& A_{11}& A_{10}& A_{9}& A_{8} &A_{7} & A_{6}& A_{5}& A_{4} & A_{3} & A_{2}& A_{1}& A_{0} \\ 1&1&1&1&1&0&0&0&1&1&1&1&1&0&0&0 \end{aligned}

If IO/\bar{M} =0; to activate \bar{G}_{2A}I/O is treated as M/M mapped I/O hence I/O has 16-bit address.

Question 4 |

There are four chips each of 1024 bytes connected to a 16 bit address bus as shown in the figure below. RAMs 1, 2, 3 and 4 respectively are mapped to addresses

0C00H-0FFFH, 1C00H-1FFFH, 2C00H-2FFFH, 3C00H-3FFFH | |

1800H-1FFFH, 2800H-2FFFH, 3800H-3FFFH, 4800H-4FFFH | |

0500H-08FFH, 1500H-18FFH, 3500H-38FFH, 5500H-58FFH | |

0800H-0BFFH, 1800H-1BFFH, 2800H-2BFFH, 3800H-3BFFH |

Question 4 Explanation:

Since the range of RAM # 1 is different in all the four options. So we will check for RAM 1 only and then the same procedure can be followed for RAM 2, 3 and 4.

So, RAM # 1 will be selected when

\begin{array}{l} S_{0}=0 \\ S_{1}=0 \\ S_{0}=A_{12}=0 \\ S_{1}=A_{13}=0 \end{array}

Now the RAM # 1 will be enable when the input of MUX is 1, or the output of AND gate is 1.

\begin{aligned} \mathrm{SO}, & A_{10}=0 \\ A_{11} &=1 \\ A_{14} &=0 \\ A_{15} &=0 \end{aligned}

\begin{array}{cccc|cccc|cccc|cccc|c} A_{15} & A_{14} & A_{13} & A_{12} & A_{11} & A_{10} & A_{9} & A_{8} & A_{7} & A_{6} & A_{5} & A_{4} & A_{3} & A_{2} & A_{1} & A_{0} & \\ \hline 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text { Start } \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \text { End } \\ \hline 0 & & &&& 8 & & & 0 & & & &0 & &&&\text { Start } \\ 0& &&& & B& & & F& & & & F & & &&\text { End } \\ \hline \end{array}

So, range of RAM # 1 is 0800H to 0BFFH

So, RAM # 1 will be selected when

\begin{array}{l} S_{0}=0 \\ S_{1}=0 \\ S_{0}=A_{12}=0 \\ S_{1}=A_{13}=0 \end{array}

Now the RAM # 1 will be enable when the input of MUX is 1, or the output of AND gate is 1.

\begin{aligned} \mathrm{SO}, & A_{10}=0 \\ A_{11} &=1 \\ A_{14} &=0 \\ A_{15} &=0 \end{aligned}

\begin{array}{cccc|cccc|cccc|cccc|c} A_{15} & A_{14} & A_{13} & A_{12} & A_{11} & A_{10} & A_{9} & A_{8} & A_{7} & A_{6} & A_{5} & A_{4} & A_{3} & A_{2} & A_{1} & A_{0} & \\ \hline 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \text { Start } \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \text { End } \\ \hline 0 & & &&& 8 & & & 0 & & & &0 & &&&\text { Start } \\ 0& &&& & B& & & F& & & & F & & &&\text { End } \\ \hline \end{array}

So, range of RAM # 1 is 0800H to 0BFFH

Question 5 |

In the circuit shown, the device connected Y5 can have address in the range

2000-20FF | |

2D00-2DFF | |

2E00-2EFF | |

FD00-FDFF |

Question 5 Explanation:

1O/\bar{M} =O to Activate G1

To select Y5, input CBA should be 101.

Possible range will be

\begin{aligned} &A_{15} &A_{14} &A_{13} &A_{12} &A_{11}&A_{10} &A_{9} &A_{8} &A_{7}\ldots&A_{0} \\ &0 &0 &1 &0 &1 &1 &0 &1 &0\ldots&0 \\ &0 &0 &1 &0 &1 &1 &0 &1 &1\ldots&1 \\ \end{aligned}

Therefore, the device can have address in the range 2D00 H -2DFF H

To select Y5, input CBA should be 101.

Possible range will be

\begin{aligned} &A_{15} &A_{14} &A_{13} &A_{12} &A_{11}&A_{10} &A_{9} &A_{8} &A_{7}\ldots&A_{0} \\ &0 &0 &1 &0 &1 &1 &0 &1 &0\ldots&0 \\ &0 &0 &1 &0 &1 &1 &0 &1 &1\ldots&1 \\ \end{aligned}

Therefore, the device can have address in the range 2D00 H -2DFF H

Question 6 |

An 8255 chip is interfaced to an 8085 microprocessor system as an I/O mapped
I/O as show in the figure. The address lines A_0 \; and \; A_1 of the 8085 are used
by the 8255 chip to decode internally its thee ports and the Control register.
The address lines A_3 \; to \; A_7 as well as the IO/\bar{M} signal are used for address
decoding. The range of addresses for which the 8255 chip would get selected is

F8H - FBH | |

F8H - FCH | |

F8H - FFH | |

F0H - F7H |

Question 6 Explanation:

As \mathrm{A}_{2} is not shown, it can be O/1.

O/P of NAND gate is 0 if A_{7} \text{ to }A_{3} \& IO / \bar{M}=1

\begin{aligned} &A_{7} &A_{6} &A_{5} &A_{4} &A_{3}\quad &A_{2} &A_{1} &A_{0} \\ \text{Starting Address }&1 &1 &1 &1 &1 &0 &0 &0 &\rightarrow \mathrm{F} 8 \mathrm{H}\\ \text{Final Address }&1 &1 &1 &1 &1 &1 &1 &1 &\rightarrow \mathrm{FFH} \end{aligned}

O/P of NAND gate is 0 if A_{7} \text{ to }A_{3} \& IO / \bar{M}=1

\begin{aligned} &A_{7} &A_{6} &A_{5} &A_{4} &A_{3}\quad &A_{2} &A_{1} &A_{0} \\ \text{Starting Address }&1 &1 &1 &1 &1 &0 &0 &0 &\rightarrow \mathrm{F} 8 \mathrm{H}\\ \text{Final Address }&1 &1 &1 &1 &1 &1 &1 &1 &\rightarrow \mathrm{FFH} \end{aligned}

Question 7 |

An I/O peripheral device shown in Fig. (b) below is to be interfaced to an 8085
microprocessor. To select the I/O device in the I/O address range D4H - D7H,
its chip-select (\overline{CS}) should be connected to the output of the decoder shown in figure (a)
below :

output 7 | |

output 5 | |

output 2 | |

output 0 |

Question 7 Explanation:

\therefore Output taken from 5th

Question 8 |

What memory address range is NOT represents by chip # 1 and chip # 2 in
the figure? A_{0} \; to \; A_{15} in this figure are the address lines and CS means chip select.

0100 - 02FF | |

1500 - 16FF | |

F900 - FAFF | |

F800 - F9FF |

Question 8 Explanation:

Chip 1

\begin{array}{cccccccccccc}A_{15} & \ldots & \ldots & A_{12} & A_{11} & A_{10} & A_{9} & A_{8} & A_{7} & \ldots & \ldots & A_{0} \\ \times & \times & \times & \times & \times & \times & 0 & 1 & 0 & 0 & 0 & 0 \\ \times & \times & \times & \times & \times & \times & 0 & 1 & 1 & 1 & 1 & 1\end{array}

Chip 2

\begin{array}{cccccccccccc}A_{15} & \ldots & \ldots & A_{12} & A_{11} & A_{10} & A_{9} & A_{8} & A_{7} & \ldots & \ldots & A_{0} \\ \times & \times & \times & \times & \times & \times & 1 & 0 & 0 & 0 & 0 & 0 \\ \times & \times & \times & \times & \times & \times & 1 & 0 & 1 & 1 & 1 & 1\end{array}

\therefore F800-F9FF cannot be the memory range for

Chip #1 and Chip #2

\begin{array}{cccccccccccc}A_{15} & \ldots & \ldots & A_{12} & A_{11} & A_{10} & A_{9} & A_{8} & A_{7} & \ldots & \ldots & A_{0} \\ \times & \times & \times & \times & \times & \times & 0 & 1 & 0 & 0 & 0 & 0 \\ \times & \times & \times & \times & \times & \times & 0 & 1 & 1 & 1 & 1 & 1\end{array}

Chip 2

\begin{array}{cccccccccccc}A_{15} & \ldots & \ldots & A_{12} & A_{11} & A_{10} & A_{9} & A_{8} & A_{7} & \ldots & \ldots & A_{0} \\ \times & \times & \times & \times & \times & \times & 1 & 0 & 0 & 0 & 0 & 0 \\ \times & \times & \times & \times & \times & \times & 1 & 0 & 1 & 1 & 1 & 1\end{array}

\therefore F800-F9FF cannot be the memory range for

Chip #1 and Chip #2

Question 9 |

The 8255 Programmable Peripheral Interface is used as described below.

(i) An A/D converter is interface to a microprocessor through an 8255. The conversion is initiated by a signal from the 8255 on Port C. A signal on Port C causes data to be stobed into Port A.

(ii) Two computers exchange data using a pair of 8255s. Port A works as a bidirectional data port supported by appropriate handshaking signals.

The appropriate modes of operation of the 8255 for (i) and (ii) would be

(i) An A/D converter is interface to a microprocessor through an 8255. The conversion is initiated by a signal from the 8255 on Port C. A signal on Port C causes data to be stobed into Port A.

(ii) Two computers exchange data using a pair of 8255s. Port A works as a bidirectional data port supported by appropriate handshaking signals.

The appropriate modes of operation of the 8255 for (i) and (ii) would be

Mode 0 for (i) and Mode 1 for (ii) | |

Mode 1 for (i) and Mode 0 for (ii) | |

Mode 2 for (i) and Mode 0 for (ii) | |

Mode 2 for (i) and Mode 1 for (ii) |

Question 10 |

An 8085 microprocessor based system uses a 4K x 8-bit RAM whose starting address is AA00. The address of the last byte in this RAM is

0FFFH | |

1000 H | |

B9FF H | |

BA00 H |

Question 10 Explanation:

12 add. lines are present 4k = 2^{12}

\begin{aligned} AA00&=1010 1010 0000 0000 \\ \text{Last Add }&= \frac{\quad 1111\quad 1111 \quad 1111\quad}{\frac{1011}{B}\quad\frac{1001}{B}\quad F\quad F} = B9FF \end{aligned}

\begin{aligned} AA00&=1010 1010 0000 0000 \\ \text{Last Add }&= \frac{\quad 1111\quad 1111 \quad 1111\quad}{\frac{1011}{B}\quad\frac{1001}{B}\quad F\quad F} = B9FF \end{aligned}

There are 10 questions to complete.