ADP3050-3.3 Analog Devices, ADP3050-3.3 Datasheet - Page 16

• Manufacturer Part Number: ADP3050-3.3
• Description: 200 Khz, 1 a Step-down High-voltage Switching Regulator
• Manufacturer: Analog Devices

ADP3050
The largest peak currents occur at the lowest input voltage. For
this design with a load current of 500 mA:
The average current diode will be equal to the load current.
An inductor with a current rating 20% greater than 0.9 A should
be used (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 will ensure that start-up and
fault conditions will not overstress the components.
For the buck-boost topology, the input voltage can be less than
the output voltage (i.e., V
able load current will be 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 V
ratings of these pins are not exceeded.
Dual Output SEPIC Regulator
For many systems, a dual polarity supply is needed. The circuit
in Figure 7 generates both a positive and a negative 5 V output
using a single magnetic component. The two inductors shown
are actually two separate windings on a single core contained in
a small, surface-mount package. The windings can be connected
in parallel or in series to be used as a single inductor for a con-
ventional buck regulator, or they can be used as a 1:1 transformer,
as in this application. One winding is used as the standard buck
inductor for the +5 V output. The second winding is used to
generate the –5 V output along with D2, C6, and C7.
These components form a SEPIC (Single Ended Primary Induc-
tance Converter) using the 1:1 coupled inductor to generate the
negative supply. When the switch is off, the voltage across the buck
winding is equal to V
generated across the second winding, which is connected to produce
1N5818
GND
D1
V
SD
*INDUCTOR IS A SINGLE CORE
IN
WITH TWO WINDINGS
COILTRONICS CTX25-4
12V
I
PEAK
IN
+ 2
=
C3
0.22 F



25 H
12
|V
+
L1*
1N4148
C1
22 F
O
OUT
+ −
12
+ V
D3
|). Make sure that the maximum voltage
IN
5
D
(V
= 4 V, V
×
D
0 5
100 F
C2
0.01 F
+
.
is the diode drop). This voltage is
1
2
3
4
C6

 +

SWITCH
BOOST
BIAS
FB
ADP3050-5
OUT



1
2
U1
25 H
L1*
1N5818
×
= –5 V), but the avail-
COMP
D2
0 375
GND
.
SD
IN
8
7
6
5
+ C7

 =

5.1k
100 F
R1
0 9
.
+5V @ 0.5A
5V @ 0.25A
+ C5
C4
1nF
100 F
A
V
V
OUT
OUT
the –5 V supply. The –5 V output would be generated even
without C6 in the circuit, but its inclusion greatly improves the
regulation of the negative output and lowers the inductor ripple
current. The total output current available for both supplies is
limited by the ADP3050 (internally limited to around 1.0 A).
Keeping load currents below 500 mA and 250 mA, for the posi-
tive and negative supplies, respectively, will ensure that current
limit is not reached under normal operation. These limits are
not interchangeable; 500 mA cannot be drawn from the –5 V
supply while drawing only 250 mA from the +5 V supply. The
maximum current available from the –5 V output is directly related
to the +5 V load current due to the fact that the +5 V output is
used to regulate both supplies. Typically, the –5 V load current
should be around one-half of the +5 V load current to ensure good
regulation of both outputs. Additionally, the –5 V output should
have a preload (the minimum current level) of 1% to 2% of the
+5 V load current. This will help maintain good regulation of the
–5 V output at light loads.
The ripple voltage of the +5 V output is that of a normal buck
regulator (as described in the Applications Information section).
This ripple voltage is determined by the inductor ripple current
and the ESR of the output capacitor. For the system below, the
positive output voltage ripple is a 30 mV peak-to-peak triangular
wave. The ripple voltage of the –5 V output is a rectangular
wave (due to the rectangular shape of the current waveform into
the –5 V output capacitor). The amplitude of this current wave-
form is approximately equal to twice the –5 V load current. For
a load current of 200 mA and an ESR of 100 mΩ, the negative
output voltage ripple is around 2
about 40 mV. The edges of this ripple waveform are quite fast,
which, along with the inductance of the output capacitor, generates
narrow spikes on the negative output voltage. These spikes can
easily be filtered out using an additional 5 µF to 10 µF bypass
capacitor close to the load (the inductance of the PC board
trace and the additional capacitor create a low-pass filter to
remove these high frequency spikes).
0.1574 (4.00)
0.1497 (3.80)
0.0098 (0.25)
0.0040 (0.10)
SEATING
PLANE
PIN 1
Dimensions shown in inches and (mm).
0.1968 (5.00)
0.1890 (4.80)
0.0500
(1.27)
BSC
8
1
OUTLINE DIMENSIONS
0.0192 (0.49)
0.0138 (0.35)
0.0688 (1.75)
0.0532 (1.35)
5
4
8-Lead SOIC
0.2440 (6.20)
0.2284 (5.80)
(R Suffix)
0.0098 (0.25)
0.0075 (0.19)
200 mA
8°
0°
0.0196 (0.50)
0.0099 (0.25)
0.0500 (1.27)
0.0160 (0.41)
100 mΩ, or
x 45°