ADP3050ARZ-3.3-RL View Datasheet(PDF) - Analog Devices
Part Name
Description
Manufacturer
ADP3050ARZ-3.3-RL Datasheet PDF : 20 Pages
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ADP3050
INVERTING (BUCK BOOST) REGULATOR
The circuit in Figure 29 shows the ADP3050 in a buck-boost
configuration that produces a negative output voltage from a
positive input voltage. This topology looks quite similar to the
buck shown in Figure 28 (except the IC and the output filter are
now referenced to the negative output instead of ground), but
its operation is quite different. For this topology, the feedback
pin is grounded and the GND pin is tied to the negative output,
allowing the feedback network of the IC to regulate the negative
output voltage.
L1
47µH
D1
1N5818
D2
1N4148
C3
0.22µF
SD
12V
VIN
GND
+ C1
22µF
1 SWITCH IN 8
C5 +
100µF
2 BOOST GND 7
3 BIAS
SD 6
4 FB COMP 5
R1
U1
5.1kΩ
ADP3050-5
C4
3.3nF
VOUT
–5V AT 0.5A
C2
0.01µF
Figure 29. Inverting (Buck-Boost) Regulator
The design procedure used for the standard buck converter
cannot be used for a buck-boost converter due to fundamental
differences in how the output voltage is generated. The switch
currents in the buck-boost are much higher than the standard
buck converter, thus lowering the available load current. To
calculate the maximum output current for a given maximum
switch current, use the following equation:
I OUT (MAX)
=
VIN
VIN
+ VOUT
×
( )
I
SW
( MAX )
−
2×
VIN
f SW × L
×
×
VOUT
VIN +
VOUT
(18)
where ISW(MAX) is the switch current limit rating of the ADP3050,
and VIN is the minimum input voltage. The inductor ripple
current is estimated using the following equation:
I RIPPLE
= VIN (MAX)
L
×
1
f SW
×
VOUT
VIN (MAX) + VOUT
(19)
Data Sheet
For the circuit in Figure 29, the maximum ripple current (at the
maximum input voltage) is
I RIPPLE
=
12
47 ×10−6
×
200
1
×10
3
× −5
12 + − 5
= 0.375 A
High ripple currents are present in both the input and output
capacitors, and their ripple current ratings must be large
enough to sustain the large switching currents present in this
topology. The capacitors should have a ripple current rating of
at least
I ≈ I × RMS(CIN , COUT )
OUT
VOUT
VIN
(20)
The peak current seen by the diode, switch, and inductor is
found by rearranging the load current equation
I PEAK
=
VIN
+ VOUT
VIN
×
I OUT
+
1
2
×
I
RIPPLE
(21)
The largest peak currents occur at the lowest input voltage. For
this design with a load current of 500 mA
I PEAK
=
12
+−
12
5
× 0.5
+
1
2
× 0.375
= 0.9
A
(22)
The average current diode is equal to the load current.
An inductor with a current rating 20% greater than 0.9 A
should be used (a rating of at least 1.2 A). Inductors and diodes
with ratings greater than 1 A should always be used, even if
the calculated peak and average currents are lower. This ensures
that start-up and fault conditions do not overstress the
components.
For the buck-boost topology, the input voltage can be less than
the output voltage, such as VIN = 4 V or VOUT = −5 V, but the
available load current is even lower. The equations given in this
section are valid for input voltages less than and greater than
the output voltage. The voltage seen by the ADP3050 is equal to
the sum of the input and output voltages (the BOOST pin sees
the sum of VIN + 2 × |VOUT|). It is important to ensure that the
maximum voltage rating of these pins is not exceeded.
Rev. C | Page 18 of 20
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