US431L/431AL

2-4

Rev. 1.2

5/11/98

APPLICATION INFORMATION

APPLICATION INFORMATION

Vo

V

Vref

R

R

Iref

R

KA

=

=

×

+









+

×

1

1

2

1

The maximum value for the biasing resistor is calcu-

lated using the following equations :

Assuming R1=2k

Ω as before,

The maximum power dissipation of the resistor is

calculated under the maximum supply voltage as

follows :

Thermal Design

The US431L is offered in the plastic TO-92 (Z) or the

surface mount SOT23-5 (L) packages. The TO-92 pack-

age has the maximum power dissipation capability of

775mW at Ta=25

°C with the derating factor of -6.2mW/

°C.The SOT23-5 package has the maximum power dis-

sipation capability of 150mW at Ta =25

°C with the

derating factor of -1.2mW/

°C.

Table below summarizes the maximum power dissipa-

tion capability of each package versus ambient tempera-

ture.

Ambient Temperature (Ta) -

°C

Pkg

25

40

50

60

70

TO 92

775mW 682mW 620mW 558mW 496mW

SOT23-5 150mW 132mW 120mW 108mW 96mW

Output Voltage Setting

The US431L can be programmed to any voltages in the

range of 1.24 to 6V with the addition of R1 and R2 exter-

nal resistors according to the following formula:

The US431L keeps a constant voltage of 1.240V be-

tween the Ref pin and ground pin. By placing a resistor

R2 across these two pins a constant current flows

through R2, adding to the Iref current and into the R1

resistor producing a voltage equal to the (1.240/R2)*R1

+ Iref * R1 which will be added to the 1.240V to set the

output voltage as shown in the above equation. Since

the Input bias current of the Ref pin is 0.5 uA max , it

adds a very small error to the output voltage and for most

applications can be ignored . For example, in a typical

5V to 3.3V application where R2=1.21k

Ω and R1=2kΩ

the error due to the Iadj is only 1mV which is about

0.03% of the nominal set point.

Figure 2 - Typical application of the US431L for program-

ming the output voltage.

not limit the input current under the minimum input

supply and maximum load and biasing current.

An example is given below on how to properly select

the biasing resistor.

Assuming :

R

V

Where :

B

MIN

MAX

=

- V

I

+ I

I

= I

+ I

V

= Minimum supply voltage

I

= Maximum load current

I

= Maximum bias current

I

= Maximum value for the minimum

cathode current spec

I

= Current through R1

KA

B

L

B

K

R

MIN

L

B

K

R

MAX

MAX

MAX

MIN

MAX

MAX

MIN

I =

3.3 - 1.24

2

mA

I

= 0.08 + 1.03 = 1.11 mA

.5 - 3.3

1.11+ 10

R = 100

R

B

B

MAX

=

=

=

103

4

108

.

R

Selecting

Ω

Ω

431app2-1.0

V

R2

Co

R1

R

US431

I

R

K

B

IN

L

(

)

P

V

Where :

P

R MAX

MAX

R MAX

B

B

(

)

(

)

=

=

- V

R

V

= Maximum supply voltage

Maximum R power dissipation

KA

B

MAX

B

2

(

)

B

(

)

=

- 3.3

mW

6

100

73

2

=