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The ALE line of an 8085 microprocessor is used to

This question was previously asked in
UGC NET 2016 Paper 3 Defence and Strategic Studies Question Paper (10-Jul-2016)
The correct answer is

latch the 8 bits of address lines AD7-AD0 into an external latch.

The problem ALE solves. The 8085 needs 16 address lines but has only 40 pins, so the designers multiplexed the lower half: the pins AD0–AD7 carry the low byte of the address during the first clock state of a machine cycle (T1) and then switch to carrying data for the rest of it. The upper byte A8–A15 has dedicated pins and stays valid throughout.

What ALE does. ALE (Address Latch Enable) goes high during T1 and falls at its end. That falling edge is used to strobe an external latch — classically a 74LS373 — which captures the low address byte before the pins change over to data:

\(AD_7\text{-}AD_0 \xrightarrow{\ \text{ALE latch}\ } A_7\text{-}A_0\)

The memory or I/O device then sees a stable, complete 16-bit address for the whole machine cycle, while the processor uses the same eight pins to transfer the data byte.

Why the other options are wrong.

Latching the output of an I/O instruction is done by the device's own chip-select and the \(\overline{WR}\) strobe, not by ALE.

Chip selects are generated by an address decoder from A15–A8 together with IO/\(\overline{M}\); ALE plays no part in enabling or disabling them.

The state of the TRAP interrupt is read with the RIM instruction; ALE has nothing to do with interrupts.

Related signals worth knowing together: IO/\(\overline{M}\) distinguishes memory from I/O, \(\overline{RD}\) and \(\overline{WR}\) time the transfer, and S0, S1 identify the machine cycle. ALE is issued once per machine cycle, so counting ALE pulses is a standard way to count machine cycles on a logic analyser.

Hence, ALE is used to latch the 8 bits of address lines AD7–AD0 into an external latch.

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