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Power and Temperature TIPL 1160 TI Precision Labs – Op Amps
Presented by Ian Williams
Prepared by Art Kay, Ian Williams and Miro Oljaca
Power Dissipation – Quiescent Current
2
PQ = quiescent power = IQ * VS
IQ = quiescent current
VS = total supply voltage
VS = VCC – VEE = (15) – (-15V) = 30V
Example: worst case over temperature
PQ = IQ ∙ VS = (6mA)(30V) = 180mW
VCC +15V
-
+
OPA827
+
VIN
VOUT
RL 1
k
VEE -15V
Power Dissipation – DC Load
3
Effective load:
RL = RLoad || (RF + R1)
RL = (1kΩ)|| (900Ω + 100Ω)
RL = 500 Ω
+15V
-15V
-
+
+
Vin
1Vdc
VOUT
RLoad 1kΩ
R1 100Ω RF 900Ω
+
5V
-
20mA
+
10V
-
10mA
10mA
+ 9V -+ 1V -
10mA10mA
Maximum DC Power Dissipation
4
0
0.02
0.04
0.06
0.08
0.1
0.12
0 2 4 6 8 10 12 14
Op
Am
p P
ow
er
Dis
ipai
ton
(W)
Vout dc Output Voltage (V)
Op Amp dc Power vs. Vout
Vcc2= 7.5V
Vcc2
4 ∙ RL= 113mW
Pdc =VoutRL
Vcc − Vout
Pdc _max = Pdc Vcc2
=Vcc
2
4 ∙ RL
DC Output Voltage (V)
Po
we
r D
issip
atio
n (
W)
Popa = Vcc − Vout ∙VoutRL
(1) Power dissipated in op amp
Popa (Vout ) =Vout ∙ VccRL
−Vout
2
RL (2) Dc output voltage
∂Popa
∂Vout=Vcc − 2 ∙ Vout
RL= 0 (3) Take the partial derivative. Set to zero and solve for maxima.
Vout =Vcc2
when ∂Popa
∂Vout= 0 (4) Solve (3) for value of Vout that yields maximum power
Pdc _max = Popa (Vcc2) =
Vcc2
4 ∙ RL (5) Substitute (4) into (2) to determine maximum dc power in Op Amp
Derivation of DC Maximum Power Transfer
5
Popa = Vcc − Vout ∙VoutRL
(1) Power dissipated in op amp
Popa (Vout ) =Vout ∙ VccRL
−Vout
2
RL (2) Dc output voltage
∂Popa
∂Vout=Vcc − 2 ∙ Vout
RL= 0 (3) Take the partial derivative. Set to zero and solve for maxima.
Vout =Vcc2
when ∂Popa
∂Vout= 0 (4) Solve (3) for value of Vout that yields maximum power
Pdc _max = Popa (Vcc2) =
Vcc2
4 ∙ RL (5) Substitute (4) into (2) to determine maximum dc power in Op Amp
Power Dissipation at AC
6
Effective Load:
RL = RLoad || (RF + R1)
RL = 500 Ω
+15V
-15V
-
++
VOUT
1VPK
VOUT
10VPK
RL 1kΩ
R1 100Ω RF 900Ω
+5V-
20mA
+
10V
-
10mA
10mA
+ 9V -+ 1V -
20mA10mA
T
Time (s)0.0 250u 500u 750u 1.0m
Vo
lta
ge
(V
)
-10.0
-5.0
0.0
5.0
10.0
T
Time (s)0.0 250u 500u 750u 1.0m
Vo
lta
ge
(V
)
-1.0
-0.5
0.0
0.5
1.0
10mA
Maximum Average AC Power Dissipation
7
0
0.02
0.04
0.06
0.08
0.1
0 2 4 6 8 10 12 14
Op
Am
p a
c A
vera
ge P
ow
er
(W)
Peak ac Output Voltage (Vout)
Op Amp ac Average Power vs. Vout
2 ∙ Vcc2
π2 ∙ RL= 91mW
2 ∙ Vccπ
= 9.55V
Pac _avg (Voutpk ) =2 ∙ Vcc ∙ Voutpk
π ∙ RL−Voutpk
2
2 ∙ RL
Pac _max_avg = Popa _avg 2 ∙ Vcc
π =
2 ∙ Vcc2
π2 ∙ RL
Po
we
r D
issip
atio
n (
W)
Peak AC Output Voltage (V)
Popa = Vcc − Vout ∙VoutRL
(1) Power dissipated in op amp
Vout (t) = Voutpk ∙ sin(ω ∙ t) (2) ac sinusoidal wave out
Pac (t) = Vcc − Voutpk ∙ sin(ω ∙ t) ∙Voutpk ∙ sin(ω ∙ t)
RL
(3) Substitute (2) into (1)
Pac (t) =VccVoutpk ∙ sin(ω ∙ t)
RL−Voutpk
2 ∙ sin2(ω ∙ t)
RL
(4) Power dissipated in op amp as a function of time
ω =2π
T
(5) Angular frequency as a function of period
Pac (t) =VccVoutpk ∙ sin(
2πT ∙ t)
RL−Voutpk
2 ∙ sin2(2πT ∙ t)
RL
(6) Substitute (5) into (4)
Pac _avg =1
T 2
VccVoutpk ∙ sin(2πT ∙ t)
RL−Voutpk
2 ∙ sin2(2πT ∙ t)
RL
T/2
0
dt
(7) Find the average power
Pac _avg (Voutpk ) =2 ∙ Vcc ∙ Voutpk
π ∙ RL−Voutpk
2
2 ∙ RL
(8) Average power as a function of peak output voltage
∂Popa _avg
∂Voutpk=2 ∙ Vccπ ∙ RL
−Voutpk
RL
(9) Take the partial derivative to find maxima. Set to zero and solve for maxima.
Pac _max _avg = Popa _avg 2 ∙ Vcc
π =
2 ∙ Vcc2
π2 ∙ RL
(10) Maximum power and the peak output voltage where max power occurs
Derivation of AC Maximum Power Transfer
8 Derivation courtesy of Miro Oljaca
Thermal Device Model – No Heat Sink
9
θJA Ambient Junction PCB
Op amp
package
θJA includes the effects of the package and PCB!
Analogous Electrical Model – No Heat Sink
10
PD
θJA
TA
TJ
TA
TJ = (PD*ΘJA) + TA
T voltage
θ resistance
P current
Temperature Rise – Maximum DC Load
11
PQ = IQ ∙ VS = 6mA ∙ 30V = 180mW
Pdc _max =Vcc
2
4 ∙ RL=
15V 2
4 ∙ (500Ω)= 113mW
Ptotal = PQ + Pdc _max = 180mW+ 113mW = 293mW
TJ = Ptotal ∙ θJA + TA = 293mW ∙ 150 W + 25 = 67.5
+15V
-15V
-
+
OPA827
+
VIN_DC
VOUT
RL 1
kΩ
R1 100Ω RF 900Ω
Absolute Maximum Ratings
12
Thermal Protection
13
50 75 100 125 1502525
30
35
40
45
Temperature (°C)
PT
AT
Cu
rre
nt (µ
A)
PTAT Current (µA) vs. Temperature
-
+
Comparator
with hysteresis
RS
PTAT
current
-
+
+VS
-VS
+VS
Disconnects power at 160⁰CReconnects power at 140⁰C
VREF
Thermal Model – Device with Heat Sink
14
θSA
Ambient
Case Junction
Heat sink
θCS
θJC
Fins
Analogous Electrical Model – Device with Heat Sink
15
θJC
θCS
θSA
TJ
TC
TS
TA
PD
TJ = PD*(θJC + θCS + θSA) + TA
PD = total power dissipation
TJ = junction temperature
TC = case temperature
TS = heat sink temperature
TA = ambient temperature
Thermal Resistance – ΘJC (Junction to Case)
16
TJ = PD*(θJC + θCS + θSA) + TA
Thermal Resistance – ΘCS (Case to Sink)
17
Typical Interface Resistances for Various Mounting
Methods with a TO-220 (interface area = 1 in2):
Thermal Joint Compound only (0.001 thick) θ = 0.056/W
Mica (0.005) and Joint Compound (0.002) θ = 0.44 /W
Series 177 Beryllium Oxide Wafers (0.062)
And joint Compound (0.002) θ = 0.13 /W
DeltaPadTM 173-9 (0.009) θ = 0.50 /W
Dry Mounting (0.001 assumed) θ = 1.2 /W
TJ = PD*(θJC + θCS + θSA) + TA
Thermal Resistance – ΘSA (Sink to Ambient)
18
Example: Aavid Thermalloy 6398BG
TJ = PD*(θJC + θCS + θSA) + TA
100
80
60
40
20
0
5
4
3
2
1
00 4 8 12 16 20
0 200 400 600 800 1000
Heat Dissipated (W)
Air Flow Velocity (Ft./Min.)
Mo
un
tin
g S
urf
ace
Te
mp
Ris
e A
bo
ve
Am
bie
nt (°
C)
Th
erm
al R
esis
tan
ce
fro
m
Su
rfa
ce
to
Am
bie
nt (°
C/W
)
Thermal Resistance in Natural Convection*
19
Device
Power
(W)
θcs Thermal
Resistance
(°C/W)
2 7.5
4 7.0
6 6.33
8 6.0
10 5.5
θSA =ΔTmountPtotal
=15
2W= 7.5 W Using Graph
100
80
60
40
20
0
5
4
3
2
1
00 4 8 12 16 20
0 200 400 600 800 1000
Heat Dissipated (W)
Air Flow Velocity (Ft./Min.)
Mo
un
tin
g S
urf
ace
Te
mp
Ris
e A
bo
ve
Am
bie
nt (°
C)
Th
erm
al R
esis
tan
ce
fro
m
Su
rfa
ce
to
Am
bie
nt (°
C/W
)
*air flow ≤ 100 ft./min
Thermal Resistance in Forced Airflow
20
Air
Velocity
(ft./min)
θcs Thermal
Resistance
(°C/W)
200 3
400 2
600 1.8
800 1.1
100
80
60
40
20
0
5
4
3
2
1
00 4 8 12 16 20
0 200 400 600 800 1000
Heat Dissipated (W)
Air Flow Velocity (Ft./Min.)
Mo
un
tin
g S
urf
ace
Te
mp
Ris
e A
bo
ve
Am
bie
nt (°
C)
Th
erm
al R
esis
tan
ce
fro
m
Su
rfa
ce
to
Am
bie
nt (°
C/W
)
Example – Calculate Total Power
21
Pdc =Vcc
2
4 ∙ RL=
(15V)2
4 ∙ (25Ω)= 2.25W Maximum dc Power
Vs = Vcc − Vee = 15V− −15V = 30V Total Supply Voltage
IQ = 25mA From OPA541 Data Sheet
PQ = IQ ∙ Vs = 25mA ∙ 30V = 0.75W Quiescent Power
Ptotal = Pdc + PQ = 2.25W+ 0.75W = 3W Total Power
+15V
-15V
-
+
OPA541
+
VIN
VOUT
RL 25Ω
22
Ptotal = 3W
θSA =ΔTmountPtotal
=22
3W= 7.33 W
Example – Calculate θSA for Given Power
100
80
60
40
20
0
5
4
3
2
1
00 4 8 12 16 20
0 200 400 600 800 1000
Heat Dissipated (W)
Air Flow Velocity (Ft./Min.)
Mo
un
tin
g S
urf
ace
Te
mp
Ris
e A
bo
ve
Am
bie
nt (°
C)
Th
erm
al R
esis
tan
ce
fro
m
Su
rfa
ce
to
Am
bie
nt (°
C/W
)
(3W, 22°C)
Example – Calculate Junction Temperature
23
θJC = 3°C/W from OPA541 data sheet (TO-220)
θCS = 0.44°C/W (Mica and joint compound)
θSA = 7.33°C/W (Heat sink specification at 3W)
TA = 25°C
PD = 3W
TJ = PD*(θJC + θCS + θSA) + TA
TJ = (3W)*(3°C/W + 0.44°C/W + 7.33°C/W) + 25°C
TJ = 57.3°C
+15V
-15V
-
+
OPA541
+
VIN
VOUT
RL 25Ω
24
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