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TL2575_10 Datasheet, PDF (18/29 Pages) Texas Instruments – 1-A SIMPLE STEP-DOWN SWITCHING VOLTAGE REGULATORS
TL2575, TL2575HV
1-A SIMPLE STEP-DOWN SWITCHING VOLTAGE REGULATORS
SLVS638B – MAY 2006 – REVISED JANUARY 2007
www.ti.com
PROCEDURE (Adjustable Output)
EXAMPLE (Adjustable Output)
Known:
VOUT(Nom)
VIN(Max) = Maximum input voltage
ILOAD(Max) = Maximum load current
1. Programming Output Voltage (Selecting R1 and R2)
Referring to Fig. 2, VOUT is defined by:
( V = V OUT
REF
1
+
R2
R1
where VREF = 1.23 V
Choose a value for R1 between 1 kΩ and 5 kΩ (use 1% metal-film
resistors for best temperature coefficient and stability over time).
( R2 = R1
VOUT
VREF
–1
Known:
VOUT = 10 V
VIN(Max) = 25 V
ILOAD(Max) = 1 A
1. Programming Output Voltage (Selecting R1 and R2)
Select R1 = 1 kΩ
R2 = 1 (10/1.23 – 1) = 7.13 kΩ
Select R2 = 7.15 kΩ (closest 1% value)
2. Inductor Selection (L1)
2. Inductor Selection (L1)
A. Calculate the "set" volts-second (E•T) across L1:
E•T = (VIN – VOUT) × ton
E•T = (VIN – VOUT) × (VOUT/VIN) × {1000/fosc(in kHz)} [V•µs]
NOTE: Along with ILOAD, the "set" volts-second (E•T) constant
establishes the minimum energy storage requirement for the
inductor.
A. Calculate the "set" volts-second (E•T) across L1:
E•T = (25 – 10) × (10/25) × (1000/52) [V•µs]
E•T = 115 V•µs
B. Using Figure 19, select the appropriate inductor code based on
the intersection of E•T value and ILOAD(Max).
C. From Table 2, choose the appropriate inductor based on the
inductor code. Parts from three well-known inductor manufacturers
are given. The inductor chosen should be rated for operation at
52-kHz and have a current rating of at least 1.15 x ILOAD(Max) to
allow for the ripple current. The actual peak current in L1 (in normal
operation) can be calculated as follows:
IL1(pk) = ILOAD(Max) + (VIN – VOUT) × ton/2L1
Where ton = VOUT/VIN × (1/fosc)
3. Output Capacitor Selection (COUT)
A. The TL2575 control loop has a two-pole two-zero frequency
response. The dominant pole-zero pair is established by COUT and
L1. To meet stability requirements, COUT must meet the following
requirement:
COUT
³
7758
VIN(Max)
(µF)
VOUT · L1(µH)
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.
B. COUT should have a voltage rating of at least 1.5 × VOUT. But if a
low output ripple voltage is desired, choose capacitors with a higher
voltage ratings than the minimum required due to their typically
lower ESRs.
B. Using Figure 19, the intersection of 115 V•µs and 1 A
corresponds to an inductor code of H470.
C. H470 → L1 = 470 µF
Choose from:
34048 (Schott)
PE-53118 (Pulse Engineering)
RL1961 (Renco)
3. Output Capacitor Selection (COUT)
A.COUT ≥ 7785 × 25/(10 × 470) [µF]
COUT ≥ 41.4 µF
To obtain an acceptable output voltage ripple →
COUT = 220 µF electrolytic
4. Catch Diode Selection (D1) (see Table 1)
4. Catch Diode Selection (D1) (see Table 1)
A. In normal operation, the catch diode requires a current rating of
at least 1.2 × ILOAD(Max). For the most robust design, D1 should be
rated for 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
1.25 × VIN(Max).
A. Pick a diode with a 3-A rating.
B. Pick a 40-V rated Schottky diode (1N5822, MBR340, 31QD04, or
SR304) or 100-V rated Fast Recovery diode (31DF1, MURD310, or
HER302)
5. Input Capacitor (CIN)
An aluminum electrolytic or tantalum capacitor is needed for input
bypassing. Locate CIN as close to VIN and GND pins as possible.
5. Input Capacitor (CIN)
CIN = 100 µF, 35 V, aluminum electrolytic
18
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