參數(shù)資料
型號(hào): MP1527
廠商: Monolithic Power Systems, Inc.
英文描述: 2A, 1.3MHz Step-Up Converter
中文描述: 第2A,1.3MHz升壓轉(zhuǎn)換器
文件頁(yè)數(shù): 11/14頁(yè)
文件大小: 279K
代理商: MP1527
MP1527
2A, 1.3MHz
Step-Up Converter
MP1527 Rev 1.8_8/31/05
Monolithic Power Systems, Inc.
11
Monolithic Power Systems
There is also a right-half-plane zero (f
RHPZ
) that
exists in all continuous mode (continuous
mode means that the inductor current does not
drop to zero on each cycle) step-up
converters. The frequency of the right half
plane zero is:
f
RHPZ
= V
IN
where L is the value of the inductor.
To stabilize the regulation control loop, the
crossover frequency (The frequency where the
loop gain drop to 0dB or gain of 1, indicated as
f
C
) should be at least one decade below the
right-half-plane zero and should be at most
75KHz. f
RHPZ
is at its lowest frequency at
maximum output load current (R
LOAD
is at a
minimum)
The crossover frequency is calculated by the
equation:
f
C
= A
VDC
*f
P1
*f
P2
/ f
Z1
or
f
C
= G
CS
*G
EA
*V
IN
*V
FB
*R3 / (2
π
*C2*V
OUT
The known values are:
G
CS
= 4.3S
G
EA
= 400
μ
S
V
FB
= 1.22V
Putting in the known constants:
f
C
= 3.3x10
-4
*V
IN
*R3/ (C2*V
OUT
If the frequency of the right-half-pane zero
f
RHPZ
is less than 750KHz, then the crossover
frequency should be 1/10 of f
RHPZ
, and
determine the compensation resistor (R3) with
equation (1). If f
RHPZ
is greater than or equal to
750KHz, set the crossover frequency to 75KHz
with equation (2).
For f
C
= f
RHPZ
/ 10, then
2
*R
LOAD
/ (2
π
*L*V
OUT
2
)
2
)
2
)
R3 = V
IN
*R
LOAD-MIN
*C2 / (10G
CS
*G
EA
*V
FB
*L)
The minimum load resistance (R
LOAD-MIN
) is
equal to the regulated output voltage (V
OUT
)
divided by the maximum load current I
LOAD-MAX
.
Substituting that into the above equation:
R3 = V
IN
*V
OUT
*C2 /(10G
CS
*G
EA
*V
FB
*L*I
LOAD-MAX
)
Putting in the known constant values:
(1) R3
48*V
IN
*V
OUT
*C2 / (L*I
LOAD-MAX
)
For f
C
= 75KHz,
f
C
= (G
CS
*G
EA
*V
IN
*V
FB
*R3) / (2
π
*C2*V
OUT
Solving for R3,
R3 = (2
π
*f
C
*C2*V
OUT
Using 75KHz for f
C
and putting in the other
known constants:
(
2) R3
2.2x10
8
*C2*V
OUT
The value of the compensation resistor is
limited to 10K
to prevent overshoot on the
output at turn-on. So if the value calculated for
R3 from either equation (1) or equation (2) is
greater than 10k
, use 10K
for R3.
Choose C3 to set the zero frequency f
Z1
to
one-fourth of the crossover frequency f
C
:
f
Z1
= f
C
/ 4
or
1 /(2
π
*C3*R3) = G
CS
*G
EA
*V
IN
*V
FB
*R3 / (8
π
*C2*V
OUT
Solving for C3:
C3 = 4*C2*V
OUT
2
)
2
/ (G
CS
*G
EA
*V
IN
*V
FB
)
2
/ V
IN
2
)
2
/ (G
CS
*G
EA
*V
IN
*V
FB
*R3
2
)
Entering the known values gives:
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