where Vripple is the peak-to-peak output voltage ripple.
Typical Applications (Continued)
C3 is necessary for continuous operation and is generally in
the 10 pF to 30 pF range.
For a full line of standard inductor values, contact Pulse En-
gineering (San Diego, Calif.) regarding their PE526XX se-
ries, or A. I. E. Magnetics (Nashville, Tenn.).
D1 should be a Schottky type diode, such as the 1N5818 or
1N5819.
A more precise inductance value may be calculated for the
Buck, Boost and Inverting Regulators as follows:
BUCK WITH BOOSTED OUTPUT CURRENT
BUCK
For applications requiring a large output current, an external
transistor may be used as shown in Figure 17. This circuit
steps a 15V supply down to 5V with 1.5A of output current.
The output ripple is 50 mV, with an efficiency of 80%, a load
regulation of 40 mV (150 mA to 1.5A), and a line regulation
of 20 mV (12V ≤ Vin ≤ 18V).
=
L
Vo (Vin − Vo)/(∆IL Vin fosc
)
BOOST
=
L
Vin (Vo − Vin)/(∆IL fosc Vo)
INVERT
=
L
Vin |Vo|/[∆IL(Vin + |Vo|)fosc]
Component values are selected as outlined for the buck
regulator with a discontinuity factor of 10%, with the addition
of R4 and R5:
where ∆IL is the current ripple through the inductor. ∆IL is
usually chosen based on the minimum load current expected
of the circuit. For the buck regulator, since the inductor cur-
rent IL equals the load current IO,
=
R4 10VBE1Bf/Ip
=
R5 (Vin − V − VBE1 − Vsat) Bf/(IL(max, DC) + IR4
)
=
∆IL 2 • IO(min)
where:
=
∆IL 140 mA for this circuit. ∆IL can also be interpreted as
VBE1 is the VBE of transistor Q1.
=
∆IL 2 • (Discontinuity Factor) • IL
Vsat is the saturation voltage of the LM1578A output transis-
tor.
where the Discontinuity Factor is the ratio of the minimum
load current to the maximum load current. For this example,
the Discontinuity Factor is 0.2.
V is the current limit sense voltage.
=
Bf is the forced current gain of transistor Q1 (Bf 30 for Fig-
The remainder of the components of Figure 15 are chosen
as follows:
ure 17 ).
=
IR4 VBE1/R4
C1 is the timing capacitor found in Figure 1.
=
Ip IL(max, DC) + 0.5∆IL
2
C2 ≥ Vo (Vin − Vo)/(8fosc VinVrippleL1)
DS008711-8
=
=
R4 200Ω
Vin 15V
=
=
R5 330Ω
Vo 5V
=
=
C1 1820 pF
Vripple 50 mV
=
=
C2 330 µF
Io 1.5A
=
=
C3 20 pF
fosc 50 kHz
=
=
L1 220 µH
R1 40 kΩ
=
=
R2 10 kΩ
D1 1N5819
=
R3 0.05Ω
Q1 = D45
FIGURE 17. Buck Converter with Boosted Output Current
BOOST REGULATOR
The boost regulator converts a low input voltage into a
higher output voltage. The basic configuration is shown in
Figure 18. Energy is stored in the inductor while the transis-
tor is on and then transferred with the input voltage to the
output capacitor for filtering when the transistor is off. Thus,
=
Vo Vin + Vin(ton/toff).
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