No. AN9416.1 Intersil Intelligent Power May 1995 Thermal Considerations In Power BiMOS Low Side Drivers (HIP0080, HIP0081, HIP0082, CA3282 and Others) Author: Wayne Austin Overview the user with a practical working knowledge of the thermal issues common to a Power Switch IC, both from a theoretical and practical approach. The Intersil HIP0080, HIP0081, HIP0082 and CA3282 Power BiMOS Low Side Drivers are designed for use in automotive and industrial switching applications over an operating temperature range of -40oC to +125oC. Each circuit has enhanced performance features that include logic level operation, overload protection and diagnostic feedback by serial interface. The focus of this Application Note is directed to the HIP0082 Power BiMOS Quad Low Side Driver but generally applies to all of the MOS Low Side Drivers. The IC interface and switching configurations are noted in the Block Diagrams below. The CA3282, HIP0081 and HIP0082 are provided in a 15 Lead Power Package with a low junction-to-case thermal resistance and a lead form option for vertical or surface mount. The chip is directly mounted on the copper heat sink tab of the IC to provide a maximum θJC that is less than 3oC/W. To begin, during switching and in the ON-state heat is generated in the drain (junction) of the MOS driver on the surface of the chip. Both transient and steady state conditions must be controlled to avoid operating temperatures in excess of the maximum junction temperature ratings, usually 150°C in plastic encapsulated chips. The temperature developed at the surface of the chip diffuses from the source of excitation and is conducted to the external surface of the IC package where adequate convection and radiation cooling are required to keep the operating junction temperature within ratings. Referring to Figure 2 as an illustrated example of the 15 Lead Single-in-Line (SIP) Package, the temperature developed at the active junction of the IC is determined by sustained dissipation over time. Heat flow from the junction-tocase follows the path of least thermal resistance. The junction-to-case thermal resistance is the complex sum of all series and parallel thermal paths that the diffusing heat takes as it flows from the chip junction to the external surface area of the IC. By design, the direct path from the die to heat sink tab and external heat sink is the primary conduit path for escaping heat. The external heat sink is cooled by convection and radiation. The designer of a power switching circuit must address many thermal related issues as a part of the design process. A key element in this process is an understanding of the temperature related parameters which are discussed here for educational information. The scope of this material includes thermal related effects on the IC resulting from the externally applied power, load conditions and high ambient temperature operation. Solutions for the problem of pulsed energy generated from inductive switching are derived as source information for the user. As such, the following information is intended to provide IN4 IN3 OUT4 OUT4 OUT5 OUT3 OUT3 OUT6 LOGIC CONTROL LOGIC LEVEL INPUTS IN2 HIP0080 HIP0081 HIP0082 IN1 LOGIC CONTROL OUT2 OUT2 OUT1 OUT1 DIAGNOSTIC SERIAL INTERFACE (SPI COMPATIBLE) CA3282 OUT8 SPI BUS INTERFACE FIGURE 1A. FIGURE 1B. FIGURE 1. SWITCHING CONFIGURATIONS OF THE QUAD AND OCTAL POWER BiMOS CIRCUITS 1 OUT7 Application Note 9416 To better understand the process of thermal conductivity, background material with thermal definitions and units related to heat are provided in Tables 1 and 2. The SI (International System)[1] symbol definitions provided in Table 1 are limited to the relevant subject material. Specific Heat, c is the heat capacity per unit mass, mk with units of Joules/kg•K and is often tabulated in calories/ gramoC units or Btu/lb•oF in reference publications. Specific Heat is commonly expressed as cp for constant pressure and cv for constant volume. Since the difference is small in solids, we may use the general expression, Electrical And Thermal Parameters CT c = ------mk To understand the process of heat flow, the following parameter definitions are provided: (EQ. 1) True specific heat can be a function of temperature and is defined by the differential expression, Temperature, T is the measure of molecular kinetic energy in a body with changes expressed in degrees Celsius or Kelvin. Thermal energy corresponds to temperature difference, ∆T as the potential for heat flow. Thermal energy units are in Btu, gram-calories or kg-calories which, by conversion, are equivalent to energy in Joules. 1 dQ c ( t ) = ------- -------m k dT (EQ. 1A) where Q is heat flow in watts (W). IC HEAT SINK TAB EXTERNAL HEAT SINK DIE ATTACH (SOLDER) IC PLASTIC 15 LEAD SIP POWER PACKAGE OF THE HIP0081, HIP0082 AND CA3282 DIE ACTIVE JUNCTION (MULTIPLE THERMAL CONDUCTION PATHS) TJ Rθ LDS IC HEAT SINK TAB TYPICAL DIMENSIONS IC SILICON DIE 20 MILS IC DIE ATTACH 1 MIL IC COPPER H. S. TAB 60 MILS TC LEADS BOND WIRES PLASTIC IC LEADS CT(LDS) Rθ PTC PD INPUT TO DIE (JUNCTION) TA CT(PTC) SI/DIE Rθ DIE CT(DIE) DIE/ATTACH H.S. TAB RθSOL Rθ TAB CT(SOL) CT(TAB) Rθ XHS CONVECTION AND RADIATION COOLING CT(XHS) EXT. H.S. PC BOARD RθJC FIGURE 2. TYPICAL HEAT SINK MOUNTING STRUCTURE OF A POWER SWITCHING IC IN THE 15 LEAD SIP POWER PACKAGE. THERMAL RESISTANCE (JUNCTION-TO-AMBIENT) IS PRIMARILY DEFINED BY THE SUM OF ALL THERMAL CON- 2 Application Note 9416 TABLE 1. INTERNATIONAL SYMBOLS (SI UNITS) UNIT SYMBOL PARAMETER ampere A Unit of electric current volt V Unit of electromotive force ohm Ω Unit of electric resistance farad F Unit of capacitance henry H Unit of inductance kilogram kg Unit of mass meter m Unit of length watt W Unit of power joule J Unit of energy (heat) Q = mk c ( T2 – T1 ) Electrical Energy, E in units of joules or watts⋅seconds is the time (t) integral of power, Unit of temperature, Celsius scale Kelvin K Unit of temperature, Kelvin scale hertz Hz Unit of frequency E= Unit of time SPECIFIC HEAT cal/gram•oC CONDUCTIVITY W/m•oC Silver 0.056 414 Copper 0.0921 381 Gold 0.031 297 Aluminum 0.226 213 Beryllia 0.286 155 Nickel 0.112 88 Silicon 0.181 See Note 1 Iron 0.108 78 Steel 0.1 46 Lead 0.03 35 5/95 Solder 0.03 35 Alumina 0.18 26 Silver Epoxy 0.15 2.3 Borosilicate Glass 0.199 0.59 0.2 0.58 Novolac Epoxy t ∫0 p( t ) dt (EQ. 4) Since Heat Flow, Q is the energy rate in joules per second which, by definition, is watts and input power is also in watts, the power supplied to an active element must reach equilibrium with the rate of heat flow being conducted or radiated away. As such, Q (power) is used as a symbol for heat flow away from a junction or device to separately distinguish this energy rate from the applied electrical Power, P at the junction. It is important to note that the temperature, ∆T must rise as a driving force to establish this equilibrium. Until equilibrium is established, the heat capacity of the body determines the transient period of change. TABLE 2. SPECIFIC HEAT AND THERMAL CONDUCTIVITY OF VARIOUS SEMICONDUCTOR MATERIALS MATERIAL (EQ. 3) Electrical Power, P in watts (joules per second) causes the heating in an electrical component. Power input is normally defined as voltage times current and instantaneous power, p(t) is instantaneous voltage times instantaneous current. °C s or Q = C T ( T 2 – T 1 ) The true three dimensional solution of heat flow is complex, requiring the use of partial differential equations and vector analysis. Linear heat flow is assumed in most cases to achieve a first order solution to a thermal problem. degree second Heat Flow, Q is analogous to electrical current and is expressed in joules per second (watt). For a body of mass, mk Heat Flow may be expressed as: When heat is generated by the output driver, it changes the parameters of the device model. As an example, for the bipolar cds Spice hybrid-π model shown in Figure 3, an added temperature node illustrates the interactive temperature effect on the parameters of the device model. [5] Power supplied to the transistor generates heat flow in the form of an equivalent current generator, Q. The gtb and gtc current sources are driven by the ∆T heat source. While similar models are not available for the Power BiMOS output devices, a device model for temperature is being developed. gµ rb rc COLLECTOR BASE gπ gm gtb go re SPICE MODEL ∆T gtc Q EMITTER NOTE: 1. Silicon Conductivity is temperature dependent. The NIST[6] bibliography reference provides a formula: k = 286/(T-100)(W/cmK) for a range of 300K to 600K. For a nominal range of junction temperatures: k ≈ 400/T (W/cmK) where T is temperature in Kelvin. Heat Capacity, CT is the ratio of heat supplied to a composite body for a corresponding temperature rise which leads to the following expression: Q C T = ------∆T THERMAL IMPEDANCE FIGURE 3. MODIFIED HYBRID-π SPICE MODEL TO SHOW THE POWER TO TEMPERATURE DRIVE SOURCE (EQ. 2) 3 Application Note 9416 direct measurement can provide θJC or θJA values for a given IC, the final application θJA is still quite dependent on the operating environment. The IC and heat sink structure should always be given a final equipment design check measurement over the operating temperature range. Thermal Conductivity and Thermal Resistance Fourier’s differential equation defines the temperature gradient for an incremental quantity of heat, dT flowing through a body in a linear unidirectional path of incremental length dx and cross-sectional area, A as: Figure 2 illustrates the Junction-to-Ambient thermal paths for the 15 Lead Single-in-Line (SIP) Power IC package. The Junction-to-Case thermal resistance, θJC , for this package is primarily the sum of the thermal resistance values for the die, die attach and tab of the IC. For an assumption of linear heat flow, each component path of the IC has a defined thermal resistance. Heat generated at the surface of the die flows through the silicon die to the copper heat sink tab and external heat sink. Heat also flows through the lead connections and through the molding of the plastic package to the surface of the IC. The combined thermal resistance paths through the plastic and leads is significantly greater than the direct die-to-tab path in the IC. (EQ. 5) Separating variables and integrating temperature, T from T1 to T2 over the heat flow path, d from x1 to x2 , gives the linear solution, kA Q = ------- ( T 1 – T 2 ) d (EQ. 5A) Thermal conductivity, k is a well documented material parameter with a definition of heat flow conducted per oC normal to the direction of heat flow through a unit distance per unit area of material. While the coefficient of thermal conductivity is a physical parameter correctly expressed in SI units as W/m • oC or (W/oC) • (m/m2), it is more commonly listed in reference publications as W/cm • oC or Btu/fthr • oF. Other methods used for maintaining junction temperature control employ the heat conducting qualities of the copper lead frame. The Power PLCC package of the HIP0080, conducts heat through the lead frame structure to the IC ground leads and then to the copper ground area on the external PC Board as illustrated in Figure 4. The θJC for this package actually refers to the Junction-to-Leads path with 14oC/W thermal resistance which is quite low compared to a standard PLCC package. For this package, the PC Board heat sink area is the most significant part of the Junction-to-Ambient, θJA thermal resistance. Thermal resistance, Rθ is more frequently used to describe the capability of heat transfer through a given body of fixed dimensions. The units of thermal resistance are oC/W. For a composite body of material, the thermal resistance equation becomes: d ∆T R θ = ------- = ------kA Q (EQ. 6) which defines thermal resistance, Rθ in relation to temperature difference and heat flow. Since ∆T represents the driving force or potential for heat flow and thermal resistance determines the rate of heat flow conduction, we may consider the equation: PC BOARD TOP SURFACE COPPER GROUND PLANE ∆T = Q R θ (EQ. 6A) as analogous to the Ohms Law equation, V = IR. This analogy is similar to a node equation voltage difference with temperature difference as the driving force. Since we cannot directly adjust temperature in an electrical circuit, temperature and heat flow are affected only by the thermal resistance and the level of power input. As such, we may directly substitute electrical power dissipation, PD for its heat equivalent, Q to establish the equation, ∆T = P D × R θ PC BOARD TOP SURFACE COPPER GROUND PLANE dT Q = – k A ------- dx DIE (TOP VIEW) (EQ. 6B) Where TJ is junction temperature and TC is case temperature, ∆T = (TJ -TC). The symbol, Rθ, is used here to represent the thermal analogy to electrical resistance. In the material that follows, thermal resistance is defined by the standard symbol now in use, which is θJC or θJA, to represent the junction-to-case or junction-to-ambient paths. Therefore, SURFACE MOUNT TO PC BOARD T J – T C = P D × θ JC (EQ. 6C) Thermal resistance of a well defined structure may be determined, given conditions of unidirectional and uniform heat flow using equation Equation 6. The thermal conductivity coefficients in Table 2 are intended as guidance to understand and improve the thermal resistance of IC heat sink structures and can be used for an approximation for the θJC of the IC package. However, the complex physical configuration of most ICs require direct measurement as the only practical way to determine thermal resistance. Even where DIE HEAT FLOW HEAT FLOW PC BOARD FIGURE 4. THE HIP0080 PLCC PACKAGE ILLUSTRATING THE COPPER LEAD FRAME CONSTRUCTION DESIGNED TO CONDUCT HEAT TO THE GROUND LEADS 4 Application Note 9416 The actual field pattern of heat flow lines is similar to that of an electric field having curving divergent lines. As heat diffuses from its source of excitation, the isothermal lines of the gradient are orthogonal to the direction of heat flow. To provide an example that illustrates how thermal resistance is determined, we can assume a simpler linear heat flow model. The direct path model of Figure 5A should give a reasonable approximation for θJC of the IC in Figure 2 by assuming linear heat diffusion in the shape of a truncated pyramid, spreading at 45o away from the active junction area of the die. Then, for an area approximation, the average length and width dimension of each material in the pyramid is increased by an amount equal to thickness. PLASTIC ENCAP. Example: Using Equation 6 for thermal resistance and the suggested model in Figure 5A, we can make the following calculations for thermal resistance. 1. For active dissipation area on the die, assume a junction area of 54 x 89 square mils and, for a die thickness of 20 mils, the heat diffuses out at 45o to increase the effective dimension in each direction by 2 x 1/2 the die thickness or 20 mils. As such, the effective die dimensions are: A = 74 x 109 = 8066 square mils or 5.204 x 10-6 square meters d = 20 mils = 50.8 x 10-5 meters Using the value for thermal conductivity for silicon from Table 2, and applying to Equation 6, we have: 5.08x10 – 4 d R θ ( Die ) = ------- = ----------------------------------- = 1.16°C ⁄ W kA 84x5.2x10 – 6 IC TAB (COPPER) 2. For the 1 mil thick Die Attach Solder, the dimensions are approximately equal to the that of the junction plus 2 times the die thickness plus 2 x 1/2 the Die Attach thickness or 40 + 1 = 41 added for each dimension. Then: A = 95 x 130 = 12350 square mils or 7.968 x 10-6 square meters d = 1 mil = 2.54 x 10-5 meters For Lead, k = 35W/m•oC Rθ (Die Attach) = 0.09o C/W. 45o (EXTERNAL HEAT SINK) HEAT FLOW DIE JUNCTION TJ CASE TC 45o 3. For the 60 mil Copper Heat Sink Tab that is an integral part of the IC, the Tab dimensions equal the active junction area dimensions plus the 45ospread in each direction or (2 x 20) + (2 x 1) + (2 x (1/2) x 60) = 102 mils A = (54 + 102) x (89 + 102) or 156 x 191 mils, A = 2.980 x 10+4 square mils = 19.22 x 10-6 square meters d = 60 mils = 1.52 x 10-3 meters k = 381 W/m•oC Rθ (Tab) = 0.21oC/W. DIE ATTACH FIGURE 5A. AN IC CROSS-SECTION USING A 45° DIFFUSION PATTERN (IN THE SHAPE OF A TRUNCATED PYRAMID) TO ILLUSTRATE BY APPROXIMATION, THE HEAT FLOW FROM THE JUNCTION TO THE CASE FOR THE IC PACKAGE SHOWN IN FIGURE 2 CASE TC JUNCTION TJ PD The calculated thermal resistance is the sum of the series thermal resistance values which are: θJC = 1.16 + 0.09 + 0.21 = 1.46oC/W. ∆X ∆T HEAT FLOW THERMAL GRADIENT SILICON DIE DIE ATTACH HEAT SINK TAB Based on measurement, the actual junction-to-case thermal resistance rating for to this package is 3oC/W. The active junction area used is the approximate dimension for one output driver of the HIP0082. With added active drivers on the chip there will not be a proportional decrease in the thermal resistance due to an overlap in the assumed 45o diffused fan-out for heat flow. EXT. HEAT SINK TAB The temperature gradient changes as heat flows from the junction to the case of the IC. The slope of the gradient changes with changes in the thermal resistance of each material. This effect is illustrated in Figure 5B. Heat Capacity and Transient Thermal Impedance Heat Capacity, CT is the amount of heat required to raise the temperature of a mass (body or component part) by 1oC. When the mass of the body is known, we may determine the specific heat (abbreviated sp. ht.). As such, the heat capacity FIGURE 5B. HEAT FLOW FROM JUNCTION-TO-CASE SHOWING THE THERMAL GRADIENT, ∆X/∆T. RESISTOR AND CAPACITOR SYMBOLS REPRESENT THERMAL RESISTANCE AND HEAT CAPACITY AS COMPONENTS IN THE HEAT FLOW PATH FIGURE 5. 5 Application Note 9416 TRANSIENT THERMAL IMPEDANCE (oC/W) of a body is calculated to provide tabulated values for specific heat using Equations 1 and 3, where: CT Q c = ------- = -------------------mk m k ( ∆T ) When gram-calorie units of specific heat, c are given in cal/gram (°C), the conversion to Joules, J is: cal sp. ht., c in --------------------------- Units = 4.186J gram ( °C ) Also, Then, 1 kilogram-calories = 4.186kJ CT = c x mk = 4.186 x (sp. ht.) x wt. where the weight, wt = Density, D x Volume, V. (See Table 2 for sp. ht. values of various materials). In Figure 5B, an electrical analogy of series resistance and shunt capacitance is used to illustrate the electrical equivalence of thermal resistance and heat capacity. A short duration power pulse at the chip junction integrates the power as heat energy which will diffuse to eventually be detected on the case (tab) of the IC. Since the RC filter circuit is an integrating network, this is a reasonable symbol analogy using the comparison of electrical current to heat flow units of joules/second or watts. 30 27 24 21 HIP0082 - 15 LEAD SIP POWER PACKAGE d = 90% d = 70% 18 15 d = 50% 12 9 d = 30% 6 3 0 0.01 d = 10% d = 2% 0.1 1 10 CYCLE PERIOD (s) 100 1000 TRANSIENT THERMAL IMPEDANCE (oC/W) FIGURE 6A. TYPICAL CHARACTERISTIC PLOT OF THE TRANSIENT THERMAL IMPEDANCE VERSUS CYCLE PERIOD (1/f) FOR THE HIP0082 QUAD LOW SIDE POWER DRIVER Calculation of the heat capacity and the thermal time constant can be made from assumptions similar to those used for the unidirectional analysis of thermal resistance. However, since the thermal capacitance is distributed in the material, a computer simulation for a ladder of small distributed RC elements should be used for an accurate result. As a first order approximation, we can assume strict unidirectional heat flow (without side diffusion) to determine the thermal time constant of the silicon die. Example: Using centimeters (cm) and grams (g) for units of length and mass, calculate the thermal time constant given a fixed volume of silicon with Area, A = x•y and die thickness, d = 20 mils = 0.0508cm. The thermal resistance is: 30 27 HIP0082 - 15 LEAD SIP POWER PACKAGE 24 21 18 15 12 9 6 3 0 0.1 1 10 PULSE DURATION (s) 100 1000 FIGURE 6B. TYPICAL CHARACTERISTIC PLOT OF THE TRANSIENT THERMAL IMPEDANCE FOR A SINGLE PULSE VERSUS PULSE DURATION FOR THE HIP0082 QUAD LOW SIDE POWER DRIVER Rθ = (1/k) cmoC/W • (d/x•y) cm/cm2. Using the formula derived value for k given Note 1, Table 2; at 120oC the conductivity, k = 0.98 W/cmoC FIGURE 6. section in an IC is generally too complex for simple calculations, evaluations can made using computer aided design, including methods used for finite element analysis and electrical circuit design, including methods used for fiite element analysis and electrical circuit design. Rθ = [(1/.98) cmoC/W] • [(d/x•y) cm/cm2] = (1.02•d/x•y) oC/W. For Heat Capacity, CT = 4.186•c • wt = 4.186 • c • D • V. Direct measurement can give us a composite transient thermal impedance curve and the thermal time constants. Power is normally applied to the IC junction as pulsed energy with variations of duty cycle and frequency. From the test data, we can derive an equivalent RC time constant by plotting the effects of transient energy change into the IC. As an example, test results for an application of the HIP0082 are illustrated in Figures 6A and 6B. The data was taken with the HIP0082 mounted on a copper PC Board. where V = (d•x•y) cm3 and silicon density, D = 2.41 g/cm3. Then, CT = [(4.186 • 0.181) J/goC] • [(2.41) g/cm3] • [(d • x • y) cm3] = (1.826•d•x•y) J/oC. Combining CT and Rθ for the thermal time constant, RθCT = [(1.02 • d/x • y) oC/W]•[(1.826•d•x•y) J/oC] = 1.862 • d2 ms. = 1.862 • (0.0508)2 = 4.81ms. In Figure 6A, Transient Thermal Impedance versus Cycle Period is plotted for a family of 6 curves with a duty cycle variation from 2%, to 90%. For low duty cycles and small cycle periods, the Transient Thermal Impedance, Zθ is very low. For very long cycle periods and large duty cycles, Zθ The actual value will be less due to the distributed capacitance effect. While the physical shape of each component 6 Application Note 9416 approaches the static thermal resistance value of θJA that includes the IC and PC Board, approximately 27oC/W. A similar curve is plotted in Figure 6B to show the transient thermal impedance versus pulse duration for a single nonrepetitive pulse. The thermal time constants are empirically derived from the log-linear slope of the curves. tion circuit. The Safe Operating Area (SOA) limit was determined by applying pulse width variation and increasing the pulse current to a point of stress. The data was taken in an IC socket without a heat sink attached. The SOA limit is noted to be lower for high ambient temperatures. Peak currents are also noted for referenceThe product of clamp voltage (90V specification maximum) times the pulse current is peak dissipation in Watts. The product of watts and milliseconds (ms) is energy in millijoules (mJ). The single pulse Energy equation is: For low duty cycle and single pulse conditions the thermal impedance is quite low at ~1Ω. This does not imply that we can have the best of both worlds by supplying all needs with short power pulses. It is important to realize that short duration pulses can inject high instantaneous peak power while the measurement of thermal impedance remains small. Localized heating does occur at the junction in proportion to the applied power level and can cause stress damage at the junction before it is detected by external monitors. Energy ≈ 90V × Current × Time The Figure 7 rating curve is based on well defined energy pulses. However, it does have good correlation to the flyback energy pulse generated when an inductive load is switched off. The collapsing energy of an inductor forces an exponential decay in the current waveform. The integrated energy of the pulse times the period of field collapse provide the basis for correlation and are discussed a later section. The transient thermal impedance time constants shown in Figures 6A and 6B are much longer than the RθCT of 2.33 ms previously calculated. The time constants represented by the change of curve slope in the range of 1s to 1000s corresponds to the IC copper tab and the external PC board. The transient thermal impedance curve of Figure 6B indicates that a single pulse width up to several seconds will not give a significant indication of heat at the chip junction. The thermal time constant of the package delays heat flow to case. Until the junction temperature rises sufficiently to force a sustained heat flow, the “apparent” thermal impedance of the package will continue to appear low. Maximum Temperature Ratings A general equation for dissipation should express the total power dissipation in a package as the sum of all significant elements of dissipation on the chip. However, in Power BiMOS Circuits very little dissipation is needed to control the logic and pre-driver circuits on the chip. The overall chip dissipation is primarily the sum of the I2R dissipation losses in each NMOS output channel, where I is the output current and R = is the channel resistance, RDSON. The dissipation, PD in each output driver is: 2 0. 5A P D = I × R DSON n PD = oC T I LIM A SO o C 5 12 OR 100 IT A SO R FO 25 F 5A 5 25 A A 0. 2 (EQ. 8) The current, I is the determined by the output load when the channel is turned ON. The channel resistance, RDSON is determined by the IC design, the level of forward gate voltage drive at Turn-ON and the on-chip temperature at the conducting output. While the effective value of RDSON may include resistance in the interconnecting metal on the chip and the bond wires of the package, this is assume to be small for the range of currents discussed here. LIM 0. 1 10 SINGLE PULSE WIDTH(ms) Pk Which, as an expression, sums the dissipation all output drivers without regard to uniformity of dissipation in each MOS channel. 12 0. 0.1 ∑ k=1 10 1 0.01 (EQ. 7) The total power dissipated for k = 1 to n output drivers: 1K 06 SINGLE PULSE ENERGY (mJ) 4A 8A 1A 2A 10K 100 Other sources of dissipation in the IC are the result of switched inductive loads. The energy stored in the Inductor is generally defined in millijoules (Watt-seconds) and, for a given current, is equal to: FIGURE 7. RATING LIMITS FOR THE HIP0082 OUTPUT CLAMP FOR ENERGY vs THE WIDTH VARIATION OF A SINGLE CURRENT PULSE INTO THE CLAMP. THE SOA LIMIT IS THE MAXIMUM SAFE LEVEL FOR SINGLE PULSE ENERGY AND SHOULD NOT BE EXCEEDED LI 2 E L = -------2 (EQ. 9) which is dissipated in the output driver when the inductive flyback pulse is clamped internally in the IC. This is discussed in a later section. Power switching output circuits with internal clamp protection require rating protection for short duration peak power pulses. As such, conditional limits are placed on the maximum energy of a single pulse. The rating curve of Figure 7 illustrates the Energy vs Time single pulse capability of the HIP0082 which has an internal drain-to-gate clamp protec- As noted in Equation 6C, the temperature rise in the package due to the power dissipation is the product of the dissi- 7 Application Note 9416 pation, PD and the θJC of the package. For a junction temperature, TJ , and a case temperature, TC; T J = T C + P D ( θ JC ) θJA = 35oC/W, determine the maximum dissipation for an ambient temperature of TA = 125oC. Using Equation 13: ( TJ – TA ) ( 150 – 125 ) P D = ------------------------ = ------------------------------ = 0.714W 35 θ JA (EQ. 10) or, given that TC should not exceed a known temperature, we may express the equation as: Dissipation Derating Curves T C = T J – P D ( θ JC ) The general solution of Equations 10 through 13 can be easily adapted to a graphical illustration as shown in the commonly used “Derating Curve” of Figure 8. The θJC rating of the HIP0082 Power Package is 3oC/W with a substantially higher value for θJA. The junction-to-case thermal resistance, θJC is based on temperature measurements at the case to external heat sink interface. Practical considerations require additional case-to-air, θCA heat sinking to achieve an overall low θJA. This is shown in the middle curve of the Figure 8 graph where the 3oC/W heat sink for θJC and 6oC/W for the external heat sink gives an overall θJA of 9oC/W. (EQ. 10A) Since this solution relates only to the package, further consideration must be given to a practical heat sink. The equation of linear heat flow assumes that the Junction-to-Ambient thermal resistance, θJA is the sum of the Junction-to-Case thermal resistance, θJC and the Case-to-Ambient thermal resistance θCA. For a power device such as illustrated in Figure 2, the Case-to-Ambient thermal resistance is the external Heat Sink. The Junction-to-Ambient thermal resistance, θJA is the sum of all thermal paths from the chip junction to the environment with an ambient temperature, TA and can be expressed as: θ JA = θ JC + θ CA It should also be noted that the “derating factor” is the reciprocal of the device thermal resistance. Dissipation Derating is commonly defined in milliwatts per oC and is done to protect the maximum junction temperature in high ambient temperature conditions. The allowed dissipation is reduced to zero when TJ = 150oC. Therefore the derating curve intercept point for PD = 0 is at the maximum junction temperature of 150oC. (EQ. 11) The Junction-to-Ambient equations equivalent to Equations 10 and 10A are: T J = T A + P D ( θ JA ) (EQ. 12) and, T A = T J – P D ( θ JA ) The junction-to-ambient thermal resistance, θJA is more correctly defined for the conditions of the application environment. Normally, a reduction in the thermal resistance can be achieved by increasing the square area size of the external heat sink such as to conduct more heat away from the package. Where the heat conducting path from the junction is through the lead frame, to the package pins, added copper ground area on PC Board is needed to reduce θJA. (EQ.12A) Where the ambient temperature and dissipation conditions are subject to user control, we can use Equation 12 and 12A to determine the junction temperature rise or maximum ambient temperature limits. When the maximum junction and ambient temperatures are known, Equation 12A gives the solution for maximum allowed IC dissipation, PD as: ( TJ – TA ) P D = -----------------------θ JA Current Limited, Maximum Dissipation Curves (EQ. 13) Achieving maximum performance from Power BiMOS Drivers with Multiple Outputs requires careful attention to the maximum current and dissipation ratings. While each output has a maximum current specification consistent with the device structure, all such devices on the chip may not be simultaneously rated to the same high current level. In such cases, the total current and dissipation on the chip must be within the maximum allowable ratings. For many applications, all output drivers on a chip are equally loaded. Given equal loading, the following discussion is intended to provide a relatively simple method to calculate the boundary conditions for maximum allowed dissipation and current. Example 1: For the HIP0082 package, θJC = 3oC/W and the worst case junction temperature, as an application design solution, should not exceed 150oC. For a given application, Equation’s. 7 and 8 determine the dissipation, PD. Assume the package is mounted to a heat sink having a thermal resistance of 6oC/W. And, for the given application, assume the dissipation is 3W and the ambient temperature (TA) is 100oC. From Equation 11, θJA is 9oC/W. The solution for junction temperature using Equation 12 is: T J = 100°C + [ 3W ( 9°C ⁄ W ) ] = 127°C Example 2: Assume for the HIP0080, θJA = 30oC/W when mounted on a PC Board with good heat sinking characteristics. Again, the worst case junction temperature, as an application design solution, should not exceed 150oC. Assume for this application, the dissipation, PD = 1.5W. Given a maximum junction temperature of 150oC, we can determine the maximum allowable ambient temperature from Equation 12A as follows: For equal current loading in each output driver, assuming saturated Turn-ON; the maximum current versus ambient temperature may be graphically plotted. To derive the curve parameters for given ratings, we need to substitute the expression for dissipation from Equation’s. 7 and 8 into the junction temperature Equation 10A. Where the currents from each output driver are equal (i.e. I = I1 = I2 .....= Ik.....= In), we have the following maximum current solution based on total dissipation: T A = 150°C – [ 1.5W ( 30°C ⁄ W ) ] = 105°C ( TJ – TC ) 2 P D = n ( P k ) = n ( I R DSON ) = ------------------------( θ JC ) Example 3: For the HIP0080 mounted on a PC Board with a 8 (EQ. 14) Application Note 9416 TJ –TC 2 P D = n x I x R DSON = -------------------( θ JC ) HIP0081, HIP0082, OR CA3282 ASSUMING THEORETICAL LIMITS AND AN INFINITE HEAT SINK, θJC = θJA = 3oC/W WHERE 14 12 MAX. OUTPUT FOR EQUAL LOADS (A) HIP0081, HIP0082 OR CA3282 WITH AN EXTERNAL 6oC/W HEAT SINK, FOR A TOTAL θJA = 9oC/W 16 DISSIPATION WATTS (W) FOR n OUTPUT DRIVERS, HIP0080 WITH θJA = 35oC/W USING PC BOARD COPPER TO HEAT SINK THE 28 LEAD PLCC PACKAGE 10 8 6 4 2 0 -50 -25 0 25 50 75 100 AMBIENT TEMPERATURE (oC) 125 150 MAX. OUTPUT DRIVE CURRENT (A) FIGURE 8. DISSIPATION DERATING CURVES SHOWING MAXIMUM DISSIPATION vs AMBIENT TEMPERATURE. THE SLOPE OF EACH CURVE IS THE RECIPROCAL OF THERMAL RESISTANCE FOR THE CONDITIONS NOTED AND THE SAFE OPERATING AREA FOR EACH CURVE IS BELOW THE LINE 2 1 HIP0082 MAX. IO 2.0 3 9.0 8.0 I = ( TJ – TC ) ------------------------j j = n ( θ JC x R DSON ) j = 1.5 7.0 j=3 6.0 5.0 j=6 4.0 j = 12 3.0 2.0 1.0 0.0 30 40 50 60 70 80 90 100 110 120 130 140 150 CASE (HEAT SINK TAB) TEMPERATURE (oC) FIGURE 9. GENERAL GRAPH SOLUTION FOR MAXIMUM OUTPUT DRIVER CURRENT, I vs CASE TEMPERATURE, TC GIVEN EQUAL LOAD CURRENTS AND A MAXIMUM JUNCTION TEMPERATURE, TJ = 150oC RDSON 0.4Ω = 1 0.5Ω = 2 4 5 0.6Ω = 3 1.5 0.8Ω = 4 HIP0081 MAX. IO 1.0Ω = 5 MAX. ALL ON CURRENT LIMIT (EQUAL CURRENT) θJC = 3oC/W, n = 4 1.0 0.5 HIP0082 OUTPUTS 3, 4 RDSON 0.57Ω AT 2A HIP0082 OUTPUTS 1, 2 RDSON 0.62Ω AT 2A HIP0081 OUTPUTS 1 - 4 RDSON 0.5Ω AT 1A HIP0081 OUTPUTS 1 - 4 RDSON 1.0Ω AT 1A (5A OUTPUTS) (2A OUTPUTS) (EA. OUTPUT) (LOW VCC OPERATION) 0.0 50 75 100 125 150 CASE (TAB) TEMPERATURE (oC) FIGURE 10. HIP0081 AND HIP0082 MAXIMUM CURRENT vs CASE TEMPERATURE, ALL 4 OUTPUTS ON WITH EQUAL CURRENT Solving for current, I gives: I = ( TJ – TC ) ----------------------------------nθ JC R DSON current. The curve coordinates are output current, I vs Case Temperature, TC. The maximum total current limits of 5A for the HIP0081 and 8A for the HIP0082 directly relate to the Absolute Maximum Ratings as defined in the datasheet. (EQ. 15) The number of output drivers ON and conducting may be from n = 1 to 4 for quad drivers such as the HIP0081 and HIP0082 or n = 1 to 8 for an octal driver such as the CA3282. Maximum temperature, dissipation and current ratings must be observed. For n equal conducting output drivers, we can now plot the maximum rated current, I vs the maximum case temperature, TC. Inductive Load Switching - Pulse Energy And Dissipation Calculations To expand on the subject of Inductive Loads, it is necessary to find the solution for energy loss that results from active switching of inductive load current, including the dissipation from the flyback pulse when the output is turned off. When the energy loss is found, duty factor and frequency variables may be applied to calculate dissipation. The denominator term in the radical of Equation 15, nθJCRDSON, is used to plot a set of normalized curves, as shown in Figure 9. Using the HIP0081 and HIP0082 as examples, Figure 10 illustrates the boundaries for temperature and current for a range of RDSON with all four outputs ON and equal output The time integral of instantaneous power is used to determine energy during Turn-ON and Turn-OFF. The indicated 9 Application Note 9416 limit the flyback voltage and is used in the CA3282, HIP0080, HIP0081 and HIP0082 Low Side Drivers. Since the power dissipated in a pulse clamped mode of operation is dissipated in the IC, a determination of the pulse energy is needed to protect the rating limits of the IC. The following solutions used here are general and may be adapted to any of the Power BiMOS IC types. solution is given in Equations 16 and 16A where the instantaneous power, P(t) = V(t)I(t) or I(t)2R, is integrated over a given time period. The answer is defined in joules (watt-seconds) and the resistance, R is the series resistance in which the energy, E is to be calculated. E t = 0 or t ∫ V( t ) I( t)dt (EQ. 16) 0 t E = 0 t 2 ∫ I( t ) The general equation for instantaneous transient current in the series loop of a step switched inductive load is: (EQ. 16A) R dt I L( t ) = I O ( 1 – A e – t / T ) 0 The general solution of Turn-ON and Turn-OFF for an RL circuit is a well documented transient differential equation. The expression for voltage is usually defined for a fixed step function but may be more complex. Given the proper values for the initial conditions, this solution applies to either Turn-ON or Turn-OFF of an inductive load. The Initial Current at t = 0 is IB and the Steady State Current, for t » T , is IO. The circuit time constant, T is L/R where R is the total series resistance in the loop and L is the inductance. Since different Turn-On and Turn-OFF equation parameters apply, the initial conditions differ for each case. For example, the loop value of R is different in the Turn-ON and Turn-OFF periods. While there may or may not be an initial current at Turn-ON, the Turn-OFF condition has an implied initial current. BATT Inductive Load Turn-On LOAD IL When the MOS input is switched ON, the I2R expression for instantaneous power may be used for the integral solution of energy loss in RDSON of the Output Driver. At Turn-ON, the general solution of energy loss in the series resistance, R; is found by substituting Equation 17 into Equation 16A. The integral equation is: L RL VZ t E RON = IL L V IL MOS TURN-ON FIGURE 11B. EQUIVALENT ON STATE CIRCUIT (EQ. 18) t 2t – --2 – ----- 2 T A T T E RON = I O R t + 2A Te – -----------e 2 BATT RL + t (EQ. 18A) 0 The current, IL(t) flows in the MOS output and, over time, dissipates power in the drain-to-source channel resistance, RDSON. Since the Equation 18A is a general solution of energy loss in the series resistance of an RL circuit, we may treat R as the sum of all series resistance in the RL loop. And, if needed to limit current in the load, additional resistance may be added in series external to the IC and Coil as part of the resistive load. L RL RDSON t 2 – --- 2 T IO 1 –A e R dt When the squared term is expanded and integrated, the solution for Turn-ON Energy in the resistance, R becomes FIGURE 11A. BATT ∫ 0 IC MOS OUTPUT DRIVER V IB A = 1 – ----- I O where For the illustrated conditions shown in Figure 11, we have a low side MOS driver with an RL load. The initial condition for current at time t = 0 may be other than zero. To assure that an inductive kick pulse (flyback) at Turn-OFF does not exceed the ratings of the MOS output driver, a zener diode feedback circuit is used to clamp the output pulse. The zener diode is used as a feedback clamp to turn on the output stage when the zener diode threshold voltage, VZ plus the gate threshold voltage, VGS(TH) is exceeded. V (EQ. 17) VCL MOS If R = RL + RDSON, the ERDSON energy loss in an IC output is the RDSON term expression, t TURN-OFF FIGURE 11C. EQUIVALENT OFF STATE CIRCUIT DURING INDUCTOR DISCHARGE t 2t – --2 – ----- 2 T A T T E RDSON = I O R DSON t + 2A Te – -----------e 2 FIGURE 11. TYPICAL MOS DRIVER OUTPUT CIRCUIT WITH A SWITCHED INDUCTIVE LOAD. AN INTERNAL ZENER DIODE VOLTAGE CLAMP IS USED FOR FLYBACK OVER-VOLTAGE PROTECTION (EQ. 19) 0 We may use Equation 19 to find the complete Turn-ON energy, ERDSON loss in each MOS output driver. Note that the time limit is not restricted and may continue to apply for the steady state ON condition. Given the following The circuit shown in Figure 11 is a common method used to 10 Application Note 9416 symbol definitions and initial conditions, the complete energy solution may now be calculated. For Figure 11, the Turn-ON circuit parameters are: IL(t) VBATT VCL L, RL T Instantaneous Inductor Load Current Output Load Power Supply Pulse Clamp Voltage, VCL = VZ + VGS(TH) Load Inductance and Resistance Time Constant, T = L/R where R = RL + RDSON IO V BATT Steady State Current, I O = -----------------for t » T R And ERDSON = 32.02 mJ. For the Short Cut method, using Equation 20A PDON = RDSONIO2 = 0.5 x (1.35)2 = 0.911W For ERDSON = (k-1.5)T PDON = 3.5TPDON = 31.89mJ. For switching conditions such as may be encountered when driving injectors and motors, total energy increases with frequency. Also, for large inductors with continuous switching, Turn-ON may not be completed before Turn-OFF; making the IB term significant and the Turn-ON Energy much larger. Initial Current flowing at t = 0. IB RDSON MOS ON State Channel Resistance PDON Steady State Dissipation, IO2RDSON Inductive Load Turn-off Expanding Equation 19, the complete expression for TurnON energy loss in the IC over time t = 0 to t is: The Turn-OFF Energy, ECLAMP in the MOS output may add a significant contribution to the power dissipation in the IC. Also, short duration pulse energy must be determined for a safe rating check. Turn-OFF current may be calculated using Equation 17, given a value for the instantaneous current, IB as an initial condition at switch-off. In the clamp mode, the MOS drain voltage is fixed at VCL and may be regarded as a fixed voltage drive source, but only while the clamp is sustained. The degree of MOS turn-on varies according to the sustaining needs of the clamp. For this reason, the I2R approach used for Turn-ON does not apply. The best approach for the TurnOFF energy solution is to use the P(t) = V(t)I(t) version of the integral power equation where V(t) = VCL during the clamp period. The Turn-OFF energy equation is: –t – 2t ----------- 2 2 T A T T A T E RDSON = P DON t + 2ATe – ----------- e – 2AT – ----------- 2 2 –t – 2t ---- -------- 2 T T A T = P DON t – 2 AT 1 – e + ----------- 1 – e 2 (EQ. 20) An approximation solution for ERDSON is impractical because the energy value will continue to increase with time. However, as the time is extended, we may assume that e –t / T terms of Equation 20 will diminish to a small value and further reduce the complexity of the Turn-ON calculation. By substituting t = kT, where k gives us a scaled value of T, and then increasing k to a large value reduces the exponential terms to zero. If the initial current, IB is zero, then A = 1. After making these substitutions in Equation 20, we have: 1 E RDSON = P DON T k – 2 + --- = ( k – 1.5 ) T PDON 2 k = 5; t E CLAMP = ∫ ( VCL I L( t ) ) dt (EQ. 21) 0 Since current must be continuous during switching, the current flowing in the inductor at switch-off becomes the Initial Current, IB in the current equation. To compensate for the switching discontinuity, a flyback voltage pulse proportional to the differential rate of current change is generated. For an unclamped pulse, (EQ. 20A) which applies for t ≥ 5T, i.e. k ≥ 5. The Equation 20A result indicates that we could have calculated the steady state energy for a kT period of time and then subtracted 1.5(TPDON) to account for the reduced energy during the transient Turn-ON interval. We may use Equation 20A for a short cut calculation, but only for time periods greater than 5T. The error is less than 1% for t ≥ 4.7T. di V ( t ) = – L ----- dt The flyback pulse may be quite large, requiring a clamp circuit for over-voltage protection. Example: Given the following data and using Equations 17 and 20, calculate IL(t) and ERDSON. For the equivalent clamp circuit shown in Figure 11, TurnOFF occurs in 3 stages: VBATT = 13.5V 1. Following turn-off, the flyback pulse quickly increases to the clamp voltage level. A small incremental loss occurs before the flyback pulse is clamped. However, this is assumed to be negligible because the Power BiMOS Output drivers typically switch off in less than 10µs. VCL = 82V L = 100mH RL = 9.5Ω 2. The flyback voltage is clamped by the output clamp circuit until the energy stored in the inductor is dissipated. RDSON = 0.5Ω R = RL+ RDSON = 9.5 + 0.5 = 10Ω 3. The output voltage settles to the steady state level, VBATT. Although not discussed here, it is possible that circulating currents may exist as a double energy transients. IO = VBATT/R = 13.5/10 = 1.35A IB = 0 Where parasitic RLC does exists, determination of the actual load circuit model may be a major problem. The designer should be aware that adding capacitance to an inductive switching circuit may create potentially damaging transients. A=1 T = L/R = 100/10 = 10ms Let t = 5T = 50ms Then,IL(t) = 1.341A at t = 5T 11 Application Note 9416 To determine the Energy delivered to the IC during TurnOFF, we are only concerned with the zener diode clamp interval. As shown in the Turn-Off model of Figure 11, we have an inductance and resistance in series, connected between the clamp voltage, VCL and the supply voltage, VBATT. And, as noted, the clamp voltage, VCL, may be treated as a voltage drive source; but only during the flyback period. When the Initial Current, IB decays to zero, the clamp energy is depleted. The energy loss in the zener diode voltage clamp is the energy dissipated in the MOS output driver. The voltage ratio expression for A assumes Turn-OFF is initiated from a steady state ON condition. To avoid limitations, it is better to stay with the current ratio expression for A. As such, the output current at switch-off is equal to the instantaneous value of the ON state current calculated using Equation 17. Given Turn-OFF boundary conditions and symbol definitions similar to Turn-ON, we have the following: When the this equation is integrated over the limits of t = 0 (at switch-OFF) to t = t, we have: IL(t) t – --- T E CLAMP = V CL I O t + A Te VBATT VCL L, RL T t0 IO IB From Equation 21, the integral equation for Clamp Energy, ECLAMP at Turn-OFF is: t E CLAMP = V CL I O ∫ ( 1 – A e – t / T ) dt Instantaneous Inductor Load Current where we must include the Initial Current, IB Output Load Power Supply which is now one of two defined voltage driver sources. Pulse Clamp Voltage, VCL = VZ+VGS(TH) and is a defined voltage drive source but only applies while IL(t) decays from the initial clamp condition at t = 0 and continues in time until IL(t) = 0. Load Inductor, L with Resistance, RL Time Constant, T = L/R where R = RL Turn-OFF time defined here as the active clamp period, t = 0+ to t = t0 where IL = 0 at t0 V t (EQ. 24A) 0 Expanding Equation 24A, we have the general solution for energy loss, ECLAMP in the time interval, t after Turn-OFF which is: t – --- T t E CLAMP = V CL I O T --- – 1 – A e T (EQ. 24B) Our primary interest is to determine the total energy loss during Turn-OFF. Given that t0 = TLn(A), the expression for total Clamp Energy at Turn-OFF is: –V BATT CL Steady State Current I O = ----------------------------------- E CLAMP = V CL I O T ( 1 – A + Ln ( A ) ) with time limited conditions, see VCL symbol definition note. Polarity is defined as positive for the VBATT supply current flow to ground Care is needed to assure that the value for A is carried out to several decimal places because the value of A may be very close to 1, causing the (1-A) term to lose accuracy. RL V BATT Initial Current I B = -----------------at t = 0. VBATT = 13.5V The direction of initial current at Turn-OFF follows that of Turn-ON conditions because current must be continuous during transient changes. The clamp voltage, VCL is typically much greater than VBATT, which gives an IO (steady state) current that is negative. However, IO does not actually reach a steady state condition because the clamp ceases to be active at t = t0 (when IL(t) goes to zero). (At Turn-OFF, to carry the integration solution for Energy past the IL(t) = 0 point will provide a fictitious answer). VCL = 82V L = 100mH, R = RL = 9.5Ω IO = (VBATT-VCL)/R = -7.21A IB = use the calculated values from Turn-ON for t = 5T, IB = 1.341A (~VBATT/R); A = 1 - (1.341/(-7.21)) = 1.18596 To simplify the Turn-OFF calculations, energy loss is assumed to begin in the voltage clamp at t = 0 when the output driver is switched OFF. Since IL(t) = 0 at t = t0 , we can set IL(t) = 0 and solve for t0 from the equation: = IO ( 1 – A e –t / T ) = 0 T = L/R = 100mH/9.5Ω = 10.526ms Then t0 = (10.526ms) Ln(1.18596) = 1.7953ms We can now calculate energy for the assumed example values. We should first note that t0 is less than T, requiring a full detailed calculation without approximations. However, unlike the Turn-ON solution which goes to a steady state condition, the Turn-OFF Energy calculation is complete when t = t0. The solution using Equation 25 is: (EQ. 17) which provides the solution, t 0 = TLn ( A ) (EQ. 22) where Ln(A) is the natural log of A. ECLAMP = (82)(-7.2)(10.5ms)(1-1.186+Ln(1.186)) = 95.9mJ For the circuit conditions illustrated in Figure 11, we can now derive our current equation for IL(t) using Equation 17 by substituting the new boundary conditions defined for Turn-OFF. To begin, we have: V CL IB A = 1 – ----- = ------------------------------------ I O V CL – V BATT (EQ. 25) Example: Using component values equal to those given for Turn-ON, and a defined initial current condition for t = 5T after Turn-ON; we have for Turn-OFF: RL (or, where otherwise defined, IB is the instantaneous current, IL(t) that is flowing at switch-off). I L( t ) (EQ. 24) 0 For the examples given above, the Turn-ON and Turn-OFF current curves are plotted in Figure 12. L and RL Energy Calculations (EQ. 23) 12 Application Note 9416 At Turn-ON, the load resistance dissipates power as an energy loss while the coil inductance stores magnetic field energy. The general equation for resistive energy loss, ER in the total series load resistance is given in Equation 16A. Since we have the same integral equation for each series component term of the loop, the Turn-ON and ON state energy loss equation for RL is: t 2t – --2 – ----- 2 T A T T E RLON = R L I O t + 2A Te – -----------e 2 As previously noted, the Turn-ON energy loss equation continues in time as a steady state condition. Therefore, Equation 27 implies that the energy loss is over the same exact time duration. At Turn-OFF, the period of integration is from switch off at t = 0 to t = t0, in which case we may then substitute t = t 0 = TLn ( A ) The solution is: t 2 1 E RLOFF = R L I O T --- ( A – 3 ) ( A – 1 ) + Ln ( A ) 2 (EQ. 26) The parameter values differ in the Turn-ON and Turn-OFF interval, as previously noted for the ERDSON and ECLAMP examples. 0 And, evaluating from t = 0 to t, we have: –t – 2t ----------- 2 2 2 T A T T A T E RLON = R L I O t + 2ATe – ----------- e – 2AT – ----------- 2 2 –t – 2t ---- -------- 2 2 T T A T = R L I O t – 2 AT 1 – e + ----------- 1 – e 2 Example: Given the same conditions used for ECLAMP TurnOFF and using Equation 28, calculate the value of ERLOFF. 2 E RLOFF = ( 13.5 ) ( – 7.2 ) ( 10.5ms ) ( Ln ( 1.186 ) – 0.1687 ) (EQ. 26A) Then, E RLOFF = 9.79mJ . At Turn-ON, the RDSON energy loss expression, ERDSON and the expression for ERLON are the same. Therefore, we can determine the Turn-ON energy loss for RL by a ratiocomparison. (Refer to Equation 19). RL E RLON = E RDSON -------------------R DSON (EQ. 28) As noted in Equation 9, the stored energy in an inductor for a given current is: 2 I E L = L ---- 2 (EQ. 27) Using the value of IB at t = 5T from the Turn-ON example, the stored inductive energy is: . TURN-OFF INDUCTOR CURRENT (A) 2 ( 1.34 ) E L = 100mH ------------------- = 89.9mJ . 2 1.8 TURN-ON In the above solutions, it was assumed that the value of the coil inductance was a fixed constant value. Inductance will vary to some degree with current and temperature. The variation will dependent on the mechanical design and the quality of the ferrite material that is an integral part of the coil. Generally, as current increases, inductance will decrease. If the core material saturates, the inductance will substantially decreased. 1.5 1.2 0.9 TURN-OFF 0.6 0.3 0.0 1.2 Where the RL load is not constant, it is possible to calculate current and energy in incremental steps. An indicator of this need would be evident in a non-exponential current rise at Turn-ON, as viewed with an oscilloscope. The inductance parameters vs current and temperature should be measured and a new time constant should be applied to each stepped increment, based on the inductance vs current data. New initial current conditions must be used for each step. 0.9 Sourced Power Supply Energy at Turn-OFF 0 5 10 15 20 25 30 35 40 45 TURN-ON, TURN-OFF TIME (ms) 50 TURN-OFF INDUCTOR CURRENT (A) FIGURE 12A. 1.8 1.5 During the discharge period, energy is sourced from the magnetic field of the inductor and from the power supply. The solution for the VBATT sourced energy, EBATT follows the same calculation method used in the clamp energy, ECLAMP solution. 0.6 t = t0 0.3 0.0 0.0 (50) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 (55) TURN-OFF TIME (EXPANDED SCALE)(ms) The expression for sourced energy from the battery is: t0 FIGURE 12B. E BATT = FIGURE 12. TRANSIENT TURN-ON AND TURN-OFF CURRENT CURVES FOR THE SWITCHED INDUCTOR LOAD EXAMPLE HAVING A TIME CONSTANT T = L/R = 10ms, VBATT = 13.5V, VCL= 82V, L = 100mH, RL = 9.5Ω AND RDSON = 0.5Ω ∫ VBATT 0 I L( t ) dt (EQ. 29) When integrated and evaluated over the limits, t = 0 to t = t0 = TLn(A), E BATT = V BATT I O T ( 1 – A + Ln ( A ) ) 13 (EQ. 29A) Application Note 9416 10000 HIP0082 SWITCHED INDUCTIVE LOAD CLAMP GRAPH PLOT OF CLAMP ENERGY VS TURN-OFF TIME FOR L = 1, 10, 100, 1,000mH; VCL = 90V, VBATT = 15V, R VARIED REFER TO TABLE 3 DATA FOR PLOT PTs 1, 2, 3 PT’s 4 - 7; PLOT 125oC, MAX. ON CURRENT POINTS OTHER DATA POINTS FOR 1A, 2A OPERATION (“ON” CURRENT REFERS TO lL (t) AT SW. OFF) NOTE: OUTPUTS 3, 4 RATED 5A MAX. OUTPUTS 1, 2 RATED 2A MAX oC T MI 1000 A SO PT9 (1.08A) SINGLE PULSE ENERGY (mJ) EA TE RA IN S HI AR T PE DO TO NO PT1(2A) 100 LI oC 5 T MI A SO 25 12 LI PT7(0.7A) PT6(1.34A) PT8 (3.87A) (1A) PT2(3A) PT5(2.5A) PT3(2A) 4A PT4(4.8A) A (I H 0m 10 5A 12 0. IN D 10 . , L 00 mH = AV C L) 25 0. mH (2A) 10 1.0 mH 0. 5A 1A 2A 10 1 0.01 0.1 1 10 100 TURN-OFF PULSE WIDTH (ms) FIGURE 13. THIS SAFE OPERATING AREA (SOA) CHART FOR THE HIP0082 ILLUSTRATES THE ESSENTIAL REQUIREMENTS FOR OPERATION WITHIN THE MAXIMUM CURRENT AND ENERGY CONSTRAINTS. IN ADDITION, INITIAL OPERATING Except for VBATT in place of VCL, this is the same integral solution as the one for ECLAMP. While noting that EBATT is sourced energy and ECLAMP is lost energy, we may solve for EBATT as a ratio of ECLAMP. external resistor load and the voltage clamp. To check our calculations, the balanced energy equation is: E BATT + E L = E RLOFF + E CLAMP Using ratio method, the EBATT equation is: V BATT E BATT = E CLAMP -------------------- V CL (EQ. 31) Filling in the numbers from the above calculation, we have ( 15.8 + 89.9 ) = ( 9.8 + 95.9 ) = 105.7 mJ. (EQ. 30) Total IC Energy Calculation and EBATT calculates to be: The total energy, loss in an output driver is the sum of the Turn-ON period loss plus the Turn-OFF loss. The calculation for ERDSON gave the RDSON Energy loss in the output driver over Turn-ON plus ON time. For t = 0 to t = 5T in the illustrated example, the loss was 32.02mJ. During the Turn-OFF time interval, the energy loss in the IC voltage clamp, 13.5 E BATT = 95.9 × ----------- = 15.79mJ 82 Turn-OFF Source and Loss Energy Check During the Turn-OFF period, the sourced power supply energy and Inductor energy should equal the energy loss in 14 Application Note 9416 ECLAMP was calculated as 95.9mJ. For each Turn-ON and Turn-OFF cycle, the total energy loss, ET in the output driver is: drive. In Figure 13 the currents shown in parenthesis are the saturated ON levels of current at the time of switch-off. An average value of the output pulse current will more closely correlate to the plotted current lines in Figure 7. If we integrate the Turn-OFF pulse current over time t0, the result is: E T = ( E RDSON + E CLAMP ) = ( 32.02 + 95.9 ) = 127.9 mJ. The illustrated energy calculation based on a Turn-ON period of 5 times the L/R time constant is an arbitrary example. The user may choose any time interval for Turn-ON. I AVCL = I O + I B ⁄ Ln ( A ) (EQ. 32) The current during the clamp interval is slightly less than one-half of IB . The ratio of IB to IAVCL is typically in the 0.46 to 0.49 range. Inductive Load Switching Dissipation Calculations Device dissipation is the energy in joules (watt-seconds) for a unit period of time. Dissipation may also be derived from the calculated energy for one switching cycle times the switching frequency, f in hertz. The Turn-ON time for t = 0 to t = 5T is 50ms and the Turn-OFF time is 1.8ms. If we assume repetitive switching with 50% ON time, the switching frequency is 10Hz. The calculated dissipation is: It is important to note that load current at Turn-OFF (shown in parenthesis) will remain the same as long as the supply voltage, VBATT and the R, L values remain the same. As such, the plotted inductance curves specifically relate to the assumed values for VBATT and VCL. While other parameters remain the same, a changing clamp voltage will generate different values for clamp energy, ECLAMP and Turn-OFF Time, t0. Increasing the clamp voltage will decrease the Turn-OFF time, t0 and cause higher dissipation for shorter periods of time, reducing the SOA margin. P D = ( E T × f ) = ( 127.9 × 10 ) = 1279 mW. To carry the results further, assume this is an HIP0082 Quad Driver with all 4 outputs switching into an equivalent load. Then, the total dissipation is 4x1.28 = 5.12W. If we use a 6oC/W external heat sink, the total θJC + θCA is 9oC/W. Using Equation 12A and 150oC for the maximum junction temperature, the solution for the maximum ambient temperature is: Sustained dissipation increases the chip junction temperature, reducing the SOA margin for safe operation. The SOA margin is reduced, as indicated by the SOA limits for 125oC operation vs 25oC operation. T A = T J – P D × θ JA = ( 150 – 5.12 × 9 ) = 104 °C. Bibliography Information [1] Reference Data for Radio Engineers, Sixth Edition Howard W. Sams & Co. Using the Inductive Switching SOA Curve Having defined the variables of inductive switching and equation solutions to calculate current and energy, we can now evaluate and tabulate the results. The key switching parameters for the Intersil family of Power BiMOS Low Side Drivers are given in Table 3. Using the IC specifications and a C Program (see the Appendix), current and energy data is generated for various points of operation. [2] Electrical Engineers’ HANDBOOK, Second Edition Fink and Christiansen, McGraw-Hill [3] Advanced Mathematics for Engineers, Third Edition H. W. Reddick and F. H. Miller [4] Heat Transfer in Electronic Equipment by K. L. Smith, Ph.D Electronics & Wireless World, Aug. 86 Figure 13 illustrates the how RL load current and energy data is applied to the HIP0082 Inductive Switching SOA Curve of Figure 7. For defined voltage conditions, a linear curve is plotted for different inductive load values. (The inductance curves are not intended as rating information.) The load resistance is varied to generate points on each inductive load curve. [5] Electrothermal Modeling of the Bipolar Transistor in cdsSpice for use in FastTrack, 8/4/94 by David T. Zweidinger, (Intersil Document) [6] NIST Special Publication 400-86, Dept. of Commerce Thermal Resistance Measurements, July 1990 F. F. Oettinger and D. L. Blackburn As an example, the Table 3 data for the HIP0082 tabulates current operation at 2A in an inductive load of 10mH. Data point PT3 is a plot of the Clamp Energy, ECLAMP = 21.14mJ and Turn-OFF time, t0 = 0.243ms. [7] Intersil Data Book, DB304.1, 1994, Intelligent Power ICs for Commercial, Industrial and Automotive Applications. [8] Lange’s Handbook of Chemistry, Twelfth Edition McGraw Hill, Editor: John A. Dean We should note as a reminder that the energy stored in the inductor (LI2/2) is not equal to the clamp energy at Turn-OFF. Additional source energy is supplied from the power supply to the clamp during the Turn-OFF interval. For the examples used here, this added contribution is small. However, if the clamp voltage level is reduced, the additional source energy will increase. [9] Transient Analysis in Electrical Engineering by Sylvan Fich Prentice-Hall, Inc. For fixed values of inductance, a plot of increasing current intersects the SOA limits to define the maximum rated output 15 Application Note 9416 TABLE 3. POWER BIMOS LOW SIDE DRIVERS - FEATURES AND INDUCTIVE SWITCHING CHARACTERISTICS HIP0080 QUAD HIP0081 QUAD HIP0082 QUAD CA3282 OCTAL 27 43 73 89 72 90 27 40 V 1 2 2A Outputs 1 and 2 5A Outputs 3 and 4 1 A 1.0Ω at 0.5A 0.5Ω at 1A 0.62Ω at 2A Outputs 1 and 2 0.57Ω at 2A Outputs 3 and 4 1.0Ω at 0.5A Ω A Minimum Output Current Limiting (Latch-OFF Threshold) 1.3 2.2 2.1A Outputs 1 and 2 5.1A Outputs 3 and 4 1.05 A Thermal Shutdown or Flag SD SD Flag None - Fault, Diagnostics, Feedback Yes Yes Yes Yes - 28 PLCC 15 SIP 15 SIP 15 SIP - 14 3 3 3 oC/W PARAMETER Output Voltage Clamp Range Min. Max. Maximum Rated DC Load Current Maximum RDSON Output Resistance Plastic Package Junction-to-Case Thermal Resistance UNITS ENERGY DATA FOR INDUCTIVE LOAD SWITCHING: Battery Supply = 15V; Clamp Voltage and RDSON at Maximum Specified Value. Data shown for 5T Turn-ON Time, then Turn-OFF. (T = L/R Time Constant) Output Voltage Clamp (Max.), VCL 43 89 90 90 90 40 V Output Operating Current, IO 0.5 1 2 2 3 0.5 A Output Drain Resistance, RDSON 1 0.5 0.6 0.6 0.6 1 Ω Load Resistance, RL 29 14.5 6.9 6.9 4.4 29 Ω Load Inductance, L 100 100 100 10 10 100 mH Turn-ON Time Const., T = L/(RL+RDSON) 3.33 6.67 13.33 1.333 2 3.33 ms Turn-OFF Time Const., T = L/(RL) 3.45 6.9 14.49 1.449 2.273 3.45 ms Turn-ON Time, t = 0 to t = 5T 16.67 33.33 66.67 6.67 10 16.67 ms Turn-OFF Time, t = t0 1.43 1.23 2.43 0.243 .366 1.57 ms Load Current at t = 5T (Turn-OFF Initial Current Condition) 0.497 0.993 1.987 1.987 2.98 0.497 A Average Current During Clamp Period 0.231 0.482 0.966 0.966 1.45 0.23 A IC Turn-ON Energy, t=0 to t = 5T; ERDSON 2.93 11.71 112.43 11.24 37.95 2.93 mJ Stored Inductor Energy, at t = 5T; EL 12.33 49.23 197.31 19.73 44.4 12.33 mJ IC Turn-OFF Clamp Energy, ECLAMP 14.23 52.6 211.38 21.14 47.78 14.41 mJ IC Turn-ON + Turn-OFF Energy 17.15 64.31 323.81 32.38 85.73 17.33 mJ Frequency for 50% duty cycle (on time) 30 15 7.5 75 50 30 Hz Dissipation, Single Output Switching 515 965 2429 2429 4286 520 mW ENERGY DISSIPATION 16 Application Note 9416 Appendix - A1: Inductive Load Calculations - C Program and Data Run POWER BiMOS INDUCTIVE SWITCHING EXAMPLE CALCULATIONS kT Steps ----1 2 3 4 5 Time in ms ----10 20 30 40 50 Load Is,Amps -----0.8534 1.167 1.283 1.325 1.341 Turn-ON Eon,mJ ------1.532 6.939 14.56 23.11 32.02 Turn-ON Eon S/C -------4.556 4.556 13.67 22.78 31.89 % Eon S/C Error ------397.5 34.34 6.152 1.438 0.3829 LOAD Input Data: HIP0082_TYPICAL_DATA Li= 100 mH, Ri= 9.5 ohms, Vbatt= 13.5 V, Vclamp= 82 V, Rd= 0.5 ohms **************************************************************************************************** V BATT Turn-ON: Output Load Current and IC/Rdson Energy Loss Tabulation Initial Current is Ib=0, Steady State Current, Io=Vbatt/(Ri+Rd) IO Is = Is(t) = Instantaneous Load Current LI Time in 10 ms steps, t=0 to t=50 ms, T=[L/(Ri+Rd)]=10 ms RI VCLAMP RD MOS (ON) IC MOS OUTPUT DRIVER k k*to/5 Steps -----0 1 2 3 4 5 Time in ms -----0 0.359 0.718 1.08 1.44 1.8 Load Current Is,Amps -----1.34 1.05 0.777 0.509 0.25 -0 Trn-OFF Energy Ec,mJ ------0 35.23 62.17 81.08 92.24 95.9 Average Current during Turn-OFF, Iavcl = (Io+Ib/lnA) = Initial Current at Switch OFF from t=5T ON State, Ib = Ratio: Iavcl to Ib, Initial ON Current = Delta Energy Inc. -----0.0 35.2 26.9 18.9 11.2 3.66 LOAD **************************************************************************************************** Turn-OFF: Output Clamp Current and Energy Tabulation Note: At t=to, Inductive Discharge Current depletes to Zero Amps V BATT Initial Current, Ib = Turn-ON step at t=5T IO Steady State Current, Io=(Vbatt-Vclamp)/Ri LI Inc.= 5 steps from t=0 to t=to= 1.7953 ms, T=(L/Ri)= 10.53 ms RI VCLAMP RD MOS (ON) IC MOS OUTPUT DRIVER 0.6514 1.3409 0.485794 **************************************************************************************************** Application Note Example Energy Source/Loss Tabulation Turn-OFF Sourced Energy, Eb(Batt.),Ei(L) from Switch OFF to t=to: Sourced Battery Energy, Eb: 15.788 mJ Stored Coil Energy at Switch-OFF, Ei: 89.901 mJ Total Sourced Turn-OFF Energy, Eb+Ei: 105.69 mJ Turn-OFF Energy Loss, Eclamp(IC), Er(Ri) from Switch OFF to t=to: Clamp Energy, Eclamp: 95.898 mJ Load Resistor Energy, Er: 9.7915 mJ Total Loss Turn-OFF Energy, Er+Eclamp: 105.69 mJ Total IC ON+OFF Energy loss, t=0 --> t= 5T --> t= 5T+to: Turn-ON Energy Loss for t=0 to t= 5T, Eon: Clamp Energy Loss at Switch Off, Eclamp: Total IC Energy Loss, Et=Eoff(0--> 5T)+Eclamp: 32.016 mJ 95.898 mJ 127.91 mJ Example Dissipation for ON/OFF cycle, t=0 --> t= 5T --> t= 5T+to: Total ON/OFF Switch Time (ton+toff)=( 50+ 1.8)= 51.8 ms For Each MOS Switching Output, Dissipation = Energy x Frequency: For repetitive switching, 50% ON time, Freq, f = 10 Hz Single Switch Dissipation, Pd = (10 x Et) = 1279 mW 17 Application Note 9416 Appendix - A2: Inductive Load Calculations - C Program and Data Run (Continued) /* Power #include #include #include #include BiMOS Application Note Inductive RL Switching Examples, <stdio.h> <sys/ieeefp.h> <floatingpoint.h> <math.h> rev 8/11/94 */ main() { char Name[10]; double R,Rd,Ri,Li,T,Vb,Vc,A,Io,Ib,t,to,tson,Is,Ison; double x,y,Eon,Eson,Eonsc,Eoff,Ei,Eb,Ec,Ecs,Eroff; int k,n=5,step=1; printf("POWER BiMOS INDUCTIVE SWITCHING EXAMPLE CALCULATIONS\n\n"); optional data entry, unix redirection from data file or type in by next 2 line prompt */ printf("Enter Data, Use only white-space or <CR> separation\n"); */ printf("Input: Name Vbatt, Vclamp, Li, Ri, Rd \n"); */ scanf("%s %lf %lf %lf %lf %lf", Name,&Vb,&Vc,&Li,&Ri,&Rd); printf("Input Data: %s\n", Name); printf("Li= %g mH, Ri= %g ohms, Vbatt= %g V, Vclamp= %g V, Rd= %g ohms\n", Li*1000, Ri, Vb, Vc, Rd); /* /* /* /* Turn ON example */ R=Ri+Rd; T=Li/R; Io=Vb/R; Ib=0; /* Assumed value for this example */ A=1-(Ib/Io); printf("********************************************************************\n"); printf("\nTurn-ON: Output Load Current and IC/Rdson Energy Loss Tabulation\n"); printf(" Initial Current is Ib=0, Steady State Current, Io=Vbatt/(Ri+Rd)\n"); printf(" Is = Is(t) = Instantaneous Load Current\n"); printf("Time in %2.2g ms steps, t=0 to t=%2.3g ms, T=[L/(Ri+Rd)]=\t%2.4g ms\n\n",step*T*1000, n*T*1000,T*1000); printf(" kT \t Time\t Load\tTurn-ON\tTurn-ON\t %% Eon\n"); printf("Steps\tin ms\tIs,Amps\tEon,mJ\tEon S/C\t S/C Error\n"); printf("-----\t-----\t------\t-------\t-------\t-------\n"); for ( k=1; k<=n; ++k ) { t=k*step*T; Is=Io*(1-A*exp(-t/T)); Eon=Rd*Io*Io*((t+2*A*T*exp(-t/T)-A*A*T/2*exp(-2*t/T))-(2*A*T-A*A*T/2)); Eonsc=(k-1.5)*T*Rd*Io*Io; printf(" %d\t% .4g\t% .4g\t% .4g\t% .4g\t% .4g\n", k,t*1000,Is,Eon*1000,Eonsc*1000,100*(EonEonsc)/Eon); if ( k == n ) { Ison=Is; Eson=Eon; tson=t; } } /* Turn OFF example */ R=Ri; T=Li/R; Io=(Vb-Vc)/R; Ib=Ison; /* From Turn-ON step data, Is at n_th step */ A=1-(Ib/Io); to=T*(log(A)); /* printf("to=%g, A= %g, T= %g, Io= %g, Ib= %g, log(A)= %g\n",to,A,T,Io,Ib,log(A)); */ printf("******************************************************************\n"); printf("\nTurn-OFF: Output Clamp Current and Energy Tabulation \n"); printf(" Note: At t=to, Inductive Discharge Current depletes to Zero Amps\n"); printf(" Initial Current, Ib = Turn-ON step at t=5T\n"); printf(" Steady State Current, Io=(Vbatt-Vclamp)/Ri\n"); 18 Application Note 9416 Appendix - A3: Inductive Load Calculations - C Program and Data Run (Continued) printf("Inc.= %d steps from t=0 to t=to=% .5g ms, T=(L/Ri)=\t% .4g ms\n\n", n,to*1000,T*1000); printf(" k \t Time\t Load \tTrn-OFF\t Delta\n"); printf("k*to/5\t in\tCurrent\tEnergy\t Energy\n"); printf(" Steps\t ms\tIs,Amps\tEc,mJ\t Inc. \n"); printf("------\t------\t------\t------\t------\n"); Ecs=0; for ( k=0; k<=5; ++k ) { t=k*to/5; Is=(Io)*(1-A*(exp(-t/T))); Ec=Vc*Io*T*((t/T)-A*(1-exp(-t/T))); x=Ec-Ecs; if ( k==0 ) printf("%d\t%2.3g\t% .3g\t % .4g\t %s\n", k, t*1000,Is,Ec*1000," 0.0"); if ( Is > 0 && k >= 1 ) { printf("%d\t%2.3g\t% .3g\t % .4g\t % .3g\n", k, t*1000,Is,Ec*1000,x*1000); Ecs=Ec; } } Is=(Io)*(1-A*exp(-to/T)); /* to verify Is=0 at t=to */ Eoff=(Vc)*Io*T*(1-A+log(A)); /* One time Energy Calculations for last row of Turn-OFF table */ printf("%d\t%2.3g\t % .3g\t % .4g\t % .3g\n", k-1,to*1000,Is,Eoff*1000,(Eoff-Ecs)*1000); Ei=0.5*Li*Ib*Ib; Eb=Eoff*Vb/Vc; Eroff=Ri*Io*Io*T*((log(A))-(2*A)+(A*A/2)+1.5); printf("\nAverage Current during Turn-OFF, Iavcl = (Io+Ib/lnA) = \t%2.4g A\n", (Io+Ib/log(A))); printf("\tRatio: Iavcl to Ib, Initial ON Current = \t%g\n", ((T/to)+1/(1-A))); printf("Initial Current at Switch OFF from t=%dT ON State, Ib =\t% .5g A\n", n*step,Ison); printf("*****************************************************************\n"); printf("\nApplication Note Example Energy Source/Loss Tabulation \n"); printf("\nTurn-OFF Sourced Energy, Eb(Batt.),Ei(L) from Switch OFF to t=to:\n"); printf(" Sourced Battery Energy, Eb: \t% .5g mJ\n", Eoff*Vb/Vc*1000); printf(" Stored Coil Energy at Switch-OFF, Ei: \t% .5g mJ\n", Ei*1000); printf(" Total Sourced Turn-OFF Energy, Eb+Ei: \t% .5g mJ\n", (Eb+Ei)*1000); printf("\nTurn-OFF Energy Loss, Eclamp(IC), Er(Ri) from Switch OFF to t=to:\n"); printf(" Clamp Energy, Eclamp: \t% .5g mJ\n", Eoff*1000); printf(" Load Resistor Energy, Er: \t% .5g mJ\n", Eroff*1000); printf(" Total Loss Turn-OFF Energy, Er+Eclamp: \t% .5g mJ\n", (Eroff+Eoff)*1000); printf("\nTotal IC ON+OFF Energy loss, t=0 --> t=%dT --> t=%dT+to: \n", n*step,n*step); printf(" Turn-ON Energy Loss for t=0 to t=%dT, Eon: \t% .5g mJ\n", n*step,Eson*1000); printf(" Clamp Energy Loss at Switch Off, Eclamp: \t% .5g mJ\n", Eoff*1000); printf(" Total IC Energy Loss, Et=Eoff(0-->%dT)+Eclamp: \t% .5g mJ\n", n*step,(Eson+Eoff)*1000); printf("\nExample Dissipation for ON/OFF cycle, t=0 --> t=%dT --> t=%dT+to:\n",n*step,n*step); printf(" Total ON/OFF Switch Time (ton+toff)=(% .3g+% .3g)= %2.4g ms\n", tson*1000, to*1000, (tson+to)*1000); printf("\nFor Each MOS Switching Output, Dissipation = Energy x Frequency:\n"); printf(" For repetitive switching, 50%% ON time, Freq, f = \t %2.3g Hz\n", 1/(2*tson)); printf(" Single Switch Dissipation, Pd = (%2.3g x Et) = \t%2.4g mW\n\n", (1/(2*tson)),(1/ (2*tson))*(Eson+Eoff)*1000); } All Intersil semiconductor products are manufactured, assembled and tested under ISO9000 quality systems certification. Intersil products are sold by description only. Intersil Corporation reserves the right to make changes in circuit design and/or specifications at any time without notice. Accordingly, the reader is cautioned to verify that data sheets are current before placing orders. Information furnished by Intersil is believed to be accurate and reliable. However, no responsibility is assumed by Intersil or its subsidiaries for its use; nor for any infringements of patents or other rights of third parties which may result from its use. No license is granted by implication or otherwise under any patent or patent rights of Intersil or its subsidiaries. For information regarding Intersil Corporation and its products, see web site http://www.intersil.com 19

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