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ENGG 1203 Tutorial

Sequential Logic (II) and Electrical Circuit (I) 22 Feb Learning Objectives

Design a finite state machine Analysis circuits through circuit laws (Ohm’s Law, KCL

and KVL) News

HW1 (Feb 22, 2013, 11:55pm) Ack.: ISU CprE 281x, HKU ELEC1008, MIT 6.111,

MIT 6.01

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Quick CheckingNOT always true

Always True

If , then

If , then

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4 5

RRR R

6 0i

2 3 4 5i i i i

2 6 3i i i

4

1 02 4

Re VR R

6 0i

32

2 4 3 5

RRR R R R

A FSM design for a Vending machine (Revisited) Vending Machine

Collect money, deliver product and change Vending machine may get three inputs

Inputs are nickel (5c), dime (10c), and quarter (25c) Only one coin input at a time Product cost is 40c Does not accept more than 50c Returns 5c or 10c back Exact change appreciated

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Solution

We are designing a state machine which output depends on both current state and inputs.

Suppose we ask the machine to directly return the coin if it cannot accept an input coin.

Input specification: I1 I2 Represent the coin inserted 00 - no coin (0 cent), 01 – nickel (5 cents), 10 – dime (10 cents),

11 – quarter (25 cents)

Output specification: C1C2P C1C2 represent the coin returned – 00, 01, 10, 11 P indicates whether to deliver product – 0, 1

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Solution

States: S1S2S3 Represent the money inside the machine now 3 bits are enough to encode the states

S00 (0 cents) – 000 S05 (5 cents) – 001 S10 – 010 S15 – 011 S20 – 100 S25 – 101 S30 – 110 S35 – 111

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Solution

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Solution

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S3511/110 S3510/011 S00 01/001 S00

11/11011/00001/00010/000

S35: Currently the machine has 35 cents e.g. 11/110 : If we insert a quarter (11), then the machine

should return one quarter and zero product (110) 35c (35 cents inside the machine now) + 25c (insert 25 cents)

= 35c (35 cents inside the machine in the next state) + 25c (return 25 cents) + 0c (return no product)

Input

Output

Next state

00/000 S35

Solution

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S3511/110 S3510/011 S0001/001 S00

11/11011/00001/00010/000

e.g. 10/011: If we insert a dime (10), then the machine should return one nickel and one product (011) 35c (35 cents inside the machine now) + 10c (insert 10 cents)

= 0c (zero cent inside the machine in the next state) + 5c (return 5 cents) + 40c (return one product)

e.g. 01/001: If we insert a nickel (01), then the machine should return zero coin and one product (001) 35c (35 cents inside the machine now) + 5c (insert 5 cents)

= 0c (zero cent inside the machine in the next state) + 0c (return zero cent) + 40c (return one product)

00/000 S35

A Parking Ticket FSM

At Back Bay garage, Don and Larry are thinking of using an automated parking ticket machine to control the number of guest cars that a member can bring. The card reader tells the controller whether the car is a member or a guest car. Only one guest car is allowed per member at a discount rate only when s/he follows out the member at the exit (within the allotted time). The second guest must pay the regular parking fees. You have been hired to implement the control system for the machine which is located at the exit.

Using your expertise on FSMs, design a FSM for the control system.

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Solution

Specifications Signals from the card reader: MEMBER and GUEST Signals from the toll booth: TOKEN (meaning one toke

received), EXP (time for discounted guest payment has expired).

Signal to the gate: OPEN. Fee: Members are free, Guest with a Member is 1

Token, Regular Guest is 2 Tokens.

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Solution

The truth table that corresponds to the FSM The state labels can be mapped to a three bit state

variable. All entries not entered below are illegal.

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Solution

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Rules Governing Currents and Voltages Rule 1: Currents flow in loops

The same amount of current flows into the bulb (top path) and out of the bulb (bottom path)

Rule 2: Like the flow of water, the flow of electrical current (charged particles) is incompressible Kirchoff’s Current Law (KCL): the sum of the currents into a

node is zero Rule 3: Voltages accumulate in loops

Kirchoff’s Voltage Law (KVL): the sum of the voltages around a closed loop is zero

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Parallel/Series Combinations of Resistance To simplify the circuit for analysis

1 2

1 2

s

s

v R i R i

v R i

R R R

1 2

1 21 2

1 2

11 1

//

p

R RR

R RR R

R R

Series

Parallel

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Voltage/Current Divider

1 2

11 1

1 2

22 2

1 2

VI

R R

RV R I V

R R

RV R I V

R R

1 2

1 2 21

1 1 1 2

12

1 2

//

//

V R R I

R R RVI I I

R R R R

RI I

R R

Voltage Divider

Current Divider

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Question: Potential Difference Assume all resistors have the same resistance,

R. Determine the voltage vAB.

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Solution

Determine VAB

We assign VG=0

2

1 2

4

3 4

5 2.5

3 1.5

A

B

RV V

R R

RV V

R R

2.5 1.5 4AB A BV V V V

For the circuit in the figure, determine i1 to i5.

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Question: Current Calculation using Parallel/Series Combinations

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Solution

21 // 2

3

2 44 //

3 7

4 253 //

7 7

40V

3Ω

4Ω 1Ω 2Ω

40V

3Ω

4Ω 2/3Ω

40V

3Ω

4/7Ω

40V 25/7Ω

(i)

(iii)

(ii)

(iv)

We apply: V = IR Series / Parallel Combinations Current Divider

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Solution

1 1

2540 11.2

7

V IR

i i A

1 2 3

2 1

3

2 13 11.2 1.62 74 3

11.2 1.6 9.6

i i i

i i A

i A

3 4 5

4

5

2 9.6 6.43

1 9.6 3.23

i i i

i A

i A

40V 25/7Ω

(v)

(vi)

(vii)

40V

3Ω

4Ω 2/3Ω

40V

3Ω

4Ω 1Ω 2Ω

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Analyzing Circuits

Assign node voltage variables to every node except ground (whose voltage is arbitrarily taken as zero)

Assign component current variables to every component in the circuit

Write one constructive relation for each component in terms of the component current variable and the component voltage

Express KCL at each node except ground in terms of the component currents

Solve the resulting equations

Power = IV = I2R = V2/R

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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,R4 = 90Ω, R5 = 100Ω

Battery: V1 = 12V, V2 = 24V, V3 = 36V

Resistor: I1, I2, …, I5 = ? P1, P2, …, P5 = ?

Question: Circuit Analysis

Step 1, Step 2

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Solution a

VN = 0

I1: M R5 V1 R1 B

I2: M V3 R3 R2 B

I4: M V2 R4 B

Step 1, Step 2

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Solution a

VM – VB = R5I1 + V1 + R1I1

I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180Step 3

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Solution a

VN – VB = R2I2 + R3I2

I2 = (VN – VB)/(R2 + R3) = – VB/30

Step 3

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Solution a

VM – VB = V2 + R4I4

I4 = (VM – VB – V2)/R4 = (12 – VB)/90

We get three relationships now (I1, I2, I4)Step 3

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Solution a

KCL of Node B: I1 + I4 + I2 = 0

(24 – VB)/180 + (12 – VB)/90 – VB/30 = 0

VB = 16/3 V Step 4, Step 5

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Solution a

I1 = (24 – VB)/180 = 14/135 A = 0.104A

I4 = (12 – VB)/90 = 2/27 A = 0.074A

I2 = – VB/30 = – 8/45 A = – 0.178AStep 5

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Solution a

P = I2R = P1 = (0.104)2 80 = 0.86528W

P4 = (0.074)2 90 = 0.49284W = VR42 / R

(6.66V, 90Ω)

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Quick CheckingNOT always true

Always True

If , then

If , then

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4 5

RRR R

6 0i

2 3 4 5i i i i

2 6 3i i i

4

1 02 4

Re VR R

6 0i

32

2 4 3 5

RRR R R R

32

Quick CheckingNOT always true

Always True

If , then

If , then

32

4 5

RRR R

6 0i

2 3 4 5i i i i

2 6 3i i i

4

1 02 4

Re VR R

6 0i

32

2 4 3 5

RRR R R R

√

√

√

√

√

(Appendix) Boolean Algebra

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(Appendix) Boolean Algebra

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(Appendix) Question: Voltage Calculation Find V2 using single loop analysis

Without simplifying the circuit Simplifying the circuit

1 2 3 1 2 32 , 2 , 2 , 1 , 2 , 4s s sV v V v V v R R R

R1

Vs1

Vs3

Vs2

R3

-R2

+

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Solution

Choose loop current

Apply KVL Replace V2 by R2I

Find V2

R1

Vs1

Vs3

Vs2

R3

-R2

+

2 1 2 3 3 1 0

2

7

s s sV R I R I R I V V

I A

2 2

4

7V R I v

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Solution

Simplify the circuit with one voltage source and one resistor

Req. = R1 + R2 + R3 = 7 ohm

Veq. = Vs1 + Vs2 + Vs3 = -2 + 2 + 2 = 2 V

I = Veq. / Req. = 2/7 A

V2 = 4/7 v Veq.

Req.