5.2b.1 Flip-Flops

Flip-Flop

A flip-flop is a basic digital memory circuit that can store one bit of state data. Its inputs are a 1-bit data value ($D$) and a one-bit clock value ($\text{clk}/\Delta$) that synchronizes the register to the system. Its output is a one-bit value $Q$:

\begin{document}
\usetikzlibrary {shapes.geometric}
\begin{tikzpicture}
   % Draw the flip-flop box
   \draw[thick] (0,0) rectangle (3,4);
   
   % Draw the input (D)
   \draw[thick, ->] (-1,3) -- (0,3) node[at start, left] {Input} node[right] {D};
 
   % Draw the output (Q)
   \draw[thick, ->] (3,3) -- (4,3) node[at start, left] {Q} node[right] {Output};
 
   % Draw the clock (clk)
   \draw[thick, scale=0.25] (1,0) -- (3,0) -- (2,{sqrt(3)}) -- cycle;
   \draw[thick, ->] (0.5,0) -- (0.5,-1) node[below]{clk};
\end{tikzpicture}
\end{document}

On the rising edge of the clock, the output $Q$ is set to the value of the input $D$ (though $D$ may be the same across multiple clock rising edges if no new value is passed). The speed at which the clock sets $Q=D$ is dependent on its clk-to-q delay, which is the time a flip-flop takes for its output to change. At all other times, the output Q remains steady. This means that the flip-flop can store data, as it will be able to maintain its value between time steps.

Note! $n$-bit register simply consists of $n$ flip-flops, each storing 1 bit of data, bundled together.

An

Examples

Plotting $Q$'s Waveform

Suppose we have the following SDS diagram of the clock and a flip-flop input $D$:

\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
   \begin{axis}[
       width=12cm,
       height=6cm,
       xlabel={Time (ns)},
       ylabel={},
       ymin=-0.5, ymax=3,
       xmin=0, xmax=28,
       xtick={0,2,4,6,8,10,12,14,16,18,20,22,24,26,28},
       ytick={0.25, 1.25, 2.25},
       yticklabels={Q, D, clk},
       grid=major,
       grid style={dotted, gray},
       xmajorgrids=true, 
       ymajorgrids=false,
       extra y ticks={0, 0.5, 1, 1.5, 2, 2.5},
       extra y tick labels={},
       extra y tick style={grid=major},
       ]
       
       % Clock signal
       \addplot[thick, black] coordinates {(0, 2) (2, 2) (2, 2.5) (4, 2.5) (4, 2) (6, 2) (6, 2.5) (8, 2.5) (8, 2) (10, 2) (10, 2.5) (12, 2.5) (12, 2) (14, 2) (14, 2.5) (16, 2.5) (16, 2) (18, 2) (18, 2.5) (20, 2.5) (20, 2) (22, 2) (22, 2.5) (24, 2.5) (24, 2) (26, 2) (26, 2.5) (28, 2.5)};
       
       % D signal
       \addplot[thick, blue] coordinates {(0, 1.5) (4, 1.5) (4, 1) (7.5, 1) (7.5, 1.5) (8.5, 1.5) (8.5, 1) (11, 1) (11, 1.5) (12, 1.5) (12, 1) (13, 1) (13, 1.5) (16, 1.5) (16, 1) (17.5, 1) (17.5, 1.5) (24, 1.5) (24, 1) (26.5, 1) (26.5, 1.5) (28, 1.5)};
       
       % Q signal
       
   \end{axis}
\end{tikzpicture}
\end{document}
 

We can then figure out the output waveform $Q$ by looking at $\text{clk}$ and $D$. Remember that the clock only syncs $D=Q$ at rising edges. In this case, at $x=kT+2$ for some $k$ and a clock period $T=4$. The flip-flop in question has a clk-to-q delay of 1 ns.

1. Since the clock signal is at $0$ until $t=2$, we are unsure of $Q$‘s value in that time frame. $Q$ may be $1$ if $D$ was $1$ before $t=0$, or $0$ otherwise.
2. ($t=2$): The clock updates, and the output syncs with the input 1 ns later due to clk-to-q delay.
3. ($t=6$): $D=0$ and the clock syncs $D=Q$ at $t=7$.
4. ($t=10$): $D=0$, so $Q$ remains at $0$.
5. ($t=14$): $D=1$, so $Q$ updates to $1$ at $t=15$.
6. ($t=18$): $D=1$ again, so there is no change to $Q$.
7. ($t=22$): $D=1$ again.
8. $t=26$: $D=0$, so $Q$ updates at $t=27$.

\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
   \begin{axis}[
       width=12cm,
       height=6cm,
       xlabel={Time (ns)},
       ylabel={},
       ymin=-0.5, ymax=3,
       xmin=0, xmax=28,
       xtick={0,2,4,6,8,10,12,14,16,18,20,22,24,26,28},
       ytick={0.25, 1.25, 2.25},
       yticklabels={Q, D, clk},
       grid=major,
       grid style={dotted, gray},
       xmajorgrids=true, 
       ymajorgrids=false,
       extra y ticks={0, 0.5, 1, 1.5, 2, 2.5},
       extra y tick labels={},
       extra y tick style={grid=major},
       ]
       
       % Clock signal
       \addplot[thick, black] coordinates {(0, 2) (2, 2) (2, 2.5) (4, 2.5) (4, 2) (6, 2) (6, 2.5) (8, 2.5) (8, 2) (10, 2) (10, 2.5) (12, 2.5) (12, 2) (14, 2) (14, 2.5) (16, 2.5) (16, 2) (18, 2) (18, 2.5) (20, 2.5) (20, 2) (22, 2) (22, 2.5) (24, 2.5) (24, 2) (26, 2) (26, 2.5) (28, 2.5)};
       
       % D signal
       \addplot[thick, blue] coordinates {(0, 1.5) (4, 1.5) (4, 1) (7.5, 1) (7.5, 1.5) (8.5, 1.5) (8.5, 1) (11, 1) (11, 1.5) (12, 1.5) (12, 1) (13, 1) (13, 1.5) (16, 1.5) (16, 1) (17.5, 1) (17.5, 1.5) (24, 1.5) (24, 1) (26.5, 1) (26.5, 1.5) (28, 1.5)};
       
       % Q signal
       \addplot[thick, red] coordinates {(3, 0.5) (7, 0.5) (7, 0) (15, 0) (15, 0.5) (27, 0.5) (27, 0) (28, 0)};
       
   \end{axis}
\end{tikzpicture}
\end{document}
 

Here it is important to reiterate that we shift $Q$‘s waveform right by 1 ns due to its clk-to-q delay.