Datasheet

( )
SEC
IN
OUT
sw
OPRI
PRI
N
V × D × 1 - D
ILpri_pospk I +
N 2 × × L
»
f
( )
IN
OMIN
SEC
sw
HSCL OUT
PRI
V × D × 1 - D
L =
N
2 × × I - I ×
N
æ ö
÷
ç
÷
ç
÷
ç
÷
÷
ç
è ø
f
( )
IN
OMAX
SEC
sw
OUT
PRI
V × D × 1 - D
L =
N
2 × × I ×
N
f
PRI
HS LS
V - 0.829V
R = R ×
0.829V
æ ö
÷
ç
÷
ç
÷
ç
÷
ç
è ø
FD
SEC OUT
PRI PRI
N V + V
=
N V
TPS55010
SLVSAV0A –APRIL 2011–REVISED JUNE 2011 www.ti.com
TURNS RATIO
The transformer turns ratio is calculated using the desired output voltage, diode voltage and the primary voltage.
Assuming a diode voltage of 0.5 V, V
OUT
of 5 V, V
PRI
of 2.2 V yields a N
PRI
:N
SEC
turns ratio of 1:2.5.
(12)
VOLTAGE FEEDBACK
Selecting 61.9 kΩ for the R
LS
, R
HS
is calculated to be 102.4 kΩ using Equation 13. Choose 100 kΩ as the
nearest standard value.
It may be necessary to adjust the feedback resistors to optimize the output voltage over the full load range.
Usually checking and setting the output voltage to the nominal voltage at 50% load, yields the best results.
(13)
SELECTING THE SWITCHING FREQUENCY and PRIMARY INDUCTANCE
The selection of switching frequency is usually a trade-off between efficiency and component size. However,
when isolation is a requirement, switching frequency is not the key variable in determining solution size. Low
switching frequency operation improves efficiency by reducing gate drive losses and MOSFET and diode
switching losses. However, a lower switching frequency operation requires a larger primary inductance which will
have more windings and higher dc resistance.
The optimal primary inductance should be selected between two inductance values, L
OMAX
and L
OMIN
. The
primary inductance should be less than L
OMAX
to maintain good efficiency and greater than Lomin to avoid the
peak switch current from exceeding the high side power switch current limit. Once the primary inductance is
selected, check against the low side current limit using the Equation 17 and the high side current limit. For this
design example, the switching frequency is selected to be 350 kHz. Using Equation 6, the resistor value is 280
kΩ. L
OMAX
and Lomin are calculated to be 3.52 µH and 1.17 µH respectively assuming a current limit of 2 A.
Selecting a primary inductance of the 2.5 µH, the positive and negative peak current are calculated as 1.204 A
and -1.99 A in the primary which do not exceed the current limits of the power switch. The rms currents can be
calculated and used to determine the power dissipation in the device.
The magnetizing ripple current is caculated as 1.41 A using Equation 18. The highside FET and lowside FET rms
currents are calculated as 0.43 A and 0.61 A, respectively using Equation 19 and Equation 20. The sum of these
currents, i.e. 1.04 A is the primary side rms current for the magnetics.
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