- 1204 Views
- Uploaded on
- Presentation posted in: General

Chap. 7 Counters and Registers

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

- Introduction
- Chap. 7의 내용
- How FFs and logic gates can be combined to produce different types of counters and registers

- Divided into 2 parts
- Part I : principles of counter operation, various counter circuit arrangement, and
representative IC counters

- Part II : counter application, types of IC register, and troubleshooting

- Part I : principles of counter operation, various counter circuit arrangement, and

- Chap. 7의 내용
- 7-1 Asynchronous(Ripple) Counters
- Asynchronous Counter : Fig. 7-1
- The FFs do not change states in exact synchronism with the applied clock pulses

- Ripple Counter
- The FFs respond one after another in a kind of rippling effect
- The terms asynchronous counter and ripple counter interchangeably

- Signal Flow(in Fig. 7-1)
- Left-to-Right : Conventional signal flow
- Right-to Left :
- FF A(rightmost) = LSB, FF D(leftmost) = MSB

- Asynchronous Counter : Fig. 7-1

We’ll break left-to-right convention,

especially in counter diagrams

- Exam. 7-1)Some time later the clock pulses are removed, and the counter FFs read 0011. How many clock pulses have occurred?
- 3 + 16 = 19 + 16 = 35 + 16 = 51 …..

- Mod Number = 2N( N : number of FF )
- Number of different states
- Fig. 7-1 : MOD-16 ripple counter ( 0000 1111)

- Number of different states
- Exam. 7-2)The counter must be able to count as many as one thousand items. How many FFs are required ?
- 10 FFs : 0 1023 ( 1001 1023은 필요 없음 )

- Frequency Division
- For any counter, the output from the last FF(the MSB) divides the input clock frequency by the MOD number of the counter
- MOD-16 Counter = Divide-by-16 Counter : Fig. 7-2

- Exam. 7-3)How many FFs are required for the MOD-60 counter?
- There is no integer power of 2 that will equal 60 : 26 = 64
- In the next section we will see how to modify the basic counter so that any MOD number can be obtained.

- 7-2 Counters with MOD Number < 2N
- Mode Number less than 2N:
- The basic counter can be modified to produce MOD numbers less than 2N by allowing the counter to skip states

- MOD-6 Counter : Fig. 7-4
- When B = C = “1”, NAND output will go “0” (few nanosecond spike or glitch)
- This glitch is very narrow and so would not produce any visible indication on LEDs
- However It could cause a problem if the B output were being used to drive other circuitry

- When B = C = “1”, NAND output will go “0” (few nanosecond spike or glitch)
- State Transition Diagram : Fig. 7-5
- Dotted line : Temporary state(=110)
- 111 state : never reached, not even temporarily

- Displaying Counter States
- Output A = “1” : Inverter output = “0” LED ON
- Output A = “0” : Inverter output = “1” LED OFF

- Exam. 7-4) a) LED status of 5, b) LED clocked by 1 kHz, c) LED will be visible for 110 in Fig.7-5
- Exam. 7-5) Determine the MOD number and the frequency at the D output of the counter in Fig.7-6(a)
- D C B A = 1 1 1 0 = 14 일 때 NAND output = 0 (Clear Input) : MOD 14
- 30 kHz/14 = 2.14 kHz

- Mode Number less than 2N:

- General Procedure (to construct MOD X Counter)
- 1) Find the smallest number of FFs such that 2N X, connect them as a counter.
If 2N = X, do not do steps 2 and 3

- 2) Connect a NAND output to the CLEAR inputs of all the FFs
- 3) Determine which FFs will be in the HIGH state at a count = X; then connect
the outputs of these FFs to the NAND inputs.

- 1) Find the smallest number of FFs such that 2N X, connect them as a counter.
- Exam. 7-6)Construct a MOD-10 (count from 0000 ~ 1001) counter : Fig. 7-6(b)
- Find the smallest number of FFs : 4 ( 24 = 16 )
- D C B A = 1 0 1 0 = 10 : D and B must be connected as the NAND gate input

- Decade Counters/BCD counters : Fig. 7-6(b) or 별도 IC
- MOD-10 Counter = Decade Counter = BCD Counter
- Count in sequence from 0000(0) to 1001(9)

- MOD-10 Counter = Decade Counter = BCD Counter
- Exam. 7-7)Construct a MOD-60 Counter : Fig. 7-7
- Find the smallest number of FFs :64 ( 26 = 64 )
- Q5 Q4 Q3 Q2 Q1 Q0 = 1 1 1 1 0 0 = 60 (32 + 16 + 8 + 4)

- 7-3 IC Asynchronous Counters
- 74LS293 Asynchronous Counter IC : Fig. 7-8
- Q0와 CP1이 연결되지 않은 이유 : 3 or 4 비트 카운터로 사용 가능

- Exam. 7-8)How the 74LS293 should be connected to operate as a MOD-16
counter : Fig. 7-9

- MR1 = MR2 = “0”, CP1 = Q0, Q3 = 10 kHz/16 = 625 Hz

- Exam. 7-9)How to wire the 74LS293 as a MOD-10 Counter : Fig. 7-10
- Q3 Q2 Q1 Q0 = 1 0 1 0 = 10 (8 + 2)

- Exam. 7-10)How to wire the 74LS293 as a MOD-14 Counter : Fig. 7-11
- Q3 Q2 Q1 Q0 = 1 1 1 0 = 14 (8 + 4 + 2)

- Exam. 7-11)Construct a MOD-60 Counter with 74LS293 : Fig. 7-12
- MOD-10 counter X MOD-6 counter = MOD-60 counter
- MOD-10 counter : Exam. 7-9
- MOD-6 counter : Q0는 사용하지 않고 3 비트(Q3 Q2 Q1)만 사용

- MOD-10 counter X MOD-6 counter = MOD-60 counter

- 74LS293 Asynchronous Counter IC : Fig. 7-8

MR1

MR2

Clear ( C0 )

- CMOS Asynchronous Counters
- 74HC4024 : MOD-128 ripple counter = CTR DIV128 ( Fig. 7-14 )
- 74HC4040 : MOD-4096 ripple counter

- MOD-8 Down Counter : Fig. 7-15
- The inverted output of each FF is connected to the CLK input of following FF.

- Ripple Counter
- The simplest type of binary counter
- Decoding Glitch : Sec. 7-12
- Propagation Delay : Fig. 7-16
- The Nth FF cannot change states until a time Nxtpd after the clock transition occurs.

- For proper counter operation
- Tclock N x tpd fmax = 1 /( N x tpd )
- Exam) 74LS112, tpd = tPHL = 24 ns
- 4 FFs : fmax = 1 / 4 x 24 ns = 10.4 MHz
- 6 FFs : fmax = 1 / 6 x 24 ns = 6.9 MHz

- Exam) 74LS112, tpd = tPHL = 24 ns

- Tclock N x tpd fmax = 1 /( N x tpd )

* FF 개수 증가

The total propagation delay

증가하고 fmax 감소

- 7-6 Synchronous(Parallel) Counter
- Synchronous/Parallel Counters
- All of the FFs are triggered simultaneously (in parallel) by the clock input pulses

- Synchronous MOD-16 Counter : Fig. 7-17
- Circuit Operation
- Each FF should have its J and K inputs connected
such that they are HIGH only when the outputs of

all lower-order FFs are in the HIGH state

- Each FF should have its J and K inputs connected

- Circuit Operation
- Advantage of Synchronous Counters over Asynchronous
- Total Delay in Synchronous Counter
- Total Delay = Single FF tpd + Single AND gate tpd
- Total delay is the same no matter how many FFs are in the counter

- Total Delay = Single FF tpd + Single AND gate tpd

- Total Delay in Synchronous Counter
- Actual ICs
- 74LS160/162, 74HC160/162 : Synchronous Decade(MOD-10) Counters
- 74LS161/163, 74HC161/163 : Synchronous MOD-16 Counters

- Exam. 7-12) (a) Determine fmax for the counter of Fig. 7-17(a) and Compare this value with MOD-16 ripple counter( FFtpd = 50 ns, AND gate tpd = 20 ns)
- Parallel Counter : fmax = 1 / ( 50 ns + 20 ns ) = 14.3 MHz
- Ripple Counter : fmax = 1 / (4 x 50 ns ) = 5 MHz

- Synchronous/Parallel Counters

A

B

C

ABC = (J = K)

Design in 7-14

p. 362

A

B

AB =( J = K)

(b) What must be done to convert this counter to MOD-32

- 5 개째 FF(25 = 32) 이 추가되며, J and K input are fed by the output of
a four input AND gate whose inputs are A, B, C, and D

(c) Determine fmax for the MOD-32 parallel counter

- FF 개수에 관계없이 14.3 MHz

- MOD-8 Parallel Up/Down Counter : Fig. 7-18
- Up Count : Up/Down = 1, AND gates 1/2 = Enabled, AND gates 3/4 = Disabled
- Down Count : Up/Down = 0, AND gates 1/2 = Disabled, AND gates 3/4 = Enabled

- Exam. 7-13)What problems might be caused if the Up/Down signal changes levels on the NGT of the clock ?
- Possible Problems : Unpredictable results of FF
- the J and K inputs change at about the same time that a NGT occurs at their CLK input.

- Effects : Predictable results of FF(No problems)
- the effects of the change in the control signal must propagate through two gates before reaching the J, K inputs(결국 다음 Clock에서 Up/Down 동작이 가능함)

- Possible Problems : Unpredictable results of FF

- 7-8 Presettable Counters
- Presettable Counter/Parallel Loading Counter
- Preset to any desired starting count either asynchronously or synchronously

- Presettable Parallel Counter with Asynchronous Preset : Fig. 7-19
- The counter is loaded with any desired count at any time
- 1) Apply the desired count to the parallel data inputs, P2, P1, and P0
- 2) Apply a Low pulse to the PARALLEL LOAD input(PL)

- PL 은 Active Low이고, 이 때 2 개 NAND Gate의 한 개 입력은 항상 1, 따라서 P에 의해 P = 1 이면 PRESET, 그리고 P = 0 이면 CLR
- Asynchronous Presetting IC Counters
- TTL : 74LS190, 191, 192, 193
- CMOS : 74HC190, 191, 192, 193

- The counter is loaded with any desired count at any time
- Synchronous Presetting
- The counter is preset on the active transition of the same clock signal
- Synchronous Presetting IC Counters
- TTL : 74LS160, 161, 162, 163
- CMOS : 74HC160, 161, 162, 163

- Presettable Counter/Parallel Loading Counter

Async presetting에서는

PRE/CLR에 의해

- 7-9 The 74LS193/HC193
- Presettable Up/Down Counter(74LS193) : Fig. 7-20
- Synchronous Counter, Asynchronous Preset, and Asynchronous Master Reset
- Terminal Count : Fig. 7-21
- TCU : 1111 0000 ( Fig. 7-22에서 t6, t7참고 )
- TCD : 0000 1111 ( Fig. 7-23에서 t9, t10참고 )

- Exam. 7-14)Determine the counter output waveforms in Fig. 7-22(a)
- Up Counter : Fig. 7-22(b)

- Exam. 7-15)Determine the counter output waveforms in Fig. 7-23(a)
- Down Counter : Fig. 7-23(b)

- Variable MOD Number Using the 74LS193
- TCD = PL : Preset to 0101(5) - Fig. 7-24
- Q2 = 5 Clock cycle : Divide the frequency by 5
- MOD-6 가 아니고 MOD-5 인 이유
- the counter gets preset back to 5 in the middle of a clock cycle

- MOD-6 가 아니고 MOD-5 인 이유
- We can vary the frequency division by changing the logic levels applied to the parallel data inputs
- A variable frequency-divider can be easily implemented by connecting switches to the parallel data inputs in Fig. 7-24

- Presettable Up/Down Counter(74LS193) : Fig. 7-20

5, 4, 3, 2, 1, 0, 5, 4, 3, 2, 1, ...

- Multistage Arrangement : Fig. 7-25
- 8 bits Up/Down Counter using two 74LS193s

- Decoding
- Electronically decode the contents of a counter and display the results
- Immediately recognizable and require no mental operations

- Electronically decode the contents of a counter and display the results
- Active-HIGH Decoding : Fig. 7-27
- At any one time only one AND gate output is HIGH

- Exam. 7-16)How many AND gates are required to decode all of the states of a MOD-32 counter? What are the inputs to the gate that decodes for 21
- MOD-32 counter has 32 possible states : 32 개 AND gate 필요
- 1 0 1 0 1(21) : E, D, C, B, A

- Active-LOW Decoding
- NAND gates are used in place of AND gates

- Exam. 7-17)Generate a control waveform which could be used to control devices such as a motor, solenoid valve, or heater.
- Control Signal Generation(On/Off control) : Fig. 7-28
- The X output is HIGH between the counts of 8 and 14 for each cycle of counter

- BCD Counter Decoding
- Decoder/Display Unit : Fig. 7-29 (refer to p. 512)

- Decoding Glitches : Fig. 7-30
- The propagation delays between FF transitions cause problems when decoding a ripple counter
- Decoder X0 = Temporary 00 state 에서 Glitch 발생 : A = B = 0 (X0=1)

- Strobing : Fig. 7-31
- A reliable method for eliminating the decoder glitches : Strobe Signal
- Decode AND gates disabled until all of the FFs have reached a stable state

- A reliable method for eliminating the decoder glitches : Strobe Signal

- Cascading BCD Counters : Fig. 7-32
- Count and Display numbers from 000 to 999
- 2 가지 Implementation
- 1) 74LS293 wired as a MOD-10 counter
- 2) 74LS90 또는 74LS192/HC192 BCD IC counter

- 7-14 Synchronous Counter Design
- Design a 3 bits Counter
- 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, … (Undesired State : 5, 6, 7) : Tab. 7-3

- Design Procedure
- 1) Determine the desired number of bits(FFs) and the desired counting sequence
- FFs = 3 개, Desired Sequence = 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, ….

- 2) Draw the state transition diagram : Fig. 7-33
- 3) Tabulate present/next state table : Tab. 7-4
- Use the state transition diagram to setup a present/next state table

- 4) Tabulate circuit excitation table : Tab. 7-5
- Add a column to this table for each J and K input by using Tab. 7-2

- 1) Determine the desired number of bits(FFs) and the desired counting sequence

- Design a 3 bits Counter

- 5) Design logic circuits to generate the levels required at each J and K input
- FF A : Fig. 7-34
- FF B : Fig. 7-35(b)
- FF C : Fig. 7-35(a)

- 6) Implement the final expressions : Fig. 7-36

- Step Motor Drive Circuit(with Direction Control) : Fig. 7-37(a)
- State Transition Diagram : Fig. 7-37(b)
- Circuit Excitation Table : Tab. 7-6
- K-map Simplification : Fig. 7-38
- Implementation : Fig. 7-39

- Shift-Register
- Transfer data left to right, or vice versa, one bit at a time(serially)
- Shift-register counters use feedback
- the output of the last FF in the register is connected back to the first FF in some way

- Ring Counter
- In most instances only a single 1 is in the register

- MOD-4 Ring Counter : Fig. 7-40
- Ring counters can be constructed for any desired MOD number
- A MOD-N ring counter used N flip-flops

- Ring counters can be constructed for any desired MOD number

- A ring counter must start off with only one FF in the 1 state and all the others in the 0 state
- Ring Counter Starter : Fig. 7-41
- 1) On power-up, the capacitor will charge up relatively slowly toward Vcc, 따라서 Inverter 1 input = 0
- 2) Inverter 1 output = 1, 따라서 Inverter 2 output = 0 until Inverter 1 input = 1
- 이때 Q3 = PRE, Q2 = Q1 = Q0 = CLR 임으로 1 0 0 0 으로 Preset 됨

- The inverted output of the last FF is connected to the input of the first FF
- 3 bits Johnson counter : Fig. 7-42
- MOD-6(six distinct states) : 000, 100, 110, 111, 011, and 001
- 50 percent duty cycle square wave at one-sixth the frequency of the clock
- MOD-N counter(N= even number) by connecting N/2 FFs
- MOD-10 Johnson Counter : 5 FF 필요

4 distinct states

- Decoding a Johnson Counter
- For a given MOD number, a Johnson counter requires only half the number of FFs that a ring counter requires
- MOD-8 Ring Counter : 8 FFs
- MOD-8 Johnson Counter : 4 FFs

- Ring Counter does not require decoding gates
- only one FF in the 1 state and all the others in the 0 state : Fig. 7-40(c) Sequence Table

- Johnson Counter requires decoding gates : Fig. 7-43
- Each decoding gate has only two inputs, even though there are three FFs in the counter
- Two of the three FFs are in a unique combination of states

- Each decoding gate has only two inputs, even though there are three FFs in the counter

- For a given MOD number, a Johnson counter requires only half the number of FFs that a ring counter requires
- IC Shift-Register Counters
- Ring/Johnson Counter는 너무 간단하게 구현 됨으로 IC가 별로 없다
- CMOS Johnson-Counter : 74HC4017, 74HC4022

- 7-16 Counter Applications : Frequency Counter
- Frequency Counter : Fig. 7-44
- Counter + Decoder/Display + AND gate
- Sampling Interval : SAMPLE pulse goes HIGH from t1 to t2
- During this sampling interval the unknown frequency pulses(fx) will pass through the AND gate and will be counted by the counter
- The accuracy of this method depends almost entirely on the duration of the sampling interval
- The sampling Interval must be very accurately controlled

- Exam. 7-18)The unknown frequency is 3792 pulses per second(pps). Determine the counter reading after a sampling interval of (a) 1 s, (b) 0.1 s, and (c)10ms
- (a) 1 s : 3792 (b) 0.1 s : 379.2, 379 or 380 (c) 0.01 s : 37.92, 37 or 38

- A method for obtaining accurate sampling interval : Fig. 7-45
- Crystal Oscillator : generate a very accurate 100-kHz waveform
- Decade Counter : divide 100-kHz frequency by 10
- Rotary Switch : select one of the decade-counter output
- Flip-Flop : 2 분주
- In position 1 : 1 Hz / 2 = 0.5 Hz
- Q = 0.5 Hz = 1 / 0.5 Hz = 2 T, 따라서 HIGH 상태의 반주기는T (Sampling Interval)

- In position 1 : 1 Hz / 2 = 0.5 Hz

- Frequency Counter : Fig. 7-44

- Exam. 7-19)The unknown input frequency is between 1, 000 pps and 9,990 pps, what is the best setting for the switch position in Fig. 7-45 with 3 BCD counter and display.
- With three BCD counter : total capacity = 000 - 999
- 0.1 s sampling interval : 100 - 999

- With three BCD counter : total capacity = 000 - 999
- Complete Frequency Counter : Fig. 7-46
- X : FF의 출력(= One-shot 의 Clock input)
- SAMPLE pulse의 2 분주이고 100ns One-shot pulse를 Trigger 시켜 Counter를 Clear 함.
- AND gate에 의해 SAMPLE pulse와 X 가 모두 HIGH 일 때만 Counter 동작

- 예제) Sampling Interval = 1 s, 그리고 unknown frequency = 237 pps 인 경우
- t1 - t2 : X = 0으로 동작하지 않음
- t2 - t3 : Counter is cleared to 0 and display 0 for 1 s ( = sample interval = 1 s )
- t3 - t4 : Counter는 1 s 동안 0 - 237 까지 Count
- t4 - t6 : 2 s 동안 237 Display

- X : FF의 출력(= One-shot 의 Clock input)

- Digital Clock operating from 60 Hz : Fig. 7-47
- Schmitt-trigger circuit : produce square pulses at the rate of 60 pps
- MOD-60 Counter : divide the 60 pps down to 1 pps

- Detailed Hour Section : Fig. 7-48
- MOD-2 ( 0 -1 ) + MOD-10 ( 0 - 9 ) Counter
- MOD-10 Counter : 74LS192(Presettable BCD Counter)
- MOD-2 Counter : JK Flip-Flop
- 12 에서 13 이 될 때 74LS192의 PL에 의해 상위 시간 = 0 (74LS112는 Clear), 그리고하위 시간 = 1( P3 P2 P1 P0 = 0 0 0 1)이 된다.
- 3 Input NAND gate 에서 상위 시간 X = 1, 그리고 하위 시간 Q1 Q0 = 11(3) 일 때 PL Enabled

- Q3 = 1, 즉 하위 시간 = 9 일 때 상위 시간 = 1 이 된다( 09 10).

- MOD-2 ( 0 -1 ) + MOD-10 ( 0 - 9 ) Counter

- 1) Parallel in/Parallel out : 74174, 74178
- 2) Serial in/Serial out : 4731B
- 3) Parallel in/Serial out : 74165
- 4) Serial in/Parallel out : 74164

- 74174 and 74HC174( 6 bit register ) : Fig. 7-49

- Exam. 7-20A)How to connect 74ALS174 so that D5 D4 D3 D2 D1 D0 ( = data input at D5 and data output at Q0 ).
- Fig. 7-50

- Exam. 7-20B)How to connect two 74ALS174 to operate as a 12 bit shift register.
- Connect the Q0 of the first IC to the D5 of the second IC.

- 4731B(CMOS Quad 64-bit) : Fig. 7-51
- 64 개 D-type Flip-Flop : 4 개가 one chip 이므로 총 256 개 FF
- Buffer circuit : Q63 (triangle symbol with no inversion bubble)
- A buffer does not change the signal’s logic level
- It is used to provide a greater output-current capability than normal

- Exam. 7-21)Delay circuit using 4731B chip : Fig. 7-52
- Q63 goes HIGH approximately 64 clock cycles after Ds input

- 8-bit parallel in/serial out register : Fig. 7-53
- Truth Table

- Exam. 7-22)Determine (a)the conditions necessary to load the register with parallel data, (b) the conditions necessary for the shifting operation
- (a) SH/LD = 0 : only Q7 will be externally available
- (b) SH/LD = 1, CP INH = 0, and PGT Clock Pulse at CP

- Exam. 7-23)What signal will appear at Q7(Q7=Ds, CP=200kHz, CP INH=0)
- 8-bit Johnson counter divided by 16 = 12.5 kHz

- 7-22 Serial in/Parallel out : 74164, 74LS164, and 74HC164
- 8-bit serial in/parallel out shift register : Fig. 7-54
- Each FF output externally accessible : Q0, Q1, …, Q7
- 2 input AND gate : one input can be used for control

- Exam. 7-24)Determine the sequence of states in Fig. 7-55(a)
(Initial Content of the 74ALS164 = 00000000)

- The correct sequence : Fig. 7-57(b)
- Q7 =1 : Temporary state
- LOW at MR ( inverted Q7 ) resets the register back to 00000000

- Q7 =1 : Temporary state

- The correct sequence : Fig. 7-57(b)

- 8-bit serial in/parallel out shift register : Fig. 7-54
- Other Register ICs
- 74194/LS194/HC194 : 4 bit bi-directional universal shift register
- 4 mode : shift left, shift right, parallel in, parallel out ( selected by 2 bit mode select code as inputs )

- 74373/LS373/HC373 : 8 bit parallel in/parallel out register
- 8 D latch with tri-state outputs : Data or Address bus buffer로 주로 사용됨
- Pin 11 : Latch Enable(LE)로 Level trigger = 1 일 때 8 개 입력 D0 - D7 이 8개 출력
Q0 - Q7으로 출력됨(따라서 Transparent Latch 라고도 함)

- Pin 11 : Latch Enable(LE)로 Level trigger = 1 일 때 8 개 입력 D0 - D7 이 8개 출력

- 8 D latch with tri-state outputs : Data or Address bus buffer로 주로 사용됨
- 74374/LS374/HC374 : 8 bit parallel in/parallel out register
- 8 edge-triggered D Flip-Flops with tri-state outputs
- Pin 11 : Clock Pulse(CP)로 Edge trigger(PGT) 일 때 373과 마찬가지로 출력됨

- 8 edge-triggered D Flip-Flops with tri-state outputs

- 74194/LS194/HC194 : 4 bit bi-directional universal shift register

- 7-24 Trouble Shooting
- Exam. 7-25)Determine the possible faults of MOD-10 counter in Fig. 7-57(a) ( The displayed waveforms of the Q output with an oscilloscope are shown in Fig. 7-57(b) )
- 이상 증상 : Q1, Q2, and Q3(except Q0) are stuck in the LOW state
- 이상 가능성 :
- 1) Q1 : internally or externally shorted to ground
- 2) MR1 : internally shorted to ground (Q1=0)
- 3) Q0 와 CP1 사이의 Open : 따라서 Q1 이 Clock Input을 받지 못함
- 4) IC 자체의 internal fault

- Exam. 7-26)위의 4 가지 이상 가능성을 조사했으나 모두 아니었으며, Fig. 7-58 과 같이 Q1에서 glitch를 발견했음. 이상 원인은 ?
- MR2 의 Open : MR2 = HIGH in TTL
- Q1의 glitch에 의해 항상 0 으로 reset 됨, 따라서 MOD-2 카운터로 동작했음( 0, 1 )

- MR2 의 Open : MR2 = HIGH in TTL
- Exam. 7-27)The displayed frequency is exactly twice in Fig. 7-46, What is the probable cause for the malfunction ?
- 3 input AND gate 의 가운데 입력이 Open 되었음 : SAMPLE pulse = HIGH:
- 따라서 t3 와 t4 동안만 sampling 되지 않고, t2 와 t4 동안 sampling 되었음.

- 3 input AND gate 의 가운데 입력이 Open 되었음 : SAMPLE pulse = HIGH:
- Exam. 7-28)The HOURS section displays in the manner shown in Tab. 7-7 . 이상 원인은 ? (p. 382, Fig. 7-48참고 )
- 74LS192의 Q3이 아니고 Q2 가 74LS112의 CLK에 잘못 연결됨
- 따라서 9 0 일 때 X = 1로 되야 하나, 7 8 일 때 X = 1로 되었음.

- 74LS192의 Q3이 아니고 Q2 가 74LS112의 CLK에 잘못 연결됨

- Exam. 7-25)Determine the possible faults of MOD-10 counter in Fig. 7-57(a) ( The displayed waveforms of the Q output with an oscilloscope are shown in Fig. 7-57(b) )

p. 333, Fig. 7-12 참고

- 7-25. Programming PLDs as Counter Circuits using Boolean Equations
- General Architecture of GAL PLD
- .D extension : F/F output is connected to the PLD output pin.

- General Architecture of GAL PLD

- Q4.D = F/F Q4의 입력 =Q4의 next state
- Exam. 7-29Design 3-bit MOD 6 Johnson Counter using GAL 16v8 (p. 373, Fig. 7-42참고 )
- Present-Next State Table : Tab. 7-8
- Q2.D = 000 or 100 or 110; Q1.D = 100 or 110 or 111; Q0.D = 110 or 111 or 011;

- Another Method : field name twisted
- D input should be 100(4) if the present state of twisted 000(0) OR…………
- CUPL file : Fig. 7-60

- Present-Next State Table : Tab. 7-8