TL2575_15 datasheet(16/32 Pages) TI

16 / 32 page

7758

OUT

(µF)

IN(Max)

L1(µH)

OUT

(

R2 = R1

– 1

OUT

REF

= V

OUT

REF

1 +

where V

= 1.23 V

REF

(

TL2575, TL2575HV

SLVS638C – JANUARY 2006 – REVISED NOVEMBER 2014

www.ti.com

Typical Application (continued)

PROCEDURE (Fixed Output)

EXAMPLE (Fixed Output)

A. In normal operation, the catch diode requires a current rating of at A. Pick a diode with 3-A rating.

least 1.2 × ILOAD(Max). For the most robust design, D1 should be

rated to handle a current equal to the TL2575 maximum switch peak

current; this represents the worst-case scenario of a continuous

short at VOUT.

B. The diode requires a reverse voltage rating of at least B. Pick 30-V rated Schottky diode (1N5821, MBR330, 31QD03, or

1.25 × VIN(Max).

SR303) or 100-V rated Fast Recovery diode (31DF1, MURD310, or

HER302).

4. Input Capacitor (CIN)

An aluminum electrolytic or tantalum capacitor is needed for input CIN = 100 μF, 25 V, aluminum electrolytic

bypassing. Locate CIN as close to the VIN and GND pins as possible.

PROCEDURE (Adjustable Output)

EXAMPLE (Adjustable Output)

Known:

VOUT(Nom)

VOUT = 10 V

VIN(Max) = Maximum input voltage

VIN(Max) = 25 V

ILOAD(Max) = Maximum load current

ILOAD(Max) = 1 A

1. Programming Output Voltage (Selecting R1 and R2)

Referring to Figure 2, VOUT is defined by:

Select R1 = 1 k

Ω

R2 = 1 (10 / 1.23 – 1) = 7.13 k

Ω

Select R2 = 7.15 k

Ω (closest 1% value)

Choose a value for R1 between 1 k

Ω and 5 kΩ (use 1% metal-film

resistors for best temperature coefficient and stability over time).

2. Inductor Selection (L1)

A. Calculate the "set" volts-second (E × T) across L1:

E × T = (VIN – VOUT) × ton

E × T = (25 – 10) × (10 / 25) × (1000 / 52) [V ×

μs]

E × T = (VIN – VOUT) × (VOUT / VIN) × {1000 / fosc(in kHz)} [V × μs]

E × T = 115 V ×

μs

NOTE: Along with ILOAD, the "set" volts-second (E × T) constant

establishes the minimum energy storage requirement for the

inductor.

B. Using Figure 17, select the appropriate inductor code based on B. Using Figure 17, the intersection of 115 V •

μs and 1 A

the intersection of E × T value and ILOAD(Max).

corresponds to an inductor code of H470.

C. The inductor chosen should be rated for operation at 52-kHz and C. H470

→ L1 = 470 μH

have a current rating of at least 1.15 x ILOAD(Max) to allow for the Choose from:

ripple current. The actual peak current in L1 (in normal operation)

34048 (Schott)

can be calculated as follows:

PE-53118 (Pulse Engineering)

IL1(pk) = ILOAD(Max) + (VIN – VOUT) × ton / 2L1

Where ton = VOUT / VIN × (1 / fosc)

RL1961 (Renco)

3. Output Capacitor Selection (COUT)

A. The TL2575 control loop has a two-pole two-zero frequency A. COUT ≥ 7785 × 25 / (10 × 470) [μF]

response. The dominant pole-zero pair is established by COUT and C

OUT ≥ 41.4 μF

L1. To meet stability requirements, COUT must meet the following

obtain

acceptable

output

voltage

ripple

→

requirement:

COUT = 220 μF electrolytic

However, COUT may need to be several times larger than the

calculated value above in order to achieve an acceptable output

ripple voltage of ~0.01 × VOUT.

Submit Documentation Feedback

Product Folder Links: TL2575 TL2575HV

TL2575 Datasheet(PDF) 16 Page - Texas Instruments

TL2575 Datasheet(HTML) 16 Page - Texas Instruments

Html Pages

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Datasheet Download

Link URL