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LT1533CS#PBF

P1-P3P4-P6P7-P9P10-P12P13-P15P16-P18P19-P20
10
LT1533
APPLICATIONS INFORMATION
WUU
U
where ∆I is the ripple current in the switch, R
CSL
and
R
VSL
are the slew resistors and f
OSC
is the oscillator
frequency.
Power dissipation P
D
is the sum of these three terms. Die
junction temperature is then computed as:
T
J
= T
AMB
+ (P
D
)(θ
JA
)
where T
AMB
is ambient temperature and θ
JA
is the package
thermal resistance. For the 16-pin SO θ
JA
is 100°C/W.
For example, with f
OSC
= 40kHz, V
IN
= 10V, 0.4A average
current and 0.1A of ripple, the maximum duty cycle is
44%. Assume slew resistors are both 17k and V
SAT
is
0.26V, then:
P
D
= 0.176W + 0.094W + 0.158W = 0.429W
In an S16 package the die junction temperature would be
43°C above ambient.
Frequency Compensation
Loop frequency compensation is accomplished by way of
a series RC network on the output of the error amplifier (V
C
pin). Referring to Figure 3, the main pole is formed by
capacitor C
VC
and the output impedance of the error
amplifier (approximately 400kΩ). The series resistor R
VC
creates a “zero” which improves loop stability and tran-
sient response. A second capacitor C
VC2
, typically one-
tenth the size of the main compensation capacitor, is
sometimes used to reduce the switching frequency ripple
on the V
C
pin. V
C
pin ripple is caused by output voltage
ripple attenuated by the output divider and multiplied by
the error amplifier. Without the second capacitor, V
C
pin
ripple is:
V
VgR
V
C PIN RIPPLE
RIPPLE m VC
OUT
=
()( )()()
125.
where V
RIPPLE
= Output ripple (V
P-P
)
g
m
= Error amplifier transconductance
R
VC
= Series resistor on V
C
pin
V
OUT
= DC output voltage
Thermal Considerations
Computing power dissipation for this IC requires careful
attention to detail. Reduced output slewing causes the part
to dissipate more power than would occur with fast edges.
However, much improvement in noise can be produced
with modest decrease in supply efficiency.
Power dissipation is a function of topology, input voltage,
switch current and slew rates. It is impractical to come up
with an all-encompassing formula. It is therefore recom-
mended that package temperature be measured in each
application. The part has an internal thermal shutdown to
prevent device destruction, but this should not replace
careful thermal design.
1. Dissipation due to input current:
PVmA
I
VIN IN
=+
11
60
where I is the average switch current.
2. Dissipation due to the drivers saturation:
P
VSAT
= (V
SAT
)(I)(DC
MAX
)
where V
SAT
is the output saturation voltage which is
approximately 0.1 + (0.4)(I), DC
MAX
is the maximum
duty cycle.
3. Dissipation due to output slew using approximations
for slew rates:
P
VI
I
R
IV
V
Rf
SLEW
IN
CSL
IN
SAT
VSL OSC
=
()
+
()
()
+
()
−
()
()
()
2
2
9
2
2
9
4
33 10
4
220 10
∆
Note if V
SAT
and ∆I are small with respect to V
IN
and I,
then:
P
IR
V
R
fVI
SLEW
CSL
IN
VSL
OSC IN
=
()( )
()
+
()
()
()
()()()
33 10 220 10
99
11
LT1533
APPLICATIONS INFORMATION
WUU
U
turns ratio of the transformer. The turns ratio must be
large enough to ensure that the transformer can put out a
voltage equal to the output voltage plus the diode under
minimum input conditions.
N
VV
DC V V
OUT F
MAX IN MIN SW
=
+
•−
()
2
()
DC
MAX
is the maximum duty cycle of each driver with
respect to the entire cycle which consists of two periods
(Q1A on and Q1B on). So the effective duty cycle is
2 • DC
MAX
. The controller, in general, determines maxi-
mum duty cycle. A 44% maximum duty cycle is a guaran-
teed value for this part.
Figure 3
V
C
PIN
1533 F03
R
VC
2k
C
VC
0.01µF
C
VC2
4.7nF
To prevent irregular switching, V
C
pin ripple should be
kept below 50mV
P-P
. Worst-case V
C
pin ripple occurs at
maximum output load current and will also be increased if
poor quality (high ESR) output capacitors are used. The
addition of a 0.0047µF capacitor on the V
C
pin reduces
switching frequency ripple to only a few millivolts. A low
value for R
VC
will also reduce V
C
pin ripple, but loop phase
margin may be inadequate.
Magnetics
Design of magnetics is dependent on topology. The fol-
lowing details the design of the magnetics for a push-pull
converter. In this converter the transformer usually stores
little energy. The following equations should be consid-
ered as the starting point to building a prototype.
Figure 4
V
IN
V
SEC
T1
D
S1
1 : N
Q1B
Q1A
L
O
C
O
V
OUT
D
S2
+
1533 F04
The following definitions will be used:
V
IN
= Input supply voltage
V
SW
= Switch-on voltage
V
OUT
= Desired output voltage
I
OUT
= Output current
f = Oscillator frequency
V
F
= Forward drop of the rectifier
Duty cycle is the major defining equation for this topology.
Note that the output L and C basically filter the chopped
voltage so duty cycle controls output voltage. N is the
Some Common Turns Ratios
V
IN
V
OUT
N
5 ±10% 12 3.6
5 ±10% 15 4.4
5 ±10% 3.3 1.1
Remember to add sufficient margin in the turns ratio to
account for IR drops in the transformer windings, worst-
case diode forward drop (V
F
) and switch-on voltage (V
SW
).
There are a number of ways to choose the inductance
value for L
O
. We suggest as a starting point that L
O
be
selected such that the converter is continuous at
I
OUT(MAX)
/4. If your minimum I
OUT
is higher than this, or
you are operating at low currents such that the IC and
components can handle higher peak currents, then use a
higher number.
Continuous operation occurs when the current in the
inductor never goes to zero. Discontinuous operation
occurs when the inductor current drops to zero before the
start of the next cycle and can occur with small inductors
and light loads. There is nothing inherently bad about
discontinuous operation, however, the converter control
and operation is somewhat different. The inductor is
smaller for discontinuous operation but the peak currents
in the switch, the transformer, the diodes, inductor and
capacitor will be higher. But for low power situations these
may not present a big constraint.
12
LT1533
APPLICATIONS INFORMATION
WUU
U
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 I
OUT
/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:
∆I
OUT
= 2 • I
OUT(MIN)
I
OUT(MIN)
= I
OUT(MAX)
/4
L
V
DC
If
O
OUT
NOM
OUT
=
−•
()
•
12
∆
The inductance of the transformer primary should be such
that L
O
, 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 L
O
. In general,
then, the inductance of the primary should be at least five
times that of L
O
. 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.
L
PRI
= 5 • L
O
/N
2
If the magnetizing current is below 100mA, then a smaller
L
O
can be used.
With the value of L
O
set, the ripple in the inductor is:
∆I
V
DC
Lf
OUT
OUT
O
=
−•
()
•
12
However, the peak inductor current is evaluated at maxi-
mum load and maximum input voltage (minimum DC).
II
I
LMAX OUT MAX
OUT MAX
=+
()
()
∆
2
The magnetizing ripple current can be shown to be:
and the peak current in the switch is:
I
SW(PEAK)
= N • I
LMAX
+ ∆I
MAG
This should be less than the 1A current limit.
In the push-pull converter the maximum switch voltage
will be 2 • (V
IN
– V
SW
) 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:
V
SW(MAX)
= 2 • V
IN
• 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:
V
IN
= 5V ±10%, V
OUT
= 12V, I
OUT(MAX)
= 150mA,
V
SW
= 0.5V, V
F
= 0.5V, f = 50kHz,
N =
+
()
−
()
=
12 0 5
2 044 45 05
355
.
•. . .
.
Round up so N = 3.6.
For continuous operation at I
OUT(MIN)
= I
OUT(MAX)
/4,
inductor ripple is:
∆I
mA
mA
OUT
==2
150
4
75•
The duty cycle for nominal input is:
DC
VV
NV V
L
mA kHz
H
NOM
OUT F
IN NOM SW
O MIN
=
+
()
−
()
=
+
()
−
()
=
=
−
()
=
2
12 0 5
236505
38 6
12 1 2 38 6
75 50
730
•
.
•. .
.%
•.%
•
()
()
µ
Off-the-shelf components can be used for this inductor.
Say we found an 800µH inductor (Coiltronics CTX200-1
for instance).
P1-P3P4-P6P7-P9P10-P12P13-P15P16-P18P19-P20

LT1533CS#PBF

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Manufacturer:
Analog Devices / Linear Technology
Description:
Switching Voltage Regulators Ultralow N 1A Sw Reg
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