Power Diodes and Switched RLC Circuits PDF

Part I Power Diodes and Rectifiers C h a p t e r 2 Power Diodes and Switched RLC Circuits After completing this chapter, students should be able to do the following: Explain th...

Part I Power Diodes and Rectifiers C h a p t e r 2 Power Diodes and Switched RLC Circuits After completing this chapter, students should be able to do the following: Explain the operating principal of power diodes. Describe the diode characteristics and its circuit models. List the types of power diodes. Explain the series and parallel operation of diodes. Derive the SPICE diode model. Explain the reverse recovery characteristics of power diodes. Calculate the reverse recovery current of diodes. Calculate the steady-state capacitor voltages of an RC circuit and the amount of stored energy. Calculate the steady-state inductor currents of an RL circuit and the amount of stored energy. Calculate the steady-state capacitor voltages of an LC circuit and the amount of stored energy. Calculate the steady-state capacitor voltage of an RLC circuit and the amount of stored energy. Determine the initial di/dt and dv/dt of RLC circuits. Symbols and Their Meanings Symbol Meaning iD, vD Instantaneous diode current and voltage, respectively i(t), iS(t) Instantaneous current and supply current, respectively ID, VD Dc diode current and voltage, respectively IS Leakage (or reverse saturation) current IO Steady-state output current IS1, IS2 Leakage (or reverse saturation) currents of diodes D1 and D2, respectively IRR Reverse recovery current trr Reverse recovery time VT Thermal voltage (continued ) 59 M02_RASH9088_04_PIE_C02.indd 59 31/07/13 3:53 PM  60   Chapter 2     Power Diodes and Switched RLC Circuits Symbol Meaning VD1, VD2 Voltage drops across diodes D1 and D2, respectively VBR, VRM Reverse breakdown and maximum repetitive voltages, respectively vR, vC, vL Instantaneous voltages across a resistor, a capacitor, and an inductor, respectively VCO, vs, VS Initial capacitor, instantaneous supply, and dc supply voltages, respectively QRR Reverse storage charge T Time constant of a circuit n Empirical emission constant 2.1 Introduction Many applications have been found for diodes in electronics and electrical engineering circuits. Power diodes play a significant role in power electronics circuits for conver- sion of electric power. Some diode circuits that are commonly encountered in power electronics for power processing are reviewed in this chapter. A diode acts as a switch to perform various functions, such as switches in rec- tifiers, freewheeling in switching regulators, charge reversal of capacitor and energy transfer between components, voltage isolation, energy feedback from the load to the power source, and trapped energy recovery. Power diodes can be assumed as ideal switches for most applications but practi- cal diodes differ from the ideal characteristics and have certain limitations. The power diodes are similar to pn-junction signal diodes. However, the power diodes have larger power-, voltage-, and current-handling capabilities than those of ordinary sig- nal ­diodes. The frequency response (or switching speed) is low compared with that of signal diodes. Inductors L and capacitors C, which are the energy storage elements, are com- monly used in power electronics circuits. A power semiconductor device is used to control the amount of energy transfer in a circuit. A clear understanding of the switch- ing behaviors of RC, RL, LC, and RLC circuits are prerequisites for understanding the operation of power electronics circuits and systems. In this chapter, we will use a diode connected in series with a switch to exhibit the characteristics of a power device and analyze switching circuits consisting of R, L, and C. The diode allows a unidirectional current flow and the switch performs the on and off functions. 2.2 Semiconductor Basics Power semiconductor devices are based on high-purity, single-crystal silicon. Single crystals of several meters long and with the required diameter (up to 150 mm) are grown in the so-called float zone furnaces. Each huge crystal is sliced into thin wafers, which then go through numerous process steps to turn into power devices. M02_RASH9088_04_PIE_C02.indd 60 31/07/13 3:53 PM  2.2  Semiconductor Basics   61 Table 2.1   A Portion of the Periodic Table Showing Elements Used in Semiconductor Materials Group Period II III IV V VI 2 B C N O Boron Carbon Nitrogen Oxygen 3 Al Si P S Aluminum Silicon Phosphorus Sulfur 4 Zn Ga Ge As Se Zinc Gallium Germanium Arsenic Selenium 5 Cd In Sn Sn Te Cadmium Indium Tin Antimony Tellurium 6 Hg Mercury Elementary Si Semiconductors Silicon Ge, Germanium Compound SiC GaAs Semiconductors Silicon carbide Gallium arsenide SiGe Silicon germanium The most commonly used semiconductors are silicon and germanium (Group IV in the periodic table as shown in Table 2.1) and gallium arsenide (Group V). Silicon materials cost less than germanium materials and allow diodes to operate at higher temperatures. For this reason, germanium diodes are rarely used. Silicon is a member of Group IV of the periodic table of elements, that is, hav- ing four electrons per atom in its outer orbit. A pure silicon material is known as an intrinsic semiconductor with resistivity that is too low to be an insulator and too high to be a conductor. It has high resistivity and very high dielectric strength (over 200 kV/ cm). The resistivity of an intrinsic semiconductor and its charge carriers that are avail- able for conduction can be changed, shaped in layers, and graded by implantation of specific impurities. The process of adding impurities is called doping, which involves a single atom of the added impurity per over a million silicon atoms. With different impurities, levels and shapes of doping, high technology of photolithography, laser cut- ting, etching, insulation, and packaging, the finished power devices are produced from various structures of n-type and p-type semiconductor layers. n-type material: If pure silicon is doped with a small amount of a Group V ele- ment, such as phosphorus, arsenic, or antimony, each atom of the dopant forms a covalent bond within the silicon lattice, leaving a loose electron. These loose electrons greatly increase the conductivity of the material. When the silicon is lightly doped with an impurity such as phosphorus, the doping is denoted as n-doping and the resultant material is referred to as n-type semiconductor. When it is heavily doped, it is denoted as n + doping and the material is referred to as n + -type semiconductor. M02_RASH9088_04_PIE_C02.indd 61 31/07/13 3:53 PM  62   Chapter 2     Power Diodes and Switched RLC Circuits p-type material: If pure silicon is doped with a small amount of a Group III ­element, such as boron, gallium, or indium, a vacant location called a hole is in- troduced into the silicon lattice. Analogous to an electron, a hole can be consid- ered a mobile charge carrier as it can be filled by an adjacent electron, which in this way leaves a hole behind. These holes greatly increase the conductivity of the material. When the silicon is lightly doped with an impurity such as boron, the doping is denoted as p-doping and the resultant material is referred to as ­p-type semiconductor. When it is heavily doped, it is denoted as p + doping and the ­material is referred to as p + -type semiconductor. Therefore, there are free electrons available in an n-type material and free holes avail- able in a p-type material. In a p-type material, the holes are called the majority carriers and electrons are called the minority carriers. In the n-type material, the electrons are called the majority carriers and holes are called the minority carriers. These carriers are continuously generated by thermal agitations, they combine and recombine in ac- cordance to their lifetime, and they achieve an equilibrium density of carriers from about 1010 to 1013/cm3 over a range of about 0°C to 1000°C. Thus, an applied electric field can cause a current flow in an n-type or p-type material. Silicon carbide (SiC) (compound material in Group IV of the periodic table) is a promising new material for high-power/high-temperature applications. SiC has a high bandgap, which is the energy needed to excite electrons from the material’s va- lence band into the conduction band. Silicon carbide electrons need about three times as much energy to reach the conduction band as compared to silicon. As a result, SiC- based devices withstand far higher voltages and temperatures than their silicon counter- parts. Silicon devices, for example, can’t withstand electric fields in excess of about 300 kV/cm. Because electrons in SiC require more energy to be pushed into the conduction band, the material can withstand much stronger electric fields, up to about 10 times the maximum for silicon. As a result, an SiC-based device can have the same dimensions as a silicon device but can withstand 10 times the voltage. Also, an SiC device can be less than a tenth the thickness of a silicon device but carry the same voltage rating. These thinner devices are faster and have less resistance, which means less energy is lost to heat when a silicon carbide diode or transistor is conducting electricity. Key Points of Section 2.2 Free electrons or holes are made available by adding impurities to the pure sili- con or germanium through a doping process. The electrons are the majority car- riers in the n-type material whereas the holes are the majority carriers in a p-type material. Thus, the application of electric field can cause a current flow in an n-type or a p-type material. 2.3 Diode Characteristics A power diode is a two-terminal pn-junction device [1, 2] and a pn-junction is normally formed by alloying, diffusion, and epitaxial growth. The modern control techniques in diffusion and epitaxial processes permit the desired device characteristics. Figure 2.1 shows the sectional view of a pn-junction and diode symbol. M02_RASH9088_04_PIE_C02.indd 62 31/07/13 3:53 PM  2.3  Diode Characteristics   63 Anode Cathode Anode Cathode p n iD iD D1 vD vD     Figure 2.1 (a) pn-junction (b) Diode symbol pn-junction and diode symbol. When the anode potential is positive with respect to the cathode, the diode is said to be forward biased and the diode conducts. A conducting diode has a relatively small forward voltage drop across it; the magnitude of this drop depends on the manu- facturing process and junction temperature. When the cathode potential is positive with respect to the anode, the diode is said to be reverse biased. Under reverse-biased conditions, a small reverse current (also known as leakage current) in the range of micro- or milliampere flows and this leakage current increases slowly in magnitude with the reverse voltage until the avalanche or zener voltage is reached. Figure 2.2a shows the steady-state v-i characteristics of a diode. For most practical purposes, a diode can be regarded as an ideal switch, whose characteristics are shown in Figure 2.2b. The v-i characteristics shown in Figure 2.2a can be expressed by an equation known as Schockley diode equation, and it is given under dc steady-state operation by ID = IS 1e VD/nVT - 12 (2.1) where ID = current through the diode, A; VD = diode voltage with anode positive with respect to cathode, V; IS = leakage (or reverse saturation) current, typically in the range 10-6    to 10-15 A; n = empirical constant known as emission coefficient, or ideality factor,      whose value varies from 1 to 2. The emission coefficient n depends on the material and the physical construction of the diode. For germanium diodes, n is considered to be 1. For silicon diodes, the pre- dicted value of n is 2, but for most practical silicon diodes, the value of n falls in the range 1.1 to 1.8. iD iD ID VBR VD 0 vD 0 vD Reverse leakage current (a) Practical (b) Ideal Figure 2.2 v - i characteristics of a diode. M02_RASH9088_04_PIE_C02.indd 63 31/07/13 3:53 PM  64   Chapter 2     Power Diodes and Switched RLC Circuits VT in Eq. (2.1) is a constant called thermal voltage and it is given by kT VT = (2.2) q where q = electron charge: 1.6022 * 10-19 coulomb 1C 2 ; T = absolute temperature in Kelvin 1K = 273 + °C 2 ; k = Boltzmann>s constant: 1.3806 * 10-23 J/K. At a junction temperature of 25°C, Eq. (2.2) gives kT 1.3806 * 10-23 * 1273 + 252 VT = = ≈ 25.7 mV q 1.6022 * 10-19 At a specified temperature, the leakage current IS is a constant for a given diode. The diode characteristic of Figure 2.2a can be divided into three regions: Forward-biased region, where VD 7 0 Reverse-biased region, where VD 6 0 Breakdown region, where VD 6 -VBR Forward-biased region. In the forward-biased region, VD 7 0. The diode cur- rent ID is very small if the diode voltage VD is less than a specific value VTD (typically 0.7 V). The diode conducts fully if VD is higher than this value VTD, which is referred to as the threshold voltage, cut-in voltage, or turn-on voltage. Thus, the threshold volt- age is a voltage at which the diode conducts fully. Let us consider a small diode voltage VD = 0.1 V, n = 1, and VT = 25.7 mV. From Eq. (2.1) we can find the corresponding diode current ID as ID = IS 1e VD/nVT - 12 = IS[e 0.1/11 * 0.02572 - 1] = IS 148.96 - 12 = 47.96 IS which can be approximated ID ≈ IS e VD/nVT = 48.96 IS, that is, with an error of 2.1%. As vD increases, the error decreases rapidly. Therefore, for VD 7 0.1 V, which is usually the case, ID W IS, and Eq. (2.1) can be approximated within 2.1% error to ID = IS 1e VD/nVT - 12 ≈ IS e VD/nVT (2.3) Reverse-biased region. In the reverse-biased region, VD 6 0. If VD is negative and  VD  W VT , which occurs for VD 6 -0.1 V, the exponential term in Eq. (2.1) becomes negligibly small compared with unity and the diode current ID becomes ID = IS 1e -VD|/nVT - 12 ≈ -IS (2.4) which indicates that the diode current ID in the reverse direction is constant and equals IS. Breakdown region. In the breakdown region, the reverse voltage is high, usu- ally with a magnitude greater than 1000 V. The magnitude of the reverse voltage may exceed a specified voltage known as the breakdown voltage VBR. With a small change M02_RASH9088_04_PIE_C02.indd 64 31/07/13 3:53 PM  2.4  Reverse Recovery Characteristics   65 in reverse voltage beyond VBR, the reverse current increases rapidly. The operation in the breakdown region will not be destructive, provided that the power dissipation is within a “safe level” that is specified in the manufacturer’s data sheet. However, it is often necessary to limit the reverse current in the breakdown region to limit the power dissipation within a permissible value. Example 2.1 Finding the Saturation Current The forward voltage drop of a power diode is VD = 1.2 V at ID = 300 A. Assuming that n = 2 and VT = 25.7 mV, find the reverse saturation current IS. Solution Applying Eq. (2.1), we can find the leakage (or saturation) current IS from -3 2 300 = IS[e 1.2/12 * 25.7 * 10 - 1] which gives IS = 2.17746 * 10-8 A. Key Points of Section 2.3 A diode exhibits a nonlinear v-i characteristic, consisting of three regions: forward biased, reverse-biased, and breakdown. In the forward condition the diode drop is small, typically 0.7 V. If the reverse voltage exceeds the breakdown voltage, the diode may be damaged. 2.4 Reverse Recovery Characteristics The current in a forward-biased junction diode is due to the net effect of majority and minority carriers. Once a diode is in a forward conduction mode and then its forward current is reduced to zero (due to the natural behavior of the diode circuit or applica- tion of a reverse voltage), the diode continues to conduct due to minority carriers that remain stored in the pn-junction and the bulk semiconductor material. The minority carriers require a certain time to recombine with opposite charges and to be neutral- ized. This time is called the reverse recovery time of the diode. Figure 2.3 shows two reverse recovery characteristics of junction diodes. It should be noted that the recov- ery curves in Figure 2.3 are not scaled and indicate only their shapes. The tailing of the recovery period is expanded to illustrate the nature of recovery although in reality t a 7 t b. The recovery process starts at t = t 0 when the diode current starts to fall from the on-state current IF at a rate of di/dt = -IF/1 t 1 - t 0 2. The diode is still conducting with a forward voltage drop of VF. The forward current IF falls to zero at t = t 1 and then continues to flow in the reverse direction because the diode is inactive and not capable of blocking the reverse current flow. At t = t 2 , the reverse current reaches a value of IRR and the diode voltage starts to reverse. After the recovery process is completed at t = t 3 , the reverse diode voltage reaches a peak of VRMS. The diode voltage passes through a transient oscillation period to complete the stored charge recovery until M02_RASH9088_04_PIE_C02.indd 65 31/07/13 3:53 PM  66   Chapter 2     Power Diodes and Switched RLC Circuits IF trr IF trr ta VF VF ta t2 t2 t t 0.25 IRR 0 t0 t1 Q1 Q2 0 t0 t1 IRR IRR tb tb VRM VRM (a) Soft recovery (b) Abrupt recovery Figure 2.3 Reverse recovery characteristics. it falls to its normal reverse operating voltage. The complete process is nonlinear and Figure 2.3 is used only to illustrate the process. There are two types of recovery: soft and hard (or abrupt). The soft-recovery type is more common. The reverse recovery time is denoted as trr and is measured from the initial zero cross- ing of the diode current to 25% of maximum (or peak) reverse current IRR. The trr consists of two components, ta and tb. Variable ta is due to charge storage in the depletion region of the junction and represents the time between the zero crossing and the peak reverse current IRR. The tb is due to charge storage in the bulk semi- conductor material. The ratio tb/ta is known as the softness factor (SF). For practi- cal purposes, one needs be concerned with the total recovery time trr and the peak value of the reverse current IRR. t rr = t a + t b (2.5) The peak reverse current can be expressed in reverse di/dt as di IRR = t a (2.6) dt Reverse recovery time trr may be defined as the time interval between the instant the current passes through zero during the changeover from forward conduction to reverse blocking condition and the moment the reverse current has decayed to 25% of its peak reverse value IRR. Variable trr is dependent on the junction temperature, rate of fall of forward current, and forward current prior to commutation, IF. Reverse recovery charge QRR is the amount of charge carriers that flows across the diode in the reverse direction due to changeover from forward conduction to re- verse blocking condition. Its value is determined from the area enclosed by the curve of the reverse recovery current. That is, QRR = Q1 + Q2. The storage charge, which is the area enclosed by the curve of the recovery current, is approximately 1 1 1 QRR = Q1 + Q2 ≅ IRRt a + IRRt b = IRRt rr (2.7) 2 2 2 M02_RASH9088_04_PIE_C02.indd 66 31/07/13 3:53 PM  2.4  Reverse Recovery Characteristics   67 or 2QRR IRR ≅ (2.8) t rr Equating IRR in Eq. (2.6) to IRR in Eq. (2.8) gives 2QRR t rrt a = (2.9) di/dt If tb is negligible as compared to ta, which is usually the case (although Figure 2.3a ­ epicts t b 7 t a), t rr ≈ t a, and Eq. (2.9) becomes d 2QRR t rr ≅ (2.10) C di/dt and di IRR = 2QRR (2.11) C dt It can be noticed from Eqs. (2.10) and (2.11) that the reverse recovery time trr and the peak reverse recovery current IRR depend on the storage charge QRR and the reverse (or reapplied) di/dt. The storage charge is dependent on the forward diode current IF. The peak reverse recovery current IRR, reverse charge QRR, and the SF are all of interest to the circuit designer, and these parameters are commonly included in the specifica- tion sheets of diodes. If a diode is in a reverse-biased condition, a leakage current flows due to the mi- nority carriers. Then the application of forward voltage would force the diode to carry current in the forward direction. However, it requires a certain time known as forward recovery (or turn-on) time before all the majority carriers over the whole junction can contribute to the current flow. If the rate of rise of the forward current is high and the forward current is concentrated to a small area of the junction, the diode may fail. Thus, the forward recovery time limits the rate of the rise of the forward current and the switching speed. Example 2.2 Finding the Reverse Recovery Current The reverse recovery time of a diode is t rr = 3 μs and the rate of fall of the diode current is di/dt = 30 A/μs. Determine (a) the storage charge QRR, and (b) the peak reverse current IRR. Solution t rr = 3 μs and di/dt = 30 A/μs. a. From Eq. (2.10), 1 di 2 QRR = t = 0.5 * 30A/μs * 13 * 10-6 2 2 = 135 μC 2 dt rr M02_RASH9088_04_PIE_C02.indd 67 31/07/13 3:53 PM  68   Chapter 2     Power Diodes and Switched RLC Circuits b. From Eq. (2.11), di IRR = 2QRR = 22 * 135 * 10-6 * 30 * 106 = 90 A C dt Key Points of Section 2.4 During the reverse recovery time trr, the diode behaves effectively as a short cir- cuit and is not capable of blocking reverse voltage, allowing reverse current flow, and then suddenly disrupting the current. Parameter trr is important for switching applications. 2.5 Power Diode Types Ideally, a diode should have no reverse recovery time. However, the manufacturing cost of such a diode may increase. In many applications, the effects of reverse recovery time is not significant, and inexpensive diodes can be used. Depending on the recovery characteristics and manufacturing techniques, the power diodes can be classified into the following three categories: 1. Standard or general-purpose diodes 2. Fast-recovery diodes 3. Schottky diodes General-purpose diodes are available up to 6000 V, 4500 A, and the rating of fast- recovery diodes can go up to 6000 V, 1100 A. The reverse recovery time varies between 0.1 μs and 5 μs. The fast-recovery diodes are essential for high-frequency switching of power converters. Schottky diodes have a low on-state voltage and a very small re- covery time, typically in nanoseconds. The leakage current increases with the voltage rating and their ratings are limited to 100 V, 300 A. A diode conducts when its anode voltage is higher than that of the cathode; and the forward voltage drop of a power diode is very low, typically 0.5 V to 1.2 V. The characteristics and practical limitations of these types restrict their applications. 2.5.1 General-Purpose Diodes The general-purpose rectifier diodes have relatively high reverse recovery time, typi- cally 25 μs; and are used in low-speed applications, where recovery time is not critical (e.g., diode rectifiers and converters for a low-input frequency up to 1-kHz applica- tions and line-commutated converters). These diodes cover current ratings from less than 1 A to several thousands of amperes, with voltage ratings from 50 V to around 5 kV. These diodes are generally manufactured by diffusion. However, alloyed types of rectifiers that are used in welding power supplies are most cost-effective and rugged, and their ratings can go up to 1500 V, 400 A. M02_RASH9088_04_PIE_C02.indd 68 31/07/13 3:53 PM  2.5   Power Diode Types   69 Figure 2.4 Various general-purpose diode configura- tions. (Courtesy of Powerex, Inc.) Figure 2.4 shows various configurations of general-purpose diodes, which basi- cally fall into two types. One is called a stud, or stud-mounted type; the other is called a disk, press pak, or hockey-puck type. In a stud-mounted type, either the anode or the cathode could be the stud. 2.5.2 Fast-Recovery Diodes The fast-recovery diodes have low recovery time, normally less than 5 μs. They are used in dc–dc and dc–ac converter circuits, where the speed of recovery is often of critical importance. These diodes cover current ratings of voltage from 50 V to around 3 kV, and from less than 1 A to hundreds of amperes. For voltage ratings above 400 V, fast-recovery diodes are generally made by diffu- sion and the recovery time is controlled by platinum or gold diffusion. For voltage rat- ings below 400 V, epitaxial diodes provide faster switching speeds than those of diffused diodes. The epitaxial diodes have a narrow base width, resulting in a fast recovery time of as low as 50 ns. Fast-recovery diodes of various sizes are shown in Figure 2.5. Figure 2.5 Fast-recovery diodes. (Courtesy of Powerex, Inc.) M02_RASH9088_04_PIE_C02.indd 69 31/07/13 3:53 PM  70   Chapter 2     Power Diodes and Switched RLC Circuits 2.5.3 Schottky Diodes The charge storage problem of a pn-junction can be eliminated (or minimized) in a Schottky diode. It is accomplished by setting up a “barrier potential” with a contact between a metal and a semiconductor. A layer of metal is deposited on a thin epitaxial layer of n-type silicon. The potential barrier simulates the behavior of a pn-junction. The rectifying action depends on the majority carriers only, and as a result there are no excess minority carriers to recombine. The recovery effect is due solely to the self- capacitance of the semiconductor junction. The recovered charge of a Schottky diode is much less than that of an equivalent pn- junction diode. Because it is due only to the junction capacitance, it is largely ­independent of the reverse di/dt. A Schottky diode has a relatively low forward ­voltage drop. The leakage current of a Schottky diode is higher than that of a pn-junction diode. A Schottky diode with relatively low-conduction voltage has relatively high leakage current, and vice versa. As a result, the maximum allowable voltage of this diode is generally limited to 100 V. The current ratings of Schottky diodes vary from 1 to 400 A. The Schottky diodes are ideal for high-current and low-voltage dc power supplies. However, these diodes are also used in low-current power supplies for increased ­efficiency. In Figure 2.6, 20- and 30-A dual Schottky rectifiers are shown. Key Points of Section 2.5 Depending on the switching recovery time and the on-state drop, the power ­diodes are of three types: general purpose, fast recovery, and Schottky. Figure 2.6 Dual Schottky center rectifiers of 20 and 30 A. 2.6 Silicon Carbide Diodes Silicon carbide (SiC) is a new material for power electronics. Its physical properties outperform Si and GaAs by far. For example, the Schottky SiC diodes manufactured by Infineon Technologies have ultralow power losses and high reliability. They also have the following features: No reverse recovery time; Ultrafast switching behavior; No temperature influence on the switching behavior. M02_RASH9088_04_PIE_C02.indd 70 31/07/13 3:53 PM  2.7   Silicon Carbide Schottky Diodes   71 iD IF SiC t Si Figure 2.7 Comparison of reverse recovery time. The typical storage charge QRR is 21 nC for a 600-V, 6-A diode and is 23 nC for a 600-V, 10-A device. The low reverse recovery characteristic of SiC diodes, as shown in Figure 2.7, has also a low reverse recovery current. It saves energy in many applications such as power supplies, solar energy conversion, transportations, and other applications such as welding equipment and air conditioners. SiC power devices enable increased effi- ciency, reduced solution size, higher switching frequency, and produce significant less electromagnetic interference (EMI) in a variety of applications. 2.7 Silicon Carbide Schottky Diodes Schottky diodes are used primarily in high frequency and fast-switching applica- tions. Many metals can create a Schottky barrier on either silicon or GaAs semi- conductors. A Schottky diode is formed by joining a doped semiconductor region, usually n-type, with a metal such as gold, silver, or platinum. Unlike a pn-junction diode, there is a metal to semiconductor junction. This is shown in Figure 2.8a and its symbol in Figure 2.8b. The Schottky diode operates only with majority carriers. There are no minority carriers and thus no reverse leakage current as in pn-junction diodes. The metal region is heavily occupied with conduction band electrons, and the n-type semiconductor ­region is lightly doped. When forward biased, the higher energy electrons in the n-region are injected into the metal region where they give up their excess energy very rapidly. Since there are no minority carriers, it is a fast- switching diode. The SiC Schottky diodes have the following features: Lowest switching losses due to low reverse recovery charge; Fully surge-current stable, high reliability, and ruggedness; Lower system costs due to reduced cooling requirements; Higher frequency designs and increased power density solutions. These devices also have low device capacitance that enhances overall system efficiency, especially at higher switching frequencies. M02_RASH9088_04_PIE_C02.indd 71 31/07/13 3:53 PM  72   Chapter 2     Power Diodes and Switched RLC Circuits Semiconductor Ohmic contact Ohmic contact Anode Cathode n-type iD Metal Anode Cathode vD (a) (b) Figure 2.8 Basic internal structure of a Schottky diode. 2.8 Spice Diode Model The SPICE model of a diode [4–6] is shown in Figure 2.9b. The diode current ID that depends on its voltage is represented by a current source. Rs is the series resistance, and it is due to the resistance of the semiconductor. Rs, also known as bulk resistance, is dependent on the amount of doping. The small-signal and static models that are generated by SPICE are shown in Figures 2.9c and 2.9d, respectively. CD is a non- linear function of the diode voltage vD and is equal to CD = dqd/dvD, where qd is the depletion-layer charge. SPICE generates the small-signal parameters from the operating point. The SPICE model statement of a diode has the general form.MODEL DNAME D (P1= V1 P2= V2 P3= V3..... PN=VN) DNAME is the model name and it can begin with any character; however, its word size is normally limited to 8. D is the type symbol for diodes. P1, P2,... and V1, V2,... are the model parameters and their values, respectively. Among many diode parameters, the important parameters [5, 8] for power switching are: IS Saturation current BV Reverse breakdown voltage IBV Reverse breakdown current TT Transit time CJO Zero-bias pn capacitance Because the SiC diodes use a completely new type of technology, the use of SPICE models for silicon diodes may introduce a significant amount of errors. The manufac- turers are, however, providing the SPICE models of SiC diodes. M02_RASH9088_04_PIE_C02.indd 72 31/07/13 3:53 PM  2.9  Series-Connected Diodes   73 A RS A ID  D1 VD ID CD  K K (a) Diode (b) SPICE model A A RS RS   VD RD CD VD ID   K K (c) Small-signal model (d) Static model Figure 2.9 SPICE diode model with reverse-biased diode. Key Points of Section 2.8 The SPICE parameters, which can be derived from the data sheet, may signifi- cantly affect the transient behavior of a switching circuit. 2.9 Series-Connected Diodes In many high-voltage applications (e.g., high-voltage direct current [HVDC] transmis- sion lines), one commercially available diode cannot meet the required voltage rating, and diodes are connected in series to increase the reverse blocking capabilities. M02_RASH9088_04_PIE_C02.indd 73 31/07/13 3:53 PM  74   Chapter 2     Power Diodes and Switched RLC Circuits iD iD  VD1 VD2 vD VD1 0 D1   vD  IS1  D2 VD2  IS (a) Circuit diagram (b) v – i characteristics Figure 2.10 Two series-connected diodes with reverse bias. Let us consider two series-connected diodes as shown in Figure 2.10a. Variables iD and vD are the current and voltage, respectively, in the forward direction; VD1 and VD2 are the sharing reverse voltages of diodes D1 and D2, respectively. In practice, the v-i characteristics for the same type of diodes differ due to tolerances in their production process. Figure 2.10b shows two v-i characteristics for such diodes. In the forward-biased condition, both diodes conduct the same amount of current, and the forward voltage drop of each diode would be almost equal. However, in the reverse blocking condition, each diode has to carry the same leakage current, and as a result the blocking voltages may differ significantly. A simple solution to this problem, as shown in Figure 2.11a, is to force equal volt- age sharing by connecting a resistor across each diode. Due to equal voltage sharing, the leakage current of each diode would be different, and this is shown in Figure 2.11b. Because the total leakage current must be shared by a diode and its resistor, iD iD VD1  VD2 vD  0 IR1 D1 VD1 R1 IS1  IS1  vD   IS2 IS2 VD2 R2 D2  IR2 IS (a) Circuit diagram (b) v – i characteristics Figure 2.11 Series-connected diodes with steady-state voltage-sharing characteristics. M02_RASH9088_04_PIE_C02.indd 74 31/07/13 3:53 PM  2.9  Series-Connected Diodes   75 Rs R1 D1 Steady- Cs Transient state voltage voltage Cs sharing Figure 2.12 sharing D2 R2 Rs Series diodes with voltage-sharing networks under steady-state and transient conditions. Is = IS1 + IR1 = IS2 + IR2 (2.12) However, IR1 = VD1/R1 and IR2 = VD2/R2 = VD1/R2. Equation (2.12) gives the rela- tionship between R1 and R2 for equal voltage sharing as VD1 VD1 IS1 + = IS2 + (2.13) R1 R2 If the resistances are equal, then R = R1 = R2 and the two diode voltages would be slightly different depending on the dissimilarities of the two v-i characteristics. The values of VD1 and VD2 can be determined from Eqs. (2.14) and (2.15): VD1 VD2 IS1 + = IS2 + (2.14) R R VD1 + VD2 = VS (2.15) The voltage sharings under transient conditions (e.g., due to switching loads, the initial applications of the input voltage) are accomplished by connecting capacitors across each diode, which is shown in Figure 2.12. Rs limits the rate of rise of the blocking voltage. Example 2.3 Finding the Voltage-Sharing Resistors Two diodes are connected in series, as shown in Figure 2.11a, to share a total dc reverse voltage of VD = 5 kV. The reverse leakage currents of the two diodes are IS1 = 30 mA and IS2 = 35 mA. (a) Find the diode voltages if the voltage-sharing resistances are equal, R1 = R2 = R = 100 kΩ. (b) Find the voltage-sharing resistances R1 and R2 if the diode voltages are equal, VD1 = VD2 = VD/2. (c) Use PSpice to check your results of part (a). PSpice model parameters of the diodes are BV = 3 kV and IS = 30 mA for diode D1, and IS = 35 mA for diode D2. Solution a. IS1 = 30 mA, IS2 = 35 mA, and R1 = R2 = R = 100 kΩ. - VD = - VD1 - VD2 or VD2 = VD - VD1. From Eq. (2.14), VD1 VD2 IS1 + = IS2 + R R M02_RASH9088_04_PIE_C02.indd 75 31/07/13 3:53 PM  76   Chapter 2     Power Diodes and Switched RLC Circuits Substituting VD2 = VD - VD1 and solving for the diode voltage D1, we get VD R VD1 = + 1I - IS1 2 2 2 S2 5 kV 100 kΩ = + 135 * 10-3 - 30 * 10-3 2 = 2750 V (2.16) 2 2 and VD2 = VD - VD1 = 5 kV - 2750 = 2250 V. b. IS1 = 30 mA, IS2 = 35 mA, and VD1 = VD2 = VD/2 = 2.5 kV. From Eq. (2.13), VD1 VD2 IS1 + = IS2 + R1 R2 which gives the resistance R2 for a known value of R1 as VD2R1 R2 =  (2.17) VD1 - R1(IS2 - IS1 2 Assuming that R1 = 100 kΩ, we get 2.5 kV * 100 kΩ R2 = = 125 kΩ 2.5 kV - 100 kΩ * (35 * 10-3 - 30 * 10-3 2 c. The diode circuit for PSpice simulation is shown in Figure 2.13. The list of the circuit file is as follows: Example 2.3 Diode Voltage-Sharing Circuit VS 1 0 DC 5KV R 1 2 0.01 R1 2 3 100K R2 3 0 100K D1 3 2 MOD1 D2 0 3 MOD2.MODEL MOD1 D (IS=30MA BV=3KV) ; Diode model parameters.MODEL MOD2 D (IS=35MA BV=3KV) ; Diode model parameters.OP ; Dc operating point analysis.END R 2 1 0.01  D1 R1 100 k  Vs 5 kV 3  D2 R2 Figure 2.13 100 k Diode circuit for PSpice simulation for Example 2.3. 0 M02_RASH9088_04_PIE_C02.indd 76 31/07/13 3:53 PM  2.10  Parallel-Connected Diodes   77 The results of PSpice simulation are NAME   D1   D2 ID –3.00E–02 ID1=–30 mA –3.50E–02 ID2=–35 mA VD –2.75E+03 VD1=–2750 V expected –2750 V –2.25E+03 VD2=–2250 V expected –2250 V REQ 1.00E+12 RD1=1 GΩ 1.00E+12 RD2=1 GΩ Note: The SPICE gives the same voltages as expected. A small resistance of R = 10 mΩ is inserted to avoid SPICE error due to a zero-resistance voltage loop. Key Points of Section 2.9 When diodes of the same type are connected in series, they do not share the same reverse voltage due to mismatches in their reverse v-i characteristics. Voltage- sharing networks are needed to equalize the voltage sharing. 2.10 Parallel-Connected Diodes In high-power applications, diodes are connected in parallel to increase the current- carrying capability to meet the desired current requirements. The current sharings of diodes would be in accord with their respective forward voltage drops. Uniform current sharing can be achieved by providing equal inductances (e.g., in the leads) or by con- necting current-sharing resistors (which may not be practical due to power losses); this is depicted in Figure 2.14. It is possible to minimize this problem by selecting diodes with equal forward voltage drops or diodes of the same type. Because the diodes are connected in parallel, the reverse blocking voltages of each diode would be the same. The resistors of Figure 2.14a help current sharing under steady-state conditions. Current sharing under dynamic conditions can be accomplished by connecting coupled inductors as shown in Figure 2.14b. If the current through D1 rises, the L di/dt across L1 increases, and a corresponding voltage of opposite polarity is induced across in- ductor L2. The result is a low-impedance path through diode D2 and the current is shifted to D2. The inductors may generate voltage spikes and they may be expensive and bulky, especially at high currents. iD iD D2 D1 D1 D2   R2 R1 vD vD   R1 R2 L2 L1 Figure 2.14 (a) Steady state (b) Dynamic sharing Parallel-connected diodes. M02_RASH9088_04_PIE_C02.indd 77 31/07/13 3:53 PM  78   Chapter 2     Power Diodes and Switched RLC Circuits Key Points of Section 2.10 When diodes of the same type are connected in parallel, they do not share the same on-state current due to mismatches in their forward v-i characteristics. Current sharing networks are needed to equalize the current sharing. 2.11 Diode Switched RC Load Figure 2.15a shows a diode circuit with an RC load. For the sake of simplicity, the di- odes are considered to be ideal. By “ideal” we mean that the reverse recovery time trr and the forward voltage drop VD are negligible. That is, t rr = 0 and VD = 0. The source voltage VS is a dc constant voltage. When the switch S1 is closed at t = 0, the charging current i that flows through the capacitor can be found from t CL 1 Vs = vR + vc = vR + i dt + vc 1t = 02 (2.18) t0 vR = Ri (2.19) With initial condition vc 1t = 02 = 0, the solution of Eq. (2.18) (which is derived in Appendix D, Eq. D.1) gives the charging current i as Vs -t/RC i1t2 = e (2.20) R The capacitor voltage vc is t CL 1 vc 1t2 = i dt = Vs 11 - e -t/RC 2 = Vs 11 - e -t/τ 2 (2.21) 0 where τ = RC is the time constant of an RC load. The rate of change of the capacitor voltage is dvc Vs -t/RC = e (2.22) dt RC and the initial rate of change of the capacitor voltage (at t = 0) is obtained from Eq. (2.22) dvc Vs ` = (2.23) dt t = 0 RC We should note that at the instant when the switch is closed at t = 0, the voltage across the capacitor is zero. The dc supply voltage VS will appear in the resistance R and the current will rise instantaneously to VS/R. That is, the initial di/dt = ∞. Note: Because the current i in Figure 2.15a is unidirectional and does not tend to change its polarity, the diode has no effect on circuit operation. M02_RASH9088_04_PIE_C02.indd 78 31/07/13 3:53 PM  2.11  Diode Switched RC Load   79 Vs i R S1 D1 i Vs 0.368 R  t0  0 t R vR vc   Vs Vs Vs   0.632 Vs   RC C vc  0 t   (a) Circuit diagram (b) Waveforms Figure 2.15 Diode circuit with an RC load. Key Points of Section 2.11 The current of an RC circuit that rises or falls exponentially with a circuit time constant does not reverse its polarity. The initial dv/dt of a charging capacitor in an RC circuit is Vs/RC. Example 2.4 Finding the Peak Current and Energy Loss in an RC Circuit A diode circuit is shown in Figure 2.16a with R = 44 Ω and C = 0.1 μF. The capacitor has an initial voltage, Vc0 = Vc 1t = 02 = 220 V. If switch S1 is closed at t = 0, determine (a) the peak diode current, (b) the energy dissipated in the resistor R, and (c) the capacitor voltage at t = 2 μs. Solution The waveforms are shown in Figure 2.16b. a. Equation (2.20) can be used with Vs = Vc0 and the peak diode current Ip is Vc0 220 IP = = = 5A R 44 V0 i S1 R t0  i R vR 0 t  vc D1 V0   C Vco vc   0 t (a) Circuit diagram (b) Waveforms Figure 2.16 Diode circuit with an RC load. M02_RASH9088_04_PIE_C02.indd 79 31/07/13 3:53 PM  80   Chapter 2     Power Diodes and Switched RLC Circuits b. The energy W dissipated is W = 0.5CV 2c0 = 0.5 * 0.1 * 10-6 * 2202 = 0.00242 J = 2.42 mJ c. For RC = 44 * 0.1 μ = 4.4 μs and t = t 1 = 2 μs, the capacitor voltage is vc 1t = 2 μs 2 = Vc0e -t/RC = 220 * e -2/4.4 = 139.64 V Note: Because the current is unidirectional, the diode does not affect circuit operation. 2.12 Diode Switched RL Load A diode circuit with an RL load is shown in Figure 2.17a. When switch S1 is closed at t = 0, the current i through the inductor increases and is expressed as di Vs = vL + vR = L + Ri (2.24) dt With initial condition i1t = 02 = 0, the solution of Eq. (2.24) (which is derived in Appendix D, Eq. D.2) yields Vs i 1t2 = 11 - e -tR/L 2 (2.25) R The rate of change of this current can be obtained from Eq. (2.25) as di Vs -tR/L = e (2.26) dt L and the initial rate of rise of the current (at t = 0) is obtained from Eq. (2.26): di Vs ` = (2.27) dt t = 0 L vL Vs S1 D1 i 0.368 Vs  t0  0 t R vR  Vs i  Is  Vs Vs R   0.632 Is L L vL R   0 t  (a) Circuit diagram (b) Waveforms Figure 2.17 Diode circuit with an RL load. M02_RASH9088_04_PIE_C02.indd 80 31/07/13 3:53 PM  2.12  Diode Switched RL Load   81 The voltage vL across the inductor is di vL 1t2 = L = Vse -tR/L (2.28) dt where L/R = τ is the time constant of an RL load. We should note that at the instant when the switch is closed at t = 0, the current is zero and the voltage across the resistance R is zero. The dc supply voltage VS will appear across the inductor L. That is, di VS = L dt which gives the initial rate of change of the current as di VS = dt L which is the same as Eq. (2.27). If there was no inductor, the current would rise instan- taneously. But due to the inductor, the current will rise with an initial slope of VS/L and the current can be approximated to i = Vs*t/L. Note: D1 is connected in series with the switch and it will prevent any negative current flow through the switch if there is an ac input supply voltage, but it is not ap- plicable for a dc supply. Normally, an electronic switch (BJT or MOSFET or IGBT) will not allow reverse current flow. The switch, along with the diode D1, emulates the switching behavior of an electronic switch. The waveforms for voltage vL and current are shown in Figure 2.17b. If t W L/R, the voltage across the inductor tends to be zero and its current reaches a steady-state value of Is = Vs/R. If an attempt is then made to open switch S1, the energy stored in the inductor 1= 0.5Li2 2 will be transformed into a high reverse voltage across the switch and diode. This energy dissipates in the form of sparks across the switch; diode D1 is likely to be damaged in this process. To overcome such a situation, a diode com- monly known as a freewheeling diode is connected across an inductive load as shown in Figure 2.24a. Note: Because the current i in Figure 2.17a is unidirectional and does not tend to change its polarity, the diode has no effect on circuit operation. Key Points of Section 2.12 The current of an RL circuit that rises or falls exponentially with a circuit time constant does not reverse its polarity. The initial di/dt in an RL circuit is Vs/L. Example 2.5 Finding the Steady-State Current and the Energy Stored in an Inductor A diode RL circuit is shown in Figure 2.17a with VS = 220 V, R = 4Ω, and L = 5 mH. The in- ductor has no initial current. If switch S1 is closed at t = 0, determine (a) the steady-state diode current, (b) the energy stored in the inductor L, and (c) the initial di/dt. M02_RASH9088_04_PIE_C02.indd 81 31/07/13 3:53 PM  82   Chapter 2     Power Diodes and Switched RLC Circuits Solution The waveforms are shown in Figure 2.17b. a. Eq. (2.25) can be used with t = ∞ and the steady-state peak diode current is VS 220 IP = = = 55 A R 4 b. The energy stored in the inductor in the steady state at a time t tending ∞ W = 0.5 L I 2P = 0.5 * 5 * 10-3552 = 7.563 mJ c. Eq. (2.26) can be used to find the initial di/dt as di VS 220 = = = 44 A/ms dt L 5 * 10-3 d. For L/R = 5 mH/4 = 1.25 ms, and t = t 1 = 1 ms, Eq. (2.25) gives the inductor current as VS 220 i1t = 1 ms 2 = 1 1 - e -tR/L 2 = * 11 - e -1/1.25 2 = 30.287 A R 4 12.13 Diode Switched LC Load A diode circuit with an LC load is shown in Figure 2.18a. The source voltage Vs is a dc constant voltage. When switch S1 is closed at t = 0, the charging current i of the capacitor is expressed as t C Lt0 di 1 Vs = L + i dt + vc 1t = 02 (2.29) dt With initial conditions i1t = 02 = 0 and vc 1t = 02 = 0, Eq. (2.29) can be solved for the capacitor current i as (in Appendix D, Eq. D.3) i Ip S1 D1  i  0 t t0 t1/2 vL t1 L vc   2Vs Vs Vs   Vs t1   LC C vc  0 t  t1 (a) Circuit diagram (b) Waveforms Figure 2.18 Diode circuit with an LC load. M02_RASH9088_04_PIE_C02.indd 82 31/07/13 3:53 PM  12.13  Diode Switched LC Load   83 C i1t2 = Vs sin ω0t (2.30) CL = Ip sin ω0t (2.31) where ω0 = 1/1LC and the peak current Ip is C Ip = Vs (2.32) CL The rate of rise of the current is obtained from Eq. (2.30) as di Vs = cos ω0t (2.33) dt L and Eq. (2.33) gives the initial rate of rise of the current (at t = 0) as di Vs ` = (2.34) dt t = 0 L The voltage vc across the capacitor can be derived as t C L0 1 vc 1t2 = i dt = Vs 11 - cos ω0 t2 (2.35) At a time t = t 1 = π1LC, the diode current i falls to zero and the capacitor is charged to 2Vs. The waveforms for the voltage vL and current i are shown in Figure 2.18b. Notes: Because there is no resistance in the circuit, there can be no energy loss. Thus, in the absence of any resistance, the current of an LC circuit oscillates and the energy is transferred from C to L and vice versa. D1 is connected in series with the switch and it will prevent any negative current flow through the switch. In the absence of the diode, the LC circuit will continue to oscillate forever. Normally, an electronic switch (BJT or MOSFET or IGBT) will not allow reverse current flow. The switch along with the diode D1 emulates the switching behavior of an electronic switch. The output of the capacitor C can be connected to other similar circuits con- sisting of a switch, and a diode connected in series with an L and a C to obtain multiples of the dc supply voltage VS. This technique is used to generate a high voltage for pulse power and superconducting applications. Example 2.6 Finding the Voltage and Current in an LC Circuit A diode circuit with an LC load is shown in Figure 2.19a with the capacitor having an initial voltage, Vc 1t = 02 = -Vc0 = V0 - 220 V; capacitance, C = 20 μF; and inductance, L = 80 μH. M02_RASH9088_04_PIE_C02.indd 83 31/07/13 3:53 PM  84   Chapter 2     Power Diodes and Switched RLC Circuits i Ip S1 t0  0 t i L vL t1/2 vc  Vc 0 D1 t1   LC   0 t C Vc0 vc Figure 2.19   Vc 0 t1 Diode circuit with an LC load. (a) Circuit diagram (b) Waveforms If switch S1 is closed at t = 0, determine (a) the peak current through the diode, (b) the conduc- tion time of the diode, and (c) the final steady-state capacitor voltage. Solution a. Using Kirchhoff’s voltage law (KVL), we can write the equation for the current i as t C Lt0 di 1 L + i dt + vc 1t = 02 = 0 dt and the current i with initial conditions of i1t = 02 = 0 and vc 1t = 02 = -Vc0 is solved as C i1t2 = Vc0 sin ω0 t CL where ω0 = 1/1LC = 106/120 * 80 = 25,000 rad/s. The peak current Ip is C 20 Ip = Vc0 = 220 = 110 A CL C 80 b. At t = t 1 = π 1LC, the diode current becomes zero and the conduction time t1 of diode is t 1 = π 1LC = π 120 * 80 = 125.66 μs c. The capacitor voltage can easily be shown to be t C L0 1 vc 1t2 = i dt - Vc0 = -Vc0 cos ω0 t For t = t 1 = 125.66 μs, vc 1t = t 1 2 = -220 cos π = 220 V. Note: This is an example of reversing the polarity of a capacitor voltage. Some ­applications may require a voltage with a polarity that is opposite of the available voltage. Key Points of Section 2.13 The current of an LC circuit goes through a resonant oscillation with a peak value of VS (C/L). The diode D1 stops the reverse current flow and the capacitor is charged to 2VS. M02_RASH9088_04_PIE_C02.indd 84 31/07/13 3:53 PM  2.14  Diode Switched RLC Load   85 2.14 Diode Switched RLC Load A diode circuit with an RLC load is shown in Figure 2.20. If switch S1 is closed at t = 0, we can use the KVL to write the equation for the load current i as CL di 1 L + Ri + i dt + vc 1t = 02 = Vs (2.36) dt with initial conditions i1 t = 0) = 0 and vc 1t = 02 = Vc0. Differentiating Eq. (2.36) and dividing both sides by L gives the characteristic equation d 2i R di i 2 + + = 0 (2.37) dt L dt LC Under final steady-state conditions, the capacitor is charged to the source voltage Vs and the steady-state current is zero. The forced component of the current in Eq. (2.37) is also zero. The current is due to the natural component. The characteristic equation in Laplace’s domain of s is R 1 s2 + s + = 0 (2.38) L LC and the roots of quadratic equation (2.38) are given by R R 2 1 s1, 2 = - { a b - (2.39) 2L C 2L LC Let us define two important properties of a second-order circuit: the damping factor, R α = (2.40) 2L and the resonant frequency,