======================SILICON_THERMAL===========================
silicon
Atomic number: 14
Atomic weight: 28.086
Density: 2329 [293 K]; 2525 [liquid at m.p.] kg m3
Molar volume: 12.06 cm3
Velocity of sound: 2200 m s1
Hardness Mineral: 6.5
Melting point: 1683K
Boiling point: 2628K
Thermal conductivity: 148 [300 k] W m1 K1
Water Standard: 2.42W
Fusion: 39.6 kJ mol1
Vaporization: 383.3 kJ mol1

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 m3
Molar volume: 12.06 cm3
Velocity of sound: 2200 m s1
Hardness Mineral: 6.5
Melting point: 1683K
Boiling point: 2628K
Thermal conductivity: 148 [300 k] W m1 K1
Water Standard: 2.42W
Fusion: 39.6 kJ mol1
Vaporization: 383.3 kJ mol1
__ 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/(T68Kdeg)_W_per_cm_K
{ L.Heasell, "The Heat Flow Problem in Silicon,"
IEEE Trans. on Elect
Dev.,ED25,12(Dec. 1978):1382}
Coeff_Therm_Expand = 2.6E6_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 m3
Molar volume: 12.06 cm3
Velocity of sound: 2200 m s1
Hardness Mineral: 6.5
Melting point: 1683K
Boiling point: 2628K
Thermal conductivity: 148 [300 k] W m1 K1
Water Standard: 2.42W
Fusion: 39.6 kJ mol1
Vaporization: 383.3 kJ mol1
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 m3
Molar volume: 12.06 cm3
Velocity of sound: 2200 m s1
Hardness Mineral: 6.5
Melting point: 1683K
Boiling point: 2628K
Thermal conductivity: 148 [300 k] W m1 K1
Water Standard: 2.42W
Specific Heat: 0.76joule/gm*C
Fusion: 39.6 kJ mol1
Thermal_Feed_back silicon (physical data)
Atomic number: 14 Atomic weight: 28.086
Density: 2329 [293 K]; 2525 [liquid at m.p.] kg m3
Molar volume: 12.06 cm3
Velocity of sound: 2200 m s1
Hardness Mineral: 6.5
Melting point: 1683K
Boiling point: 2628K
Thermal conductivity: 148 [300 k] W m1 K1
Coefficient of linear thermal expansion: K1
Water Standard: 2.42W
Fusion: 39.6 kJ mol1
Vaporization: 383.3 kJ mol1
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 m3
Molar volume: 12.06 cm3
Velocity of sound: 2200 m s1
Hardness Mineral: 6.5
Melting point: 1683K
Boiling point: 2628K
Thermal conductivity: 148 [300 k] W m1 K1
Water Standard: 2.42W
Specific Heat: 0.76joule/gm*C
Fusion: 39.6 kJ mol1
Vaporization: 383.3 kJ mol1
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 =11,664e18 m3 =11.66e15 m3
mass = 28.2e12 Kg
Spectif heat = 760 J/KgK
C_therm = 21,431e12 J/K = 21e9 J/K
R_therm*Ctherm = 186 X 21.4e9j/K = 39.8e6
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.