Rainbow Electronics MAX1847 User Manual
Page 13
MAX1846/MAX1847
High-Efficiency, Current-Mode,
Inverting PWM Controller
______________________________________________________________________________________
13
2)
Higher frequencies allow the use of smaller value
(hence smaller size) inductors and capacitors.
3)
Higher frequencies consume more operating
power both to operate the IC and to charge and
discharge the gate of the external FET. This tends
to reduce the efficiency at light loads.
4)
Higher frequencies may exhibit lower overall effi-
ciency due to more transition losses in the FET;
however, this shortcoming can often be nullified
by trading some of the inductor and capacitor size
benefits for lower-resistance components.
5)
High-duty-cycle applications may require lower
frequencies to accommodate the controller mini-
mum off-time of 0.4µs. Calculate the maximum
oscillator frequency with the following formula:
Remember that V
OUT
is negative when using this formula.
The oscillator frequency is set by a resistor, R
FREQ
,
connected from FREQ to GND. The relationship
between f
OSC
(in Hz) and R
FREQ
(in
Ω) is slightly non-
linear, as illustrated in the Typical Operating
Characteristics. Choose the resistor value from the
graph and check the oscillator frequency using the fol-
lowing formula:
External Synchronization (MAX1847 only)
The SYNC input provides external-clock synchroniza-
tion (if desired). When SYNC is driven with an external
clock, the frequency of the clock directly sets the
MAX1847’s switching frequency. A rising clock edge
on SYNC is interpreted as a synchronization input. If
the sync signal is lost, the internal oscillator takes over
at the end of the last cycle, and the frequency is
returned to the rate set by R
FREQ
. Choose R
FREQ
such
that f
OSC
= 0.9 x f
SYNC
.
Choosing Inductance Value
The inductance value determines the operation of the
current-mode regulator. Except for low-current applica-
tions, most circuits are more efficient and economical
operating in continuous mode, which refers to continu-
ous current in the inductor. In continuous mode there is
a trade-off between efficiency and transient response.
Higher inductance means lower inductor ripple current,
lower peak current, lower switching losses, and, there-
fore, higher efficiency. Lower inductance means higher
inductor ripple current and faster transient response. A
reasonable compromise is to choose the ratio of induc-
tor ripple current to average continuous current at mini-
mum duty cycle to be 0.4. Calculate the inductor ripple
with the following formula:
Then calculate an inductance value:
L = (V
IN(MAX)
/ I
RIPPLE
) x (D
MIN
/ f
OSC
)
Choose the closest standard value. Once again, remem-
ber that V
OUT
is negative when using this formula.
Determining Peak Inductor Current
The peak inductor current required for a particular out-
put is:
I
LPEAK
= I
LDC
+ (I
LPP
/ 2)
where I
LDC
is the average DC input current and I
LPP
is
the inductor peak-to-peak ripple current. The I
LDC
and
I
LPP
terms are determined as follows:
where L is the selected inductance value. The satura-
tion rating of the selected inductor should meet or
exceed the calculated value for I
LPEAK
, although most
coil types can be operated up to 20% over their satura-
tion rating without difficulty. In addition to the saturation
criteria, the inductor should have as low a series resis-
tance as possible. For continuous inductor current, the
power loss in the inductor resistance (P
LR
) is approxi-
mated by:
P
LR
~ (I
LOAD
x V
OUT
/ V
IN
)
2
x R
L
where R
L
is the inductor series resistance.
I
I
V
V
V
V
V
I
V
V
V
V
V
L f
V
V
LDC
LOAD
OUT
D
IN MIN
SW
LIM
LPP
IN MIN
SW
LIM
OUT
D
OSC
OUT
D
=
Ч
+
(
)
=
(
)
Ч
+
(
)
Ч
Ч
+
(
)
−
−
−
−
−
−
−
(
)
(
)
I
I
V
V
V
V
V
V
V
V
RIPPLE
LOAD MAX
IN MAX
SW
LIM
OUT
D
IN MAX
SW
LIM
=
Ч
Ч
+
(
)
(
)
−
−
−
−
−
0 4
.
(
)
(
)
(
)
f
R
R
OSC
FREQ
FREQ
=
Ч
(
)
+
Ч
(
)
Ч
Ч
(
)
Ч
(
)
−
−
−
−
1
5 21
10
1 92
10
4 86
10
7
11
19
2
.
.
.
f
V
V
V
V
V
V
V
V
t
OSC MAX
IN MIN
SW
LIM
IN MIN
SW
LIM
OUT
D
OFF MIN
(
)
(
)
(
)
(
)
=
+
Ч
−
−
−
−
−
1