Datasheet
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 at the external FET, which tends to reduce the
efficiency at light loads.
4) Higher frequencies may exhibit lower overall efficiency
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-resis-
tance components.
5) High-duty-cycle applications may require lower fre-
quencies to accommodate the controller minimum
off-time of 0.4µs. Calculate the maximum oscillator
frequency with the following formula:
IN(MIN) SW LIM
OSC(MAX)
IN(MIN) SW LIM OUT D
OFF(MIN)
V VV
f
V VVV V
1
t
−−
−−−
=
+
×
Remember that V
OUT
is negative when using this formula.
When running at the maximum oscillator frequency
(f
OSCILLATOR
) and maximum duty cycle (D
MAX
), do
not exceed the minimum value of D
MAX
stated in the
Electrical Characteristics table. For designs that exceed
the D
MAX
and f
OSC(MAX)
, an autotransformer can reduce
the duty cycle and allow higher operating frequencies.
The oscillator frequency is set by a resistor, RFREQ,
which is connected from FREQ to GND. The relationship
between fOSC (in Hz) and RFREQ (in W) is slightly nonlin-
ear, as illustrated in the Typical Operating Characteristics.
Choose the resistor value from the graph and check the
oscillator frequency using the following formula:
( ) ( ) ( )
( )
2
7 11 19
OSC
FREQ FREQ
1
f
5.21 10 1.92 10 R 4.86 10 R
−− −
−
=
× +× × × ×
External Synchronization (MAX1847 only)
The SYNC input provides external-clock synchroniza-
tion (if desired). When SYNC is driven with an exter-
nal 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 inductor
ripple current to average continuous current at mini-
mum duty cycle to be 0.4. Calculate the inductor ripple
with the following formula:
( )
( )
RIPPLE
LOAD(MAX) IN(MAX) SW LIM OUT D
IN(MAX) SW LIM
I
0.4 I V V V V V
V VV
−−−
−−
=
×× +
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 output
is:
I
LPEAK
= I
LDC
+ (I
LPP
/ 2)
where I
LDC
is the average DC inductor current and I
LPP
is the inductor peak-to-peak ripple current. The I
LDC
and
I
LPP
terms are determined as follows:
( )
( )
(
)
LOAD
LDC
MAX
SW LIM MAX
IN MIN
LPP
OSC
I
I
1 D
V V V x D
I
L x f
=
−
−−
=
where L is the selected inductance value. The satu-
ration 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-
www.maximintegrated.com
Maxim Integrated
│
13
MAX1846–MAX1847 High-Efciency, Current-Mode,
Inverting PWM Controller