RC4194D 데이터 시트보기 (PDF) - Fairchild Semiconductor

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RC4194D

Dual Tracking Voltage Regulators

Fairchild Semiconductor 

RC4194

PRODUCT SPECIFICATION

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 mW 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:

TJ = TA + PDqJ – A

Equation 1

where

TJ = junction temperature (°C)

TA = ambient air temperature (°C)

PD = power dissipated by device (W)

qJ-A = thermal resistance from junction to ambient

air (°C/W)

The power dissipated by the voltage regulator can be detailed

as follows:

PD = (VIN – VOUT) ´ IO + VIN ´ IQ

Equation 2

where

VIN = input voltage

VOUT = regulated output voltage

IO = load current

IQ = quiescent current drain

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:

TJ at thermal shutdown = 150°C

TA = 125°C

qJ-A = 41.6°C/W, K (TO-66) pkg.

VIN = 40V

VOUT = 30V

IQ = 1 mA + 75 mA/VOUT x 30V

= 3.25 mA*

qJ – A

=

T----J----–----T----A--

PD

PD = T---q-J---J-–--–--T-A---A--

= (VIN – VOUT) ´ IO + VIN ´ IQ

Solve for IO,

IO = -q---J---–---A----(--TV---J--I--–N----T-–---A-V-----O----U---T----) – (---V----VI--N---I--N–-----´V----OI---Q-U----T---)-

IO

=

4---11---5.--6-0--°-°--CC----/-W-–----1---´2---5-1---°0--C--V---

–

-4--0-----´-----3---.-2---5-----´-----1---0---–--3-

10

= 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, qJ-A can be broken into the following compo-

nents:

qJ-A = qJ-C + qC-S + qS-A

where

qJ-C = junction-to-case thermal resistance

qC-S = case-to-heatsink thermal resistance

qS-A = heatsink-to-ambient thermal resistance

———————————————

*The current drain will increase by 50mA/VOUT on positive side and 100mA/VOUT on negative side

9

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