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
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