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參數(shù)資料
| 型號： | AAT1160 |
| 廠商： | Advanced Analog Technology,lnc. |
| 英文描述： | Step Up Current Mode PWM Converter with 4+1 Operational Amplifiers |
| 中文描述： | 升壓電流模式PWM轉(zhuǎn)換器，具有4 1運算放大器 |
| 文件頁數(shù)： | 10/14頁 |
| 文件大小： | 372K |
| 代理商： | AAT1160 |
Advanced Analog Technology, Inc.
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Advanced Analog Technology, Inc
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Page 10 of 14 V 1.0
AAT1160
Operation Information:
Setting the Output Voltage:
Output voltage is set by using a feedback pin and a
resistor divider; equation is as follows:
V =1.265 (1+R1/R2)
Soft-Start:
The soft-start function is embedded with a typical
setting of 14ms
Operation Frequency Setting:
Pin4 (FREQ) can set PWM operation to low with
f
=600kHz, or set PWM operation to high with
f
=1.2 MHz.
OSC
OSC
Diode Selection:
Schottky diode for the boost regulator must be
selected correctly depending on the output voltage
and the output current. The diode must have a
reverse voltage equal to or greater than the output
voltage. In addition, the diode current must
exceed the switch current limit as a lower forward
voltage will increase efficiency.
Inductor Selection:
I
=
IN
I
+
V *D/(2*L*
L(peak)
S
f )
D: PWM duty
V /
V =1/ (1-D)
Maximum current of the inductor must be greater
than
L(peak)
I
The Compensation:
Pin17 (EO) series with
compensation, usually range as 1k
, 100pF
≤
C
C
≤
6,800pF. Standard formulas are
only for reference purpose, as PCB has a lot of
noise or parasitic capacitance and resistance, it
often requires on board adjustment.
The key steps for step-up compensation are as
follows:
Transconductance (from FB to CC),
Current-sense amplifier transresistance,
R &
C for
≤
C
R
≤
200k
g (105
μ
S)
R
CS
(0.275V/A)
For continuous conduction, the right-half-plane
zero frequency (
RHPZ
f
) is given by the following:
f
=
V (1-D)
2
/(2
*L*
duty cycle=1-(
V /
V ), L is the inductance value,
and
LOAD
I
is the maximum output current. Typical
target, crossover (
RHPZ
LOAD
I
) where D=the
C
f ), is 1/6 of the RHPZ.
For example, if oscillation frequency is assumed to
be
OSC
f
=500kHz,
V =2.5V,
I
=0.5A, then
LOAD
R
L=4.7
μ
H, then:
f
=5(2.5/5)
2
/(2
*4.7*10
-6
*0.5)=84.65kHz
If
C =(
V /
V )(
LOAD
R
/
R
1.25/5)(10/0.275)*[105
μ
S/(6.28*14kHz)](1/4)=2.7
nF
Now selected
C
R
requirements. For example, if 4% of transient
droop is allowed, input to error amplifier moves
V =5V, and
. If we select
OUT
=10
RHPZ
C
f =14kHz, C
C
would be calculated
as:
CS
)(
g /2
π
*
C
f )(1-D)/2=(
has fulfilled transient-droop
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