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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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