Data Sheet ADP3050
Rev. C | Page 19 of 20
Dual Output SEPIC Regulator
For many systems, a dual polarity supply is needed. The circuit
in Figure 30 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 conven-
tional buck regulator, or they can be used as a 1:1 transformer,
as in this application. The first 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.
00125-030
C1
22µF
C2
0.01µF
1
2
3
4
8
7
6
5
U1
ADP3050-5
V
IN
C3
0.22µF
D1
GND
12V
+
L1*
25µH
V
OUT
+5V AT 0.5A
R1
5.1kΩ
C4
1nF
D3
1N4148
+
C5
100µF
SD
*INDUCTOR IS A SINGLE CORE
WITH TWO WINDINGS
COILTRONICS CTX25-4
–5V AT 0.25A
V
OUT
+
C6
100µF
L1*
25µH
+
C7
100µF
D2
1N5818
SWITCH
BOOST
BIAS
FB
IN
GND
SD
COMP
Figure 30. Dual Output +5 V and −5 V Regulator
These components form a single-ended primary inductance
converter (SEPIC) 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
O
+ V
D
(V
D
is the diode drop).
This voltage is generated across the second winding, which is
connected to produce the −5 V supply. The −5 V output is
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
positive and negative supplies, respectively, ensures that the
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 helps 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 Figure 30, 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 waveform 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 approximately 2 × 200 mA × 100 mΩ, or about
40 mV. The edges of this ripple waveform are quite fast. Along
with the inductance of the output capacitor, it 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 PCB trace and the
additional capacitor create a low-pass filter to remove these
high frequency spikes).
