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ENGG 1203 Tutorial
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
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Sequential Logic (II) and Electrical Circuit (I)
22 Feb
Learning Objectives



Analysis circuits through circuit laws (Ohm’s Law, KCL
and KVL)
News
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Design a finite state machine
HW1 (Feb 22, 2013, 11:55pm)
Ack.: ISU CprE 281x, HKU ELEC1008, MIT 6.111,
MIT 6.01
1
Quick Checking
NOT
Always
always True
true
R2
If
R4

R3
, then
i6  0
R5
i2  i3  i4  i5
i2  i6  i3
e1 
R4
 R2  R4 
V0
If i6  0, then
R2
 R2  R4 

R3
 R3  R5 
2
A FSM design for a Vending machine
(Revisited)

Vending Machine


Collect money, deliver product and change
Vending machine may get three inputs
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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
3
Solution
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
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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
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C1C2 represent the coin returned – 00, 01, 10, 11
P indicates whether to deliver product – 0, 1
4
Solution
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States: S1S2S3
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Represent the money inside the machine now
3 bits are enough to encode the states
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S00 (0 cents) – 000
S05 (5 cents) – 001
S10 – 010
S15 – 011
S20 – 100
S25 – 101
S30 – 110
S35 – 111
5
Solution
6
Solution
Next state
Input
Output
11/110
11/000
01/000
10/000


S35
11/110  S35
10/011  S00
01/001  S00
00/000  S35
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)
7
Solution
11/110
11/000
01/000
10/000
S35
11/110  S35
10/011  S00
01/001  S00
00/000  S35

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)
8
A Parking Ticket FSM
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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.
9
Solution
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Specifications
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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.
10
11
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.
12
Solution
13
Rules Governing Currents and
Voltages

Rule 1: Currents flow in loops

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Rule 2: Like the flow of water, the flow of electrical
current (charged particles) is incompressible

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The same amount of current flows into the bulb (top path)
and out of the bulb (bottom path)
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
14
Parallel/Series Combinations of
Resistance

To simplify the circuit for analysis
v  R1i  R2i
v  Rs i
Series
 Rs  R1  R2
Rp 
1
1
R1
 R1 // R2
 1
R2
R1 R2

R1  R2
Parallel
15
Voltage/Current Divider
 Voltage
Divider
I
V
R1  R2
V1  R1 I 
R1
V
R1  R2
V2  R2 I 
R2
V
R1  R2
Current 
Divider
V   R1 // R2  I
I1 
R2
V R1 // R2

I
I
R1
R1
R1  R2
R1
I2 
I
R1  R2
16
Question: Potential Difference

Assume all resistors have the same resistance,
R. Determine the voltage vAB.
17
Solution
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
Determine VAB
We assign VG=0
R2
VA  5 
 2.5V
R1  R2
R4
VB  3 
 1.5V
R3  R4
VAB  VA  VB  2.5   1.5  4V
18
Question: Current Calculation using
Parallel/Series Combinations
For the circuit in the figure, determine i1 to i5.
19
Solution

We apply:


(i)
3Ω

4Ω
40V
1Ω
V = IR
Series / Parallel Combinations
Current Divider
2Ω
(ii)
2
1 // 2  
3
(iii)
3Ω
40V
3Ω
4Ω
40V
2/3Ω
2
4
4 //   
3
7
4/7Ω
(iv)
40V
25/7Ω
4
25
3 //   
7
7
20
Solution
(vi)
40V
(vii)
3Ω
(v)
25/7Ω
4Ω
40V
2/3Ω
3Ω
4Ω
40V
V  IR
40  i1
1Ω
2Ω
i1  i2  i3
25
 i1  11.2 A
7
i2 
2
3 i   1  11.2   1.6 A
1
 
4 2
7
3
i3  i4  i5
i3  11.2  1.6  9.6 A
i4  2  9.6   6.4 A
3
i5  1  9.6   3.2 A
3
 
 
21
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
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Power = IV = I2R = V2/R
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22
Question: Circuit Analysis
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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,
Step 1, Step 2
R4 = 90Ω, R5 = 100Ω
Battery: V1 = 12V, V2 = 24V, V3 = 36V
Resistor: I1, I2, …, I5 = ?
P1, P2, …, P5 = ?
23
Solution a
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VN = 0
I1: M  R5  V1  R1  B
I2: M  V3  R3  R2  B
I4: M  V2  R4  B
Step 1, Step 2
24
Solution a
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
VM – VB = R5I1 + V1 + R1I1
I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180
Step 3
25
Solution a
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VN – VB = R2I2 + R3I2
I2 = (VN – VB)/(R2 + R3) = – VB/30
Step 3
26
Solution a
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VM – VB = V2 + R4I4
I4 = (VM – VB – V2)/R4 = (12 – VB)/90
We get three relationships now (I1, I2, I4)
Step 3
27
Solution a
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KCL of Node B: I1 + I4 + I2 = 0
(24 – VB)/180 + (12 – VB)/90 – VB/30 = 0
Step 4, Step 5
 VB = 16/3 V
28
Solution a
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I1 = (24 – VB)/180 = 14/135 A = 0.104A
I4 = (12 – VB)/90 = 2/27 A = 0.074A
I2 = – VB/30 = – 8/45 A = – 0.178A
Step 5
29
Solution a
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P = I2R =
P1 = (0.104)2 80 = 0.86528W
P4 = (0.074)2 90 = 0.49284W = VR42 / R
(6.66V, 90Ω)
30
Quick Checking
NOT
Always
always True
true
R2
If
R4

R3
, then
i6  0
R5
i2  i3  i4  i5
i2  i6  i3
e1 
R4
 R2  R4 
V0
If i6  0, then
R2
 R2  R4 

R3
 R3  R5 
31
Quick Checking
NOT
Always
always True
true
R2
If
R4

R3
, then
i6  0
√
R5
i2  i3  i4  i5
i2  i6  i3
e1 
R4
 R2  R4 
√
V0
If i6  0, then
R2
 R2  R4 

R3
√
 R3  R5 
√
√
32
(Appendix) Boolean Algebra
33
(Appendix) Boolean Algebra
34
(Appendix) Question: Voltage
Calculation

Find V2 using single loop analysis


Without simplifying the circuit
Simplifying the circuit
Vs1  2v,Vs 2  2v,Vs 3  2v, R1  1, R2  2, R3  4
Vs2
R1
R2
Vs1
+
Vs3
R3
35
Solution

Choose loop current
Vs2
R1
R2
Vs1
+
Vs3

Apply KVL

R3
Replace V2 by R2I
Vs 2  R1I  R2 I  R3 I  Vs 3  Vs1  0
2
I  A
7

Find V2
4
V2   R2 I  v
7
36
Solution



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
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
Req.
V2 = 4/7 v
Veq.
37