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TL3577 датащи(PDF) 13 Page - Texas Instruments

номер детали TL3577
подробное описание детали  100-kHz CURRENT-MODE SIMPLE STEP-UP/FLYBACK SWITCHING REGULATOR
PDF  23 Pages
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производитель  TI [Texas Instruments]
домашняя страница  http://www.ti.com
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TL3577 датащи(HTML) 13 Page - Texas Instruments

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V
+ V – V
OUT
F
IN
V
– V
OUT
IN
V
+ V – V
OUT
F
SAT
V
OUT
≈
I
LOAD
1 – D
V – V
IN
SAT
D
L
100,000
•
I
LOAD
∆I
IND
1 – D
2
+
I
LOAD
∆I
IND
1 – D
2
+
V
+ V
OUT
F
V
– V
OUT
SAT
I
LOAD
I
LOAD
∆I
IND
1 – D
2
+
I
LOAD
I
D
V
LOAD
IN
• •
1 – D
50 (1 – D)
+
0.25
Ω
(
(
2
D
1. Output Voltage Section
2. Inductor Selection (L)
TL3577
100-kHz CURRENT-MODE SIMPLE STEP-UP/FLYBACK SWITCHING REGULATOR
SLVS633 – OCTOBER 2006
Table 1. Step-up Regulator Formulas(1)
Duty cycle
D
Average inductor current
IIND(AVG)
Inductor current ripple
∆I
IND
Peak inductor current
IIND(PK)
Peak switch current
ISW(PK)
Switch voltage when off
VSW(OFF)
Diode reverse voltage
VR
Average diode current
ID(AVG)
Peak diode current
ID(PK)
Power dissipation
PD
(1)
VF = forward-biased diode voltage, ILOAD = output load
First, determine if the TL3577 can provide these values of VOUT and ILOAD(max) when operating with the minimum
value of VIN. The upper limits for VOUT and ILOAD(max) are given by the following equations.
VOUT ≤ 60 V and
VOUT ≤ 10 × VIN
ILOAD(max) ≤ (2.1 A × VIN(min))/VOUT
These limits must be greater than or equal to the values specified in this application.
Resistors R1 and R2 are used to select the desired output voltage. These resistors form a voltage divider and
present a portion of the output voltage to the error amplifier, which compares it to an internal 1.23-V reference.
Select R1 and R2 such that:
R1/R2 = (VOUT/1.23 V) – 1
A. Preliminary Calculations
To select the inductor, the calculation of the following three parameters is necessary:
Dmax, the maximum switch duty cycle (0 ≤ D ≤ 0.9):
Dmax = VOUT + VF – VIN(min)/VOUT + VF – 0.6 V
where, typically, VF = 0.5 V for Schottky diodes and VF = 0.8 V for fast-recovery diodes.
E
• T, the product of volts • time that charges the inductor:
E
• T = D
max × (VIN(min) – 0.6V)10
6/100,000 Hz (V
µs)
IIND,DC, the average inductor current under full load:
IIND,DC = (1.05 × ILOAD(max))/(1 – Dmax)
13
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