.BIB+HYHELP.AID &LAYER1A.AIDLAYER2A.AID?LAYER3A.AIDgLAYER4A.AIDLAYER5A.AIDMENULAY2.AIDMENULAY3.AIDMENULAY4.AIDMENULAY5.AIDXSTR1.AID FET1.AID FET2.AID `DIODE1.AID HELPXSTR.AIDHELPDIOD.AIDHELPFET1.AIDqHELPFET2.AIDKTABLE.AID 3THERM0.AID YTHERM1.AID THERM2.AID THERM3.AID 9THERM4.AID THERM5.AID THERM7.AID TTHERM9.AID THERM10.AID THERM11.AIDU THERM12.AID THERM14.AID THERM15.AID: THERM17.AID THERM18.AID THERM19.AID~ THERM20.AID THERM21.AID2 THERM22.AIDs THERM23.AID THERM24.AID THERM25.AIDi THERM26.AID THERM27.AID P O Hewlett-PackardO O!Hybrid Circuit Thermal CalculatorO OAppCAD Pͻ;This program analyzes a typical thermal problem encountered =in the design of hybrid (MIC) circuits. An active device and =it's mounting system up to a heat sink make up the equivalent =of a series R network. In a MIC amplifier, for example, this Gmight include: a transistor, the die attach preform, a substrate, backside of the chip is the heat source. If this is not valid=for your application, choose "custom" with = .001. The 1st @layer is then the semiconductor with junction size of Chip widthGand Chip length. =.001 is also useful for a general heat source. F [Esc] G Return to the Thermal Analysis main menu. Depending on the number G of layers used, the Esc key may need to be pressed several times. F Note: : If Power = 1000 mW, then each T is the for that layer.  [ Press any key ] ͼͻ!HYHELP P. O Hewlett-PackardO O!Hybrid Circuit Thermal CalculatorO OAppCAD Pͻ Device: Power dissipated = p mW  Theta (jc) = p$ C/watt Temperature (jn) = p C  Chip width = p (  Chip length = p  Ŀ    device T = p         Material 1:   Thickness = p Kth = p T = p   K  2   3 5  3    3   3    3   3    3  Spreading Angle% Mounting Surface Temperature = p C      = p45 Overall Thermal Resistance = p /Watt   PͼP Units: Kth = watts/cmC Temperature = C Dimensions = mils (.001") P[F1]=Help [F2]=Compute [F4]=Add Layer [F6]=Device [F7]=Material [Esc]=Main !HYHELP THETATOT RL?  99999  .1 * DEVICE EL * PDISS RL>  9999  .1 Power dissipated in device junction in milliwatts TEMPJN &RL>  9999 -273 * DELTA1 RL F  9999  .0001 * DELTA6 RL F  9999  .0001 * KTH1 &RL 9  9.99  0 Thermal conductivity of material THICK1 RL )  9999  0 Thickness of material MATERIAL1 EL * TEMPMOUNT RL?  999 -273 Temperature of mounting (heatsink) surface THETADEV &RL  9999  .001 Thermal resistance of device - junction to chip backside WIDTHCHIP &RL  9999  .001 Width of chip bottom (<= Length)LENGTHCHIP&RL  9999  .001 Length of chip bottom (>= Width)ANGLE IL Z Heat spreading angle: 90 deg = columnar; 45 deg = typical; 5 deg MIN L K   Material 2:   Thickness = p Kth = p T = p L '  REMLINE BAS!HYHELP DELTA2 RL F  9999  .0001 * KTH2 &RL 9  9.99  0 Thermal conductivity of material THICK2 RL )  9999  0 Thickness of material MATERIAL2 EL * L K   Material 3:   Thickness = p Kth = p T = p L '  REMLINE BAS!HYHELP DELTA3 RL F  9999  .0001 * KTH3 &RL 9  9.99  0 Thermal conductivity of material THICK3 RL )  9999  0 Thickness of material MATERIAL3 EL $L L  Material 4:   Thickness = p Kth = p T = p K '  REMLINE BAS!HYHELP DELTA4 RL F  9999  .0001 * KTH4 &RL9  9.99  0 Thermal conductivity of material THICK4 RL)  9999  0 Thickness of material MATERIAL4 EL d L K   Material 5:   Thickness = p Kth = p T = p L '  REMLINE BAS!HYHELP DELTA5 RL F  9999  .0001 * KTH5 &RL9  9.99  0 Thermal conductivity of material THICK5 RL)  9999  0 Thickness of material MATERIAL5 EL  B  Material 1:    Material 2: (#$!HYHELP  b  Material 1:    Material 2:    Material 3: CODER BAS!HYHELP    Material 1:    Material 2:    Material 3:    Material 4: MO3 BAS!HYHELP    Material 1:    Material 2:    Material 3:    Material 4:    Material 5: BRARY EXE!HYHELP P O Hewlett-PackardOO(Thermal Resistance of Bipolar TransistorO OAppCAD PĿ!> Ŀ    Material: pSiliconp  > ? >>     Units:  pmicrometers (um)  > ? l >>  !> ?  >^ B > >  !> ?> ^ ͻ>>  > p^ >>  Length =    Power Dissipated> ?> ^ ͼ>> p    in junction = p mW  > ?  >^ E > >    > ? ? ? >     Temperature of chip > ?  w  >      backside = p   C> ? p? >     >            Chip Thickness = p   Width = p     F    Thermal Resistance = p C/Watt  Temperature rise = p C    (approximate)   Junction Temperature = p C  Thermal Resistance = p C/Watt Junction Temperature = p C  (exact solution)   (corrected Kth vs. Temp) PP [F1]=Help [F2]=Calculate [F3]=Units [F4]=Material [F5]=Drawing [Esc]=Exit !HELPXSTR MATERIAL EL* UNITS EL* POWER RR   99999  .01 Average power dissipated in the transistor junction TEMPBACK RR   300  25 Temperature of back of the chip, e.g.,collector mounting (>= 25 deg) THICKNESS RR  999  .0001 Thickness of the transistor chip WIDTHC RR5  9999  .001 Width of transistor chip (less than or equal to Length) LENGTHC RR D  9999  .0001 Length of transistor chip (greater than or equal Width) WIDTHJ RR 1  999  .001 Width of active juction area (less than or equal to Length) LENGTHJ RR&  999  .001 Length of active juction area (greater than or equal to Width) THETAQ RR  9999.9  .01 * THETAE RR  9999.9  .01 * DELTATEMP RRC  9999.9  .01 * TEMPJ RRC  9999.9  .01 * TEMPCORR RRC  9999.9  .01 * P" O Hewlett-PackardOO%Thermal Resistance of Single Gate FETO OAppCAD PĿN  Material: pGallium Arsenide  ^ Ŀ   Units:  pmicrometers (um)  ^^^ $^ _Gate Length ^ x ^  ! ^^^  !^^ ^S۳ _ _^    Power Dissipated^^ ^GĴD x ^ _ ^     in channel = p mW ^^ ^S۳ _ _^    !^^_ Gate Width^    Temperature of chip^^^    backside = p   C ^^^   ^^^   Chip Thickness = p   ^    !'+ F    Thermal Resistance = p C/Watt  Channel Temperature = p C     ,  Temperature rise = p C    Channel Temperature = p C  ' (corrected Kth vs. Temp) PP [F1]=Help [F2]=Calculate [F3]=Units [F4]=Material [F5]=Drawing [Esc]=Exit D7A AID!HELPFET1 MATERIAL EL* UNITS EL* POWER RR   99999  .01 Average power dissipated in the transistor junction TEMPBACK RR   999 -273 Temperature of the back of the chip (collector mounting surface) THICKNESS RR  999  .0001 Thickness of the transistor chip THETAQ RR  9999.9  .01 * DELTATEMP RR  9999.9  .01 * TEMPJ RRC  9999.9  .01 * TEMPCORR RRC  9999.9  .01 * GATEW RR 9  99999  .001 Width of gate GATEL RR 8  99999  .001 Length of gate P O Hewlett-PackardOO'Thermal Resistance of Multiple Gate FETO OAppCAD PĿD   Material: pGallium Arsenide  ^"Ŀ  Units:  pmicrometers (um)  ^_ Number of gates =^ x ^^$^^^D^^! ^^^>^^ ^ ^^^ _ Gate Width^ Power Dissipated^^ ^ۺۺۺۺ _^   in channel = p mW ^^ ~S^ۺۺ~S^ۺ~S^ x ^ !^^ ^ۺۺۺۺ _^  Temperature of chip^^ ^ͼ^ ^  backside = p   C ^ x ^ _ ^ _^ x ^  ^ _ Gate Spacing^_ Gate Length^  Chip Thickness = p ^^ ^! ^"  + F    Thermal Resistance = p C/Watt  Channel Temperature = p C     ,  Temperature rise = p C  Channel Temperature = p C  ( (corrected Kth vs. Temp) PP [F1]=Help [F2]=Calculate [F3]=Units [F4]=Material [F5]=Drawing [Esc]=Exit u [!HELPFET2 MATERIAL EL* UNITS EL* POWER RR   99999  .01 Average power dissipated in the transistor junction TEMPBACK RR   999 -273 Temperature of the back of the chip (collector mounting surface) THICKNESS RR  999  .0001 Thickness of the transistor chip THETAQ RR  9999.9  .01 * DELTATEMP RR  9999.9  .01 * TEMPJ RRC  9999.9  .01 * TEMPCORR RRC  9999.9  .01 * GATEW RR >  99999  .001 Width of gate GATEL RR =  99999  .001 Length of gate GATESPACE RR )  99999  .001 Spacing between gates GATES IL>c Number of gates (must be > 1) P O Hewlett-PackardOO!Thermal Resistance of Round DiodeOOAppCAD PĿN Junction Diameter = p  Power Dissipated = p  mW *Temp (chip back) = p  C   / Ŀ  Ŀ  p                Thickness  p    p        WidthLength    H    Material: pSiliconp +    Units:  pmicrometers (um) "  ' Temperature rise = p C     Thermal Resistance = p%       (approximate) (C/Watt)Junction Temperature = p  C   !*  Thermal Resistance = p Junction Temperature = p  C     (exact solution)  (C/Watt)$(with Kth vs. Temp. correction)  '    PP [F1]=Help [F2]=Calculate [F3]=Units [F4]=Material [F5]=Drawing [Esc]=Exit !HELPDIOD DIAMETER RR   999  .00001 Diameter of active junction area of diode THICKNESS RR    999  .0001 Thickness of chip WIDTHC RR   999  .0001 Width of chip (must be equal to or less than Length) LENGTHC RR 8  999  .0001 Length of chip (must be greater to or equal to Width) POWER RR ?  9999.9  .0001 Average power dissipated in the junction area UNITS EL* MATERIAL EL* TEMPBACK RR?  300  25 Temperature of back of chip, e.g.,mounting surface  (>= 25 deg) THETAQ RR  9999  .001 * THETAE RR  9999  .001 * DELTATEMP RRC  999  .001 * TEMPJ RRC  999  .01 * TEMPCORR RRC  999  .01 * P O Hewlett-PackardOO(Thermal Resistance of Bipolar TransistorO OAppCAD PĿI This module calculates the thermal resistance of a bipolar transistorF based on the methods of Linsted and Surty (Ref. 2). The basics of G this calculation are described in the Tutorial section beginning onE Page 15. In addition to this exact calculation, an approximation > based on a 45 spreading angle is also made for reference.NE The power dissipated in the transistor and the temperature of the E backside of the chip (die attach surface) are used to compute the D temperature rise from the mounting surface to the junction. The G junction temperature is calculated both with and without the effectF of decreasing thermal conductivity of Si or GaAs with temperature. N [F3] Gtoggles dimensional units between microns, millimeters, and mils.   F [F4]  based on a 45 spreading angle is also made for reference.NE The power dissipated in the diode and the temperature of the back H of the chip (die attach surface) are used to compute the temperatureA rise from the mounting surface to the junction. The junctionA temperature is calculated both with and without the effect ofC decreasing thermal conductivity of Si or GaAs with temperature. N [F3] Gtoggles dimensional units between microns, millimeters, and mils.   F [F4]  25 while glass    B is < 0.01 watts/(cm-deg C). The thermal conductivity of Silicon   H (Si) is about 1.46 at 27 deg C, and Gallium Arsenide (GaAs) is about    H 0.44 at the same temperature. A complicating factor in semiconductor    H temperature calculations is that the thermal conductivity of both Si    H and GaAs decrease as temperature increases. It turns out, however,    H that there is a fairly simple way of taking this into consideration.    H   ACurrently there are two systems for defining Kth, the thermal    0 conductivity, and the dimensions are different.   H   #The one to be used in this note is:   H    Kth = watts/(cm * degC)$   H  LP Page 9 ͼrmal PB D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ LK  'The second system defines K as follows: K 'K' = (cal * cm)/(sec * cm * cm * deg C) K  The two are related by:) K Kth = 4.18 * K'Eq. 4 K  Equation 1 is usually given as:! K Delta T = P(F/Kth * A)Eq. 3a K  where: Rth = F/Kth * AEq. 5 K  "and Rth is the THERMAL RESISTANCE. K   H   H  LP Page 10 ͼ The tPz D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  G K K  BThe dimensions of thermal resistance are then degrees C per watt  C (deg C/W). Therefore, for columnar heat flow, the formula for  F thermal resistance, Eq.5, is very simple. Unfortunately, not all L thermal resistance calculations are this simple. Many configurations  L involve what is known as heat spreading, and this is the subject of   the next section.6 K K M M M M M M M M   H  LP Page 11 ͼesistance,PT D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I SPREADING HEAT FLOW L  CTo explain thermal spreading, an example from Reference 2 will  G be used. It takes the form of a small circular heat source on the M top of a circular cylinder an illustration of which you will find on  = the next page (13). If a2 > 5 * a1 and h > 5 * a1, then: L Rth = 1/(pi * Kth * a1)Eq. 6  L  CIn this example the heat spreads out in three dimensions and in  N the ultimate limit, takes the form of a hemisphere. Also the thermal  H resistance becomes independent of how thick the material is as long 3 as the inequalities given above are satisfied. M M M M M   H  LP Page 12 ͼPQ D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I GENERAL HEAT FLOW PROBLEM L  CPure spreading is seldom met in semiconductor systems. Although  E the heat flow is three dimensional, the boundary conditions have D specific limits. The equations on page 6 are not easy to solve  C without use of a computer (that is, if you don't want to spend  E several weeks solving one problem). The first modern attempt at M solving this type of problem was made by Linsted and Surty (Ref. 2)  B in 1972. They published their results in the form of several  M normalized curves for specific dimensional ratios. Their final form  H for the solution of the heat flow equation is for square geometries H only although the curves cover ten ratios. In this note the method N is the same as L&S although the results differ slightly, probably due  > to the manner in which the infinite series are evaluated. M  8The general approach to setting up and solving heat flow  B equations will now be discussed briefly. For a more thorough  G discussion of heat problems in general, see Ref 1. This, however,   7 does not cover the semiconductor problem specifically. LP Page 14 ͼP D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I  9The method of determining heat flow in a semiconductor is  ' illustrated on the next page (16).(L   is assumed not to be limited by chip size. Both of these G approximations are valid for modern FET designs. Once the thermal F resistance, Rth, has been found the temperature can be evaluated. E For the static case (power is CW) the temperature of the channel  - (FET) or junction (bipolar or diode) is:! M T = (P * Rth degrees) + TaEq. 7  M   1where Ta is the ambient or heat sink temperature.  LP Page 17 ͼ`T]END P D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  THERMAL TIME CONSTANT N =It was mentioned earlier that the input power could vary with M time. When this occurs, there is a time delay before the heat rises  C or flows out of the system. Thus, in pulsed applications, the  M thermal time constant of the device can be of great importance. For  M example, power transistors are often used in a pulsed mode where the  E peak output power can be increased, because the average power is M reduced. This can be very effective as long as the pulse length is  M short compared to the thermal time constant of the transistor and its  M mount. If the pulse length is equal to or greater than the thermal  H time constant, the maximum junction temperature will be the same as E with CW power. When the thermal time constant is calculated, it  H should include the chip and its mounting system. It has been found N that the thermal time constant has a useful electrical analog in the  H RC line. Therefore the total thermal time constant is equal to the G sum of the time constants of the chip and its mounting components. H In most systems, the chip time constant dominates this calculation. F The next figure on page 19 shows how the RC line and the multiple    thermal time constants compare.)  LP Page 18 ͼrlier that theP O Hewlett-PackardO OThermal Time ConstantOO AppCAD Pͻ FĿ   Thermal Model:4  Ŀ Rth-1Rth-2Rth-3\    P //\/\/\/\/\/\/\/\/\Ĵ\     \   ,Cth-1 Cth-2 Cth-3 \   !#Ĵ\   ?\   , T(t) = P * Rth (4/ exp 1.5) * (t/)    F  FĿ   Electrical Analog:0   R1 R2 R3    1o/\/\/\/\/\/\/\/\/\o   Ŀ    Io     C1 C2 C3        1oo    I(t) = Io * exp(t/RC)    F P Page 19 ͼ MPc D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I  9The thermal time constant has two components, the thermal  B resistance, which has already been discussed, and the thermal  D capacitance. To calculate the thermal capacitance, two material  = constants are needed, the density and the specific heat. L   In general:7 L Cth = rho * c * F * AEq. 8  L  )where: rho = density in grams/cubic cm ,c = specific heat in watt-seconds/gr * deg C  F = die thickness in cm A = area in square cm! M  5The thermal RC product is Rth * Cth so for the simple  columnar case:9 M 3Rth * Cth = (F/Kth * A) * (rho * c * F * A) Eq. 9    = rho * c * F * F/Kth  LP Page 20 ͼP D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I L  For Silicon: rho = 2.33 gr/cc# c = 0.7 watt * sec/gr * deg C L  For GaAsrho = 5.31 gr/cc# c = 0.35 watt * sec/gr * deg C L  5The thermal time constant, tau, is not Rth * Cth but:  L tau = [4/(pi * pi)] * Rth * Cth M  %which is about (2.5 * Rth * Cth) and, M .T(tau) = P * Rth * 4 * (1/pi^1.5) * (t/tau)^.5 = 0.718 * P * Rth * (t/tau)^.5 Eq.10  M  =Thus the analog between the transmission line is not exact in H all respects. Temperature is the analog of current in the RC line.   I  LP Page 21 ͼ* pi)] * P D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I F The following equations compute the junction temperature rise vs. / time, with the current into an RC network: L  Thermal:&T(t) = P * Rth(4/pi^1.5) * (t/tau)^0.5  L   RC circuit:I = Io * e^(t/RC) L  ANote that the temperature response varies as the square root  of the t/tau ratio.3 L  9For Si: tau = 0.453 * F^2 (at 300 deg K) Eq. 11  M  9For GaAs: tau = 1.71 * F^2 (at 300 deg K) Eq. 12  M  DIn a multiple time constant system (page 18), the time constants  H do add as in an RC network. Usually the chip time constant will be F the longest, because Rth is largest. Also the resistances add in  series.B   I  LP Page 22 ͼPs D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I !MEASUREMENT OF THERMAL RESISTANCE L  CThis note will not go into the details of measurement, but the  ; three methods in common use will be briefly described. L  Junction temperature method:, L  ;The voltage across a forward-biased junction depends on the A materials in the junction, the current, and the temperature.  M Initially the device is placed in an oven at a known temperature and  N current (emitter-base, for the bipolar and gate-source for the FET).  N The voltage is measured at at least three temperatures to establish a  H temperature plot. During the actual test the bipolar transistor is N pulsed from the low-current, or measuring mode, to a high current or  C heating mode. A sample-and-hold circuit measures the junction  H voltage from which the temperature is derived. The high current DC 9 input (which must include the duty-cycle) gives Pin. M   I  LP Page 23 ͼribed.P- D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I  Then,> L Rth = T/(Pin * % duty cycle).  L  CFor the FET, the device is usually tested at Idss, and the gate  D pulsed into forward bias during the measurement interval. This  M method gives an average temperature. This is the method used by HP  M Microwave Semiconductor Division for measurement of Rth for bipolar  P transistors and device test cells used to characterize Si & GaAs MMICs. L  Liquid crystal method: * M  7Before the device is tested it is coated with a special  F solution which dries to a crystalline layer on the surface of the N transistor. When the device is biased on, distinctive colors can be  F observed under a microscope set up with cross-polarizing filters. F Each color indicates a certain temperature range. This method is 2 considered very good for finding "hot spots".   I  LP Page 24 ͼusuallyP D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I L L  Infrared method: 3 L  CThe transistor is placed on the stage of a infrared microscope  D and the device is biased to its normal operating condition. An  M infrared sensitive diode in the microscope scans the surface of the  M transistor, the intensity of the reading indicating the temperature.  C This test involves the use of expensive equipment and requires  M special training to get accurate results. The minimum resolution is   about 0.0002 inches.5 M M M M M M M   I  LP Page 25 ͼp) P D:THERMO Hewlett-PackardOOThermal AnalysisOO AppCAD Pͻ  I L SUMMARY AND CONCLUSION L