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SC104EVB

Micro Power Constant-Current DC-DC Converter

Semtech Corporation 

SC104

POWER MANAGEMENT

Applications Information

Component Selection - Introduction

Referring to the 6 LED typical schematic below, there are

three components that depend upon the application that

need to be determined:

RSET - this resistor sets the output current for the device

RLIM - this resistor sets the peak inductor current

L - the output inductor

All the other components can be mostly generalized and

are addressed below the following design steps.

VIN = 3V to 5V

Vf = 0.35V

VCE(SAT) = 0.25V

thus DC = 0.87

Since this value is greater than the guaranteed minimum

value for maximum duty cycle, the device will be operating

in discontinuous mode to provide the desired output. Note

that the duty cycle does not depend upon the output

current, and that unless the output to input ratio is low,

the device will usually need to be in discontinuous mode,

so we will cover that first (Step 1 through Step 5).

Continuous mode calculations start at Step 6.

IOUT ADJUST

IOUT = 15mA

LED4

LED5

LED6

LED3

LED2

LED1

RSET

23.2R

COUT

0.47uF

U1

1

ADJ

2

FB

3

GND

4

OUT

SC104

8

EN

7

LIM

6

IN

5

LX

D1

CIN

4.7uF

ENABLE

L1

12uH

RLIM

7.50k

Step 1: Continuous or Discontinuous?

The first thing to do when designing with the SC104 is to

determine whether the output inductor will be operating

in continuous mode (where the inductor current does not

drop to zero while the device is switching) or discontinuous

mode (where the inductor current drops to zero while

switching). This determination can be made simply by

calculating the required duty cycle needed for the target

output voltage, and comparing it to the guaranteed

minimum value for the maximum duty cycle from the

Step 2: Calculating the Inductor for Discontinuous Mode

Having determined that we need to be operating in

discontinuous mode, we next need to calculate the

maximum inductor value allowed that will permit the part

to output the correct power. The maximum discontinuous

inductor value, L(D) is given by:

( ( ) ) ( ( ) ) L(D)

=

t2

ON(MIN)

•

VIN

•

VIN − VCE(SAT )

2 • 1.4 • VOUT • IOUT • t + t ON(MIN) OFF(MIN)

•

VOUT − VCE(SAT ) + Vf

VOUT − VIN + Vf

Where:

t = minimum switch on-time = 1.8µs

ON(MIN)

IOUT = required output current

tOFF(MIN) = minimum switch off-time = 0.6µs

Using our 6 LED example:

IOUT = 15mA

Electrical Characteristics on Page 3. %DC(MIN) = 70% (or 0.7

duty). If DC is greater than 0.7 then discontinuous mode

is required. The required duty cycle is calculated as

follows:

( ) DC = VOUT − VIN + Vf

( ) VOUT − VCE(SAT) + Vf

thus L(D) = 14.4µH

Selecting the next lower standard value gives us

L(D) = 12µH. Of course a lower value inductor may be used

if desired, but may not necessarily be the most efficient

choice.

Where:

VOUT = output voltage, the sum of the total LED (max.)

forward voltage drop at the required output voltage

plus the feedback voltage, 0.35V.

VIN = minimum input voltage

Vf = Schottky diode (D1) forward voltage drop

V = power switch saturation voltage

CE(SAT)

Using the 6 LED example above:

VOUT = (6 * 3.475) + 0.35 = 21.2V

V = 3V

IN

Step 3: Calculating the Current Limit Required with this

Inductor for Discontinuous Mode

Having determined the inductor value we are going to use,

we next need to calculate the current limit required to

meet the necessary output power. The discontinuous

mode current limit, I , is given by:

LIM(D)

( ) ILIM(D) =

VIN − VCE(SAT)

L(D)

• tON(MIN)

 2005 Semtech Corp.

6

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