Input Output Interfacing Questions and Answers

Answer: B

The keyed answer is memory mapped IOs and IO mapped IOs.

In the 8085, the \(IO/\overline{M}\) line tells external circuits whether the address on the bus belongs to the I/O space or to memory. It is high for IN and OUT instructions, which address ports in the separate I/O space (I/O mapped I/O), and low for memory operations, which include devices placed in the memory address space (memory mapped I/O).

It plays no part in telling keyboard input from memory, serial from parallel communication, or synchronous from asynchronous transfers.

Answer: B

Both (A) and (R) are true but (R) is not the correct explanation of (A).

In the 8085, the \(\mathrm{IO}/\overline{\mathrm{M}}\) signal tells the system whether the current cycle is an I/O or a memory operation: low for memory, high for I/O. So it is used when accessing memory, and the assertion is true.

The reason is also true: in memory-mapped I/O the ports are given memory addresses and are accessed with memory instructions. But that is a different scheme. In memory-mapped I/O this pin is not needed to tell ports from memory, so the reason does not explain why the pin exists.

Answer: B

The sequence is (ii), (iv), (iii), (i): address enable, start of conversion, end of conversion, data ready.

The microprocessor first places the ADC's address on the bus and enables it, which selects the converter (and, for a multiplexed ADC, the channel). It then sends a start-of-conversion pulse. When the converter finishes it raises its end-of-conversion signal, and the microprocessor then reads the converted data, which is now ready on the data lines.

The other options start the conversion before the ADC is selected, or put end of conversion before the start, which cannot happen.

Answer: B

This is the common anode configuration.

Each LED's anode is tied to the positive supply, through a current-limiting resistor, and its cathode to a microcontroller port pin. When the pin is driven low, current flows from the supply through the LED into the pin, so the port sinks the LED current, about 8 mA here. Microcontrollers such as the 8051 can sink far more current than they can source, because their port outputs are strong only when pulling low, which is why LEDs are wired this way.

In a common cathode connection the cathodes go to ground and the port would have to source the current. 'Driver' and 'buffer' configurations use an external transistor or IC to supply the current, so the port would not be sinking it directly.

Answer: B

The ULN2003 is needed for increasing the current.

An 8051 port pin can supply or sink only a few milliamperes, far too little to drive a stepper motor winding that needs hundreds of milliamperes. The ULN2003 contains seven Darlington transistor pairs, each able to sink about 500 mA, with built-in suppression diodes for inductive loads. The port drives the Darlington inputs and the Darlingtons carry the winding current.

Switching the windings on and off in sequence is done by the 8051's program; the ULN2003 only amplifies the current. 'Increasing the power' is less precise, since the supply voltage is unchanged and the driver's role is current gain. The ULN2003 has no short-circuit shut-off.