参数资料
| 型号： | LT1946EMS8E#TRPBF |
| 厂商： | Linear Technology |
| 文件页数： | 7/12页 |
| 文件大小： | 0K |
| 描述： | IC REG BOOST 1.5A 8MSOP |
| 标准包装： | 2,500 |
| 类型： | 升压（升压） |
| 输出数： | 1 |
| 输入电压： | 2.45 V ~ 16 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 2.7MHz |
| 电流 - 输出： | 1.5A |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-TSSOP，8-MSOP（0.118"，3.00mm 宽）裸露焊盘 |
| 包装： | 带卷 (TR) |
| 供应商设备封装： | 8-MSOP-EP |
LT1946
APPLICATIO S I FOR ATIO
Compensation—Theory
Like all other current mode switching regulators, the
LT1946 needs to be compensated for stable and efficient
g mp
C OUT
R L
V OUT
operation. Two feedback loops are used in the LT1946: a
fast current loop which does not require compensation,
and a slower voltage loop which does. Standard Bode plot
analysis can be used to understand and adjust the voltage
V C
R C
R O
g ma
+
–
1.250V
REFERENCE
R1
feedback loop.
C C
R2
1946 F04
As with any feedback loop, identifying the gain and phase
contribution of the various elements in the loop is critical.
Figure 4 shows the key equivalent elements of a boost
converter. Because of the fast current control loop, the
power stage of the IC, inductor and diode have been
replaced by the equivalent transconductance amplifier
g mp . g mp acts as a current source where the output current
is proportional to the V C voltage. Note that the maximum
output current of g mp is finite due to the current limit in the
IC.
From Figure 4, the DC gain, poles and zeroes can be
calculated as follows:
C C : COMPENSATION CAPACITOR
C OUT : OUTPUT CAPACITOR
g ma : TRANSCONDUCTANCE AMPLIFIER INSIDE IC
g mp : POWER STAGE TRANSCONDUCTANCE AMPLIFIER
R C : COMPENSATION RESISTOR
R L : OUTPUT RESISTANCE DEFINED AS V OUT DIVIDED BY I LOAD(MAX)
R O : OUTPUT RESISTANCE OF g ma
R1, R2: FEEDBACK RESISTOR DIVIDER NETWORK
Figure 4. Boost Converter Equivalent Model
The Current Mode zero is a right half plane zero which can
be an issue in feedback control design, but is manageable
with proper external component selection.
Using the circuit of Figure 1 as an example, the following
table shows the parameters used to generate the Bode plot
shown in Figure 5.
Output Pole: P1=
2
2 ? π ? R L ? C OUT
Table 3. Bode Plot Parameters
Parameter Value
R L 18.6
Units
?
Comment
Application Specific
Error Amp Pole: P2 =
1
2 ? π ? R O ? C C
C OUT
R O
20
10
μ F
M ?
Application Specific
Not Adjustable
? g ma O mp ? L
1.25 V IN R
2
ESR Zero: Z 2 =
1
Error Amp Zero: Z1=
2 ? π ? R C ? C C
DC GAIN: A = ? R ? g
2
V OUT
1
2 ? π ? ESR ? C OUT
C C
R C
V OUT
V IN
g ma
g mp
L
f S
470
49.9
8
3.3
40
5
5.4
1.2
pF
k ?
V
V
μ mho
mho
μ H
MHz
Adjustable
Adjustable
Application Specific
Application Specific
Not Adjustable
Not Adjustable
Application Specific
Not Adjustable
V IN ? R L
2 ? π ? V OUT ? L
High Frequency Pole: P3 > S
RHP Zero: Z3 =
2
2
f
3
From Figure 5, the phase is 120 ° when the gain reaches
0dB giving a phase margin of 60 ° . This is more than
adequate. The crossover frequency is 25kHz, which is
about three times lower than the frequency of the right half
plane zero Z2. It is important that the crossover frequency
be at least three times lower than the frequency of the RHP
zero to achieve adequate phase margin.
1946fb
7
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