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参数资料
| 型号： | ADP1850DP-EVALZ |
| 厂商： | Analog Devices Inc |
| 文件页数： | 21/32页 |
| 文件大小： | 0K |
| 描述： | EVAL BOARD FOR ADP1850DP |
| 标准包装： | 1 |
| 主要目的： | DC/DC，步降 |
| 输出及类型： | 1，非隔离 |
| 输出电压： | 1.09V |
| 电流 - 输出： | 50A |
| 输入电压： | 10 ~ 15 V |
| 稳压器拓扑结构： | 降压 |
| 频率 - 开关： | 300kHz |
| 板类型： | 完全填充 |
| 已供物品： | 板 |
| 已用 IC / 零件： | ADP1850DP |
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Data Sheet
LOOP COMPENSATION (SINGLE PHASE
OPERATION)
As with most current mode step-down controller, a transcon-
ADP1850
At the crossover frequency, the open-loop transfer function is
unity or 0 dB, H (f CROSS ) = 1. Combining Equation 1 and
Z COMP ( f CROSS ) = ?
? G × G
? ? C OUT × V OUT
? ?
?
?
ductance error amplifier is used to stabilize the external voltage
loop. Compensating the ADP1850 is fairly easy; an RC compen-
sator is needed between COMPx and AGND. Figure 33 shows
? 2 π × f CROSS
? m CS
? ?
? ? V REF
?
?
(5)
the configuration of the compensation components: R COMP ,
The zero produced by R COMP and C COMP is
C COMP , and C C2 . Because C C2 is very small compared to C COMP ,
to simplify calculation, C C2 is ignored for the stability
compensation analysis.
f ZERO =
1
2 π R COMP × C COMP
(6)
ADP1850
At the crossover frequency, Equation 4 can be shown as
f CROSS 2 + f ZERO 2
C C2
COMPx
R COMP
C COMP
G m
0.6V
FBx
Z COMP ( f CROSS ) = R COMP ×
f CROSS
Combining Equation 5 and Equation 7 and solving for
(7)
AGND
R COMP gives
f CROSS + f ZERO
? 2 π × f CROSS
? G × G
? ? C OUT × V OUT
? × ?
? ?
? (8)
?
Figure 33. Compensation Components
The open loop gain transfer function at angular frequency, s, is
given by
R COMP =
f CROSS
2
2
× ?
? m CS
? ? V REF
?
?
H ( s ) = G m × G CS ×
× Z COMP ( s ) × Z FILTER ( s )
f CROSS =
f ZERO =
= SW
V REF
V OUT
where:
G m is the transconductance of the error amplifier, 500 μS.
G CS is the tranconductance of the power stage.
Z COMP is the impedance of the compensation network.
Z F ILTER is the impedance of the output filter.
(1)
Choose the crossover and zero frequencies as follows:
f SW
12
f CROSS f
4 48
Substituting Equation 2, Equation 9, and Equation 10 into
(9)
(10)
G CS =
? 2 π × f CROSS ? ? C OUT × V OUT
R COMP = 0 . 97 × A CS × R DSON ? ?
? × ?
? ?
? (11)
?
V REF = 0.6 V.
G CS with units of A/V is given by
1
A CS × R DSON _ MIN
(2)
? G m ? ? V REF
where:
G m is the transconductance of the error amplifier, 500 μS.
?
?
G CS =
C COMP =
π R COMP × f CROSS
where:
A CS is the current sense gain of either 3 V/V, 6 V/V, 12 V/V, or
24 V/V set by the gain resistor between DLx and PGNDx.
R DSON_MIN is the low-side MOSFET minimum on resistance.
If a sense resistor, R S , is added in series with the low-side FET,
then G CS becomes
1
A CS × ( R DSON _ MIN + R S )
Because the zero produced by the ESR of the output capacitor is
not needed to stabilize the control loop, assuming ESR is small
the ESR is ignored for analysis. Then Z FILTER is given by
A CS is the current sense gain of 3 V/V, 6 V/V, 12 V/V, or 24 V/V.
R DSON is on resistance of the low-side MOSFET.
V REF = 0.6 V.
And combining Equation 6 and Equation 10 yields
2
(12)
Note that the previous simplified compensation equations for
R COMP and C COMP yield reasonable results in f CROSS and phase
margin assuming that the compensation ramp current is ideal.
Varying the ramp current or deviating the ramp current from
ideal can affect f CROSS and phase margin.
Z FILTER =
× C COMP ≤ C C 2 ≤ × C COMP
1
sC OUT
Because C C2 is small relative to C COMP , Z COMP can be simplified to
(3)
And lastly, set C C2 to
1 1
20 10
(13)
Z COMP = R COMP +
1
sC COMP
=
1 + sR COMP × C COMP
sC COMP
(4)
Rev. A | Page 21 of 32
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