7-Segment Decoder Driver and Display

Introduction

In this experiment you will demonstrate the operation of a decoder-driver circuit that accepts a binary or BCD input code and generates the 7 segment display signals to produce the numbers 0 through 9 and other characters.

Figure 1 shows a simplified block of the 74LS48 BCD to 7-Segment Decoder. The 74LS48 contains three main block circuits, a 7-segment decoder, a driver and a system of basic memory units. The basic memory unit is often called a latch or a flip-flop. The decoder outputs drive an encoder circuit made up of OR gates that generate the 7-segment code necessary to display the digits 0 through 9 and the letters a through f. The output devices are current driver transistors that supply the proper current to th e segments in the driver.

Figure 1

Equipment needed

Microprocessor power supply
Logic Probe
74LS00
74LS02
74LS04
74LS48 (46)
7-Segment LED

Figure 2

Part 1. -- Set-Reset flip-flop

1. Wire the latch circuit shown in fiqure 2. The Set (A) and Reset (B) are the inputs and C (L1) and C (L2) are the outputs.
2. Apply power to the circuit and create a truth table for S and R Inputs and C and C outputs.
3. Wire the latch circuit shown in figure 3. Repeat steps 1. and 2. for circuit 3. These new outputs are labeled D and D. Why do we call this circuit a basic memory unit? What happens to the outputs when S and R both 0? Refer to the textbook (Katz) for a discussion of flip-flops (chapter 6).

Part 2. -- (7-Segment Decoder-Driver and Display)

1. Construct the circuit shown in figure 4. Use the TTL handbook to verify the correct conections. The pin connections for the 74LS48 and the 7-Segment Display are shown in fiqure 5.
2. Calculate the value of the resistor between the 74LS48 and the 7-seg LED.
3. Apply power to the circuit. Create a truth table for figure 4. Do the LEDs L1-L4 which output the binary word agree with the output of the 7-Segment LED? What does the 7-Segment LED read in binary states 1010-1111? What do you think the LT, RBI and BI/ RBO pins do?

Figure 3

Figure 4

Figure 5

Check the MAN74A documentation.

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DISPLAY 7 SEGMENT

   7-Segment Displays
Truth Table to Decode Binary into 7-Segment

Decimal Number		Inputs					Outputs
		D	C	B	A		a	b	c	d	e	f	g
0 1 2 3 45 6 7 8 9		0 0 0 0 00 0 0 1 1	0 0 0 0 11 1 1 0 0	0 0 1 1 00 1 1 0 0	0 1 0 1 01 0 1 0 1		1 0 1 1 01 1 1 1 1	1 1 1 1 10 0 1 1 1	1 1 0 1 11 1 1 1 1	1 0 1 1 01 1 0 1 1	1 0 1 0 00 1 0 1 0	1 0 0 0 11 1 0 1 1	0 0 1 1 11 1 0 1 1

 


Karnaugh Mapping for Segment a:
The Truth Table:

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

Boxing ones:

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

Red cells are represented by B; this is a 4x2 supercell.

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

Green cells are represented by D; this is a 2x4 supercell.
So far then we have:

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

where grey boxes indicate cells included in more that one supercell. The shaded boxes are represented by B+D.
Focusing on the remaining cells with ones:

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

Blue cells are represented by A^.C; this is a 2x2 supercell.

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

Cyan cells are represented by A'^.C'; this is a 2x2 wrap-around supercell.

	BA	00	01	11	10
DC\
00		1	0	1	1
01		0	1	1	1
11		x	x	x	x
10		1	1	x	x

All cells are represented by B + D + A^.C + A'^.C', as shown for segment a.
Karnaugh Mapping for Segment c:
The Truth Table:

	BA	00	01	11	10
DC\
00		1	1	1	0
01		1	1	1	1
11		x	x	x	x
10		1	1	x	x

Boxing zeroes:

	BA	00	01	11	10
DC\
00		1	1	1	0
01		1	1	1	1
11		x	x	x	x
10		1	1	x	x

The red cells are represented by C'^.B^.A'; of course the complement is required for the final realization. Thus (C'^.B^.A')' is the combination of gates that is realized from the Karnaugh mapping of zeroes. Application of De Morgan's theorem results in the combination of gates shown for segment c:
(C'^.B^.A')' = C + B' + A

   Decoder for segment a

D C B A a 0 1 2 3 45 6 7 8 9 0 0 0 0 00 0 0 1 1 0 0 0 0 11 1 1 0 0 0 0 1 1 00 1 1 0 0 0 1 0 1 01 0 1 0 1 1 0 1 1 01 1 1 1 1

   Decoder for segment b

		D	C	B	A		b
0 1 2 3 45 6 7 8 9		0 0 0 0 00 0 0 1 1	0 0 0 0 11 1 1 0 0	0 0 1 1 00 1 1 0 0	0 1 0 1 01 0 1 0 1		1 1 1 1 10 0 1 1 1

   Decoder for segment c

D C B A c 0 1 2 3 45 6 7 8 9 0 0 0 0 00 0 0 1 1 0 0 0 0 11 1 1 0 0 0 0 1 1 00 1 1 0 0 0 1 0 1 01 0 1 0 1 1 1 0 1 11 1 1 1 1

BA  00   01   11   10  DC\  00  1 1 1 0  01  1 1 1 1  11  x x x x  10  1 1 x x Karnaugh

   Decoder for segment d

D C B A d 0 1 2 3 45 6 7 8 9 0 0 0 0 00 0 0 1 1 0 0 0 0 11 1 1 0 0 0 0 1 1 00 1 1 0 0 0 1 0 1 01 0 1 0 1 1 0 1 1 01 1 0 1 1

   Decoder for segment e

		D	C	B	A		e
0 1 2 3 45 6 7 8 9		0 0 0 0 00 0 0 1 1	0 0 0 0 11 1 1 0 0	0 0 1 1 00 1 1 0 0	0 1 0 1 01 0 1 0 1		1 0 1 0 00 1 0 1

	BA	00	01	11	10
DC\
00		1	0	0	1
01		0	0	0	1
11		x	x	x	x
10		1	0	x	x

 

   Decoder for segment f

		D	C	B	A		f
0 1 2 3 45 6 7 8 9		0 0 0 0 00 0 0 1 1	0 0 0 0 11 1 1 0 0	0 0 1 1 00 1 1 0 0	0 1 0 1 01 0 1 0 1		1 0 0 0 11 1 0 1 1      Decoder for segment g D C B A g 0 1 2 3 45 6 7 8 9 0 0 0 0 00 0 0 1 1 0 0 0 0 11 1 1 0 0 0 0 1 1 00 1 1 0 0 0 1 0 1 01 0 1 0 1 0 0 1 1 11 1 0 1 1

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PERCOBAAN 3
DISPLAY 7 SEGMEN

TUJUAN:
1. Siswa memahami rangkaian interface mikrokontroller dengan 7 segmen
2. Siswa dapat memahami program assembly untuk menampilkan data ke 7 segment
3. Siswa memahami beberapa instruksi assembly dasar, MOV, Setb, Clr, dan waktu tunda.

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