TL3577 Datasheet, PDF (13/21 Pages) Texas Instruments – 100-kHz CURRENT-MODE SIMPLE STEP-UP/FLYBACK SWITCHING REGULATOR

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TL3577
100-kHz CURRENT-MODE SIMPLE STEP-UP/FLYBACK SWITCHING REGULATOR
SLVS633 – OCTOBER 2006
Table 1. Step-up Regulator Formulas⁽¹⁾
Duty cycle
Average inductor current
Inductor current ripple
Peak inductor current
Peak switch current
Switch voltage when off
Diode reverse voltage
Average diode current
Peak diode current
Power dissipation
D
I_IND(AVG)
∆I_IND
I_IND(PK)
I_SW(PK)
V_SW(OFF)
V_R
I_D(AVG)
I_D(PK)
P_D
V_OUT + V_F – V_{IN ≈} V_OUT – V_IN
V_OUT + V_F – V_SAT
V_OUT
I_LOAD
1–D
V_IN – V_{SAT •}
D
L
100,000
I_{LOAD +} ∆I_IND
1–D 2
I_{LOAD +} ∆I_IND
1–D 2
V_OUT + V_F
V – V _OUT
SAT
I_LOAD
I_{LOAD +} ∆I_IND
1–D 2
( ^{0.25 Ω}
I_LOAD
2
D
+
I_LOAD
•
D
•
V_IN
1–D
50 (1 – D)
(1) V_F = forward-biased diode voltage, I_LOAD = output load
First, determine if the TL3577 can provide these values of V_OUT and I_LOAD(max) when operating with the minimum
value of V_IN. The upper limits for V_OUT and I_LOAD(max) are given by the following equations.
V_OUT ≤ 60 V and
V_OUT ≤ 10 × V_IN
I_LOAD(max) ≤ (2.1 A × V_IN(min))/V_OUT
These limits must be greater than or equal to the values specified in this application.
1. Output Voltage Section
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 = (V_OUT/1.23 V) – 1
2. Inductor Selection (L)
A. Preliminary Calculations
To select the inductor, the calculation of the following three parameters is necessary:
D_max, the maximum switch duty cycle (0 ≤ D ≤ 0.9):
Dmax = V_OUT + V_F – V_IN(min)/V_OUT + V_F – 0.6 V
where, typically, V_F = 0.5 V for Schottky diodes and V_F = 0.8 V for fast-recovery diodes.
E • T, the product of volts • time that charges the inductor:
E • T = D_max × (V_IN(min) – 0.6V)10⁶/100,000 Hz (Vµs)
I_IND,DC, the average inductor current under full load:
I_IND,DC = (1.05 × I_LOAD(max))/(1 – D_max)
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