======================SILICON_THERMAL=========================== silicon Atomic number: 14 Atomic weight: 28.086 Density: 2329 [293 K]; 2525 [liquid at m.p.] kg m-3 Molar volume: 12.06 cm3 Velocity of sound: 2200 m s-1 Hardness Mineral: 6.5 Melting point: 1683K Boiling point: 2628K Thermal conductivity: 148 [300 k] W m-1 K-1 Water Standard: 2.42W Fusion: 39.6 kJ mol-1 Vaporization: 383.3 kJ mol-1 -------------------------------------------------------------------------------------- thermal resistance 1mil cube geometry of Silicon? Therm_Conduct 148 W/(m*K) 148uW/(um*K) 3.76mW/(mil*K) which means ... __ 3.76mWatt /\ -> ______ Delta Temp / /| If L =1mil, A = 1mil^2, and 3.76mWatt 1 deg_K /L / | Delta Temp = 1 deg_K -> /_____/ | ________________________________ |__ A | / | R_th = 266 deg_K/W for 1mil^3 | | /\ | / |________________________________| |/____|/ / R_th = Rho*L/A 3.76mWatt Rho = 266mil*deg_K/W If you have a single transistor which is powered at 5mA*5V =25mW such that its temperature is uniform over half shell 1 mil in radius, shell temperature? 25mW of power | - - - - |- - - - _ -|- _ - - - - - \|/ - - - - - - _- V -_ - - _ - - - - - _____________________ - - - - -_ /1mil/ | | | - - - - -/_|_/ / / | - - - - _/ 2mil__/ _/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/ _/ - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ The thermal resistance can be modeled in terms of half shells of silicon. Given that the thermal gradients will follow this pattern, the following gives the area of a half sphere. A = 2*PI()*R^2 If you look at a 1mil thick shell which is 1mil away from the heat source, its thermal resistance should be about one sixth the thermal resistance of a 1mil cube. And thermal resistance should drop with the square of the distance since because of R_th ={1_mil*(266 deg_K/W)/6.28}/R_mils^2 R_th =(42 deg_K/W)*1_mil/R_mils^2 At 5mA current at 5volts supply, the temperature gradient works out very close to 1 deg_C. ___________________________________ | delta_Temp_1mil_@25mW = 1.05deg_K | |___________________________________| >From 1 mils to 15mils, the total resistance and thermal gradient follows.. R_th = (42 deg_K/W)*( 1/1mils + 1/2mils + 1/3mils ..etc) T_delta =( 1 +0.25 +0.11 +0.06 +etc) =1.58 deg_K 25mW*(63deg_K/W) = 1.58deg_K ____________________________________ | 25mW of power from 1mil to 15mils | |sees a thermal R of 63 degree/watt | |____________________________________| | - - - - |- - - - _ -|- _ - - - - - \|/ - - - - - - _- V -_ - - _ - - - - - _____________________ - - - - -_ /1mil/ | | | - - - - -/_|_/1.6C/ / | - - - - _/ 2mil__/.6C_/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/ .3C _/ .2C - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ Once you exceed the thickness of the wafer (15mils), the thermal resistance of die bonded to metal is of course is much lower. It is common practice to layout a die such that anything sensitive to thermal gradients is on isothermal lines with anything that can dissipate power. It also does not hurt to put somes distance (say 15mils) between them. Cross coupling is also preferred if possible. ___________________________________________________ | ---___ | | \__ | | \_ ISOTHERMAL LAYOUT | | \ | | \ \ | | \ \ | | +1.6deg_C \ \ | | ___ |_| | | |5mA| |T| | | |@5V|<- 15mils ->|_| Input | | |___| |T| transistors | | Output |_| | | transistors / /<--5mdeg_C gradient | | / /<---across 1mil | | / / | | _/ 0.5mdeg_C*3uV/deg_C =1.5nV | | _/ | | ___/ | |___________________________________________________| If the input TC is about 3uV/deg_C, and if the isothermal layout reduces the gradient by 1/10th, the thermal feedback will induce 1.8nV of thermal induced offset. Compare this TC to the effects of thermalcouples on the PC board. Their TC can be as high as 15uV/deg_C compared to the 1nV/deg_C example given above. It is common for this thermal couple error to have an impact around the one Hertz frequency range. When there is a need for microvolt accuracy, there are a lot more things to consider. -------------------------------------------------------------------------------------- silicon Atomic number: 14 Atomic weight: 28.086 Density: 2329 [293 K]; 2525 at m.p. kg m-3 Molar volume: 12.06 cm3 Velocity of sound: 2200 m s-1 Hardness Mineral: 6.5 Melting point: 1683K Boiling point: 2628K Thermal conductivity: 148 [300 k] W m-1 K-1 Water Standard: 2.42W Fusion: 39.6 kJ mol-1 Vaporization: 383.3 kJ mol-1 __ Delta Temp /\ ______ L / /| Therm_Conduct =1.48 W/(cm*K) / / | /_____/ | R = Rho*L/A |__ A | / | /\ | / Rho = 266mil*K/W |/____|/ / Watt 180mW of power sees a thermal R of 70 degree/watt | - - - - |- - - - _ -|- _ - - - - \|/ - - - - - _- V -_ - - _ - - - - - 12C____-________________ - - - - -_ /1mil/ | | | - - - - -/_|_/ 3C / / | - - - - _/ 2mil__/2C _/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/1.4C _/ 1C - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ -------------------------------------------------------------------------------------- Thermal_Feed_back_Measured Temperature vs mil (180mW) _________________________________________________ | | 100C |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ | | | 10C |_XXXXXX_ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ | | XXXXXXXXX | 1C |_ _ |_ _ |_ _ |_ _XXXXXXXX _ |_ _ |_ _ |_ _ |_ _ | | XXXXXXXX | .1C |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ XXXXXX |_ _ | | | .01C |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ | 0mil 2mil 4mil 6mil 8mil 10mil Thermal Test ^ 13.2v SUPPLY VOLTAGE /|\ ____|____ | | | _|_ 200 Ohms / / _ \ 1uA \ \/ \/ | 4VpK / /\_/\ V __ | \___/ 1mVpK |__| _|_ _|_ __ in output1| | | |______ |__| __ |___| |___| | output2 | |__ | | | tau 100us __ _| |_ _|_ | |____|' NPN NPN `|___| | |__| |`-> <-'| |___| |_________| | _|_ KP101 7mils /// 180mW of power sees a thermal R of 70 degree/watt | - - - - |- - - - _ -|- _ - - - - \|/ - - - - - _- V -_ - - _ - - - - - 12C____-________________ - - - - -_ /1mil/ | | | - - - - -/_|_/ 3C / / | - - - - _/ 2mil__/2C _/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/1.4C _/ 1C - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ 180mW of power sees a thermal R of 70 degree/watt | | Area_Sphere = 4*PI()*R^2 \|/ ___________________V_________________ | | | \_ 1mil_/ | | | | \ \ \___/ / / | \ \_ \__ 2mil__/ _/ / \ \ \_____/ / / \_ \__ 3mil __/ _/ \_ \_________/ _/ \__ 4mil __/ \_ \_____________/ _/ \__ 5mil __/ \_______________/ -------------------------------------------------------------------------------------- Si density 2330 kg_per_m^3 mole volume = 12.06_cm^3 28.09_gm_per_mole Thermal_conductive 150_W_per_m_K 350/(T-68Kdeg)_W_per_cm_K { L.Heasell, "The Heat Flow Problem in Silicon," IEEE Trans. on Elect Dev.,ED-25,12(Dec. 1978):1382} Coeff_Therm_Expand = 2.6E-6_per_K CpH= 20.0_J_per_Mol_K = .76_J_per_gm_K Cv = 3*R {for most solids} = 25.031_J_per_Mol_K = 1.126_J_per_gm_K { approximated by the Dulong_Petit Law } Cv = for most minerals is about 1.3_J_per_gm_K Electrical Thermal Ohms Volts/current C_Deg/Watt Cap Volts/Q C_Deg/Joules Voltage Volts Temperature_C_Deg Current Amps Heat[energy]_Watts ___ 215 54 24 13 8.6 6 4.4 3.3 | |__/\ ___/\ __/\ _/\ __/\ __/\ __/\ __/\ __.. |___| \/ | \/| \/| \/ | \/| \/| \/| \/| Heat is _|_ _|_ _|_ _|_ _|_ _|_ _|_ _|_ Like a ___ ___ ___ ___ ___ ___ ___ ___ Current |1.3n |5.2n |12n |21n |33n |47n |64n |83n source _|_ _|_ _|_ _|_ _|_ _|_ _|_ _|_ /// /// /// /// /// /// /// /// 5um 10um 15um 20um 25um 30um 35um 40um ___ 1000 20um 107.4 | |___/\ _____________________/\ _____ |___| \/ | \/ | Atten 1/9.3 _|_ 34n | RC = 3.4us ___ | _|_ _|_ /// /// ___ 1000 40um 34 | |___/\ __________________/\ _____ |___| \/ | \/ | Attenuation 1/31 _|_ 180n | RC = 6us ___ | _|_ _|_ /// /// ___ 1000 60um 7.1 | |___/\ _________________/\ _____ |___| \/ | \/ | Attenuation 1/139 _|_ 1010n | RC = 7.5us ___ | _|_ _|_ /// /// Simulated/Measured Sanity Checked Attenuation and time delay 1000 ....................................... | . . . . | | . . . . | 100 |.......................................| | . . 6us . 7.4us | . . . . | 10 |.......................................| | 3.4us . . . | | . . . . | 1 |_______________________________________| | . . . . | | . . . . | 1/10 |......Atten............................| | . . . . | | . Atten . | 1/100 |.......................................| | . . . . Atten | . . . . | -1.5mV|_______________________________________| 10u 20u 30u 40u 50u 60u PNPL number ----------------------Package_Thermals------------------ silicon (physical data) Atomic number: 14 Atomic weight: 28.086 Density: 2329 [293 K]; 2525 at m.p. kg m-3 Molar volume: 12.06 cm3 Velocity of sound: 2200 m s-1 Hardness Mineral: 6.5 Melting point: 1683K Boiling point: 2628K Thermal conductivity: 148 [300 k] W m-1 K-1 Water Standard: 2.42W Fusion: 39.6 kJ mol-1 Vaporization: 383.3 kJ mol-1 First find the thermal mass (thermal capacitance) of a 200mil by 200mil by 15mil thick die. __________ / /| 200mils / / / 200mils*25.4u =>.5cm / / / 15mils*25.4u =>.038cm /_________/ / 15mils|_________|/ 200mils Silicon Volume 0.5cm*0.5cm*.038cm=.0095cc_Si Thermal Mass 0.0095cc*2.42Water= .0229gm Thermal CAP 0.0229gm*0.76joule/gm*C =.0174J/deg_C 1/60 of a Joule raise chip 1deg_C. The package in still air can have a thermal resistance around 200deg_C/W. So there is nothing stopping the modeling of the packaged silicon as a thermal RC. Applying 25mW for 1sec is 25mJoules. You would expect the die temperature to increase 1.47deg_C over that time. 1sec / __ 200deg_C/W _/ ____| |_________/\ /\___ _|_ |__| _|_ \/ | / _ \ ___ | Tau = R*C \/ \/ ^ 25mW | 17.4mJ/deg_C| /\_/\ /|\ | | \___/ | _|_ _|_ _|_ /// /// /// Tau_still_air = 200deg_C/W * 17.4mJ/deg_C = 3.5seconds Of course any air movement across the package will greatly influence this time constant. -----------------------Bench_Check_Thermal_gadients------------------ 180mW of power sees a thermal R of 70 degree/watt | - - - - |- - - - _ -|- _ - - - - \|/ - - - - - _- V -_ - - _ - - - - - 12C____-________________ - - - - -_ /1mil/ | | | - - - - -/_|_/ 3C / / | - - - - _/ 2mil__/2C _/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/1.4C _/ 1C - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ Temperature vs mil (180mW) _________________________________________________ | | 100C |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ | | | 10C |_XXXXXX_ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ | | XXXXXXXXX | 1C |_ _ |_ _ |_ _ |_ _XXXXXXXX _ |_ _ |_ _ |_ _ |_ _ | | XXXXXXXX | .1C |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ XXXXXX |_ _ | | | .01C |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ |_ _ | 0mil 2mil 4mil 6mil 8mil 10mil ___________________________________ ^TEMP | Temperature_180mW vs distance_mil | /|\ |___________________________________| * . . 10C |.*............................................ | * . . | * . . | * . . | * . . | * . . | * . . | * . . | * . 1C |...................*.......................... | . * . | . * . | . * X . | . * . | . * . | . * . | . * . | . * 0.1C |.....................................*....... | . . * | . . * | . . * |______________________________________________\ / 0mil 4mil 8mil Distance Thermal Test ^ 13.2v SUPPLY VOLTAGE /|\ ____|__________ | | | _|_ 200 Ohms / |20mA / _ \ 1uA \ V \/ \/ | 4VpK / /\_/\ V __ |180mW_pp \___/ 1mVpK =1/2deg_C |__| _|_ _|_ __ in output1| |13deg_C_pp | |______ |__| __ |___| |___| | output2 | |__ | | | tau 100us __ _| |_ _|_ | |____|'NPN <-7mils-> NPN`|___| | |__| |`-> <-'| |___| |_______________| | _|_ KP101 /// ------------------------Self_Heating_Bipolar----------------------- Base Emitter Collector __|__ __|__ __|__ ______|_ _|____|_ _|_____|_ _|__________________ | | | P+ | | N+ | | | N+ | | P+Field | | \ \_____/Base\_____/ / \_____/ | | | \__________________/ NWELL | PWELL | | _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ | _ _ _ _ _ _ | |/ N BURIED LAYER \|/ P buryLay \| \ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _/ \ _ _ _ _ _ _/ As the bipolar geometries keep getting smaller in size, the effective thermal resistance that the transistor's power should see should get larger. R_th = 266 deg_K/W for 1mil^3 180mW of power sees a thermal R of 70 degree/watt | - - - - |- - - - _ -|- _ - - - - \|/ - - - - - _- V -_ - - _ - - - - - 12C____-________________ - - - - -_ /1mil/ | | | - - - - -/_|_/ 3C / / | - - - - _/ 2mil__/2C _/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/1.4C _/ 1C - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ The thermal resistance of a half shell will be 266deg_K/W*mil*delta_radius/(2*PI*Rad^2).From a spread sheet, the relationship is as such. ^ R_th Thermal resistance emitters /|\ sees versus emitter radius | . . | * . . | * . . | * . . 10K degC/W |.......*............................. | * . . | * . . | * . 1000degC/W |...............*...................... | . * . | . * . | . * . 100degC/W |.......................*............. | . * . | . * . | . * . 10degC/W |...............................*..... | . . * | . . * | . . * |____________________________________________\ / 1u 100u Distance So how much mismatch error does this produce? ^ 5V /|\ ___|_______ _|_ | / _ \ 1uA | a 1u radius \/ \/ | | means 1000deg_C/W /\_/\ V | \___/ | 1uA*5V*1000 | | is 5mdeg_C |_____ | |_ | _| NPN`|__|__|'NPN <-'| |`-> |___________| | _|_ 1uA => 5mdeg_C /// 100uA => 1/2deg_C => 1mV 1mA => 5deg_C -------------------------------------------------------------------------------------- as costly as germanium. Package_Thermals silicon (physical data) Atomic number: 14 Atomic weight: 28.086 Density: 2329 [293 K];2525 [at m.p.] kg m-3 Molar volume: 12.06 cm3 Velocity of sound: 2200 m s-1 Hardness Mineral: 6.5 Melting point: 1683K Boiling point: 2628K Thermal conductivity: 148 [300 k] W m-1 K-1 Water Standard: 2.42W Specific Heat: 0.76joule/gm*C Fusion: 39.6 kJ mol-1 Thermal_Feed_back silicon (physical data) Atomic number: 14 Atomic weight: 28.086 Density: 2329 [293 K]; 2525 [liquid at m.p.] kg m-3 Molar volume: 12.06 cm3 Velocity of sound: 2200 m s-1 Hardness Mineral: 6.5 Melting point: 1683K Boiling point: 2628K Thermal conductivity: 148 [300 k] W m-1 K-1 Coefficient of linear thermal expansion: K-1 Water Standard: 2.42W Fusion: 39.6 kJ mol-1 Vaporization: 383.3 kJ mol-1 thermal resistance for a 1mil cube geometry of Silicon? Therm_Conduct = 148 W/(m*K) = 148uW/(um*K) = 3.76mW/(mil*K) which means ... __ 3.76mWatt /\ -> ______ Delta Temp / /| If L =1mil,A = 1mil^2,and 3.76mWat 1 deg_K /L / | Delta Temp = 1 deg_K -> /_____/ | ________________________________ |__ A | / | R_th = 266 deg_K/W for 1mil^3 | | /\ | / |________________________________| |/____|/ / R_th = Rho*L/A 3.76mWatt Rho = 266mil*deg_K/W If you have a single transistor which is powered at 5mA*5V =25mW such that its temperature is uniform over a half shell 1 mil in radius temperature? 25mW of power | - - - - |- - - - _ -|- _ - - - - - \|/ - - - - - - _- V -_ - - _ - - - - - _____________________ - - - - -_ /1mil/ | | | - - - - -/_|_/ / / | - - - - _/ 2mil__/ _/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/ _/ - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ The thermal resistance can be modeled in terms of half shells of silicon. Given that the thermal gradients will follow this pattern, the following gives the area of a half sphere. A = 2*PI()*R^2 If you look at a 1mil thick shell which is 1mil away from the heat source, its thermal resistance should be about one sixth the thermal resistance of a 1mil cube. And thermal resistance should drop with the square of the distance since because of R_th ={1_mil*(266 deg_K/W)/6.28}/R_mils^2 R_th =(42 deg_K/W)*1_mil/R_mils^2 At 5mA current at 5volts supply, the temperature gradient works out very close to 1 deg_C. ___________________________________ | delta_Temp_1mil_@25mW = 1.05deg_K | |___________________________________| >From 1 mils to 15mils, the total resistance and thermal gradient follows.. R_th = (42 deg_K/W)*( 1/1mils + 1/2mils + 1/3mils ..etc) T_delta =( 1 +0.25 +0.11 +0.06 +etc) =1.58 deg_K 25mW*(63deg_K/W) = 1.58deg_K ____________________________________ | 25mW of power from 1mil to 15mils | |sees a thermal R of 63 degree/watt | |____________________________________| | - - - - |- - - - _ -|- _ - - - - - \|/ - - - - - - _- V -_ - - _ - - - - - _____________________ - - - - -_ /1mil/ | | | - - - - -/_|_/1.6C/ / | - - - - _/ 2mil__/.6C_/ / - - - /\__|__/ / / - - - /_ |3mil __/ _/ - - / \__|____/ .3C _/ .2C - /\_ 4mil __/ - / \___|______/ _/ /\__ |5mil __/ / \____|_______/ Once you exceed the thickness of the wafer (15mils), the thermal resistance of die bonded to metal is of course is much lower. It is common practice to layout a die such that anything sensitive to thermal gradients is on isothermal lines with anything that can dissipate power. It also does not hurt to put somes distance (say 15mils) between them. Cross coupling is also preferred if possible. ___________________________________________________ | ---___ | | \__ | | \_ ISOTHERMAL LAYOUT | | \ | | \ \ | | \ \ | | +1.6deg_C \ \ | | ___ |_| | | |5mA| |T| | | |@5V|<- 15mils ->|_| Input | | |___| |T| transistors | | Output |_| | | transistors / /<--5mdeg_C gradient | | / /<---across 1mil | | / / | | _/ 0.5mdeg_C*3uV/deg_C =1.5nV | | _/ | | ___/ | |___________________________________________________| If the input TC is about 3uV/deg_C, and if the isothermal layout reduces the gradient by 1/10th, the thermal feedback will induce 1.8nV of thermal induced offset. Compare this TC to the effects of thermalcouples on the PC board. Their TC can be as high as 15uV/deg_C compared to the 1nV/deg_C example given above. It is common for this thermal couple error to have an impact around the one Hertz frequency range. When there is a need for microvolt accuracy, there are a lot more things to consider. -------------------------------------------------------------------------------------- Package_Thermals silicon (physical data) Atomic number: 14 Atomic weight: 28.086 Density: 2329 [293 K];2525 [at m.p.] kg m-3 Molar volume: 12.06 cm3 Velocity of sound: 2200 m s-1 Hardness Mineral: 6.5 Melting point: 1683K Boiling point: 2628K Thermal conductivity: 148 [300 k] W m-1 K-1 Water Standard: 2.42W Specific Heat: 0.76joule/gm*C Fusion: 39.6 kJ mol-1 Vaporization: 383.3 kJ mol-1 thermal mass (thermal capacitance) of a 200mil by 200mil by 15mil thick die. __________ / /| 200mils / / / 200mils*25.4u =>.5cm / / / 15mils*25.4u =>.038cm /_________/ / 15mils|_________|/ 200mils 0.5cm*0.5cm*.038cm=.0095cc_Si Silicon Volume .0095cc*2.42Water= .0229gm Thermal Mass 0.0229gm*0.76joule/gm*C =.0174J/deg_C In other words, it takes about 1/60 of a Joule to raise the chip temperature 1deg_C. The package in still air can have a thermal resistance around 200deg_C/W. So there is nothing stopping the modeling of the packaged silicon as a thermal RC. Applying 25mW for 1sec is 25mJoules. You would expect the die temperature to increase 1.47deg_C over that time. 1sec / __ 200deg_C/W _/ ____| |_________/\ /\___ _|_ |__| _|_ \/ | / _ \ ___ | Tau = R*C \/ \/ ^ 25mW | 17.4mJ/deg_C| /\_/\ /|\ | | \___/ | _|_ _|_ _|_ /// /// /// Tau_still_air = 200deg_C/W * 17.4mJ/deg_C = 3.5seconds Of course any air movement across the package will greatly influence this time constant. -------------------------------------------------------------------------------------- It appears to be possible to accurately measure the self heating of transistors in the Perfect 10 process. The thermal resistance of silicon oxide is about 100 times larger than silicon. Each transistor is a small volume of silicon inside a its own tub of oxide. ________ ________________________ _____________________ |..| |..| |..| n40x5x1w |..| NPN transistor in |..|________________________|..| oxide tub |..............................| |______________________________| When power is applied to a transistor, the thermal resistance to the ambient is going to be define by the silicon oxide. The tiny volume of silicon mass is going to behave like a thermal capacitor. From the electrical to thermal analogy, power is like current and temperature is like voltage. ___ Power_Watts | |Temp _ _ |___| / \/ \ | R_therm ____/ _/\_\\______|_/\ _____ | \ \/ // | \/ | | \_/\_/ _|_ | | ___ C_therm | _|_ _|_ _|_ /// /// /// So when you apply a pulse of power to a transistor, you would expect the kind of response that you would expect from the circuit above. Because transitors have a large temperature coefficient , it was possible to view the RC transient response using the circuit shown below. 100 ^ VCC 10V ___ /_\ ____/\ _____ | C |_____| | \/ | |___| | | ____| 16mA*8.4V = 134mW _|_ | | / \ Pulse _|_ _| \___/ -0.3V | B |___|' n40x5x1w | +2.5V |___| |`-> _____----- _|_ 10KHz |___ <= 47mV pk => _-- Vbe heats up! /// | | _|_ 18uS tau | E | Delta 23C/134mW |___| 100 assume -2mV/C Rth =170degC/W |____/\ ______ \/ _|_ 0->1.6V /// The base of the transistor is switched on and off. When first on, the transistor will start raising the temperature of the tiny volume of silicon at a rate set by the amount of power and the small mass of silicon which makes up the transistor. Utimately, the temperature gets high enough such that all the power is flowing out of the thermal resistor and the temperature levels off. In the circuit above, it appears to be possible to monitor everything and see the emitter base voltage decrease in voltage as the transitor heats up. Using a digital oscilloscope, it is possible to tranfer the scope wave curve into a spread sheet where a curve fit method is applied. By applying 134mW, it appears the transistor's vbe changed 47mV with a RC time of 18us. This appears to correspond to the transistor seeing a thermal resistance of 170degC/W. The next step is to see how close this responce comes to what can be calculated. ________ ________________________ _____________________ |..| |..| |..| n40x5x1w |..| NPN transistor in |..|________________________|..| oxide tub |..............................| |______________________________| Assume oxide is 1u thick all around Assume Silicon tranistor is 4u deep The dimension of the transistor are shown above. I am also showing all the assumptions. Using the simplest method to calculate the thermal resistance of the oxide is as follows. transitor dimensions = 36 X 81 = 2916u2 = 4.52mil2 perimeter = 234u X 4u = 936u2 = 1.45mils2 Therm_Conduct_SiO2 1.4 W/(m*K) __ Delta Temp /\ ______ L / /| / / | /_____/ | R_therm = Rho*L/A |__ A | / | /\ | / Rho = 28,200 mil*K/W |/____|/ / Watt Rth = Rho*L/A = (28,200mil*K/W)*(1/25.4)/(4.5+1.45) = 186 degC/W The measured thermal resistance appears to be about 170 degC/W. Given the method of calculations, the two numbers are pretty close. The mass of silicon is nexted calculated and this leads to the thermal capacitance and thermal time constant. density = 2.42 gm/cm3v = 2.42e+6 gm/m3 = 2.42e+3 Kgm/m3 volume = 4u X 36u X 81u =11,664e-18 m3 =11.66e-15 m3 mass = 28.2e-12 Kg Spectif heat = 760 J/Kg-K C_therm = 21,431e-12 J/K = 21e-9 J/K R_therm*Ctherm = 186 X 21.4e-9j/K = 39.8e-6 The measured time constant looks pretty solid at 18us. So the the calculated time constant appears to be a factor of two higher. Again, the calcuation methods were simple and all the assumptions have been stated. But the agreement between the calculations and measurements appear to be close enough to suggest the measurements really are showing the thermal behavior in this process. A min geometry transistor is about 20 smaller in die size. In the test circuit, 16mA was the collector current. This says that currents under 1mA could effectively lower the emitter base voltage of a minimum transitor by as much as 50mV. The min geometries also have an emitter resistance of about 40->70 ohms. This degeneration should help. Inside a output power transistor, all the silicon should be close to the same temperature. So current hogging may not be too bad of a problem. The fun comes when matching the vbe of an output transistor to one that is not drawing as much power.