DESIGN APPLICATION NOTE --- AN027 Calculating Junction Temperature from Thermal Resistance DETERMINING THE JUNCTION TEMPERATURE APPLICATION FROM DEVICENOTE THERMAL RESISTANCE EA-101542 Rev A FOR PACKAGED SEMICONDUCTOR DEVICES To illustrate the thermal resistance of a packaged power semiconductor device, the analogy of an electrical resistor network will be used. In the electrical resistor model the electrical resistance is defined by the potential difference (voltage) across the resistor divided by the current through that resistor. In a thermal resistance the thermal potential difference (temperature) divided by the thermal current (heat) through the thermal resistor defines the thermal resistance, RTh. Figure 1 will be used to illustrate the electrical resistor equivalent circuit for a semiconductor device. Starting from the heat source (transistor junction), heat may transport through two paths. In the first path heat transfers from the transistor junction, through the molding compound by conduction, and then to the air surrounding the device by convection. In the second path, which is parallel with the first, heat flows from the junction of the device through the lead, through the PCB, into the chassis by conduction and finally to the air surrounding by convection. The second path is the primary focus of calculating the junction temperature since the majority of heat generated in the device transports through this path. The following example illustrates how to calculate the junction temperature of a device given its thermal resistance. The first consideration involves how the device will be used in operation. If a significant amount of power is output as RF energy, that energy is not being dissipated in the device and needs to be subtracted from the dissipated power in the calculation of junction temperature. To account for this, the effective power (Peff) dissipated is calculated by adding the DC and RF input powers (PDC & PRFin) and subtracting the RF output power (PRFout). Peff = PDC + PRFin − PRFout If the input and the output RF power are very small compare to the DC power, Peff can be simplified to: Peff = PDC The remaining steps are illustrated in the following example of determining the junction temperature of a device, given its operating parameters and thermal resistance. Junction High Thermal Resistance Path. Usually neglected Dominant Path R Th, j - l R Th, j - p Package Surface or Exposed Backside Ground PCB R Th, p - a Air R Th, l - p Ground Lead R Th, p - c Chassis Firgure 1: Thermal Resistance Equivalent Circut The information provided herein is believed to be reliable at press time.RFMD assumes no responsibility for inaccuracies or omissions.RFMD assumes no responsibility for the use of this information, and all such information shall be entirely at the user’s own risk. Prices and specifications are subject to change without notice. No patent rights or licenses to any of the circuits described herein are implied or granted to any third party.RFMD does not authorize or warrant anyRFMD product for use in life-support devices and/or systems. Copyright 2002 RFMD All worldwide rights reserved. 522 Almanor Ave., Sunnyvale, CA 94085 Phone: (800) SMI-MMIC 1 http://www.RFMD.com EAN-101542 Rev B DESIGN APPLICATION NOTE --- AN027 Calculating Junction Temperature from Thermal Resistance Example 1 Given the following information, calculate the junction temperature of a NGA-589 device. Tchassis = 70oC where Tchassis is the temperature of the chassis that the PCB is attached to P = 1mW P RFin V = 50mW RFout = 5.0V I dev = 80mA dev RTh, j - l = 100oC/W where RTh, j - l is the thermal resistance from the junction to the lead of the device ∆T Lead-PCB = 10oC where ∆T Lead-PCB is the temperature difference between the lead of the device and the PCB ∆T PCB-Chassis = 5oC where ∆T PCB-Chassis is the temperature difference between PCB and the Chassis Solution: Step 1: Calculate Effective Dissipated Power Peff = PDC + PRFin − PRFout (1) Peff = [(5.0V ) * (0.080 A)] + (0.001W ) − (0.050W ) Peff = 0.351W Step 2: Setup Junction Temperature Equation TJunction = ∆TJunct −Chassis + TChassis (2) ∆TJunct−Chassis = ∑∆Ti = ∑(RTh,i ) * (Peff ) (3) i i where i is each thermal resistance from Fig. 1 Step 3: Calculate each ∆Ti from Fig. 1 ∆TJunct − Lead = ( RTh , j −l ) * ( Peff ) from (3) ∆TJunct−Lead = (100o C / W ) * (0.351W ) PCB ∆TJunct−Lead = 35.1o C Screw NGA-589 Lead Step 4: Solve Equation (3) ∆TJunct−Chassis = ∆TJunct−Lead + ∆TLead−PCB + ∆TPCB−Chassis ∆TJunct−Chassis = (35.1+ 10 + 5)o C Chassis Figure 2: Cross Section of a PCB Assembly ∆TJunct−Chassis = 50.1o C Step 5: Solve Equation (2) TJunction = ∆TJunct−Chassis + TChassis TJunction = 50.1o C + 70 o C TJunction = 120 .1o C 522 Almanor Ave., Sunnyvale, CA 94085 Phone: (800) SMI-MMIC 2 http://www.RFMD.com EAN-101542 Rev B