ADP1610
The regulator loop gain is
DIODE SELECTION
The output rectifier conducts the inductor current to the output
capacitor and load while the switch is off. For high efficiency,
minimize the forward voltage drop of the diode. For this reason,
Schottky rectifiers are recommended. However, for high voltage,
high temperature applications, where the Schottky rectifier
reverse leakage current becomes significant and can degrade
efficiency, use an ultrafast junction diode.
VFB
VIN
VOUT VOUT
(14)
×GMEA × ZCOMP ×GCS × ZOUT
AVL
=
×
where:
A
V
V
V
G
VL is the loop gain.
FB is the feedback regulation voltage, 1.230 V.
OUT is the regulated output voltage.
IN is the input voltage.
Make sure that the diode is rated to handle the average output
load current. Many diode manufacturers derate the current
capability of the diode as a function of the duty cycle. Verify
that the output diode is rated to handle the average output load
current with the minimum duty cycle. The minimum duty cycle
of the ADP1610 is
MEA is the error amplifier transconductance gain.
Z
COMP is the impedance of the series RC network from COMP to
GND.
G
CS is the current sense transconductance gain (the inductor
current divided by the voltage at COMP), which is internally set
by the ADP1610.
VOUT −VIN−MAX
DMIN
=
(12)
VOUT
Z
OUT is the impedance of the load and output capacitor.
where VIN-MAX is the maximum input voltage.
To determine the crossover frequency, it is important to note
that, at that frequency, the compensation impedance (ZCOMP) is
dominated by the resistor, and the output impedance (ZOUT) is
dominated by the impedance of the output capacitor. So, when
solving for the crossover frequency, the equation (by definition
of the crossover frequency) is simplified to
Table 6. Schottky Diode Manufacturers
Vendor
Motorola
Diodes, Inc.
Sanyo
Phone No.
Web Address
602-244-3576
805-446-4800
310-322-3331
www.mot.com
www.diodes.com
www.irf.com
VFB
V
1
(15)
IN
| A | =
×
× GMEA× RCOMP×G ×
=1
VL
CS
LOOP COMPENSATION
VOUT VOUT
2π × fC ×COUT
The ADP1610 uses external components to compensate the
regulator loop, allowing optimization of the loop dynamics for a
given application.
where:
fC is the crossover frequency.
COMP is the compensation resistor.
Solving for RCOMP
The step-up converter produces an undesirable right-half plane
zero in the regulation feedback loop. This requires compensat-
ing the regulator such that the crossover frequency occurs well
below the frequency of the right-half plane zero. The right-half
plane zero is determined by the following equation:
R
,
2π × fC ×COUT ×VOUT ×VOUT
VFB ×VIN × GMEA × GCS
(16)
(17)
R COMP
=
2
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
VIN
VOUT
RLOAD
2π×L
FZ (RHP) =
×
(13)
For VFB = 1.23, GMEA = 100 µS, and GCS = 2 S,
2.55×104 × fC ×COUT ×VOUT ×VOUT
RCOMP
=
where:
FZ(RHP) is the right-half plane zero.
VIN
Once the compensation resistor is known, set the zero formed
by the compensation capacitor and resistor to one-fourth of the
crossover frequency, or
R
LOAD is the equivalent load resistance or the output voltage
divided by the load current.
To stabilize the regulator, make sure that the regulator crossover
frequency is less than or equal to one-fifth of the right-half
plane zero and less than or equal to one-fifteenth of the
switching frequency.
2
(18)
CCOMP
=
π× fC ×RCOMP
where CCOMP is the compensation capacitor.
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