Power and Temperature

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1 Power and Temperature TIPL 1160 TI Precision Labs Op Amps Presented by Ian Williams Prepared by Art Kay, Ian Williams and Miro Oljaca

2 Power Dissipation Quiescent Current V IN V CC 15V - V EE -15V OPA87 RL 1k V OUT P Q = quiescent power = I Q * V S I Q = quiescent current V S = total supply voltage V S = V CC V EE = (15) (-15V) = 30V Example: worst case over temperature P Q = I Q V S = (6mA)(30V) = 180mW

3 Power Dissipation DC Load 1V - R 1 100Ω 9V - R F 900Ω 10mA - 10mA 15V 5V - 0mA 10mA V OUT Effective load: R L = R Load (R F R 1 ) R L = (1kΩ) (900Ω 100Ω) R L = 500 Ω Vin 1Vdc -15V 10mA 10V - R Load 1kΩ 3

4 Maximum DC Power Dissipation P dc = V out V R cc V out L P dc _max = P V cc dc = V cc 4 R L Power Dissipation (W) V cc = 7.5V V cc 4 R L = 113mW DC Output Voltage (V) 4

5 Derivation of DC Maximum Power Transfer P opa = V cc V out V out P opa = V cc V out V out (1) (1) Power dissipated in in op op amp amp R L R L P opa (V out ) = V out V cc V out P opa (V out ) = V out V cc V out () () Dc Dc output voltage R L R L P P opa opa R L R L = V cc V out = V cc V out = = 0 0 (3) (3) Take the the partial derivative. Set Set to to zero zero and and solve for for maxima. V V out out R L R L V out = V cc V out = V cc when P P opa opa = = 0 0 (4) (4) Solve (3) (3) for for value of of Vout that that yields maximum power V V out out P dc _max = P opa ( V cc ) = V cc P dc _max = P opa ( V cc (5) (4) into () to dc in Op Amp ) = V cc (5) Substitute (4) into () to determine maximum dc power in Op Amp 4 4 R L R L 5

6 Power Dissipation at AC 1V - R 1 100Ω 9V - R F 900Ω Effective Load: R L = R Load (R F R 1 ) R L = 500 Ω T 1.0 Voltage (V) V OUT 1V PK - 15V 5V - -15V 0mA 10mA 0mA 10mA 10mA V OUT 10V PK 10mA R L 1kΩ 10V - T10.0 Voltage (V) u 500u 750u 1.0m Time (s) u 500u 750u 1.0m Time (s) 6

7 Maximum Average AC Power Dissipation P ac_avg (V outpk ) = V cc V outpk π R L P ac_max _avg = P opa _avg V cc π V outpk R L = V cc π R L Power Dissipation (W) V cc π = 9.55V V cc π R L = 91mW Peak ac Output Voltage (Vout) Peak AC Output Voltage (V) 7

8 Derivation of AC Maximum Power Transfer P opa = V cc V out V out R L (1) Power dissipated in op amp V out (t) = V outpk sin (ω t) () ac sinusoidal wave out P ac (t) = V cc V outpk sin (ω t) V outpk sin (ω t) R L (3) Substitute () into (1) P ac (t) = V ccv outpk sin (ω t) R L V outpk ω = π T P ac (t) = V ccv outpk sin ( π T t) V outpk sin ( π T t) R L R L P ac_avg = 1 T/ V cc V outpk sin ( π T t) V outpk sin ( π T t) T 0 R L sin (ω t) (4) Power dissipated in op amp as a function of time R L R L dt (5) Angular frequency as a function of period (6) Substitute (5) into (4) (7) Find the average power P ac_avg (V outpk ) = V cc V outpk V outpk (8) Average power as a function of peak output voltage π R L R L P opa _avg = V cc V outpk (9) Take the partial derivative to find maxima. Set to π R L R L zero and solve for maxima. V outpk P ac_max _avg = P opa _avg V cc π = V cc (10) Maximum power and the peak output voltage where π R max power occurs L Derivation courtesy of Miro Oljaca 8

9 Thermal Device Model No Heat Sink θ Op amp JA Ambient Junction package PCB θ JA includes the effects of the package and PCB! 9

10 Analogous Electrical Model No Heat Sink T J P D θ JA T A T A T J = (P D *Θ JA ) T A T voltage θ resistance P current 10

11 Temperature Rise Maximum DC Load R 1 100Ω V IN_DC R F 900Ω 15V - OPA87-15V RL 1kΩ V OUT P Q = I Q V S = 6mA 30V = 180mW P dc _max = V cc 4 R L = 15V 4 (500Ω) = 113mW P total = P Q P dc _max = 180mW 113mW = 93mW T J = P total θ JA T A = 93mW 150 W 5 =

12 Absolute Maximum Ratings 1

13 Thermal Protection 45 PTAT Current (µa) vs. Temperature Disconnects power at 160⁰C V S Reconnects power at 140⁰C PTAT current V S PTAT Current (µa) R S V REF - Comparator with hysteresis V S Temperature ( C) 13

14 Thermal Model Device with Heat Sink Case θ JC Junction θ CS Heat sink Fins θ SA Ambient 14

15 Analogous Electrical Model Device with Heat Sink T J P D θ JC θ CS θ SA T C T S T A T J = P D *(θ JC θ CS θ SA ) T A P D = total power dissipation T J = junction temperature T C = case temperature T S = heat sink temperature T A = ambient temperature 15

16 Thermal Resistance Θ JC (Junction to Case) T J = P D *(θ JC θ CS θ SA ) T A 16

17 Thermal Resistance Θ CS (Case to Sink) Typical Interface Resistances for Various Mounting Methods with a TO-0 (interface area = 1 in ): Thermal Joint Compound only (0.001 thick) θ = /W Mica (0.005) and Joint Compound (0.00) θ = 0.44 /W Series 177 Beryllium Oxide Wafers (0.06) And joint Compound (0.00) θ = 0.13 /W DeltaPad TM (0.009) θ = 0.50 /W Dry Mounting (0.001 assumed) θ = 1. /W T J = P D *(θ JC θ CS θ SA ) T A 17

Thermal Resistance Θ SA (Sink to Ambient) Example: Aavid Thermalloy 6398BG Mounting Surface Temp Rise Above Ambient ( C) 0 00 400 600 800 1000 100 5 80 60 40

18 Thermal Resistance Θ SA (Sink to Ambient) Example: Aavid Thermalloy 6398BG Mounting Surface Temp Rise Above Ambient ( C) Air Flow Velocity (Ft./Min.) Heat Dissipated (W) Thermal Resistance from Surface to Ambient ( C/W) T J = P D *(θ JC θ CS θ SA ) T A 18

19 Thermal Resistance in Natural Convection* Mounting Surface Temp Rise Above Ambient ( C) Air Flow Velocity (Ft./Min.) Heat Dissipated (W) *air flow 100 ft./min Thermal Resistance from Surface to Ambient ( C/W) Device Power (W) θ cs Thermal Resistance ( C/W) θ SA = ΔT mount P total = 15 W = 7.5 W 19

20 Thermal Resistance in Forced Airflow Air Flow Velocity (Ft./Min.) Mounting Surface Temp Rise Above Ambient ( C) Thermal Resistance from Surface to Ambient ( C/W) Air Velocity (ft./min) θ cs Thermal Resistance ( C/W) Heat Dissipated (W) 0

21 Example Calculate Total Power P dc = V cc 4 R L = (15V) 4 (5Ω) =.5W Maximum dc Power 15V V s = V cc V ee = 15V 15V = 30V I Q = 5mA Total Supply Voltage From OPA541 Data Sheet - OPA541 V OUT P Q = I Q V s = 5mA 30V = 0.75W P total = P dc P Q =.5W 0.75W = 3W Quiescent Power Total Power V IN -15V R L 5Ω 1

22 Example Calculate θ SA for Given Power Air Flow Velocity (Ft./Min.) Mounting Surface Temp Rise Above Ambient ( C) (3W, C) Thermal Resistance from Surface to Ambient ( C/W) P total = 3W θ SA = ΔT mount P total = 3W = 7.33 W Heat Dissipated (W)

23 Example Calculate Junction Temperature θ JC = 3 C/W from OPA541 data sheet (TO-0) θ CS = 0.44 C/W (Mica and joint compound) θ SA = 7.33 C/W (Heat sink specification at 3W) T A = 5 C P D = 3W T J = P D *(θ JC θ CS θ SA ) T A T J = (3W)*(3 C/W 0.44 C/W 7.33 C/W) 5 C T J = 57.3 C 15V - OPA541 V OUT V IN -15V R L 5Ω 3

24 Thanks for your time! Please try the quiz. 4

AN An Analysis for Power Dissipation of LDO Application with High Power. Hawk Chen. Thermal topology of LDO: Introduction:

AN An Analysis for Power Dissipation of LDO Application with High Power. Hawk Chen. Thermal topology of LDO: Introduction: An Analysis for Power Dissipation of LDO Application with High Power Hawk Chen When an LDO is operating, its on-chip devices dissipate power as a heat way. Heat flow from a higher to a lower temperature