Heatsinking – Fairchild DUAL TRACKING VOLTAGE REGULATORS RC4194 User Manual

Page 9

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RC4194

PRODUCT SPECIFICATION

9

If a small signal silicon diode is used, it will clamp the nega-
tive output voltage at about +0.55V. A Schottky barrier or
germanium device would clamp the voltage at about +0.3V.
Another cure which will keep the negative output negative at
all times is the 1 m

W resistor connected between the +15V

output and the Comp- terminal. This resistor will then sup-
ply drive to the negative output transistor, causing it to satu-
rate to -1V during the brownout.

Heatsinking

Voltage Regulators are power devices which are used in a
wide range of applications.

When operating these devices near their extremes of load
current, ambient temperature and input-output differential,
consideration of package dissipation becomes important to
avoid thermal shutdown at 175

°C. The RC4194 has this fea-

ture to prevent damage to the device. It typically starts
affecting load regulation approximately 2

°C below 175°C.

To avoid shutdown, some form of heatsinking should be used
or one of the above operating conditions would need to be
derated.*

The following is the basic equation for junction temperature:

Equation 1

where

T

J

= junction temperature (

°C)

T

A

= ambient air temperature (

°C)

P

D

= power dissipated by device (W)

q

J-A

= thermal resistance from junction to ambient

air (

°C/W)

The power dissipated by the voltage regulator can be detailed
as follows:

Equation 2

where

V

IN

= input voltage

V

OUT

= regulated output voltage

I

O

= load current

I

Q

= quiescent current drain

T

J

T

A

P

D

q

J

A

+

=

P

D

V

IN

V

OUT

(

) I

O

V

IN

I

Q

´

+

´

=

Let’s look at an application where a user is trying to deter-
mine whether the RC4194 in a high temperature environ-
ment will need a heatsink.

Given:

T

J

at thermal shutdown = 150

°C

T

A

= 125

°C

q

J-A

= 41.6

°C/W, K (TO-66) pkg.

V

IN

= 40V

V

OUT

= 30V

I

Q

= 1 mA + 75

mA/V

OUT

x 30V

= 3.25 mA*

Solve for I

O

,

= 60 mA – 13 mA ~ 47 mA

If this supply current does not provide at least a 10% margin
under worst case load conditions, heatsinking should be
employed. If reliability is of prime importance, the multiple
regulator approach should be considered.

In Equation 1,

q

J-A

can be broken into the following compo-

nents:

q

J-A

=

q

J-C

+

q

C-S

+

q

S-A

where

q

J-C

= junction-to-case thermal resistance

q

C-S

= case-to-heatsink thermal resistance

q

S-A

= heatsink-to-ambient thermal resistance

q

J

A

T

J

T

A

P

D

------------------

=

P

D

T

J

T

A

q

J

A

------------------

=

V

IN

V

OUT

(

) I

O

V

IN

I

Q

´

+

´

=

I

O

T

J

T

A

q

J

A

V

IN

V

OUT

(

)

-------------------------------------------------

V

IN

I

Q

´

V

IN

V

OUT

(

)

-----------------------------------

=

I

O

150

°C 125°C

41.6

°C/W 10V

´

-----------------------------------------

40

3.25

´

10

3

´

10

----------------------------------------

=

———————————————
*The current drain will increase by 50

m

A/V

OUT

on positive side and 100

m

A/V

OUT

on negative side

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