APPLICATION NOTES Thermal Resistivity Table Simplifies Temperature Calculations How much power can a microwave device handle before it overheats? Here is a good approximation of the technique for determining "hot spots" of different structures. Reprinted with permission from Microwaves/RF Quick estimates of temperature rises in conduction heat flow situations can be made easily when it is realized that the steady state Fourier heat flow equation is the mathematical equivalent of Ohms Law for electrical circuits with temperature differences replacing voltage differences and heat flow (expressed in calories, per second, B.T.U. per hour or watts) replacing current flow. In usual form the one dimensional steady state flow equation is: Q = K (A/L) ΔT (1) where Q = heat flow K = thermal conductivity A = cross section of heat flow path L = path length ΔT = temperature drop along path of length, L. Inverting the relationship to: ΔT = (1/K) (L/A) Q then (1/K) equals thermal resistivity and (1/K) (L/A) equals thermal resistance. (By anology to V = (ρ) (L/A) I and V = R . I). By preparing a table of thermal resistivity (1/K) in units of °C – inches/W for the materials encountered in microwave work, temperature drops can be readily estimated in series heat conduction situations. The application of these concepts can best be illustrated by the following two examples. EXAMPLE PROBLEM 1. EXAMPLE PROBLEM 2. Figure 1 shows a thin film resistor in a microwave integrated circuit. How hot will it get with 2 W dissipated over its 0.1 x 0.2 in. surface (100 W/in.2)? A miniature rod resistor terminates a coaxial line as shown in Fig. 2. Estimate the temperature rise at the front end of the resistor relative to a point on the outer shell when 5 W is being dissipated: Electrical engineers are usually introduced to heat flow either as studies in physics or problems to be solved by a mechanical engineer. Heat is shown to flow via the mechanisms of radiation, convection and conduction—each with their own equations and coefficients. The subject is made more confusing by the use of c.g.s. units in physics and British Thermal Units in engineering. Most of the important temperature calculations faced by a microwave engineer result from the need to determine how hot a particular point in a structure will get when power is being dissipated. While the ultimate heat sink usually expels its energy to the outside world by free or forced convection to air or liquid, the internal circuit elements, whose temperature rises usually set a limit on power handling capability, most often are conduction cooled. Radiation plays a relatively insignificant role below 300°C. ΔT1–4 =ΔT1–2 + ΔT2–3+ ΔT3–4 ΔT1–2 (in ceramic) = (2.13)(0.025)(2) = 5.3°C (0.1) (0.2) ΔT2–3 (in grease) = ΔT3–4 (in aluminum) = 46 (0.002)(2) = 9.2°C (0.1) (0.2) ΔT1–5 =ΔT1–2+ ΔT2–3+ ΔT3–4 + ΔT4–5 Since the power is uniformly dissipated along the resistor, ΔT1–2 can be estimated by assuming that the full 5 W travels 1/2 the length of the resistor to the mid plane of the end wall: ΔT1–2 = (0.23)(0.125)(2) = 2.9°C (0.1) (0.2) (0.24) (0.020 + 0.031) (5) = 48.7°C π(0.020)2 ΔT2–3 (solder joint) = Therefore, ΔT1–4 = 5.3 + 9.2 + 2.9 = 17.4°C Adding to this an estimate of maximum absolute temperature of 80°C for the aluminum ground plane leads to an estimated 100°C for the maximum resistor temperature. This is a conservative estimate since it neglects transverse heat spreading, but it serves to illustrate the ease and value of quick estimates. The major temperature drop occurs at the grease interface between the ceramic and the metal. Changing from Aluminum Oxide to Beryllium Oxide with 1/10 the thermal resistivity wouldn’t help in this situation as much as working to minimize the interface resistance. Thin Film resistor as heat source in microwave integrated circuit RESISTOR 0.1X0.2[2,5X5,1] SILICONE GREASE LAYER 0.002[0,05] COPPER CIRCUIT HEAT ALUMINA CERAMIC 0 025[0 64] THICK ALUMINUM GROUND PLANE 0.125[3,18] THICK (0.78) (0.005) (5) = 3.2°C π (0.040) (0.062) For the radial path 3-4, use an average diameter of 0.120 for the area calculation and a path length of 0.080. ΔT3–4 = ΔT4–5 = Then (0.11) (0.080) (5) = 1.9°C π (0.120) (0.062) (0.11) (0.040 + 0.031) (5) = 1°C π (0.062) (0.2) ΔT1–5 = 48.7 + 3.2 + 1.9 +1 = 54.8°C A review of the results reveals that the manner of estimating the area and length of the metal paths and the solder joint are relatively unimportant since they contribute only 11 percent to the total ΔT. For this configuration the greatest potential reductions in temperature drop can be made by increasing the resistor diameter or reducing its length. Resistor as heat source in coaxial line 0.040[1,02] DIAMETER BeO ROD T5 0.062 [1,575] T1 T4 T3 0.2 [5,1] T2 0.040[1,02] SOLDER JOINT 0.062 [1,575] COPPER CHELL Figure. 2 Figure 1 KEY: Inches [Millimeters] .XX ±.03 .XXX ±.010 [.X ±0.8 .XX ±0.25] 300 Dino Drive, Ann Arbor, MI 48103 888-244-6638 or 734-426-5553 • FAX 734-425-5557 www.aeroflex.com/inmet • [email protected] REV 3/09 APPLICATION NOTES Ta ble 1. Thermal Resistivity (R) – Typical Values R in °C - inches Watt conversion from thermal conductivity K to R R= R= 22.8 BTU - ft for K in K hr - ft2 - °F 0.094 cal - cm for K in K sec - cm2 - °C Material R Diamond 0.06 Lead Silver 0.10 Indium Copper 0.11 Boron nitride Material R 1.14 2.1 Material R Quartz 27.6 Glass (7740) 34.8 (isotropic) 1.24 Silicon thermal grease 46 Gold 0.13 Aluminum 0.23 Alumina ceramic 2.13 Water 63 Beryllia ceramic 0.24 Kovar 2.34 Mica (avg) 80 Molybdenum 0.27 Silicon carbide 2.3 Polyethylene 120 Brass 0.34 Steel (300 series) 2.4 Teflon 190 Silicon 0.47 Nichrome 3.00 Nylon 190 Platinum 0.54 Carbon 5.7 Silicone Rubber 190 Tin 0.60 Ferrite 6.3 P.P.O. 205 Nickel 0.61 Pyroceram (9606) Polystyrene 380 Eutectic lead tin solder 11.7 Epoxy—high 0.78 conductivity 24 Mylar 1040 Air 2280 KEY: Inches [Millimeters] .XX ±.03 .XXX ±.010 [.X ±0.8 .XX ±0.25] 300 Dino Drive, Ann Arbor, MI 48103 888-244-6638 or 734-426-5553 • FAX 734-425-5557 www.aeroflex.com/inmet • [email protected] REV 3/09