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參數(shù)資料
| 型號： | ADDC02808PB |
| 廠商： | Analog Devices, Inc. |
| 英文描述： | 28 V, 200 W Pulsed DC/DC Converter with Integral EMI Filter |
| 中文描述： | 28五，200瓦脈沖直流/直流轉換器積分EMI濾波器 |
| 文件頁數(shù)： | 10/20頁 |
| 文件大?。?/td> | 287K |
| 代理商： | ADDC02808PB |
ADDC02808PB
REV. A
–10–
8.1
7.4
–200
800
–100
V
O
0
100
200
300
400
500
600
700
8.0
7.9
7.8
7.7
7.6
7.5
TIME –
m
s
Figure 25. Predicted Response for 24 A Step Load Change,
di/dt = 12 A/
μ
s, with Factory Set Internal Compensation
Optimized for C
LOAD
= 4,000
μ
F and R
ESR
= 2.5 m
RESPONSE AT END OF PULSE
The previous section describes how the ADDC02808PB con-
verter responds to the positive step change in load current that
occurs at the beginning of a power pulse. This section will
discuss the converter’s response at the end of the power pulse
when the load current is abruptly returned to a small value.
Figures 26-29 show the converter’s measured output voltage as
the load current is stepped from 25 A down to 4 A, 2 A, 1 A, and
0.1 A, respectively. The load capacitance is 1,000
μ
F with
100mV
10
0%
100
90
100
m
s
V
O
Figure 26. Output Voltage Transient Response to a 25 A
to 4 A Step Change in Load, di/dt/ = 12 A/
μ
s, with 1,000
μ
F
Load Capacitance (R
ESR
= 10 m
)
100mV
10
0%
100
90
100
m
s
V
O
Figure 27. Output Voltage Transient Response to a 25 A
to 2 A Step Change in Load, di/dt/ = 12 A/
μ
s, with 1,000
μ
F
Load Capacitance (R
ESR
= 10 m
)
R
ESR
= 10 m
. The di/dt is 12 A/
μ
s. As can be seen, the peak
deviations for these curves are close to each other and com-
parable to the negative deviation shown in Figure 6 for a simi-
larly sized positive step change in load current.
100mV
10
0%
100
90
100
m
s
V
O
Figure 28. Output Voltage Transient Response to a 25 A
to 1 A Step Change in Load, di/dt/ = 12 A/
μ
s, with
1,000
μ
F Load Capacitance (R
ESR
= 10 m
)
100mV
10
0%
100
90
1ms
V
O
Figure 29. Output Voltage Transient Response to a 25 A
to 0.1 A Step Change in Load, di/dt/ = 12 A/
μ
s, with
1,000
μ
F Load Capacitance (R
ESR
= 10 m
)
What is different about these curves is the settling time. Once
the converter’s output voltage rises above nominal, the con-
verter cannot help to discharge the load capacitor. It can only
reduce its output current to zero; it cannot draw a negative
current. As such, the time it takes to bring the output voltage
back down to its nominal value depends on the load current
during the low load portion of the cycle. The rate at which the
output voltage falls to its nominal value is the load current
divided by the load capacitance (including the 150
μ
F capaci-
tance that is inside the converter). The smaller the load current,
the longer it takes to get the output voltage back to its nominal
value.
During the time that the output voltage is too high, the integra-
tor in the converter’s feedback circuitry is continuing to ramp
out of range. As the output voltage then falls below its nominal
value, it must have an undershoot error to bring the integrator
back into range. As can be seen from these figures, the lower
the load current, the longer the output voltage remains too
high, and the longer and the greater the output voltage under-
shoot is.
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