LT1533C View Datasheet(PDF) - Linear Technology
Part Name
Description
MFG CO.
'LT1533C' PDF : 20 Pages View PDF
LT1533
APPLICATIONS INFORMATION
For continuous operation the inductor ripple current must
be less than twice the output current. The worst case for
this is at maximum input (lowest DC) but we will evaluate
at nominal input since the IOUT/4 is somewhat arbitrary.
Note when both inputs are off, inductor current splits
between outputs and the diode common goes to 0V.
Looking at the inductor current during off time, output
ripple current is:
∆IOUT = 2 • IOUT(MIN)
IOUT(MIN) = IOUT(MAX)/4
( ) LO
=
VOUT
1− 2•
∆IOUT
DCNOM
•f
The inductance of the transformer primary should be such
that LO, when reflected into the primary, dominates the
input current. In other words, we want the magnetizing
current of the transformer small with respect to the
current going through the transformer to LO. In general,
then, the inductance of the primary should be at least five
times that of LO. This ensures that most of the power will
be passed through the transformer to the load. It also
increases the power capability of the converter and
reduces the peak currents that the switch will see.
LPRI = 5 • LO/N2
If the magnetizing current is below 100mA, then a smaller
LO can be used.
With the value of LO set, the ripple in the inductor is:
( ) VOUT 1− 2 •DC
∆IOUT =
LO • f
However, the peak inductor current is evaluated at maxi-
mum load and maximum input voltage (minimum DC).
ILMAX
=
IOUT(MAX)
+
∆IOUT(MAX)
2
The magnetizing ripple current can be shown to be:
∆IMAG
=
VOUT +
N • LPRI
VF
•f
and the peak current in the switch is:
ISW(PEAK) = N • ILMAX + ∆IMAG
This should be less than the 1A current limit.
In the push-pull converter the maximum switch voltage
will be 2 • (VIN – VSW) plus a small amount (10%) for
leakage spikes. Because voltage is slew-controlled, the
spikes will be less than normal. So, maximum switch
voltage is:
VSW(MAX) = 2 • VIN • 1.1
This should be below the maximum rated switch voltage.
So, given the turns ratio, primary inductance and current,
the transformer can be designed. As an example:
VIN = 5V ±10%, VOUT = 12V, IOUT(MAX) = 150mA,
VSW = 0.5V, VF = 0.5V, f = 50kHz,
( )( ) N =
12 + 0.5
= 3.55
2 • 0.44 4.5 − 0.5
Round up so N = 3.6.
For continuous operation at IOUT(MIN) = IOUT(MAX)/4,
inductor ripple is:
∆IOUT
=
2•
150mA
4
=
75mA
The duty cycle for nominal input is:
( )( ) DCNOM =
2•N
VOUT + VF
VIN(NOM) − VSW
( )( ) = 12 + 0.5 = 38.6%
2 • 3.6 5 − 0.5
( ) 12 1− 2 • 38.6%
LO(MIN) = 75mA • 50kHz = 730µH
Off-the-shelf components can be used for this inductor.
Say we found an 800µH inductor (Coiltronics CTX200-1
for instance).
12
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