Datasheet

DI
L
+
V
IN
V
OUT
L
ǒ
V
OUT
* V
IN
Ǔ
f
I
Lmax
+
V
IN
* V
OUT
V
IN
0.8
I
OUT
)
DI
L
2
L +
V
IN
V
OUT
DI
L
ǒ
V
OUT
* V
IN
Ǔ
f
I
Lavg
+
V
IN
* V
OUT
V
IN
0.8
I
OUT
TPS63700
SLVS530C –SEPTEMBER 2005–REVISED JUNE 2013 www.ti.com
For example, if an output voltage of –5 V is needed and a resistor of 150 kΩ has been chosen for R2, a 619-kΩ
resistor is needed to program the desired output voltage.
Inductor Selection
An inductive converter normally requires two main passive components for storing energy during the conversion.
An inductor and a storage capacitor at the output are required.
The average inductor current depends on the output load, the input voltage (VIN), and the output voltage VOUT.
It can be estimated with Equation 2, which shows the formula for the inverting converter.
(2)
with:
I
Lavg
= average inductor current
An important parameter for choosing the inductor is the desired current ripple in the inductor.
A ripple current value between 20% and 80% of the average inductor current can be considered as reasonable,
depending on the application requirements. A smaller ripple reduces the losses in the inductor, as well as output
voltage ripple and EMI. But in the same way, the inductor becomes larger and more expensive.
Keeping those parameters in mind, the possible inductor value can be calculated using Equation 3.
(3)
with:
ΔI
L
= peak-to-peak ripple current
f = switching frequency
L = inductor value
With the known inductor current ripple, the peak inductor value can be approximated with Equation 4. The peak
current through the switch and the inductor depends also on the output load, the input voltage (VIN), and the
output voltage (VOUT). To select the right inductor, it is recommended to keep the possible peak inductor current
below the current-limit threshold of the power switch. For example, the current-limit threshold of the TPS63700
switch for the inverting converter is nominally 1000 mA.
(4)
with:
I
LMAX
= peak inductor current
With Equation 5, the inductor current ripple at a given inductor can be approximated.
(5)
Care has to be taken for the possibility that load transients and losses in the circuit can lead to higher currents as
estimated in Equation 4. Also, the losses caused by magnetic hysteresis losses and copper losses are a major
parameter for total circuit efficiency.
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