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參數(shù)資料
| 型號(hào)： | CS5166HGDW16 |
| 廠商： | ZF Electronics Corporation |
| 英文描述： | 5-Bit Synchronous CPU Controller with Power-Good and Current Limit |
| 中文描述： | 5位同步CPU，具有功率控制器，準(zhǔn)確和電流限制 |
| 文件頁數(shù)： | 18/22頁 |
| 文件大?。?/td> | 435K |
| 代理商： | CS5166HGDW16 |
Figure 25: 30A load waveforms.
Figure 26: 15A load transient waveforms.
Figure 26 shows supply response to a 15A load step with a
30A/μs slew rate. The V
2
TM
control loop immediately
forces the duty cycle to 100%, ramping the current in both
inductors up. A voltage spike of 136mV due to output
capacitor impedance occurs. The inductive component of
the spike due to ESL recovers within several microseconds.
The resistive component due to ESR decreases as inductor
current replaces capacitor current.
The benefit of adaptive voltage positioning in reducing the
voltage spike can readily be seen. The differences in DC
voltage and duty cycle can also be observed. This particu-
lar transient occurred near the beginning of regulator off-
time, resulting in a longer recovery time and increased
voltage spike.
Output Inductor
The inductor should be selected based on its inductance,
current capability, and DC resistance. Increasing the induc-
tor value will decrease output voltage ripple, but degrade
transient response.
Inductor Ripple Current
Ripple current =
Example: V
IN
= +5V, V
OUT
= +2.8V, I
LOAD
= 14.2A, L =
1.2μH, Freq = 200KHz
Ripple current =
= 5.1A
Output Ripple Voltage
V
RIPPLE
= Inductor Ripple Current
×
Output Capacitor
ESR
Example:
V
IN
= +5V, V
OUT
= +2.8V, I
LOAD
= 14.2A, L = 1.2μH,
Switching Frequency = 200KHz
Output Ripple Voltage = 5.1A
×
Output Capacitor ESR
(from manufacturer’s specs)
ESR of Output Capacitors to limit Output Voltage Spikes
ESR =
This applies for current spikes that are faster than regula-
tor response time. Printed Circuit Board resistance will add
to the ESR of the output capacitors.
In order to limit spikes to 100mV for a 14.2A Load Step,
ESR = 0.1/14.2 = 0.007
Inductor Peak Current
Peak Current = Maximum Load Current +
(
)
Example: V
IN
= +5V, V
OUT
= +2.8V, I
LOAD
= 14.2A, L = 1.2μH,
Freq = 200KHz
Peak Current = 14.2A + (5.1/2) = 16.75A
A key consideration is that the inductor must be able to
deliver the Peak Current at the switching frequency with-
out saturating.
Response Time to Load Increas
e
(limited by Inductor value unless Maximum On-Time is
exceeded)
L
×
I
OUT
(V
IN
-V
OUT
)
Response Time =
Example: V
IN
= +5V, V
OUT
= +2.8V, L = 1.2μH, 14.2A
change in Load Current
Response Time =
= 7.7μs
1.2μH
×
14.2A
(5V-2.8V)
Ripple
Current
2
V
OUT
I
OUT
[(5V-2.8V)
×
2.8V]
[200KHz
×
1.2μH
×
5V]
[(V
IN
- V
OUT
)
×
V
OUT
]
(
Switching Frequency
×
L
×
V
IN
)
Trace 1 Output voltage ripple
Trace 2 Buck regulator #1 inductor switching node
Trace 3 Buck regulator #2 inductor switching node
Trace 1 Output voltage ripple
Trace 2 Buck regulator #1 inductor switching node
Trace 3 Buck regulator #2 inductor switching node
Application Information: continued
18
C
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