AL8400

0.2V LINEAR LED DRIVER CONTROLLER

AL8400

Document number: DS35115 Rev. 1 - 2

6 of 12

www.diodes.com

December 2010

© Diodes Incorporated

There are two important equations for the circuit:

LED circuit path:

Control drive circuit path

The maximum total LED voltage plus the reference voltage determines the minimum supply voltage. Substituting into

equation 1 yields:

REF

CEsat

LED

min

CC

V

V

V

V

+

+

=

max

B

min

OUT

REF

max

BE

min

CC

max

B

I

I

V

V

V

R

+

−

−

=

min

B

max

OUT

REF

min

BE

max

CC

min

B

I

I

V

V

V

R

+

−

−

=

min

FE

LED

max

B

h

I

I

=

current gain of the transistor.

max

FE

LED

min

B

h

I

I

=

the transistor.

Finally, the bipolar selection is also influenced by the maximum power dissipation

Since this determines the package choice (

θ

Bipolar Example

The driver is required to control 2 series connected LEDs at 150mA ±10%, each having a forward voltage of 3V minimum and

3.6V maximum. Hence the minimum operating supply voltage is 3.6*2 + 0.2 = 7.4V. The actual supply voltage given is

8V ±5%, i.e. 7.6V minimum. We will use the DNLS320E bipolar transistor (Q2).

The DNLS320E datasheet shows:

The datasheet graph shows a very slow variation at his current, so a value of 500 is appropriate.

Then

500

150

B

=

= 0.3mA

is approximately 0.8V at -55°C.

From these the maximum allowed bias resistor value is:

0003

.

0

0003

.

0

2

.

0

8

.

0

6

.

7

B

+

−

−

=

= 11k

Ω

1200

150

B

=

= 0.125mA