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VHDL 0
INTRODUCTION TO
VHDL
K H Wong
khwong@cse
3943-8397,
Room 907 SHB-Engineering building
http://www.cse.cuhk.edu.hk/~khwong/www2/ceng3430/ceng3430.html
VHDL 0 (v.6A) : Introduction 1
CENG3430 Rapid Prototyping of Digital Systems
• You will learn:
• The hardware description language VHDL
• Techniques to build a Logic system e.g. building blocks of a
Central Processing Unit (CPU)
• High speed logic circuits analysis: time delay estimation, testing,
power supply stability, etc.
1 entity and2 is port (a,b : in std_logic;
2 c : out std_logic);
3 end and2
4 architecture and2_arch of and2
5 begin
6 c <=a and b;
7 end and2_arch
Write VHDL code, then
it will generate the
hardware chip automatically
Example: A VHDL AND-gate Program
VHDL 0 (v.6A) : Introduction 2
A QUICK RUN
THROUGH
Overview
VHDL 0 (v.6A) : Introduction 3
Overview
•What is VHDL used for?
• To design
• Hardware systems (an industrial standard)
• Microprocessors: Arm7 etc.
• Design new Digital systems: e.g. mobile phone, camera chips
VHDL 0 (v.6A) : Introduction 4
Motivations
• Learn to design digital systems.
• Provide knowledge for you to :
• Design products:
• Robots controllers, media players, portable games, mobile phones.
• Advanced examples
• Image processing
• Computer vision
• Super computer
• Start a business.
VHDL 0 (v.6A) : Introduction 5
Examples of digital system design
• Mass products
• Media players
• Mobile phones
• Novel products
• Wearable computer
• Robots
• Research
• Real time edge detection for
computer vision
VHDL 0 (v.6A) : Introduction 6
www.cnn.com/.../06/10/mars.rover/index.html
To learn
• Design digital processing components using
programmable logic
• Two existing Methods
• (a) Schematic, (too complicated
• But is suitable to describe the top level design like a data flow block diagram
• (b) Language (e.g. VHDL--Very-High-Speed-Integrated-Circuits
Hardware Description Language): Each module in the schematic
can be written in VHDL.
VHDL 0 (v.6A) : Introduction 7
1 entity and2 is port (a,b : in std_logic;
2 c : out std_logic);
3 end and2
4 architecture and2_arch of and2
5 begin
6 c <=a and b;
7 end and2_arch
DIGITAL DESIGN
Work Flow
VHDL 0 (v.6A) : Introduction 8
Digital Design Work Flow
• Idea generation
• Drafting on paper
• Design the chip (use VHDL)
• Test
• Manufacturing production line design
• Quality control
VHDL 0 (v.6A) : Introduction 9
1 entity and2 is port (a,b : in std_logic;
2 c : out std_logic);
3 end and2
4 architecture and2_arch of and2
5 begin
6 c <=a and b;
7 end and2_arch
WE USE IN OUR LAB
Hardware: FPGA (Field Programmable Gate Array)
The hardware can be reprogrammable , so you can change your design
rapidly and easily with no additional hardware manufacturing cost.
Software: VHDL (Very-High-Speed-Integrated-
Circuits Hardware Description Language)
VHDL 0 (v.6A) : Introduction 10
Re-programmable Hardware: FPGAField
Programmable GateArray
• So what is inside an FPGA
• IOB=Input/Output block
• CLB=Configurable Logic
block (static ram based)
• Change the CLBs to get
the desired functions
VHDL 0 (v.6A) : Introduction 11
From http://www.alldatasheet.co.kr/datasheet-
pdf/pdf_kor/49173/XILINX/XCS10-3PC84C.html
Inside a CLB (Configurable Logic block )
• The CLB is a fixed design but you can change the logic
function for generating output from input G1-G4 by
reprogramming the bits in the logic function lookup table.
• This will change the overall logic function of the CLB
• Re-programming the logic table
VHDL 0 (v.6A) : Introduction 12
http://www.design-reuse.com/news_img/20100913_1.gif
http://pldworld.biz/html/technote/pldesignline/bobz-02.gif
CLB
FPGA CLB
(Configurable Logic block )
Software: to program an FPGA
•
VHDL 0 (v.6A) : Introduction 14
1 entity and2 is port (a,b : in std_logic;
2 c : out std_logic);
3 end and2
4 architecture and2_arch of and2
5 begin
6 c <=a and b;
7 end and2_arch
Use a schematic: (Top level
design to merge modules)
Use a language VHDL (for each
module)
or/and
VHDL 0 (v.6A) : Introduction 15
1 entity and2 is port (a,b : in std_logic;
2 c : out std_logic);
3 end and2
4 architecture and2_arch of and2
5 begin
6 c <=a and b;
7 end and2_arch
Development cycle
VHDL language
Schematic (diagram)
Timing simulation
•
VHDL 0 (v.6A) : Introduction 16
Summary of VHDL
• For hardware Design
• Parallel language (not sequential)
• Different! (not the same as C++ or Java)
• VHDL is the industrial standard for CE.
VHDL 0 (v.6A) : Introduction 17
An example: “And” gate in VHDL
• 1 entity and2 is port (a,b : in std_logic;
• 2 c : out std_logic);
• 3 end and2
• 4 architecture and2_arch of and2
• 5 begin
• 6 c <=a and b;
• 7 end and2_arch
VHDL 0 (v.6A) : Introduction 18
a
b
c
The chip
C<=a and b
COMPUTER
ENGINEERING MARKET
and VHDL
VHDL 0 (v.6A) : Introduction 19
TSMC(TaiwanSemicon.ManufacturingComp.)
台灣積體電路製造股份有限公司
http://www.tsmc.com
• From Wiki:
• Has the largest asset in Taiwan stock market,
• One of the World's largest dedicated
independent semiconductor foundry.
• Products: Apple iphone6 plus A8-cpu
• Relation to VHDL
• Design ideaWrite VHDL  TSMC chips
VHDL 0 (v.6A) : Introduction 20
Huawei Technologies Co. Ltd
http://www.huawei.com/en/
• From wiki:
• Telecom equipment manufacture
• China large private company--http://money.163.com 500 (2011-08-
25)
• Products: the second-largest supplier of mobile
telecommunications infrastructure equipment in the
world (after Ericsson).
VHDL 0 (v.6A) : Introduction 21
References
• See course web page
• Digital Systems Design Using VHDL, Charles H. Roth
(first or second edition)
• Rapid Prototyping of Digital Systems, by Hamblen,
James etal. Springer 2008. (read_online)
• Digital Design: Principles and Practices, 4/E John F.
Wakerly, Prentice Hall.
• High-Speed Digital Design: A Handbook of Black
Magic by Howard W. Johnson and Martin Graham
Prentice Hall.
• BOOKBOON (Free text books)
• http://www.alldatasheet.com/
VHDL 0 (v.6A) : Introduction 22
APPENDIX
VHDL 0 (v.6A) : Introduction 23
Major companies , a comparison in 2011
(from wiki)
•
Company Boeing Nestle Honda Toyota Ford HS
BC
Len
ovo
BP Sony
Revenue
US Billion
68.5 125 120 235 128 98.
9
21.
59
308
.9
86.64
Asset 64.3 126 125 370 166 24
54
10.
71
272
.2
155.9
4
Profit
US Billion
3.3 39 1.39 5.07 6.56 13.
15
0.2
73
3.3 2.96
VHDL 0 (v.6A) : Introduction 24
(from wiki) Wiki: 2009年資本額約新台幣2,589.6億元,市值約1兆6,000億元,為台灣股市中市值最大的公司。
http://money.163.com中国民营企业500强榜单发布,华为第一(2011-08-25)
•
Company Apple IBM Microsoft Intel HP TSMC
台灣積
體電路
(largest
asset in
Taiwan
stock
market)
Huawei
华为
(Telecom
equipm’t,
China
large
private
company)
Revenue
US Billion
65.23 99 69.94 43.6 99.87 13.98 21.8
Asset 75.1 113.5 108.7 63.2 124.5 20.43 Not
known
Profit
US Billion
14.01 14 23.15 11.46 14.83 5.55 2.67
VHDL 0 (v.6A) : Introduction 25
TRI-STATE LOGIC
A revision:’
The concept of tri-state logic is essential in
computer design, so we want to revise these
techniques before we move on.
VHDL 0 (v.6A) : Introduction 26
Appendix 1:Tri-state logic
**At the float state, the wire is cut
•
VHDL 0 (v.6A) : Introduction 27
Input
Output
enable (OE)
Output
Input OE (input) Output
0 0 Z(Float)
1 0 Z(Float)
0 1 0
1 1 1
Tri-state equivalent circuit
(using output connect/cut view)
•
VHDL 0 (v.6A) : Introduction 28
Input
Output
enable (OE)
Output
Input
Output
enable (OE)
Output
OE=1, switch close
OE=0, switch open
Same as
Alternatively: we can treat the Tri-state equivalent
circuit using the Rout impedance view
•
VHDL 0 (v.6A) : Introduction 29
Input
Output
enable (OE)
Output
Input
Output
enable (OE)
Output
When OE=1, Rout= small, (e.g. 50 Ω )
When OE=0, Rout=infinity (e.g. 10 MΩ)
Same as
Rout
OE (output enable)
controls the value
of Rout
Tri-state equivalent
circuit :
Rout impedance view
to explain the
concept of tri-stat
A tri-state
circuit diagram
Student ID: ___________,Date:_____________
Name: _______________
Exercise0.1:Tri-state logic with pull up resistor
•
VHDL 0 (v.6A) : Introduction 30
Input1
Output-Enable (OE)
Output
5V
10K
Input1 Output-Enable
OE (input)
Output
0 0 ? ___
1 0 ? ___
0 1 ? ___
1 1 ? ___
**At float the wire
is cut
Exercise0.2: Use Rout ( Impedance view) to explain the result of
exercise 0.1
•
VHDL 0 (v.6A) : Introduction 31
Input1 Output-
Enable
OE
(input)
Output Equivalent Rout
(10M or 50)
Draw equivalent circuit
and find output Voltage
0 0 ? _1__ ? ?
1 0 ? _1__ ? ?
0 1 ? _0__ ? ?
1 1 ? _1__ ? ?
Resistance view
Exercise 0.3
Application 1 of Tri-state logic:
Input/Output pin
• OE1 controls the traffic.
• Fill in the cells with ‘?’.
VHDL 0 (v.6A) : Introduction 32
Directional
control(OE1)
A
B
A Output
Enable
OE1 (input)
B
0 0 ?
1 0 ?
? 1 0
? 1 1
Exercise 0.4
Application 2 of Tri-state logic:
Transceivers for I/O data pins
• When T =1, A->B; T
controls the traffic,
• when /OE=1, IO pins
A,B are disabled
• Fill in the cells with ‘?’.
•
VHDL 0 (v.6A) : Introduction 33
A
B
/OE
T
T A Output
Enable
/OE1
(input)
B Which
controls
which
1 0 0 ? ?
0 1 0 ? ?
? ? 1 Float ?
? Float 1 ? ?
All data-lines are transceiver buffers
• A good controller will enable the CPU to
• read/write RAM, and read ROM
•
VHDL 0 (v.6A) : Introduction 34
CPU data lines
transceivers
ROM
data
lines
RAM
data
lines transceivers
transceivers
/OE1, T1
/OE2, T2
/OE3,
T3
Exercise 0.5 : List OE1,2,3 and T1,2,3 for
the followings cases
• a) CPU writes to RAM:
• /OE1=___ , /OE2___, /OE3=___, T1___, T2=____, T3_____
• b) CPU reads from ROM
• /OE1=___ , /OE2___, /OE3=___, T1___, T2=____, T3_____
• c) CPU reads from RAM
• /OE1=___ , /OE2___, /OE3=___, T1___, T2=____, T3_____
VHDL 0 (v.6A) : Introduction 35
CPU data lines
transceivers
ROM
data
lines
RAM
data
lines transceivers
transceivers
/OE1, T1
/OE2, T2
/OE3,
T3
A
B
A B
B A
Exercise 0.6
Application 3 of Tri-state logic:
Selectionofcontrolsignal(resolvedlogic)
• Output depends on Input_A if OE is _?___
• Output depends on Input_B if OE is _?___
• Discuss the operation of this circuit.
VHDL 0 (v.6A) : Introduction
36
Input_B
Input_A Output
OE
Exercise 0.7
• Fill in ‘?’. Is it a “nor-gate” or an “or-gate”?
• Discuss the operation of this circuit.
• Answer :
VHDL 0 (v.6A) : Introduction 37
Output
OE2
5V
10K
OE1
0V
0V
OE1 OE2 Output
0 0 ?
1 0 ?
0 1 ?
1 1 ?

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dokumen.tips_vhdl-0-introduction-to-vhdl.ppt

  • 1. VHDL 0 INTRODUCTION TO VHDL K H Wong khwong@cse 3943-8397, Room 907 SHB-Engineering building http://www.cse.cuhk.edu.hk/~khwong/www2/ceng3430/ceng3430.html VHDL 0 (v.6A) : Introduction 1
  • 2. CENG3430 Rapid Prototyping of Digital Systems • You will learn: • The hardware description language VHDL • Techniques to build a Logic system e.g. building blocks of a Central Processing Unit (CPU) • High speed logic circuits analysis: time delay estimation, testing, power supply stability, etc. 1 entity and2 is port (a,b : in std_logic; 2 c : out std_logic); 3 end and2 4 architecture and2_arch of and2 5 begin 6 c <=a and b; 7 end and2_arch Write VHDL code, then it will generate the hardware chip automatically Example: A VHDL AND-gate Program VHDL 0 (v.6A) : Introduction 2
  • 3. A QUICK RUN THROUGH Overview VHDL 0 (v.6A) : Introduction 3
  • 4. Overview •What is VHDL used for? • To design • Hardware systems (an industrial standard) • Microprocessors: Arm7 etc. • Design new Digital systems: e.g. mobile phone, camera chips VHDL 0 (v.6A) : Introduction 4
  • 5. Motivations • Learn to design digital systems. • Provide knowledge for you to : • Design products: • Robots controllers, media players, portable games, mobile phones. • Advanced examples • Image processing • Computer vision • Super computer • Start a business. VHDL 0 (v.6A) : Introduction 5
  • 6. Examples of digital system design • Mass products • Media players • Mobile phones • Novel products • Wearable computer • Robots • Research • Real time edge detection for computer vision VHDL 0 (v.6A) : Introduction 6 www.cnn.com/.../06/10/mars.rover/index.html
  • 7. To learn • Design digital processing components using programmable logic • Two existing Methods • (a) Schematic, (too complicated • But is suitable to describe the top level design like a data flow block diagram • (b) Language (e.g. VHDL--Very-High-Speed-Integrated-Circuits Hardware Description Language): Each module in the schematic can be written in VHDL. VHDL 0 (v.6A) : Introduction 7 1 entity and2 is port (a,b : in std_logic; 2 c : out std_logic); 3 end and2 4 architecture and2_arch of and2 5 begin 6 c <=a and b; 7 end and2_arch
  • 8. DIGITAL DESIGN Work Flow VHDL 0 (v.6A) : Introduction 8
  • 9. Digital Design Work Flow • Idea generation • Drafting on paper • Design the chip (use VHDL) • Test • Manufacturing production line design • Quality control VHDL 0 (v.6A) : Introduction 9 1 entity and2 is port (a,b : in std_logic; 2 c : out std_logic); 3 end and2 4 architecture and2_arch of and2 5 begin 6 c <=a and b; 7 end and2_arch
  • 10. WE USE IN OUR LAB Hardware: FPGA (Field Programmable Gate Array) The hardware can be reprogrammable , so you can change your design rapidly and easily with no additional hardware manufacturing cost. Software: VHDL (Very-High-Speed-Integrated- Circuits Hardware Description Language) VHDL 0 (v.6A) : Introduction 10
  • 11. Re-programmable Hardware: FPGAField Programmable GateArray • So what is inside an FPGA • IOB=Input/Output block • CLB=Configurable Logic block (static ram based) • Change the CLBs to get the desired functions VHDL 0 (v.6A) : Introduction 11 From http://www.alldatasheet.co.kr/datasheet- pdf/pdf_kor/49173/XILINX/XCS10-3PC84C.html
  • 12. Inside a CLB (Configurable Logic block ) • The CLB is a fixed design but you can change the logic function for generating output from input G1-G4 by reprogramming the bits in the logic function lookup table. • This will change the overall logic function of the CLB • Re-programming the logic table VHDL 0 (v.6A) : Introduction 12 http://www.design-reuse.com/news_img/20100913_1.gif http://pldworld.biz/html/technote/pldesignline/bobz-02.gif CLB FPGA CLB (Configurable Logic block )
  • 13. Software: to program an FPGA • VHDL 0 (v.6A) : Introduction 14 1 entity and2 is port (a,b : in std_logic; 2 c : out std_logic); 3 end and2 4 architecture and2_arch of and2 5 begin 6 c <=a and b; 7 end and2_arch Use a schematic: (Top level design to merge modules) Use a language VHDL (for each module) or/and
  • 14. VHDL 0 (v.6A) : Introduction 15 1 entity and2 is port (a,b : in std_logic; 2 c : out std_logic); 3 end and2 4 architecture and2_arch of and2 5 begin 6 c <=a and b; 7 end and2_arch Development cycle VHDL language Schematic (diagram)
  • 15. Timing simulation • VHDL 0 (v.6A) : Introduction 16
  • 16. Summary of VHDL • For hardware Design • Parallel language (not sequential) • Different! (not the same as C++ or Java) • VHDL is the industrial standard for CE. VHDL 0 (v.6A) : Introduction 17
  • 17. An example: “And” gate in VHDL • 1 entity and2 is port (a,b : in std_logic; • 2 c : out std_logic); • 3 end and2 • 4 architecture and2_arch of and2 • 5 begin • 6 c <=a and b; • 7 end and2_arch VHDL 0 (v.6A) : Introduction 18 a b c The chip C<=a and b
  • 18. COMPUTER ENGINEERING MARKET and VHDL VHDL 0 (v.6A) : Introduction 19
  • 19. TSMC(TaiwanSemicon.ManufacturingComp.) 台灣積體電路製造股份有限公司 http://www.tsmc.com • From Wiki: • Has the largest asset in Taiwan stock market, • One of the World's largest dedicated independent semiconductor foundry. • Products: Apple iphone6 plus A8-cpu • Relation to VHDL • Design ideaWrite VHDL  TSMC chips VHDL 0 (v.6A) : Introduction 20
  • 20. Huawei Technologies Co. Ltd http://www.huawei.com/en/ • From wiki: • Telecom equipment manufacture • China large private company--http://money.163.com 500 (2011-08- 25) • Products: the second-largest supplier of mobile telecommunications infrastructure equipment in the world (after Ericsson). VHDL 0 (v.6A) : Introduction 21
  • 21. References • See course web page • Digital Systems Design Using VHDL, Charles H. Roth (first or second edition) • Rapid Prototyping of Digital Systems, by Hamblen, James etal. Springer 2008. (read_online) • Digital Design: Principles and Practices, 4/E John F. Wakerly, Prentice Hall. • High-Speed Digital Design: A Handbook of Black Magic by Howard W. Johnson and Martin Graham Prentice Hall. • BOOKBOON (Free text books) • http://www.alldatasheet.com/ VHDL 0 (v.6A) : Introduction 22
  • 22. APPENDIX VHDL 0 (v.6A) : Introduction 23
  • 23. Major companies , a comparison in 2011 (from wiki) • Company Boeing Nestle Honda Toyota Ford HS BC Len ovo BP Sony Revenue US Billion 68.5 125 120 235 128 98. 9 21. 59 308 .9 86.64 Asset 64.3 126 125 370 166 24 54 10. 71 272 .2 155.9 4 Profit US Billion 3.3 39 1.39 5.07 6.56 13. 15 0.2 73 3.3 2.96 VHDL 0 (v.6A) : Introduction 24
  • 24. (from wiki) Wiki: 2009年資本額約新台幣2,589.6億元,市值約1兆6,000億元,為台灣股市中市值最大的公司。 http://money.163.com中国民营企业500强榜单发布,华为第一(2011-08-25) • Company Apple IBM Microsoft Intel HP TSMC 台灣積 體電路 (largest asset in Taiwan stock market) Huawei 华为 (Telecom equipm’t, China large private company) Revenue US Billion 65.23 99 69.94 43.6 99.87 13.98 21.8 Asset 75.1 113.5 108.7 63.2 124.5 20.43 Not known Profit US Billion 14.01 14 23.15 11.46 14.83 5.55 2.67 VHDL 0 (v.6A) : Introduction 25
  • 25. TRI-STATE LOGIC A revision:’ The concept of tri-state logic is essential in computer design, so we want to revise these techniques before we move on. VHDL 0 (v.6A) : Introduction 26
  • 26. Appendix 1:Tri-state logic **At the float state, the wire is cut • VHDL 0 (v.6A) : Introduction 27 Input Output enable (OE) Output Input OE (input) Output 0 0 Z(Float) 1 0 Z(Float) 0 1 0 1 1 1
  • 27. Tri-state equivalent circuit (using output connect/cut view) • VHDL 0 (v.6A) : Introduction 28 Input Output enable (OE) Output Input Output enable (OE) Output OE=1, switch close OE=0, switch open Same as
  • 28. Alternatively: we can treat the Tri-state equivalent circuit using the Rout impedance view • VHDL 0 (v.6A) : Introduction 29 Input Output enable (OE) Output Input Output enable (OE) Output When OE=1, Rout= small, (e.g. 50 Ω ) When OE=0, Rout=infinity (e.g. 10 MΩ) Same as Rout OE (output enable) controls the value of Rout Tri-state equivalent circuit : Rout impedance view to explain the concept of tri-stat A tri-state circuit diagram
  • 29. Student ID: ___________,Date:_____________ Name: _______________ Exercise0.1:Tri-state logic with pull up resistor • VHDL 0 (v.6A) : Introduction 30 Input1 Output-Enable (OE) Output 5V 10K Input1 Output-Enable OE (input) Output 0 0 ? ___ 1 0 ? ___ 0 1 ? ___ 1 1 ? ___ **At float the wire is cut
  • 30. Exercise0.2: Use Rout ( Impedance view) to explain the result of exercise 0.1 • VHDL 0 (v.6A) : Introduction 31 Input1 Output- Enable OE (input) Output Equivalent Rout (10M or 50) Draw equivalent circuit and find output Voltage 0 0 ? _1__ ? ? 1 0 ? _1__ ? ? 0 1 ? _0__ ? ? 1 1 ? _1__ ? ? Resistance view
  • 31. Exercise 0.3 Application 1 of Tri-state logic: Input/Output pin • OE1 controls the traffic. • Fill in the cells with ‘?’. VHDL 0 (v.6A) : Introduction 32 Directional control(OE1) A B A Output Enable OE1 (input) B 0 0 ? 1 0 ? ? 1 0 ? 1 1
  • 32. Exercise 0.4 Application 2 of Tri-state logic: Transceivers for I/O data pins • When T =1, A->B; T controls the traffic, • when /OE=1, IO pins A,B are disabled • Fill in the cells with ‘?’. • VHDL 0 (v.6A) : Introduction 33 A B /OE T T A Output Enable /OE1 (input) B Which controls which 1 0 0 ? ? 0 1 0 ? ? ? ? 1 Float ? ? Float 1 ? ?
  • 33. All data-lines are transceiver buffers • A good controller will enable the CPU to • read/write RAM, and read ROM • VHDL 0 (v.6A) : Introduction 34 CPU data lines transceivers ROM data lines RAM data lines transceivers transceivers /OE1, T1 /OE2, T2 /OE3, T3
  • 34. Exercise 0.5 : List OE1,2,3 and T1,2,3 for the followings cases • a) CPU writes to RAM: • /OE1=___ , /OE2___, /OE3=___, T1___, T2=____, T3_____ • b) CPU reads from ROM • /OE1=___ , /OE2___, /OE3=___, T1___, T2=____, T3_____ • c) CPU reads from RAM • /OE1=___ , /OE2___, /OE3=___, T1___, T2=____, T3_____ VHDL 0 (v.6A) : Introduction 35 CPU data lines transceivers ROM data lines RAM data lines transceivers transceivers /OE1, T1 /OE2, T2 /OE3, T3 A B A B B A
  • 35. Exercise 0.6 Application 3 of Tri-state logic: Selectionofcontrolsignal(resolvedlogic) • Output depends on Input_A if OE is _?___ • Output depends on Input_B if OE is _?___ • Discuss the operation of this circuit. VHDL 0 (v.6A) : Introduction 36 Input_B Input_A Output OE
  • 36. Exercise 0.7 • Fill in ‘?’. Is it a “nor-gate” or an “or-gate”? • Discuss the operation of this circuit. • Answer : VHDL 0 (v.6A) : Introduction 37 Output OE2 5V 10K OE1 0V 0V OE1 OE2 Output 0 0 ? 1 0 ? 0 1 ? 1 1 ?