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Dueck R.Digital design with CPLD applications and VHDL.2000

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480

C H A P T E R 1 0 •

2digit.gdf count_8.vhd sev_segv.vhd

State Machine Design

If the debouncer is working properly, the seven-segment display should advance by one each time pb_in is pressed. If the debouncer is not working, the display will change by

with each switch press.

ﬁles for the debouncer and test circuit components are supplied this book in the folder drive:\Student Files\Chapter 10\. To use these ﬁles, create a symbol for each one (File menu; Project; Set Project to Current File;

then File menu; Create Default Symbol) and draw the Graphic Design File of Figure 10.26. Alternatively, you can instantiate each ﬁle as a component in a VHDL design entity

(all components are designed in VHDL) and connect them together with internal signals.

Behaviorally Designed Switch Debouncer

We can also design a switch debouncer by using a behavioral state machine description in VHDL. In order to do so, we need to deﬁne the operation of the circuit with a state diagram, as in Figure 10.27.

FIGURE 10.27	pb-in, pb-out/pb-out
State Diagram for a Behaviorally	01/1
Designed Switch Debouncer	01/1
	10/0
	s0
	00

		00/0
		11/1
		11/1
01/1	s1
01/1	01
10/0	01
10/0

		00/0
01/1		11/1
		11/1
	s2
10/0	s2
10/0	02
	02		00/1
		00/0	00/1
		00/0	11/0
		11/1	11/0
01/1		11/1
01/1	s3
10/0	s3
	03

Transitions between states are determined by comparing pb_in and pb_out. If they are the same (00 or 11), the machine advances to the next state; if they are different (01 or 10), the machine reverts to the initial state, s0. At any point in the state diagram (including state s3, the last state), the machine will reset if pb_in and pb_out are different, indicating a bounce on the input.

If pb_in and pb_out are the same for four clock pulses, the input is deemed to be stable. Only at this point will the output change to its opposite state.

N O T E

In the shift register–based debouncer, the circuit advanced to the next state if the shift register input and output were different and reset if they were the same. This might appear to be opposite to our behavioral description, but it is not if you look carefully. The shift register debouncer circuit inverts pb_in before applying the signal to the serial input of the shift register. Therefore, viewed from the circuit input and output terminals, rather than at the shift register input and output, the description is the same in both cases.

10.4 • Switch Debouncer for a Normally Open Pushbutton Switch

481

The VHDL code corresponding to the behavioral description of the switch debouncer is given next. The only output change is speciﬁed on the transition from state s3 to s0 when pb_in pb_out. Since no change is allowed at any other time, no other output state needs to be speciﬁed.

--dbc_behv.vhd

--Behavioral deﬁnition of a switch debouncer LIBRARY ieee;

USE ieee.std_logic_1164.ALL;

ENTITY dbc_behv IS

PORT(

clk, pb_in : IN STD_LOGIC; pb_out : BUFFER STD_LOGIC);

END dbc_behv;

ARCHITECTURE debounce of dbc_behv IS

TYPE sequence IS (s0, s1, s2, s3);

SIGNAL state: sequence;

dbc_behv.vhd dbc_behv.scf

IF (clk‘EVENT and clk=‘1’) THEN

CASE state IS

WHEN s0=> IF (pb_in = pb_out) THEN state <= s1;

ELSE

state <= s0; END IF;

WHEN s1=> IF (pb_in = pb_out) THEN state <= s2;

ELSE

state <= s0; END IF;

WHEN S2=> IF (pb_in = pb_out) THEN state <= s3;

ELSE

state <= s0; END IF;

WHEN s3=> IF (pb_in = pb_out) THEN state <= s0;

pb_out <= not pb_out;

ELSE

state <= s0; END IF;

WHEN others => state <= s0;

END CASE;

END IF;

END PROCESS;

END debounce;

Figure 10.28 shows a simulation of the behaviorally-designed switch debouncer. State s1 through s3 are of too short a duration to show properly on the simulation, so further details of the simulation are shown in Figures 10.29 and 10.30.

482 C H A P T E R 1 0 • State Machine Design

FIGURE 10.28

Simulation of a Behaviorally Designed Switch Debouncer

FIGURE 10.29

Simulation Detail (Behaviorally Designed Switch Debouncer)

FIGURE 10.30

Simulation Detail (Behaviorally Designed Switch Debouncer)

Note that the behaviorally designed switch debouncer does not have a built-in clock divider. If we were to use the circuit on the Altera UP-1 board, we would need to include a divide-by-216 counter to the circuit, as shown in Figure 10.31.

SECTION 10.4 REVIEW PROBLEM

10.4What is the fastest acceptable clock rate for the shift register portion of the debouncer in Figure 10.24 if the pushbutton switch bounces for 15ms?

COUNT_16

clock

INPUT

clkdiv[15..0]

2digit@83

clk

Q[15..0]

VCC

DBC_BEHV

clkdiv15

clk

pb_out

2digit@51

INPUT

pb_in

VCC

COUNT_8

clk

Q[7..0]

q[7..0]

2digit@52

reset

INPUT

reset

VCC

FIGURE 10.31

Using a Behaviorally Designed Debouncer with a 16-bit Clock Divider

VCC

SEV_SEGV

a d3 b

d2	c
	c
d1	d
	d
d0	e
	e
	f
	g

SEV_SEGV

	a
d3	b
	b
d2	c
	c
d1	d
	d
d0	e
	e
	f
	g

OUTPUT

dp1

OUTPUT

dp2

OUTPUT

2digit@68

2digit@79

2digit@58

2digit@60

2digit@61

2digit@63

2digit@64

2digit@65

2digit@67

2digit@69

2digit@70

2digit@73

2digit@74

2digit@76

2digit@75

2digit@77

483

484 C H A P T E R 1 0 • State Machine Design

	0/00
	start	in1/out1, out2
	start
X /01	000
X /01	1/00
pulse2	wait1	1/00
100	001

X /10	0/00
	0/00
pulse1	wait2
011	010
	1/00
	0/00

FIGURE 10.32

State Diagram for a Two-pulse Generator

10.5 Unused States in State Machines

In our study of counter circuits in Chapter 9, we found that when a counter modulus is not equal to a power of two there were unused states in the counter’s sequence. For example, a mod-10 counter has six unused states, as the counter requires four bits to express ten states and the maximum number of 4-bit states is sixteen. The unused states (1010, 1011, 1100, 1101, 1110, and 1111) have to be accounted for in the design of a mod-10 counter.

	The same is true of state machines whose number of states does not equal a power of
	two. For instance, a machine with ﬁve states requires three state variables. There are up to
	eight states available in a machine with three state variables, leaving three unused states.
	Figure 10.32 shows the state diagram of such a machine.
	Unused states can be dealt with in two ways: they can be treated as don’t care states,
	or they can be assigned speciﬁc destinations in the state diagram. In the latter case, the
	safest destination is the ﬁrst state, in this case the state called start.

EXAMPLE 10.4	Redraw the state diagram of Figure 10.32 to include the unused states of the machine’s
	state variables. Set the unused states to have a destination state of start. Brieﬂy describe
	the intended operation of the state machine.
	Solution Figure 10.33 shows the revised state diagram.
	The machine begins in state start and waits for a HIGH on in1. The machine then
	makes a transition to wait1 and stays there until in1 goes LOW again. The machine goes to
	wait2 and stays there until in1 goes HIGH and then makes an unconditional transition to
	pulse1 on the next clock pulse. Until this point, there is no change in either output.
	The machine makes an unconditional transition to pulse2 and makes out1 go HIGH.
	The next transition, also unconditional, is to start, when out1 goes LOW and out2 goes
	HIGH. If in1 is LOW, the machine stays in start. Otherwise, the cycle continues as above.
	In either case, out2 goes LOW again.

10.5

• Unused States in State Machines

485

FIGURE 10.33

unused3

Example 10.4

X /00

0/00

State Diagram for Two-pulse

111

Generator Showing Unused

unused2

States

110

X /00

in1/out1, out2

start

unused1

X /00

000

X /01

1/00

101

pulse2

wait1

1/00

100

001

	X /10	0/00
		0/00
	pulse1	wait2
	011	010
		1/00
		0/00
	Thus the machine waits for a HIGH-LOW-HIGH input sequence and generates a pulse
	sequence on two outputs.

EXAMPLE 10.5	Use classical state machine design techniques to implement the state machine described in
	the modiﬁed state diagram of Figure 10.33. Draw the state machine as a Graphic Design
	File in Max PLUS II and create a simulation to verify its function.

Solution Table 10.6 shows the state table of the state machine represented by Figure 10.33.

Table 10.6 State Table for State Machine of

Figure 10.33

Present		Next
State	Input	State	Outputs

Q2Q1Q0	in1	Q2Q1Q0	out1	out2

000	0	000	0	0
000	1	001	0	0
001	0	010	0	0
001	1	001	0	0

010	0	010	0	0
010	1	011	0	0
011	0	100	1	0
011	1	100	1	0

100	0	000	0	1
100	1	000	0	1
101	0	000	0	0
101	1	000	0	0

110	0	000	0	0
110	1	000	0	0
111	0	000	0	0
111	1	000	0	0

486 C H A P T E R 1 0 • State Machine Design

Figure 10.34 shows the Karnaugh maps used to simplify the next-state equations for the state variable ﬂip-ﬂops. The output equations can be simpliﬁed by inspection.

The next-state and output equations for the state machine are:

Q0 in1					Q0 in1					Q0 in1
Q2 Q1	00	01	11	10	Q2 Q1	00	01	11	10	Q2 Q1	00	01	11	10
00	0	0	0	0	00	0	0	0	1	00	0	1	1	0
01	0	0	1	1	01	1	1	0	0	01	0	1	0	0
11	0	0	0	0	11	0	0	0	0	11	0	0	0	0
10	0	0	0	0	10	0	0	0	0	10	0	0	0	0
			D2					D1					D0

FIGURE 10.34

Example 10.5

K-Maps for Two-pulse Generator

D2 Q2Q1Q0

D1 Q2Q1Q0 Q2Q1Q0in1 D0 Q2Q0in1 Q2Q1in1 out1 Q2Q1Q0

out2 Q2Q1Q0

Figure 10.35 shows the Graphic Design File schematic for the state machine. Figure 10.36 shows the MAX PLUS II simulation waveforms.

We can monitor the state variables in the MAX PLUS II simulation ﬁle by adding a group of waveforms for the buried nodes q2, q1, and q0. These are shown on the simulation as q[2..0].Q, meaning the Q outputs of the ﬂip-ﬂops named q2, q1, q0.

To add the buried nodes, select Enter Node from SNF from the Node menu in the simulator window. In the dialog box shown in Figure 10.37, check the box that says All, and click on List. Select the nodes q2.Q, q1.Q, and q0.Q from the Available Nodes and Groups and transfer them to the Selected Nodes and Groups. ClickOK. Select the three new waveforms and from the Node menu, select Group. Click OK in the resulting dialog box.

487

INPUT
in1
		q2	q1		q0
		NOT	NOT	NOT	NOT
					AND3
DFF					d2
DFF
d2	PRN	q2			AND3
D	Q				AND3
	CLRN				OR2
	CLRN				d1
					AND4
DFF
d1	PRN	q1
D	Q
					AND3
	CLRN				OR2
					d0
					AND3
DFF
d0	PRN	q0
D	Q
INPUT
clk
	CLRN
					AND3
					OUTPUT
					out1
					AND3
					OUTPUT
					out2

FIGURE 10.35

Example 10.5

Two-pulse Generator

488 C H A P T E R 1 0 • State Machine Design

FIGURE 10.36

Example 10.5

Simulation of a Two-pulse

Generator (GDF)

FIGURE 10.37

Adding Buried Nodes to a

Simulation

EXAMPLE 10.6	Write the VHDL code required to implement the two-pulse generator described in Exam-
	ples 10.4 and 10.5. Create a MAX PLUS II simulation to verify the operation of the
	design. Based on your examination of the simulations for the VHDL design and the GDF
	design of the previous example, how do the two designs differ in their operation? What is
	the reason for the difference?
	Solution The VHDL code for the state machine in design entity two_pulse.vhd fol-
	lows. The unused states are accounted for in the others clause.
	-- two_pulse.vhd
	LIBRARY ieee;
	USE ieee.std_logic_1164.ALL;
	ENTITY two_pulse IS

PORT(
clk, in1	: IN	STD_LOGIC;
output	: OUT	STD_LOGIC_VECTOR (1 to 2));
END two_pulse;

ARCHITECTURE a OF two_pulse IS

TYPE SEQUENCE IS (start, wait1, wait2, pulse1, pulse2);

SIGNAL pulse_state : SEQUENCE;

BEGIN

PROCESS(clk)

BEGIN

IF (clk‘EVENT and clk = ‘1’) THEN

CASE pulse_state IS

two_pulse.vhd two_pulse.scf two_pulse.rpt

10.38

10.6

of a Two-pulse (VHDL)

10.6		• Unused States in State Machines	489
WHEN start =>
IF in1 = ‘0’ THEN
pulse_state		<= start;
output		<= “00”;
ELSIF in1 = ‘1’ THEN
pulse_state <= wait1;
output		<= “00”;
END IF;
WHEN wait1 =>
IF in1 ‘0’ THEN
pulse_state		<= wait2;
output		<= “00”;
ELSIF in1 = ‘1’ THEN
pulse_state <= wait1;
output		<= “00”;
END IF;
WHEN wait2 =>
IF in1 = ‘0’ THEN
pulse_state		<= wait2;
output		<= “00”;
ELSIF in1 = ‘1’ THEN
pulse_state <= pulse1;
output		<= “00”;
END IF;
WHEN pulse1 =>
pulse_state	<= pulse2;
output	<= “10”;
WHEN pulse2 =>
pulse_state	<= start;
output	<= “01”;
WHEN others =>
pulse_state	<= start;
output	<= “00”;

END CASE;

END IF;

END PROCESS;

END a;

Figure 10.38 shows the MAX PLUS II simulation of the state machine.

If you closely examine the simulation waveforms in Figures 10.36 and 10.38, you will note that the pulse outputs in Figure 10.38 (VHDL design) occur one clock cycle later than they do in Figure 10.36 (graphical design). This is because the VHDL compiler has synthesized each output with a D ﬂip-ﬂop, as we did for the single-pulse circuit in Figure

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