1 Final Control Introduction Thermal Circuits –Thermal resistance –Heat Sinks –2N3904 example –OPA512 example JFET/MOSFET Review Summary

2 Types of Heat Sinks Ball Grid Array For: Intel AMD PowerPC Surface Mount, Wave Solderable, Snap-down

3 System Parameters Type of System Parameter QuantityPotentialTime ElectricalChargeEMF or Voltage Second LiquidVolumePressureSecond GasMassPressureSecond ThermalHeatTemperatureSecond MechanicalDistanceForceSecond

4 Defining Resistance Type of SystemChange inQuantity Parameter DescriptionName (SI Units) ElectricalCharge xfrd. per sec. Current (amperes) LiquidVolume Xfrd. per sec. Flow Rate (Cubic meters per second) GasGas xfrd. per sec. Flow Rate (kilograms per second) ThermalHeat xfrd. per sec. Flow rate of energy (joules per second) MechanicalDistance per sec. Velocity (meters per second)

5 Thermal Definitions RθRθ TJTJ TATA TCTC R θJC R θCA T= T J -T A J = junction C=case A=ambient P D = power dissipated PDPD

6 A simple thermal circuit P D = 120 mW =.12 W TJTJ TATA R θJA = C/W T = 24 0 C = T J – T A = P D *R θJA T J (max) = C Max power: BJT ~ P CE FET ~ P DS SCR ~ P AK

7 A complete thermal circuit with heat sink PDPD TJTJ TCTC TATA TSTS R θJC R θCS R θSA R θJC = Thermal Resistance from Junction to Case T R θSA = Thermal Resistance from Case to ambient R θCS = Thermal Resistance from Case to sink T

8 Common Cases

9 Mounting a heat sink

10 Mounting a TO-3 Heat Sink

11 2N3904 BJT Specs

12 BJT Amplifier Circuit: Spec says max. ambient temperature will be C Example 1: R θJA for the 2N3904 is C/W, V C = 10 volts V E = 1.05 volts V CE = 9 volts P D = 9*20 = 180 mW P D = 180 mW =.18 W TJTJ TATA R θJA T = 36 0 C = T J – T A = P D *R θJA T J (max) = C TO-92 Style Case

13 Adding a heat sink TO-92 Heat Sink From Digikey: R θSA = 64 0 C per Watt PDPD TJTJ TCTC TATA TSTS R θJC R θCS R θSA T From 2N3904 spec: R θJC = C per Watt From general: R θCS = 5 0 C per Watt R θTotal = C/Watt T = T J -T A = C P D = 180 mW =.18 W T J (max) = C

14 OPA Absolute Max Internal power Dissipation = 125 Watts

15 OPA512 Problem OPA512, 125W, 15A R θJA = 30 0 C/ Watt R θJC =.9 0 C/ Watt P D = 9*20 = 180 mW TJTJ TATA R θJA P D = (T J – T A )/R θJA = 110/30 = 3.67Watts T J (max) = CT A = 40 0 C Need 30 watt AC amplifier for an audio application. The max temp will be F First look at no heat sink: TJTJ TCTC TATA TSTS R θJC =.9 R θCS R θSA T 30 = (T J – T A )/R T =110/R T R T = C/Watt total So R θCS +R θSA = =2.77

16 TO3 Mounting Kit Mounting a TO-3 Heat Sink

17 Pentium 4 Heat Sink

18 Athlon Heat Sink

19 Power Derating Curve C 5 W 25 0 C 5W 125 = 25 0 C/WattR θJC =

20 JFET & MOSFET Review

21 IRL630 HEXFET Switching Power MOSFET R θJC = C/Watt (max) R θCS =.5 0 C/Watt (flat, greased surface) R θJA = 62 0 C/Watt (max)

22 Another Heat Sink Problem R θJC = C/Watt (max) R θCS =.5 0 C/Watt (flat, greased surface) R θJA = 62 0 C/Watt (max) Max continuous drain current = 9 amps R ds =.4Ω Max power dissipated = C Linear derating factor =.59 W/ 0 C = 2.02 WPD=PD= C 25 0 C PDPD T TJTJ TCTC TATA TSTS R θJC =1.7 R θCS =.5 R θSA T So the max power that can be dissipated is: R θJC +R θCS +R θSA = =6.2 0 C/W and P D =125/6.2=20.2W Best TO-220 at Digikey is: R θSA =4 0 C/W

23 CMOS: C omplementary Metal- Oxide Semiconductor CMOS is a widely used type of semiconductor. CMOS semiconductors use both NMOS (negative polarity) and PMOS (positive polarity) circuits. Since only one of the circuit types is on at any given time, CMOS chips require less power than chips using just one type of transistor.semiconductortransistor

24 Summary Thermal Circuits –Thermal resistance –Heat Sinks –2N3904 example –OPA512 example JFET/MOSFET Review –IRL630 Heat Sink example Next –Industrial Electronics SCR TRIAC DIAC –Stepping Motors