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1
Prof. N. Sujatha Phone: 6513 8079 Email: [email protected] Office: N1.3-B4-07
BG 2104: W1-W7
Basic concepts and basic circuit analysis Principles of semiconductor devices Diodes and applications (self-learning) Bipolar Junction Transistor (BJT)
Contents CA (15%)
Assignment on electric circuits
2
Electrical circuits and analysis
An electrical system
3
Basic electrical quantities
• Electric charge (Q) – units of Coulombs (C)
– quantized in units of charge on electron e = -1.610-19 C - How many electrons would build up 1C charge?
+ - -
4
• Current (I)
– rate of flow of electric charge (water pipe analogy) – I = dQ/dt – units of Ampere = 1C/sec (A)
Basic electrical quantities
What motivates the flow of charges / electrons (here)?
5
Electric potential energy: is the energy stored in a system due to accumulation of charges, due the interaction between charges
Basic electrical quantities
charge
Electric potential Electric potential energy / unit charge
6
•Voltage (V) difference in the electric potential between 2 points (nodes) units of Volts (V)
Basic electrical quantities
7 Voltaic cell- diagrammatic representation
Electron flow Conventional flow (used in this course)
Basic electrical quantities
–conventional flow: define the current direction as the direction of flow of positive charges (high potential low potential, opposite to the direction of electron flow. 8
Voltage sources
An ideal voltage source delivers a constant voltage regardless of the current it produces. Its internal resistance is zero.
Current sources
An ideal current source delivers a constant current regardless of the output voltage. Its internal resistance is infinity.
+ –
9
Symbol Convention
ID (VD) means DC current (voltage) only.
id (vd) means AC current (voltage) only.
iD (vD) means DC and AC current (voltage) in superposition.
Voltage source
Current source
10
When a current is constant with time, we say that we have direct current, abbreviated as dc. On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as ac.
Similarly, we have dc and ac voltage.
Direct and alternating signal
11
.
Examples of DC and AC currents versus time
I
12
AC currents (or voltages) can have various waveforms
13
0 V 0 V
+ 5 V
V
+ 5 V
- 5 V
AC signal can be superimposed upon a DC signal (value)
14
The basic electrical system: What else?
+
-
Charge flow rate
Voltage Load
15
Types of loads
1. Resistive 2. Capacitive 3. Inductive 4. Combination (practical)
Passive
Active Amplifiers
16
Resistance • Resistance (R)
-units of Ohms (Ω) -conductance G = 1/R in units of Siemens (S)
• Resistors: - is a two-terminal electronic component that resists the flow of current, producing a voltage drop between its terminals in accordance with Ohm's law.
- typical range 1Ω to100MΩ - types: Carbon (cheapest) Metal film (more stable, better accuracy) Wire wound (most accurate and most expensive)
17
Ohm’s Law
For a resistor R, as shown below, the voltage drop from point a to b, V=Vab=Va-Vb is given by:
The power dissipated by the resistor is:
RVRIVIP /22 ===
IRV =
18
Resistor- Value?
19
Equivalent Resistance: in series
21
122313
RRI
VVI
VVI
VVRR toteq
+=
−+
−=
−==
R1
R3 R2 Req
Req = R1 + R2 + R3
A series circuit is a single path for electric current through all of its components.
What if multiple resistances in the circuit?
20
Equivalent Resistance: in parallel
At point A, I=I1+I2
21
21
212112
1212
/)(/)( RRRR
RVVRVVVV
IVVRR toteq +
=−+−
−=
−==
Req R3 R2 R1
321
1111RRRReq
++=
A parallel circuit has multiple paths for current components to pass separately. A parallel circuit provides the same voltage across all its branchs.
21
Equivalent Resistance: combinations of series and parallel
22
Voltage divider- A series resistor circuit connected to a voltage source
+
==21
22 RR
RVVV inout
In general, ∑
==i
xinRout R
RVVVx
V1
V2
Vin = V1+ V2
+ -
Vin
23
Current divider
Before you move on to the next slide, can you try to solve for i1 and i2 using Ohm’s law and equivalent resistance? 24
Current divider
21
212211 RR
RRiRiRiRiv seqs +====
2121
12
21
21 iii
RRRii
RRRii sss +=
+=
+=
25
Current divider
IS R2 V
+
–
R1 I1 I2
R3 I3
j
parSR R
RII
j=∑=+++=
iMpar RRRRR11111
21
In general,
26
Circuit diagrams • A circuit diagram represents the interconnections between electrical units in a circuit.
• A circuit diagram is just one representation of the actual circuit, among many possible diagams that can equally represent the circuit.
• A circuit diagram do not reflect the actual physical lengths, sizes, shapes, placement of the electrical units.
• A line (or wire) in a circuit diagram does not necessarily indicates the physical existence of a connecting wire. It simply indicates that the points connected by this line in the diagram is electrically “identical”, i.e., they assumes the same voltage. • A node in diagram is where three or more circuit components are connected together.
27
What are the differences between these two circuits? Please think carefully before move to the next slide.
Circuit diagrams
28
Circuit diagrams Every circuit can be drawn in different ways without changing the physics of the circuit. In fact, the two circuits shown are identical !! Examine complicated diagrams and redraw them to clearly show how the elements are interconnected.
What is the voltage across R4?
29
Circuit diagrams
R4 is shorted-out as if it doesn’t exist, therefore, no voltage across it; and no current passing through it.
30
Simple circuit Analysis
Simple circuit can be analyzed simply by: • represent the circuit with a clear diagram
• identify parallel and series combinations of resistors
• replace such combinations by their equivalent resistors
• continue until the circuit is reduced to a straightforward parallel or series circuit
• finally use Ohm’s law, or the voltage and current divider rules, to find the individual resistor voltages and currents if required.
31
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
+ – 10V
+
–
V1 +
–
V3 +
–
V2
Simple circuit analysis: an example
Find V1, V2, and V3
32
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
10V +
–
V1 +
–
V3 +
–
V2
Find an equivalent resistance for the network with V1 across it, then find V1
using a voltage divider.
+ –
Simple circuit analysis: an example
33
1kΩ
1kΩ
10V +
–
V1
V5k1k1
k1V101 =Ω+Ω
Ω=V
+ –
Simple circuit analysis: an example
34
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
10V +
–
5V +
–
V3 +
–
V2
Find an equivalent resistance for the network with V2 across it, then find V2.
+ –
Simple circuit analysis: an example
35
1kΩ
2kΩ
1kΩ
10V +
–
5V +
–
V2 1kΩ
V5.2k1k1
k1V52 =Ω+Ω
Ω=V
+ –
Simple circuit analysis: an example
36
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
10V +
–
5V +
–
V3 +
–
2.5V
V25.1k1k1
k1V5.23 =Ω+Ω
Ω=V
+ –
Simple circuit analysis: an example
37
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
+ – 10V
+
–
V1 +
–
V3 +
–
V2
Simple circuit analysis: an example
Mark the currents I, I1, I2, I3 and I4
38
I1= 5 A I2=2.5 A I3= 2.5 A I4 = 1.25 A
39
• Simple circuit analysis can only be applied for a simple circuit. Often, circuit is a complicated network even after simplification.
• Simple circuit analysis cannot be applied systematically using computer program. • Therefore, network analysis based on Kirchoff’s current and voltage laws usually need to be applied.
Is simple circuit analysis applicable in all cases?
40
Kirchoff’s current law (KCL) Current law: for any node of the circuit, the sum of the currents flowing into the node is equal to the sum of the currents flowing out of the node. Or the algebraic (net) sum of current flow is zero, if you assign inward and outward current with different sign.
∑∑ =outin
II
I1=I2+I3
KCL
think of water: what goes into a pipe junction must come out
Because charge cannot build up at a node.
41
Kirchoff’s voltage law (KVL)
Voltage law: The algebraic sum of all the voltages across individual components around any loop of a circuit is zero.
v1+v2+v3+v4=0 KVL
think of gravity: what goes up must come down.
∑ =loop iV 0
Because electrical potential is conservative.
42
Network analysis In a circuit diagram: • A branch is one path through which the same current flow through all the components in series in this path.
• A loop is a closed path in the circuit
• A node or junction is a point where three or more circuit components are joined.
43
Network analysis Two methods for calculating currents and voltages in a circuit based on KCL and KVL: (1) Nodal method; (2) Mesh (or loop) current method.
• In all methods, usually it is a good idea to replace simple series and parallel combinations of resistances by their equivalent resistance to simplify the circuit. • Choose one node as a reference (ground) node, which is considered to have zero voltage. Make a convenient choice of ground to simplify the analysis. A good choice is the node with the most branches, or a node which can immediately give you another node voltage (e.g., directly connect to a voltage source).
44
Nodal method
• Choose ground node, identify all nodes and assign node voltages to all nodes (other than reference node). They are the unknowns.
• First assume (arbitrarily) a current direction in each branch of the circuit (draw an arrow) and define a current symbol (e.g., I1). If there is a current source in the branch, the branch current is then specified by the current output from this source.
• A KCL equation is written for each node in terms of nodal voltages.
• Number of equations equals number of independent nodes (k-1 for k total nodes).
• Complete the analysis by working out branch currents
If a current in your solution comes out negative, it just means that it’s in the opposite direction from the one you initially assume. So don’t worry about the actual current directions before you start solving.
45
Nodal method: an example
Assume: R1=R2=R3=1Ω; R4=R5=2Ω V1 = 1V V2 = 2V VA = 0.6 V and VB = 0.8 V
46
Nodal method: another example with current sources
500Ω
500Ω I1
V1 V2
Ω+
Ω−
=500500
1211
VVVI
KCL at node 1
500Ω
500Ω
1kΩ
500Ω
500Ω I1
3 mA
1 2 3
V1 V2 V3
I2 4mA
i1
i2
i3
i4
i5
i1
i2
47
Nodal method: another example with current sources
500Ω
1kΩ
500Ω V2 V3 V1
Ω−
+Ω
=Ω
−500k1500
32221 VVVVV
KCL at node 2
Ω=
Ω−
+500500
3322
VVVI500Ω
500Ω
I2
V2 V3 KCL at node 3
V1 = 1.33V, V2=1.17V, V3=1.58V
i2
i3
i4
i4
i5
48
Mesh (loop) current method • Choose ground node and identify meshes (loops).
• Assign a current (imaginary loop current) to each mesh (they are unknowns). By convention, define all mesh currents in the clockwise direction.
• Write a KVL equation in terms of mesh currents for each mesh.
• Complete the analysis by working out the branch currents. (actual branch current = algebraic sum of relevant imaginary loop currents at a given branch).
1kΩ
1kΩ
1kΩ
V1 V2 I1 I2 + –
+ –
49
Mesh (loop) current method: voltages across resistors
R
I1
+ – VR
VR = I1 R
R
I1
+ – VR I2
VR = (I1 - I2 ) R When R is shared by two loops: Actual current through R = algebraic sum of relevant imaginary loop currents following through R 50
Mesh (loop) current method: an example
If, Vbb=1V, All Ri = 1 ohm, I1 = 0.5 A I2 = I3 = 0.25A Actual current passing R2 is I1-I2=0.25 A (downwards direction)
51
1kΩ
2kΩ
2kΩ
12V 4mA
2mA
I0
+ –
Mesh (loop) current method: another example with current sources
52
• The current sources will have whatever voltage necessary produce the nominal current which cannot be determined immediately. We can’t use KVL around the loop because we don’t know the voltages across the current sources.
Mesh (loop) current method: another example with current sources
I1 I2
I3 1kΩ
2kΩ
2kΩ
12V 4mA
2mA
I0
+ –
What to do?
53
• The 4mA current source sets I2: I2 = -4 mA
• The 2mA current source sets a constraint on I1 and I3:
I1 - I3 = 2 mA • We have two equations and three
unknowns. Where is the third equation?
Mesh (loop) current method: another example with current sources
Notice the negative sign. Why?
54
1kΩ
2kΩ
2kΩ
12V
4mA
2mA
I0 I1 I2
I3
+ –
Apply KVL to supermesh
12V - I3*2kΩ - (I3 - I2)*1kΩ - (I1 - I2)*2kΩ = 0
ground node
Supermesh is highlighted by the dashed yellow line and excludes current sources.
Solution: I1 = 1.2 mA; I2 = -4 mA; I3 = -0.8 mA; I0 = I1 -I2 = 5.2 mA 55
Linear superposition principle
For any linear circuit containing more than one independent voltage source or current source, the total current in any part of the circuit equals the algebraic sum of the currents produced by each source separately. To isolate the current from a particular source
• replace all other voltage sources by short circuits • replace all other current sources by open circuits
56
Linear superposition principle: example 1
ARR
Vi s
41
483'
212 =
+=
+=
Find I2 using superposition
ARR
Rii s
34
4882''
21
12 =
+⋅
=+
=
''' 222 iii +=57
2kΩ 1kΩ
2kΩ 12V
I0
2mA
4mA – +
Linear superposition principle: example 2
Solve for I0?
58
2kΩ 1kΩ
2kΩ
I’0
2mA
I’0 = -4/3 mA why?
• 2mA source contribution
Linear superposition principle: example 2
59
2kΩ 1kΩ
2kΩ
I’’0
4mA
I’’0 = 0 why?
• 4mA source contribution
Linear superposition principle: example 2
60
2kΩ 1kΩ
2kΩ 12V
I’’’0
– +
I’’’0 = -4 mA why?
• 12V source contribution
Linear superposition principle: example 2
61
I’0 = -4/3 mA I’’0 = 0
I’’’0 = -4 mA
I0 = I’0+ I’’0+ I’’’0 = -16/3 mA
• Final result
Linear superposition principle: example 2
As a practice, can you analyze this circuit using nodal or mesh current analysis? 62
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
+ – 10V
+
–
V1 +
–
V3 +
–
V2
Varying load
Estimation of load currents for changing loads
63
Thevenin equivalent circuit
Thévenin's theorem for electrical networks states that any combination of sources and resistors with two terminals is electrically equivalent to a single voltage source Vth in series with a single resistor Rth. For single frequency AC systems the theorem can also be applied to general impedances (AC resistance).
• Vth = open circuit voltage at the output port • Rth = Vth / (short circuit current at the output port)
Rth also equals to the equivalent resistance looking into the circuit while replace voltage sources with shorts and current sources with open circuits. 64
The Thevenin idea is most useful when one considers two circuits or circuit elements, with the first circuit’s output providing the input for the second circuit. In the figure, the output of the first circuit (A), consisting of Vth and Rth, is fed to the second circuit element (B), which consists simply of a load resistance (RL) to ground. This simple configuration represents, in a general way, a very broad range of analog electronics.
Thevenin equivalent circuit
65
Circuit loading
)/(1 Lth
th
Lth
Lthout RR
VRR
RVV+
=
+
=
Thus, we should try to keep the ratio Rth/RL small in order to avoid the loss in voltage due to loading the circuit . A maximum ratio of 1/10 is often used as a rule of thumb in circuit design. 66
1kΩ
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
+ – 10V
+
–
V1 +
–
V3 +
–
V2
Varying load
Draw the Thevenin equivalent circuit for the below circuit
67
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
+ – 10V
+
–
+
–
+
–
To find Thevenin voltage
?
3.64V
68
1kΩ
2kΩ
1kΩ
2kΩ
1kΩ
+
–
+
–
+
–
Draw the Thevenin equivalent circuit for the below circuit
For finding Thevenin resistance
1.91 kΩ
69
To find the load current through 1kΩ
3.64V
1.91 kΩ
1 kΩ
Load current = 3.64 V /(1.91+1) kΩ =1.25 mA Voltage across 1 kΩ resistor = 1.25V Note that you have obtained the same result using simple circuit analysis! (Refer slide number 36) But, Thevenin circuit is useful for computing load current in a variable load condition; its only a one time effort! 70
Norton equivalent circuit
Norton's theorem for electrical networks states that any network of sources and resistors can be replaced by a single current source in parallel with a single resistor. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors.
• Int = short circuit current at the output port • Rnt = (open circuit voltage at the output port) / Int Rnt also equals to the equivalent resistance looking into the circuit while replace voltage sources with shorts and current sources with open circuits. (Rnt= Rth).
71
• Short A and B, calculate the current: Int=Vin/R1 • Now replace voltage sources with shorts and current sources with open circuits. • Evaluate the resistance across points A and B "looking back" into the circuit. Rnt=R1*R2/(R1+R2)
A
B
A
B
nt nt
Example: Norton equivalent circuit
72
nteqth IRV =eq
thnt R
VI =
Vth
Rth
+ –
Int Rnt
nttheq RRR ==
Thevenin equivalent vs Norton equivalent
73
Revisiting types of loads
1. Resistive 2. Capacitive 3. Inductive 4. Combination (practical)
Passive
Active Amplifiers
74
Capacitor • Stores charge on electrodes (parallel conductive plates separated by insulator)
Q=VC
• Capacitance measure in units of farads (F = 1 C/volt)
• Range of typical values (1pF to 1000 μF)
• Capacitor types ceramic (pF) mylar (nF) electrolytic (μF) 75
Capacitor
For DC voltages, no current passes through a capacitor. It “blocks DC”. When a time varying potential is applied, the current passing through the capacitor is:
The impedance (or AC resistance) of a capacitor is:
fCjCjZC πω 2/1/1 ==
Where ω is the frequency. ZC 0 when ω ∞
dtdVCI =
76
in series
Equivalent Capacitance
in parallel
77
Inductor
• Energy stored in magnetic field
• Measure in units of henries (H)
• Typical range (1μH to 1H)
78
Inductor
LjZL ω=
It passes DC current, and resist AC current. The impedance of an inductor is:
79
in series
Equivalent Inductance
in parallel
neq LLL
L 111
21
+++=
neq LLLL +++= 21
80
Semiconductor Devices
The semiconductor devices we will focus in this course behave very differently from resistors, capacitors, and inductors that are familiar to you. Semiconductors all regulate, in some manner, the amount of current that can pass through them in a given direction.
81