Download EE101-Lect2- Basic Concepts for Electronic and Electric

Synergy of Design and Analysis
Circuit Analysis
•Check against specifications
•If specs are met, done
•Otherwise, improve/modify the
design until specs are met
Product Specifications
Circuit Design
•Circuit Structure (topology)creative part
•Determination of
Parameter Values (R, L, C,
transistor sizes, etc.) –
mathematical optimization
EE101 Fall 2012 Lect 2- Kang
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Electronic Circuits consisted of R, L, C, Independent
Sources, Dependent Sources, Transistors, OP Amps, etc.
.
Circuit Design involves (1) Determination of a Structure/Topology for
Right Components, and then (2) Determining Component Values
EE101 Fall 2012 Lect 2- Kang
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Design an Ohmmeter that measures the
Resistance (Rx) of Resistors
Measure using
(E/I- Ri), where
Ri is the internal
resistance in
series.
Ifs = Im when Rx = 0
Using an
Ammeter
EE101 Fall 2012 Lect 2- Kang
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EE101Lect2- Basic Concepts for Electronic and
Electric Circuits
Linearity vs. Nonlinearity: Most models are nonlinear but linear models are
conveniently used.
Resistors: v= f(i)= i3 is nonlinear, and v= 2i is linear.
The diode equation is nonlinear, i =𝐾(𝑒 𝑉 -1), but for small voltages near v=0, i=Kv
(linear, like a K mho conductor) – conductance = 1/(resistance)
Most linear models are approximation for small signal cases. Analog Circuits are for
small signals and often called Linear Circuits.
For linear relationship, superposition principle applies.
For v=v1 + v2, i=K(v1+v2)= Kv1+Kv2= i1+i2.
What does this mean? For two different sources, you can solve the circuit equation
one at a time by assuming the other source is zero, and then add the solutions at the
end.
But for diode with i=K(𝑒 v1+v2 −1), the superposition principle cannot applied
(because the equation is not linear).
EE101 Fall 2012 Lect 2- Kang
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Current Through R3?
EE101 Fall 2012 Lect 2- Kang
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Simple Example of Resistive Circuits
Problem:
SOLUTION:
Let i3 (downward direction) be the current through
R3.
Let i2 (downward direction) be the current through
R2.
Let i1 (left to right direction) be the current through
R1.
Then i1=i2+i3
(Eq.1)
Vs (=90V) = VR1+ VR2(=VR3)=10 (i2+i3)+60 X i3 (Eq.2)
We are mainly interested in iR3. So how do we
express i2 in terms of i3?
[Q] Find is the current through R3
[Q] Find the power dissipated in R3
The voltages across both R2 and R3 are same.
VR2=VR3 and thus i2=60 X i3/30=2 X i3
(Eq.3)
From Eqs. 2 and 3, 90=10(3 X i3)+60 X i3=90 X i3, thus
i3=1 Amp. (Ans)
The power dissipated in R3 is
P3= V3x i3= (60x1) x 1= 60 Watt. (ans)
This is like a 60W light bulb case, although real-life
light bulbs are for AC (alternating current) of 110120V, whose average value is close to 90V DC.
EE101 Fall 2012 Lect 2- Kang
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Some Observations (KVL, KCL)
• Two branches (R2 and R3) in parallel have the same voltage
VR2=VR3
This can be restated with generalization as “the sum of voltages in a loop is
zero”,
(𝑣𝑜𝑙𝑡𝑎𝑔𝑒𝑠 𝑖𝑛 𝑎 𝑙𝑜𝑜𝑝) = 0
This is called Kirchhoff’s Voltage Law (KVL)
• The Sum of the currents at a node is zero.
(what goes into a node comes out, that is
𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛𝑤𝑎𝑟𝑑 = ∑𝑐𝑢𝑟𝑟𝑒𝑛𝑡 out𝑤𝑎𝑟𝑑
i1=i2+i3
Another way of saying this is
(𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠 𝑖𝑛𝑡𝑜 𝑎 𝑛𝑜𝑑𝑒) = 0
This is called Kirchhoff’s Current Law (KCL)
EE101 Fall 2012 Lect 2- Kang
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Some Basic Mathematics
• i(t)=dq(t)/dt
•
𝑡
𝑡0
q(t)= −∞ 𝑖(τ) 𝑑τ= −∞ 𝑖(τ)dτ
𝑡
= q0 + 𝑡𝑜 𝑖 𝜏 𝑑τ
• 𝑒𝑥 = 1 +
𝑥
𝑥2
+
1!
2!
+
𝑥3
3!
+
𝑡
𝑖(τ)dτ
𝑡0
+ ⋯ , −∞ < 𝑥 < ∞
• 𝑒 𝑖𝜔𝑡 = cos𝜔𝑡 + i sin𝜔𝑡, where i= −1
EE101 Fall 2012 Lect 2- Kang
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Examples of Dependent Sources
• A device model for Bipolar Junction
Transistor (BJT) shown in Fig. 3.40 of our
textbook.
C
B +
VBE
iB
B
C
VBE
-
βiB
E
E
EE101 Fall 2012 Lect 2- Kang
Diamond symbol
stands for dependent
sources. In this
example, it is currentcontrolled current
source (CCCS) whose
magnitude depends on
other current, iB.
In contract
independence sources
are marked in circle or
by the battery (DC)
symbol.
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Examples of Dependent Sources (continued)
• Voltage-controlled voltage source (VCVS)
Let us study the case of an operational amplifier
(OP Amp) shown in Fig. 5.4 of our textbook.
Ro
a
a
_
c
b
+
b
c
_
+
vd Ri _
+
A vd
Ri = input resistance, usually very large
= almost infinite (∞).
Ro = output resistance, usually very small.
A = amplification factor, usually very large in thousands
or even more.
EE101 Fall 2012 Lect 2- Kang
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How doe we handle dependent sources?
EE101 Fall 2012 Lect 2- Kang
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A Resistive Circuit with both Independent and
Dependent Sources
We can apply KCL to find v1, v2, v3:
Node 1: is+2ix=(v1-v2)/R1= is+2 (v3-v2)/R3
Node 2: (v1-v2)/R1 + (v3-v2)/R3=v2/R2
Node 3: 3 (v3-v2)/R3= -v3/R4
Here the values of is, R1, R2, R3, R4 are known.
(1)
(2)
(3)
So, we have 3 (independent) equations with 3 unknowns of v1, v2, v3
and thus can find node voltages. (matrix determinant method can be used.)
After finding these voltages, we can also find currents (ik) through all branch
Elements and power dissipationEE101
in each
element
x ik).
Fall 2012
Lect 2- (Pk=vk
Kang
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.
EE101 Fall 2012 Lect 2- Kang
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EE101 Fall 2012 Lect 2- Kang
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