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 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:
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:
Equation 2
where
VIN = input voltage
VOUT = regulated output voltage
IO = load current
IQ = quiescent current drain
TJTAPDqJA–
+=
PDVIN VOUT
–()IOVIN IQ
´+´=
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*
Solve for IO,
= 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
qJA–
TJTA
–
PD
------------------=
PD
TJTA
–
qJA–
------------------=
VIN VOUT
–()IOVIN IQ
´+´=
IO
TJTA
–
qJA–VIN VOUT
–()
-------------------------------------------------VIN IQ
´
VIN VOUT
–()
-----------------------------------–=
IO150°C 125°C–
41.6°C/W 10V´
----------------------------------------- 40 3.25´10 3–
´
10
----------------------------------------–=
———————————————
*The current drain will increase by 50mA/VOUT on positive side and 100mA/VOUT on negative side