LT1576/LT1576-5
U
W U U
APPLICATIONS INFORMATION
withlargeinductors.Outputripplevoltageforthepositive-
to-negative converter will be much higher than a buck
converter. Ripple current in the output capacitor will also
be much higher. The following equations can be used to
calculateoperatingconditionsforthepositive-to-negative
converter.
This duty cycle is close enough to 50% that IP can be
assumed to be 1.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
VOUT (R2) should be set to approximately 5k. R1 is
calculated from:
Maximum load current:
V
V
(
)(
)
IN OUT
R2 V
− 1.21
(
)
OUT
I −
V
V −
0.35
(
)(
)
P
OUT IN
R1=
2 V
+ V f L
(
)( )( )
OUT
IN
1.21
I
=
MAX
V
+V − 0.35 V
+ V
F
(
)(
)
OUT
IN
OUT
INDUCTOR VALUE
IP = Maximum rated switch current
VIN = Minimum input voltage
VOUT = Output voltage
VF = Catch diode forward voltage
0.35 = Switch voltage drop at 1.5A
Unlike buck converters, positive-to-negative converters
cannot use large inductor values to reduce output ripple
voltage. At 200kHz, values larger than 75µH make almost
no change in output ripple. The graph in Figure 16 shows
peak-to-peak output ripple voltage for a 5V to –5V con-
verter versus inductor value. The criteria for choosing the
Example: with VIN(MIN) = 5.5V, VOUT = 5V, L = 30µH,
VF = 0.5V, IP = 1.5A: IMAX = 0.6A. Note that this equation
does not take into account that maximum rated switch
current (IP) on the LT1576 is reduced slightly for duty
cyclesabove50%. Ifdutycycleisexpectedtoexceed50%
(input voltage less than output voltage), use the actual IP
value from the Electrical Characteristics table.
150
5V TO –5V CONVERTER
OUTPUT CAPACITOR’S
ESR = 0.1Ω
120
DISCONTINUOUS
I
= 0.1A
90
60
30
0
LOAD
DISCONTINUOUS
= 0.25A
I
Operating duty cycle:
LOAD
V
OUT + VF
V − 0.3 + VOUT + VF
DC =
CONTINUOUS
IN
I
> 0.38A
LOAD
(This formula uses an average value for switch loss, so it
may be several percent in error.)
0
15
30
45
60
75
INDUCTOR SIZE (µH)
1576 F16
With the conditions above:
Figure 16. Ripple Voltage on Positive-to-Negative Converter
5 + 0.5
5.5 − 0.3 + 5 + 0.5
DC =
= 51%
24
Linear [ Linear ]
