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EE70 Review.

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Presentation on theme: "EE70 Review."— Presentation transcript:

1 EE70 Review

2 Electrical Current

3 Circuit Elements An electrical circuit consists of circuit elements such as voltage sources, resistances, inductances and capacitances that are connected in closed paths by conductors

4 Reference Directions The voltage vab has a reference polarity that is positive at point a and negative at point b.

5 Reference Directions

6 Reference Directions “uphill: battery” “downhill: resistor”
Energy is transferred when charge flows through an element having a voltage across it.

7 Power and Energy Watts Joules

8 Current is flowing in the passive configuration
Reference Direction Current is flowing in the passive configuration If current flows in the passive configuration the power is given by p = vi If the current flows opposite to the passive configuration, the power is given by p = -vi

9 Dependent Sources

10 Resistors and Ohm’s Law
b The units of resistance are Volts/Amp which are called “ohms”. The symbol for ohms is omega: 

11 Resistance Related to Physical Parameters
 is the resistivity of the material used to fabricate the resistor. The units of resitivity are ohm-meters (-m)

12 Power dissipation in a resistor

13 Kircohoff’s Current Law (KCL)
The net current entering a node is zero Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node

14 Kircohoff’s Current Law (KCL)

15 Series Connection

16 Kircohoff’s Voltage Law (KVL)
The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit.

17 Kircohoff’s Voltage Law (KVL)

18 Parallel Connection KVL through A and B: -va+vb = 0  va = vb
KVL through A and C: -va - vc = 0  va = -vc

19 Equivalent Series Resistance

20 Equivalent Parallel Resistance

21 Circuit Analysis using Series/Parallel Equivalents
Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source. Redraw the circuit with the equivalent resistance for the combination found in step 1.

22 Voltage Division Of the total voltage, the fraction that appears across a given resistance in a series circuit is the ratio of the given resistance to the total series resistance.

23 Current Division For two resistances in parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to the sum of the two resistances.

24 Node Voltage Analysis

25 Node Voltage Analysis

26 Mesh Current Analysis

27 Mesh Current Analysis

28 Thévenin Equivalent Circuits

29 Thévenin Equivalent Circuits

30 Thévenin Equivalent Circuits

31 Thévenin Equivalent Circuits

32 Thévenin Equivalent Circuits

33 Norton Equivalent Circuits

34 Norton Equivalent Circuits

35 Source Transformations

36 Maximum Power Transfer

37 Superposition Principle
The superposition principle states that the total response is the sum of the responses to each of the independent sources acting individually. In equation form, this is

38 Superposition Principle

39 Superposition Principle
Current source open circuit

40 Superposition Principle
Voltage source short circuit Req

41 Voltage-Amplifier Model
The input resistance Ri is the equivalent resistance see when looking into the input terminals of the amplifier. Ro is the output resistance. Avoc is the open circuit voltage gain.

42 Voltage Gain Ideally, an amplifier produces an output signal with identical waveshape as the input signal, but with a larger amplitude.

43 Current Gain

44 Power Gain

45 Operational Amplifier

46 Summing Point Constraint
Operational amplifiers are almost always used with negative feedback, in which part of the output signal is returned to the input in opposition to the source signal.

47 Summing Point Constraint
In a negative feedback system, the ideal op-amp output voltage attains the value needed to force the differential input voltage and input current to zero. We call this fact the summing-point constraint.

48 Summing Point Constraint
Verify that negative feedback is present. Assume that the differential input voltage and the input current of the op amp are forced to zero. (This is the summing-point constraint.) Apply standard circuit-analysis principles, such as Kirchhoff’s laws and Ohm’s law, to solve for the quantities of interest.

49 The Basic Inverter

50 Applying the Summing Point Constraint

51 Inverting Amplifier

52 Summing Amplifier

53 Non-inverting Amplifiers

54 Voltage Follower

55 Capacitance

56 Capacitances in Parallel

57 Capacitances in Series

58 Inductance

59 Inductance The polarity of the voltage is such as to oppose the change in current (Lenz’s law).

60 Series Inductances

61 Parallel Inductances

62 Magnetic flux produced by one coil links the other coil
Mutual Inductance Fields are aiding Fields are opposing Magnetic flux produced by one coil links the other coil

63 Discharge of a Capacitance through a Resistance
KCL at the top node of the circuit: iC iR

64 Discharge of a Capacitance through a Resistance
We need a function vC(t) that has the same form as it’s derivative. Substituting this in for vc(t)

65 Discharge of a Capacitance through a Resistance
Solving for s: Substituting into vc(t): Initial Condition: Full Solution:

66 Discharge of a Capacitance through a Resistance
To find the unknown constant K, we need to use the boundary conditions at t=0. At t=0 the capacitor is initially charged to a voltage Vi and then discharges through the resistor.

67 Discharge of a Capacitance through a Resistance

68 Charging a Capacitance from a DC Source through a Resistance

69 Charging a Capacitance from a DC Source through a Resistance
KCL at the node that joins the resistor and the capacitor Current into the capacitor: Current through the resistor:

70 Charging a Capacitance from a DC Source through a Resistance
Rearranging: This is a linear first-order differential equation with contant coefficients.

71 Charging a Capacitance from a DC Source through a Resistance
The boundary conditions are given by the fact that the voltage across the capacitance cannot change instantaneously:

72 Charging a Capacitance from a DC Source through a Resistance
Try the solution: Substituting into the differential equation: Gives:

73 Charging a Capacitance from a DC Source through a Resistance
For equality, the coefficient of est must be zero: Which gives K1=VS

74 Charging a Capacitance from a DC Source through a Resistance
Substituting in for K1 and s: Evaluating at t=0 and remembering that vC(0+)=0 Substituting in for K2 gives:

75 Charging a Capacitance from a DC Source through a Resistance

76 DC Steady State In steady state, the voltage is constant, so the current through the capacitor is zero, so it behaves as an open circuit.

77 DC Steady State In steady state, the current is constant, so the voltage across and inductor is zero, so it behaves as a short circuit.

78 DC Steady State The steps in determining the forced response for RLC circuits with dc sources are: 1. Replace capacitances with open circuits. 2. Replace inductances with short circuits. 3. Solve the remaining circuit.

79 RL Transient Analysis

80 RL Transient Analysis

81 RC and RL Circuits with General Sources
First order differential equation with constant coefficients Forcing function

82 RC and RL Circuits with General Sources
The general solution consists of two parts.

83 The particular solution (also called the forced response) is any expression that satisfies the equation. In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution.

84 The homogeneous equation is obtained by setting the forcing function to zero.
The complementary solution (also called the natural response) is obtained by solving the homogeneous equation.

85 Integrators and Differentiators
Integrators produce output voltages that are proportional to the running time integral of the input voltages. In a running time integral, the upper limit of integration is t .

86

87 Differentiator Circuit

88 Second–Order Circuits
Differentiating with respect to time:

89 Second–Order Circuits
Dampening coefficient Define: Undamped resonant frequency Forcing function

90 Solution of the Second-Order Equation

91 Solution of the Complementary Equation

92 Solution of the Complementary Equation
Roots of the characteristic equation: Dampening ratio

93 1. Overdamped case (ζ > 1). If ζ > 1 (or
equivalently, if α > ω0), the roots of the characteristic equation are real and distinct. Then the complementary solution is: In this case, we say that the circuit is overdamped.

94 2. Critically damped case (ζ = 1). If ζ = 1 (or
equivalently, if α = ω0 ), the roots are real and equal. Then the complementary solution is In this case, we say that the circuit is critically damped.

95 natural frequency is given by:
3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently, if α < ω0), the roots are complex. (By the term complex, we mean that the roots involve the square root of –1.) In other words, the roots are of the form: in which j is the square root of -1 and the natural frequency is given by:

96 For complex roots, the complementary solution is of the form:
In this case, we say that the circuit is underdamped.

97 Circuits with Parallel L and C
We can replace the circuit with it’s Norton equivalent and then analyze the circuit by writing KCL at the top node:

98 Circuits with Parallel L and C

99 Circuits with Parallel L and C

100 Circuits with Parallel L and C
This is the same equation as we found for the series LC circuit with the following changes for :

101 Complex Impedances-Inductor

102 Complex Impedances-Inductor

103 Complex Impedances-Capacitor

104 Complex Impedances-Capacitor

105 Impedances-Resistor

106 Impedances-Resistor

107 Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path. The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

108 Power in AC Circuits For >0 the load is called “inductive” since Z=jL for an inductor For <0 the load is “capacitive” since Z=-j/C for a capacitor

109 Load Impedance in the Complex Plane

110 Power for a General Load
If the phase angle for the voltage is not zero, we define the power angle : Power angle:

111 AC Power Calculations Average Power: Reactive Power: Apparent Power:

112 Power Triangles Average power Average power Reactive power
Apparent power

113 Thevenin Equivalent Circuits
The Thevenin equivalent for an ac circuit consists of a phasor voltage source Vt in series with a complex impedance Zt

114 Thevenin Equivalent Circuits
The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit. We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.

115 Thevenin Equivalent Circuits
The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.

116 Norton Equivalent Circuit
The Norton equivalent for an ac circuit consists of a phasor current source In in parallel with a complex impedance Zt

117 Maximum Average Power Transfer
The Thevenin equivalent of a two-terminal circuit delivering power to a load impedance.

118 Maximum Average Power Transfer
If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance. If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.

119 Transfer Functions The transfer function H(f ) of the two-port filter is defined to be the ratio of the phasor output voltage to the phasor input voltage as a function of frequency:

120 First-Order Low Pass Filter
Half power frequency

121 First-Order Low Pass Filter

122 First-Order Low Pass Filter
For low frequency signals the magnitude of the transfer function is unity and the phase is 0. Low frequency signals are passed while high frequency signals are attenuated and phase shifted.

123 Magnitude Bode Plot for First-Order Low Pass Filter

124 A horizontal line at zero for f < fB /10.
2. A sloping line from zero phase at fB /10 to –90° at 10fB. 3. A horizontal line at –90° for f > 10fB.

125 First-Order High-Pass Filter

126 First-Order High-Pass Filter

127 First-Order High-Pass Filter

128 Bode Plots for the First-Order High-Pass Filter

129 For resonance the reactance of the inductor and the capacitor cancel:
Series Resonance For resonance the reactance of the inductor and the capacitor cancel:

130 Series Resonance Quality factor QS

131 Series Resonant Band-Pass Filter

132 Series Resonant Band-Pass Filter

133 At resonance ZP is purely resistive:
Parallel Resonance At resonance ZP is purely resistive:

134 Parallel Resonance Quality factor QP

135 Parallel Resonance Vout for constant current, varying the frequency

136 Second Order Low-Pass Filter

137 Second Order Low-Pass Filter

138 Second Order Low-Pass Filter

139 Second Order High-Pass Filter
At low frequency the capacitor is an open circuit At high frequency the capacitor is a short and the inductor is open

140 Second Order Band-Pass Filter
At low frequency the capacitor is an open circuit At high frequency the inductor is an open circuit

141 Second Order Band-Reject Filter
At low frequency the capacitor is an open circuit At high frequency the inductor is an open circuit

142 First-Order Low-Pass Filter
A low-pass filter with a dc gain of -Rf/Ri

143 First-Order High-Pass Filter
A high-pass filter with a high frequency gain of -Rf/Ri

144 Flux Linkages and Faraday’s Law
Magnetic flux passing through a surface area A: For a constant magnetic flux density perpendicular to the surface: The flux linking a coil with N turns:

145 Faraday’s Law Faraday’s law of magnetic induction:
The voltage induced in a coil whenever its flux linkages are changing. Changes occur from: Magnetic field changing in time Coil moving relative to magnetic field

146 Lenz’s Law Lenz’s law states that the polarity of the induced voltage is such that the voltage would produce a current (through an external resistance) that opposes the original change in flux linkages.

147 Lenz’s Law

148 Magnetic Field Intensity and Ampère’s Law

149 Ampère’s Law The line integral of the magnetic field intensity around a closed path is equal to the sum of the currents flowing through the surface bounded by the path.

150 Magnetic Field Intensity and Ampère’s Law

151 Magnetic Circuits In many engineering applications, we need to compute the magnetic fields for structures that lack sufficient symmetry for straight-forward application of Ampère’s law. Then, we use an approximate method known as magnetic-circuit analysis.

152 magnetomotive force (mmf) of an N-turn current-carrying coil
Analog: Voltage (emf) reluctance of a path for magnetic flux Analog: Resistance Analog: Ohm’s Law

153 Magnetic Circuit for Toroidal Coil

154 A Magnetic Circuit with Reluctances in Series and Parallel

155 A Magnetic Circuit with Reluctances in Series and Parallel

156 Mutual inductance between coils 1 and 2:
Self inductance for coil 1 Self inductance for coil 2 Mutual inductance between coils 1 and 2:

157 Total fluxes linking the coils:
Mutual Inductance Total fluxes linking the coils:

158 Currents entering the dotted terminals produce aiding fluxes
Mutual Inductance Currents entering the dotted terminals produce aiding fluxes

159 Circuit Equations for Mutual Inductance

160 Ideal Transformers

161 Ideal Transformers

162 Mechanical Analog d1 d2

163 Impedance Transformations

164 Semiconductor Diode

165 Shockley Equation

166 Load-Line Analysis of Diode Circuits
Assume VSS and R are known. Find iD and vD

167 Load-Line Analysis of Diode Circuits

168 Ideal Diode Model The ideal diode acts as a short
circuit for forward currents and as an open circuit with reverse voltage applied. iD > 0 vD < 0  diode is in the “on” state vD < 0 ID = 0  diode is in the “off” state

169 Assumed States for Analysis of Ideal-Diode Circuits
1. Assume a state for each diode, either on (i.e., a short circuit) or off (i.e., an open circuit). For n diodes there are 2n possible combinations of diode states. 2. Analyze the circuit to determine the current through the diodes assumed to be on and the voltage across the diodes assumed to be off.

170 3. Check to see if the result is consistent with the assumed state for each diode. Current must flow in the forward direction for diodes assumed to be on. Furthermore, the voltage across the diodes assumed to be off must be positive at the cathode (i.e., reverse bias). 4. If the results are consistent with the assumed states, the analysis is finished. Otherwise, return to step 1 and choose a different combination of diode states.

171 Half-Wave Rectifier with Resistive Load
The diode is on during the positive half of the cycle. The diode is off during the negative half of the cycle.

172 Half-Wave Rectifier with Smoothing Capacitor
The charge removed from the capacitor in one cycle:

173 Full-Wave Rectifier The capacitance required for a full-wave rectifier is given by:

174 Full-Wave Rectifier

175 Clipper Circuit

176 NPN Bipolar Junction Transistor

177 Bias Conditions for PN Junctions
The base emitter p-n junction of an npn transistor is normally forward biased The base collector p-n junction of an npn transistor is normally reverse biased

178 Equations of Operation
From Kirchoff’s current law:

179 Equations of Operation
Define  as the ratio of collector current to emitter current: Values for  range from 0.9 to with 0.99 being typical. Since: Most of the emitter current comes from the collector and very little (1%) from the base.

180 Equations of Operation
Define  as the ratio of collector current to base current: Values for  range from about 10 to 1,000 with a common value being   100. The collector current is an amplified version of the base current.

181 The base region is very thin
Only a small fraction of the emitter current flows into the base provided that the collector-base junction is reverse biased and the base-emitter junction is forward biased.

182 Common-Emitter Characteristics
vBC vCE

183 Common-Emitter Input Characteristics

184 Common-Emitter Output Characteristics

185 Amplification by the BJT
A small change in vBE results in a large change in iB if the base emitter is forward biased. Provided vCE is more than a few tenth’s of a volt, this change in iB results in a larger change in iC since iC=iB.

186 Common-Emitter Amplifier

187 Load-Line Analysis of a Common Emitter Amplifier (Input Circuit)

188 Load-Line Analysis of a Common Emitter Amplifier (Output Circuit)

189 Inverting Amplifier As vin(t) goes positive, the load line moves upward and to the right, and the value of iB increases. This causes the operating point on the output to move upwards, decreasing vCE  An increase in vin(t) results in a much larger decrease in vCE so that the common emitter amplifier is an inverting amplifier

190 PNP Bipolar Junction Transistor
Except for reversal of current directions and voltage polarities, the pnp BJT is almost identical to the npn BJT.

191 PNP Bipolar Junction Transistor

192 NMOS Transistor

193 NMOS Transistor

194 Operation in the Cutoff Region

195 Operation Slightly Above Cut-Off
By applying a positive bias between the Gate (G) and the body (B), electrons are attracted to the gate to form a conducting n-type channel between the source and drain. The positive charge on the gate and the negative charge in the channel form a capacitor where:

196 Operation Slightly Above Cut-Off
For small values of vDS, iD is proportional to vDS. The device behaves as a resistance whose value depends on vGS.

197 Operation in the Triode Region

198

199 Load-Line Analysis of a Simple NMOS Circuit

200 Load-Line Analysis of a Simple NMOS Circuit

201 Load-Line Analysis of a Simple NMOS Circuit

202 CMOS Inverter

203 Two-Input CMOS NAND Gate

204 Two-Input CMOS NOR Gate


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