ECE 205: Intro Elec & Electr Circuits


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1 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Version 1.00 Created by Charles Feng
2 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 1 Contents 1 Introductory Matter International System of Units (SI) Current and Voltage Power DC Analysis Passive Elements (Resistors) Series Parallel Active Elements Ohm s Law Kirchoff s Laws Nodal Analysis Mesh Analysis Superposition Thevenin s and Norton s Theorems Transient Analysis Capacitors and Inductors FirstOrder Networks Step Response SecondOrder Networks Semiconductor Devices Diodes Simple Diode Model Zener Diode Metal Oxide Semiconductor Field Effect Transistors (MOSFET) nchannel MOSFET pchannel MOSFET Loads for Enhancement Drivers nmos Inverter with Resistive Load nmos Inverter with Depletion Load
3 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide CMOS Inverter Bipolar Junction Transistors (BJT) BJT Circuit Analysis ResistorTransistor Logic (RTL) DiodeTransistor Logic (DTL) TransistorTransistor Logic (TTL) Emitter Coupled Logic (ECL) Operational Amplifiers AC SteadyState Analysis Sinusoidal Signals Complex Numbers Phasors Impedance and Admittance Phasor Circuits AC Steady State Power
4 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 3 1 Introductory Matter 1.1 International System of Units (SI) Basic units these units represent the six fundamental concepts in physics. 1. Length: m (meter) 2. Time: s (second) 3. Mass: kg (kilogram) 4. Charge: C (coulomb) 5. Temperature: K (Kelvin) 6. Luminous intensity: cd (candela) Derived units these units can be represented as a combination of basic units but exist for simplicity. 1. Force: N (newton) 2. Energy: J (joule) 3. Power: W (watt) 4. Current: A (ampere) 5. Voltage: V (volt) 6. Resistance: Ω (ohm) 7. Conductance: S (siemens) 8. Capacitance: F (faraday) 9. Inductance: H (henry) 10. Frequency: Hz (hertz) 1.2 Current and Voltage Current is defined as the time rate of charge flow. Conventionally, it represents the motion of positive charges. It has both magnitude and direction. i a b In the above diagram, if i > 0, then the current is going from a to b. If i < 0, the current is going from b to a. The little arrow under the i is called the reference direction. The voltage of an element, also known as its potential difference, measures the work done in moving a unit charge through the element from one terminal to the other. Voltages have both magnitudes and polarities. V a b Terminal a is a voltage of V higher or lower than terminal b, or V = V a V b. If V is positive, then V a > V b ; if V is negative, then V b > V a. Whenever the reference direction for the current in an element is in the direction of the reference voltage drop across the element, use a positive sign in any express that relates v to i. Otherwise, use a negative sign. We will see an example of this when we look at power. 1.3 Power Power is defined as the rate at which energy is dissipated. It can be written in several ways, depending on what factors are known: p = vi = i 2 R = v2 R
5 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 4 Since the formula for power has both v and i in it, we must apply the passive sign convention. When the reference direction of the current is in the direction of the reference voltage drop, we will use the formula p = vi; when it is in the direction of the reference voltage rise, we will use the formula p = vi. If p > 0, that means that energy is dissipated; if p < 0, that means that energy is delivered. 2 DC Analysis 2.1 Passive Elements (Resistors) A passive element cannot deliver power or store energy. It can only absorb and dissipate energy. Examples of passive elements include resistors, inductors, and capacitors. Passive elements can be in series (same current), in which they share a common node that has no other currents entering or leaving it, or in parallel (same voltage), in which they form a loop containing no other elements. For two elements in series, the current is the same and the total voltage is the sum of the individual voltages of the elements. Two parallel elements have the same voltage and a total current of the sum of each of the parallel currents. To put it in a more understandable format, v s = v 1 + v 2, i s = i 1 = i 2, v p = v 1 = v 2, and i p = i 1 + i 1. Extreme cases for resistors are short circuit, when R = 0 and so v = Ri = 0, and open circuit, when R = and so i = v R 0. On the figure below, on left is a short circuit and on right is an open circuit. v = 0 i = 0 The maximum power transfer theorem states that if you have a voltage source in series with a resistor R and a variable load resistor R L, maximum power is delivered to the load resistor when R L = R Series R 1 R 2 a + v 1 + v 2 b Resistors in series: The total resistance is R s = R 1 R 2 v and v 2 = R 2 i = v. R 1 + R 2 R 1 + R 2 n R j. To find the voltage on each resistor, use voltage division: v 1 = R 1 i = j=1
6 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Parallel + v R 1 i 1 R 2 i 2 Resistors in parallel: i 1 = v R 1 = The total resistance is R 2 i and i 2 = v R 1 = i. R 1 + R 2 R 2 R 1 + R 2 1 R p = n j=1 1 R j. To find the current on each resistor, we use current division: 2.2 Active Elements Active elements are basically those that aren t passive. They include batteries, generators, etc. There are two main types of sources: voltage sources and current sources. Moreover, each of these two sources has two subtypes: independent and dependent. Independent sources are not affected by the voltage or current through the source; dependent sources are affected by either one. In the below diagram, from left to right: independent voltage source, independent current source, dependent voltage source, and dependent current source. Both dependent voltage and current sources can be controlled by either voltage or current through the source. 2.3 Ohm s Law Basically Ohm s Law states that v = Ri and i = Gv, where R is resistance and G is conductance. Passive sign convention must be used when applying Ohm s Law. 2.4 Kirchoff s Laws We will begin by listing several definitions: 1. A node is a point at which two or more circuit elements are connected. Note that nodes can include more than one vertex as long as there is nothing between the vertices. 2. A branch is a twoterminal circuit element connected between two nodes. 3. A loop is a closed path through the circuit in which no node is crossed more than once. 4. A mesh is a loop that does not contain with it another loop.
7 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 6 In the below diagram, d, e, f, g, h, j, and abc are nodes. There are seven resistors and two sources and so there are = 9 branches. The loops are abeda, abcjhgfda, et cetera; the meshes are abeda, degfd, and bcjhgeb. a b c R 1 i(t) R 6 d e v(t) R 2 R 3 j R 7 f g h R 4 R 5 Kirchoff s current law states that the algebraic sum of all the currents entering/leaving any node in a circuit is zero. If the reference direction of the current leaves the node, use a positive sign for the current; if it enters the node, use a negative sign. Kirchoff s voltage law states that the algebraic sum of all the voltages around any mesh or loop in a circuit is zero. If there is a reference voltage drop, use a positive sign; if there is a reference voltage rise, use a negative sign. 2.5 Nodal Analysis The reference node is chosen to be the node to which the largest number of branches is connected. It is usually referred to as ground and we will consider its voltage to be zero. The node voltage is defined as the voltage rise from the reference node to a nonreference node. For a circuit with n nodes, the number of independent node voltages is n 1. If the circuit only has current sources, do the following: 1. Find the number of nodes n. Choose one as the reference node as defined above. Label the node voltages with respect to the reference node. Also, label the branch currents. 2. Apply KCL at each of the nonreference nodes. 3. Express the currents in terms of node voltages using Ohm s law. 4. Solve fot he node voltages. If the circuit also has voltage sources, form supernodes. A supernode is a voltage source and its two terminals. Apply KCL at each of the nodes and supernodes except the one containing the reference node. 2.6 Mesh Analysis This type of analysis makes use of Kirchoff s voltage law. A mesh/loop current is a fictional current that exists only in the perimeter of a mesh/loop. mesh/loop, and all the elements in a mesh/loop have a common mesh/loop current. It flows continously around the The reference directions of the mesh/loop currents can be chosen arbitrarily; if the wrong direction is chosen then the current will turn out to be negative. The number of independent mesh/loop currents in a circuit is equal to B N + 1 where B is the number of branches and N is the number of nodes.
8 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 7 The branch current is the actual current through an element. It can be expressed in terms of the mesh/loop currents; if the mesh/loop currents overlap on the branch then the branch current is just their algebraic sum or difference, depending on their reference directions. For circuits containing only voltage sources, do the following: 1. Determine how many independent mesh/loop currents there are in the circuit. 2. Assign and label the mesh/loop currents and voltage across circuit elements. 3. Apply KVL for each mesh/loop to obtain (B N + 1) equations containing (B N + 1) independent mesh/loop currents. 4. Solve for the mesh/loop currents to find the branch currents and voltages. For circuits containing current sources as well, make sure you choose your mesh/loop currents so that there is only one current going through each current source. Then, apply KVL around the meshes/loops that don t contain current sources. 2.7 Superposition If there are multiple independent sources in a circuit, then the total current or voltage for any element is equal to the sum of all the individual currents or voltages produced by each independent source acting alone. To use superposition, do the following: 1. Choose one independent source. 2. Kill or zero all the other independent sources. For voltage sources, set v = 0 to make a short circuit. For current sources, set i = 0 to make an open circuit. 3. Analyze the circuit with one source. Find the current and voltage on each of the circuit elements. 4. Choose another source. Repeat steps (2) and (3). 5. Sum the individual currents and voltages caused by each source. Note that superposition can only be used if there are only independent sources in the circuit. It cannot be used for dependent sources. Moreover, we cannot use superposition directly to calculate power. 2.8 Thevenin s and Norton s Theorems The purpose of these theorems is to replace a linear circuit seen at a pair of terminals with an equivalent circuit containing only a single independent source and a single resistor with the intent of leaving the voltage and current at the terminals the same. Thevenin s theorem allows for a voltage source and a resistor in series; Norton s theorem allows for a current source and a resistor in parallel. In the figure below, on left is a Thevenin circuit and on right is a Norton circuit. R T a a v T i N R N b b We can easily convert between the Thevenin voltage v T and resistance R T and the Norton current i N and resistance R N using the following formulas: v T = R N i N and R T = R N.
9 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 8 The Thevenin voltage v T of a circuit is equal to v oc, the open circuit voltage across the two terminals a and b. The Norton current is equal to i sc, the short circuit current through a and b when the two terminals are shorted. Then, to find R T and R N, we can use one of two methods: (1) R T = R N = v T = v oc, or (2) R T = R N = the equivalent i N i sc resistance seen looking into the two terminals with all the independent sources in the circuit killed (shorting voltage sources and turning current sources into open circuits). Note that if there are dependent sources in the circuit, R T and R N must be derived using the first method. 3 Transient Analysis 3.1 Capacitors and Inductors A capacitor is a passive element that opposes change in voltage. The voltage across a capacitor cannot change instantaneously; it must be continuous. At steadystate, a capacitor acts like an open circuit no current flows through it and dv/dt = 0. For a capacitor, charge is related to voltage by the formula q = Cv where C is the capacitance, measured in farads. Current is related to voltage by the formula i = dq dt = C dv. This equation can be solved for v to find the voltage across a capacitor, which dt is v(t) = v t i(τ) dτ. Finally, the energy stored in a capacitor is W C (t) = 1 C 2 Cv2 (t) = 1 q2 qv = 2 2C. For capacitors in series, 0 1 C s = N j=1 1 C j and for series in parallel, C p = N C j. An inductor is a passive element that opposes change in current. The current across an inductor cannot change instantaneously and must be continuous. At steadystate, an inductor acts like a short circuit, with no voltage difference at its two terminals and di/dt = 0. In an inductor, flux (measured in Webers) is related to current by the formula λ = Li where L is the inductance, measured in henrys. Voltage is related to current by the formula v(t) = L di dt, which can be solved for i to get i(t) = i t v(τ) dτ. The L energy stored in an inductor is W L (t) = 1 2 Li2. For inductors in series, L s = N L j, and for inductors in parallel, j=1 j=1 1 L p = N j=1 1 L j. 0
10 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide FirstOrder Networks An RC circuit is a circuit containing a voltage source, a resistor, and a capacitor. By applying KVL, we see that dv C dt + 1 RC v C = v S RC. At steadystate, v C = v S. R v S i(t) C + v C (t) An RL circuit is a circuit containing a voltage source, a resistor, and an inductor. By applying KVL, we see that di L dt + R L i L = v S L. At steadystate, i L = v S R. R v S i (t) L L + v L (t) The general case for circuits containing one energystorage element is dx + ax = f(t) where a = 1/RC for RC circuits, a = R/L dt for RL circuits, f(t) is the forcing function, and X(t) = v C (t) or i L (t). The solution of this equation is called the complete response, or X(t) = X c (t) + X p (t). X c (t) is the complementary solution of the homogeneous equation and is called the natural or source free response. It occurs when f(t) = 0. Solving for X c, we get X c (t) = Ae t/τ where τ = 1/a. Note that τ = RC for an RC circuit and L/R for an RL circuit. X p (t) is a particular solution of the nonhomogeneous equation and is called the steady state or fixed response. Its formula is given by dx p dt + ax p = f(t). When f(t) is a constant, we get a special case where X p = K. Therefore, the complete response for f(t) being constant is: X(t) = K + Ae t/τ τ is refered to as the time constant, and τ = RC for an RC circuit and τ = L/R for an RL circuit. It is measured in seconds and is the time required for the natural response to decay by a factor of e. If τ is large, things happen slowly; if τ is small, things happen quickly. This gives us equations for RC and RL circuits: v C (t) = K + Ae t/rc i L (t) = K + Ae tr/l To solve the above equations, do v C ( ) which makes the equation v C = K which gives you K, and then do v C (0 + ) which makes the equation v C = K + A so you can solve for A. The same method applies also for i L (t). If there is more than one capacitor/inductor, try to reduce them if they are in series/parallel; otherwise, use node/mesh analysis to obtain a set of differential equations. If there is more than one resistor/source, use Thevenin s theorem to reduce the circuit into one source and one resistor.
11 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Step Response If v S (t) is equal to v 1 for t < 0 and v 2 for t > 0, then we come up to a problem called step response. For an RC circuit, our step response formulas are v C (t) = v 1 for t < 0 and v C (t) = v 2 (v 2 v 1 )e t/τ for t > 0. Also, i(t) = 0 for t < 0 and i(t) = v 2 v 1 R e t/τ for t > 0. For an RL circuit, our step response formulas are i L (t) = i 1 for t < 0 and i L (t) = i 2 (i 2 i 1 )e t/τ for t > 0. Also, v L (t) = 0 for t < 0 and i(t) = v 2 v 1 R e t/τ for t > SecondOrder Networks For a series RLC circuit containing a voltage source v S (t), a resistor R, an inductor L and a capacitor C, we have the formula 4 Semiconductor Devices 4.1 Diodes Simple Diode Model L d2 i dt 2 + R di dt + dv C dt The simple diode model states that when the voltage across the diode is less than the diode turnon voltage, v F 0.7 (for silicon diodes), the diode is off and current through it is zero (open circuit). When the current is above zero, then the voltage across the diode is equal to v F. v AC + = dv S dt To analyze diode circuits, do the following: i 1. Guess on/off for each diode. If on, then v AC = v F = 0.7 V. If off, then i = Analyze the circuit, Find i for each on diode and v AC for each off diode. 3. Look for contradictions: for example, if you guessed that a diode was on yet its current i 0; or if you guessed a diode was off but its v AC v F. 4. If there is a contradiction, modify your guess. The power dissipated by a diode is equal to p = v ac i = { 0.7i (on) 0 (off) There are a couple other models, the diode equation model, the ideal model, and the piecewise linear model (of which the simple diode model is a special case) but we won t go into detail on them Zener Diode Diodes can undergo what is called reverse breakdown in which the diode converts to a set voltage called the breakdown voltage, v R, which is independent of i. This condition happens when v AC = v R or v CA = v R. If the power dissipation is small, no
12 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 11 damage is done to the diode under breakdown conditions. For most common diodes, the breakdown voltage is 100 to 1000 V; however, for Zener diodes, v R = V. This special property of Zener diodes make it suitable for use in reverse breakdown conditions, in which they act as a constant voltage source or a regulator. They are used upsidedown to force the current backwards through the diode in order to take advantage of the constant v R. 4.2 Metal Oxide Semiconductor Field Effect Transistors (MOSFET) Transistors are a threeterminal device used for signal amplification (as opposed to diodes, which are used for rectifying, regulating, and switching). We will talk about two types of transistors: MOSFET (metal oxide semiconductor FET) and BJT (bipolar junction transistor) nchannel MOSFET i D i D D D G S G S i G i S i G i S Above left: enhancementmode nmos. Above right: depletionmode nmos. There are three currents going into and out of the transistor: the gate, i G, the drain, i D, and the source, i S. Between the drain and source is the base, but it is shorted with the source and so will be considered part of the latter. A MOSFET transistor has a certain threshold voltage denoted v T N. Enhancement mode MOSFETs always have v T N > 0. There are three different states the transistor can be in: 1. If v GS < v T N then the transistor is in cutoff mode and i D = If v T N v GS v DS + v T N then the transistor is on and in saturation mode for which i D = k N 2 (v GS v T N ) If v GS > v DS + v T N then the transistor is on and in nonsaturation or triode mode for which i D = k N 2 (2(v GS v T N ) v DS )v DS. In order to solve a MOSFET problem, firstly assume that it is in saturation and solve for i D. Then, check if v T N v GS v DS +v T N ; if not, change your guess and check again. For a depletionmode MOSFET, v T N < 0 and so when v GS = 0 current still flows. Depletionmode MOSFETs are usually only used as loads for enhancementmode MOSFETS.
13 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide pchannel MOSFET i G G S i S i G G S i S D D i D i D Above left: enhancementmode pmos. Above right: depletionmode pmos. pmos is very similar to nmos except that everything is backwards. Our three regions of operation are: 1. If v SG < v T P then the transistor is in cutoff mode and i D = If v T P v SG v SD + v T P then the transistor is on and in saturation mode for which i D = k P 2 (v SG v T P ) If v SG > v SD + v T P then the transistor is on and in nonsaturation or triode mode for which i D = k P 2 (2(v SG v T P ) v SD )v SD Loads for Enhancement Drivers Resistive Load Saturated Load NonSaturated Load Depletion Load v DD v DD v GG v DD v DD i L R L i D1 i D1 i D1 i D D v out i D2 v out i D2 v out i D2 v out G S G S G S G S i G i S i G i S i G i S i G i S Active loads (saturated, nonsaturated, depletion) are generally better than resistive loads because transistors take up less space than resistors in a circuit. Depletion load is the best because it doesn t cutoff at v DD v T N (as does the saturated load) nor does it require another power supply (as does the nonsaturated load). For all the loads, the driver equations are the same as outlined above for enhancementmode nmos systems. A resistive load is formed simply by an nmos logic and a resistor. The load equation i L = v DD v out R L. A saturated load is, as its name suggests, always saturated since v GS1 = v DS1 < v DS + v T N1. Its load equations are v GS1 = v DS1 = v DD v out and i DN1 = k N1 2 (v DD v out v T N1 ) 2.
14 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 13 A nonsaturated load is always nonsaturated since v GG v out > v DD v out + v T N1. Its load equations are v GS1 = v GG v out, v DS1 = v DD v out, and i DN1 = k N1 2 (2(v GG v out v T N1 )(v DD v out ) (v DD v out ) 2 ). A depletion load is formed by a depletionmode nmos in addition to the enhancement driver. v GS1 = 0 > v T N1, and its load equations are v DS1 = v DD v out and two equations for i DN1 : 1. If v out > v DD + v T N1 : i DN1 = k N1 2 (2(0 v T N1)(v DD v out ) (v D D v out ) 2 ) 2. If v DD + v T N1 > v out : i DN1 = k N1 2 (0 v T N1) 2 The load is always on since nmos Inverter with Resistive Load v out v out = v DD cutoff saturation v out = v TN2 linear = v TN2 As you can see in the above graph, there are three zones and two boundary points: 1. When < v T N2, the MOSFET is cutoff and v out = 12 V. 2. The first boundary point occurs when = v T N2 = 2 V. At this point, v out = 12 V. 3. When v out + v T N2 > > v T N2, the MOSFET is in saturation. In this zone, set i L = i DN2 in order to solve for v out in terms of. 4. The second boundary point occurs when = v out + v T N2. 5. When > v out + v T N2, the MOSFET is nonsaturated. Set i L = i DN2 in order to solve for v out in terms of.
15 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide nmos Inverter with Depletion Load v out v out = v TN1 v out = v out = v DD + v DD v TN1 T2 nonsat T2 sat T1 off T1 sat T1 nonsat = v TN1 There are four regions of operation: 1. When v T N1, T1 is off and T2 is nonsaturated. Furthermore, i D1 = 0 = i D2 and v out = v DD. 2. When v T N1 v out +v T N1 and v out v DD +v T N2, T1 is saturated and T2 is nonsaturated. Therefore, to solve for v out in terms of, write a quadratic equation with i D1 = i D2 using the respective formulas for saturation and nonsaturation. 3. When V T N1 v out + v T N1 and v T N1 v out v DD + v T N2, both T1 and T2 are saturated. 4. When v out > v T N1, T1 is nonsaturated and T2 is nonsaturated CMOS Inverter v DD G S T2 D i DP v out G D S i DN T1 A CMOS inverter contains both an nchannel and a pchannel MOSFET.
16 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 15 A CMOS inverter has five regions of operation, as outlined below: Region nmos driver pmos driver 1 cutoff nonsaturated 2 saturated nonsaturated 3 saturated saturated 4 nonsaturated saturated 5 nonsaturated cutoff v out v out = + v TP v out = v out = v DD + v DD v TN1 1 2 v out = v TN 3 = v TN 4 5 = v DD v TN 4.3 Bipolar Junction Transistors (BJT) C i C i B B E i E A BJT is a semiconductor device containing three doped regions of ptype and ntype materials, resulting in two pn junctions. There are two types of BJTs: npn and pnp; we will only discuss npn BJTs here. There are three different modes of operation for BJTs:
17 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 16 C C C B B B E Cutoff Forward Active Saturated E E When a BJT is cutoff, i C = i B = i E = 0 and V BE < V BE(on) = 0.7 V. When a BJT is forwardactive, i C = βi B, i E = i B + i C = (1 + β)i B, V BE = V BE(on) = 0.7 V, and V CE > V CE(sat) = 0.2 V. When a BJT is in saturation, i C βi B, i E = i B + i C (1 + β)i B, V BE = V BE(on) = 0.7 V, and V CE = V CE(sat) = 0.2 V. 4.4 BJT Circuit Analysis Our method is as follows: 1. Determine whether it is on or off by finding V BE. 2. Assume it is forwardactive. Then i C = βi B and i E = (1 + β)i B. Now apply KVL around the BE circuit and solve for i B to find i C andi E. Finally, apply KVL around the CE circuit to find V CE ; if it is greater than 0.2 V then our assumption is correct and the transistor is forwardactive. 3. If our assumption is not correct, the BJT is saturated. V CE = 0.2; we apply KVL around the BE circuit to find one equation in terms of i B and i E = i B + i C. Then we apply KVL around the CE circuit to find another equation in terms of i C and i E. We solve for i B and i C which in turn gives us i E. The power of a BJT is p = i B V BE + i C V CE. 4.5 ResistorTransistor Logic (RTL) In an RTL inverter, we have three different regions of operation: 1. When V in < 0.7 V, the BJT is cutoff and so i C = 0 and V out = V CC. 2. When V in 0.7 V then the BJT is on. Assume forwardactive and check if V CE > 0.2 V; if not, it is saturated. 4.6 DiodeTransistor Logic (DTL) This is a system comprised of diodes and BJTs. When solving these problems, remember the simple diode model and the BJT firstorder model. 4.7 TransistorTransistor Logic (TTL) This is a system comprised of only BJTs.
18 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Emitter Coupled Logic (ECL) This is fast because none of the transistors are ever saturated; they are all either in cutoff or are forwardactive. There are three sections of an ECL gate: the input (current steering network), the output (emitter follower), and a constant current source. Basically, an ECL works by steering a current from the constant current source through either the left side or the right side of the input section, depending on whether the input voltage is above or below a reference voltage. The emitter follower output section provides an output that is always V BE(on) less than the collector voltage in the input section. Usually ECLs are made with two transistors, one of which will be cutoff and one which will be forwardactive depending on the input voltage. 4.9 Operational Amplifiers i V+ V out i V Essentially all we have to know is the ideal op amp model, in which the current going into the op amp (i) is equal to the current going out which is equal to zero (virtual open principle) and V + = V (virtual short principle). Based on these rules, we can analyze a variety of op amp circuits. 5 AC SteadyState Analysis 5.1 Sinusoidal Signals Any periodic function can be presented by a sum of sinusoidals. The sinusoidal functions for voltage and current are the following: v(t) = V m cos (ωt + θ v ) i(t) = I m cos (ωt + θ i ) The period of the above functions is T = 2π ω and the frequency is 1/T. To convert between sine and cosine, remember the following equations: sin (ωt + 90 ) = cos (ωt) sin (ωt ± 180 ) = sin (ωt) cos (ωt 90 ) = sin (ωt) cos (ωt ± 180 ) = cos (ωt) A sinusoidal function may be represented in the form A cos (ωt) + B sin (ωt). In order to convert this into our standard form, use the formula ( ) B A 2 + B 2 cos (ωt θ), where θ = tan 1 and the quadrant of θ is determined by the location of the point A (A, B). 5.2 Complex Numbers One can represent complex numbers in one of three forms: 1. Rectangular: A = a + jb where j = 1
19 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Exponential polar: A = re jθ 3. Polar: A = r θ The real part of a complex number is equal to a = R(A) = r cos (θ). The imaginary part is equal to b = I(A) = r sin (θ). The magnitude is r = A = ( ) b a 2 + b 2 and the argument is θ = tan 1. a 5.3 Phasors A phasor is a complex number whose magnitude is the amplitude of the sinusoid and whose angle is the angle of the sinusoid. The two sinusoids we are worrying about are and their phasors are v = V m cos (ωt + θ) = R(V m e jθv e jωt ) i = I m cos (ωt + θ) = R(I m e jθi e jωt ) ( ) ( ) V = Vm θ v v = R(V e or jωt ) I = I m θ i i = R(Ie jωt. ) Note that phasors are defined by the cosine function so if a sine function is given we must convert it to cosine using one of the rules seen in the previous section. For resistors, our equation is V = RI or V m θ v = RI m θ i. V and I are in phase, or θ v = θ i. For inductors, our equation is V = jωli or V m θ v = ωli m θ i V leads I by 90, or θ v = θ i For capacitors, or equation is I = jωcv, or I m θ i = ωcv m θ i V lags I by 90, or θ v = θ i Impedance and Admittance Impedance is defined as the opposition to a sinusoidal electric current. It is equal to Z = V I. In its polar form, Z = Z the total reactance. θ z = V m I m θ v θ i. In rectangular form, we have Z = R + jx where R is the total resistance and X is We have four formulas to convert between polar and rectangular forms: R = Z cos (θ z ), X = Z sin (θ z ), Z = ( ) R 2 + X 2, and X θ z = tan 1. R The impedances of resistors, inductors, and capacitors are defined as follows: 1. Resistor: V = RI, Z R = R (real number) 2. Inductor: V = jωli, Z L = jωl = jx L (imaginary number) 3. Capacitor: I = jωcv, Z C = 1 jωc = jx C (imagine number) X L and X C are the inductive reactance and capacitive reactance respectively. In general, Z is a function of ω. The unit of impedance is Ω if we use H for inductors and F for capacitors. The admittance is defined as Y = 1 Z = I V = Y θ Y and is measured in siemens. When put into rectangular form, we get the expression Y = G + jb where G is the conductance and B is the susceptance.
20 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 19 In series, the total impedance is equal to the sum of the individual impedances: Z s = equal to the harmonic mean of the individual impedances: We can also use voltage division: And current division (where Y = 1 Z ): 1 Z p = V 1 = n j=1 1 Z j. Z 1 Z 1 + Z 2 V n Z j. In parallel, the total impedance is j=1 I 1 = Y 1 Y 1 + Y 2 I = Z 2 Z 1 + Z 2 I I 2 = Y 2 Y 1 + Y 2 I = Z 1 Z 1 + Z 2 I 5.5 Phasor Circuits To solve a phasor circuit, do the following: 1. Replace resistors, inductors, and capacitors with their impedance equivalents in terms of ω. Replace v and i with the phasors V and I. This step yields a phasor circuit. 2. Write the phasor equation using KCL, KVL, etc. 3. Solve for unknown phasors. 4. Convert phasors to timedomain sinusoidal responses. Note that you can use KCL, KVL, nodal analysis, mesh/loop analysis, and superposition to solve phasor circuits. In circuits with sources of multiple frequencies, you must apply superposition, then find the phasor response for each frequency, and finally convert the phasor responses to sinusoidal responses and add them to get the total response. If the frequency is given in hertz, the units for ω is radians per second and therefore ω = ν where ν is the hertz frequency of the 2π source. Thevenin s and Norton s theorems can be applied to a phasor circuit as long as there is only one radian frequency ω. In this case, V T = V oc, I N = I sc, and Z T = Z N = V oc I sc. 5.6 AC Steady State Power The instantaneous power of a phasor circuit is defined as p(t) = v(t) i(t) = V m I m cos (ωt + θ v ) cos (ωt + θ i ) = V mi m (cos (θ v θ i ) + cos (2ωt + θ v + θ i )) 2
21 ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 20 The average power of a phasor circuit is equal to P = 1 T t0+t t 0 = V mi m 2 = V mi m 2 p(t) dt cos (θ v θ i ) cos θ z = 1 V I cos (arg V arg I) 2 = 1 2 I2 m Z cos θ z = 1 2 Im2 R(Z) Note that there are special equations for average power for resistors, inductors and capacitors: 1. Resistor: P R = 1 2 RI2 m = V 2 m 2R m 2. Inductor: P L = 0 3. Capacitor: P C = 0 Ideal capacitors and inductors are lossless elements. We can also find rootmeansquare (RMS) or effective values for our periodic wave forms. For a DC circuit, P = Ieff 2 R = V eff 2 R circuit: P = 1 t0+t i 2 R dt = 1 t0+t v 2 T T t 0 For an AC circuit, I eff = I rms = t 0 since the current and voltage do not fluctuate. Similarly, we can set up integrals for an AC R dt. 1 t0+t i T 2 dt and V eff = V rms = t 0 For i(t) = I m cos (ωt + θ i ), I rms = Im 2 which is independent of ω or θ i. 1 t0+t v T 2 dt. t 0 The average power of a circuit is equal to P = V mi m 2 cos (θ v θ i ) = V rms I rms cos (θ v θ i ).
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