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4-bit PISO shift register working
Short answer
The 4-bit PISO register contains D flip-flops with input selectors controlled by SH/LD'. Setting SH/LD'=0 loads parallel data in one clock; SH/LD'=1 enables right-shift, sending bits out serially at QD LSB first. For example, loading 1011 yields serial output 1,1,0,1 after four clocks.
Draw the circuit diagram and explain the working principle of 4-bit parallel in serial out (PISO) shift register.
A 4-bit parallel-in serial-out (PISO) shift register stores a 4-bit word applied on four parallel lines in one clock and then shifts it out bit by bit on a single serial output. It uses four D flip-flops, a common clock and a SHIFT/LOAD' control that, through a 2-to-1 AND-OR selector at each D input, chooses between parallel data and the output of the previous stage. The working is illustrated with the word 1011.
Circuit
PA PB PC PD
| | | |
+-----+ +-----+ +-----+ +-----+
0->| MUX | +-->| MUX | +-->| MUX | +-->| MUX |
+-----+ | +-----+ | +-----+ | +-----+
| | | | | | |
+-----+ | +-----+ | +-----+ | +-----+
|D QA|-+ |D QB|-+ |D QC|-+ |D QD|-->
|> | |> | |> | |> | Serial
+-----+ +-----+ +-----+ +-----+ out
MUX = 2 AND + 1 OR: D = SH.(Q of left FF) + SH'.P
SH/LD' and CLK are common to all stages.
One stage (e.g. stage B):
QA ---------->+-----+
SH/LD' --+--->| AND |--+
| +-----+ | +----+
| +->| OR |--> DB
| +-----+ +->| |
+NOT>| AND |--+ +----+
PB ---------->+-----+
DB = SH.QA + SH'.PB
- SH/LD' = 0 (load): the lower AND gates are enabled, so . At one clock edge all four bits are stored at the same time.
- SH/LD' = 1 (shift): the upper AND gates are enabled, so (and ). Each clock moves the data one place right and the next bit appears at the serial output QD.
Storing and retrieving 1011
Parallel inputs: PA PB PC PD = 1 0 1 1 (PA = MSB, PD = LSB).
- Store (clock 1, SH/LD' = 0): all bits load together: QA QB QC QD = 1011. The LSB (1) is already at the output QD.
- Retrieve (clocks 2, 3, 4, SH/LD' = 1): each clock shifts right and a 0 enters QA. QD shows 1, 0, 1 in turn.
- Serial output after clocks 1 to 4: 1, 1, 0, 1, i.e. 1011 read LSB first. One clock loads, and n - 1 = 3 more clocks deliver the remaining bits.
| CLK | SH/LD' | QA | QB | QC | QD | Serial out |
|---|---|---|---|---|---|---|
| 0 (initial) | - | 0 | 0 | 0 | 0 | - |
| 1 | Load | 1 | 0 | 1 | 1 | 1 |
| 2 | Shift | 0 | 1 | 0 | 1 | 1 |
| 3 | Shift | 0 | 0 | 1 | 0 | 0 |
| 4 | Shift | 0 | 0 | 0 | 1 | 1 |
| 5 | Shift | 0 | 0 | 0 | 0 | 0 |
Timing diagram
1 2 3 4 5
CLK _|‾|_|‾|_|‾|_|‾|_|‾|
SH/LD' ___|‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾|
QA _|‾‾‾|______________
QB _____|‾‾‾|__________
QC _|‾‾‾|___|‾‾‾|______
QD _|‾‾‾‾‾‾‾|___|‾‾‾|__
Working principle in short: load in parallel, then shift right. A 4-bit word needs 1 load clock plus 3 shift clocks to transmit all bits. PISO registers are used for parallel-to-serial conversion in data transmission.
✓ Answer checked by Claude Opus 5.5 (high effort)
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