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| 型号： | LTC1702IGN#TR |
| 厂商： | Linear Technology |
| 文件页数： | 19/36页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM VM 24-SSOP |
| 标准包装： | 2,500 |
| 系列： | PolyPhase® |
| PWM 型： | 电压模式 |
| 输出数： | 2 |
| 频率 - 最大： | 750kHz |
| 占空比： | 93% |
| 电源电压： | 3 V ~ 7 V |
| 降压： | 是 |
| 升压： | 无 |
| 回扫： | 无 |
| 反相： | 无 |
| 倍增器： | 无 |
| 除法器： | 无 |
| Cuk： | 无 |
| 隔离： | 无 |
| 工作温度： | -40°C ~ 85°C |
| 封装/外壳： | 24-SSOP（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
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LTC1702
APPLICATIO N S I N FOR M ATIO N
( ) . ? 0 5 + ( ) 16 . ? 0 +
– 2 . 18 7 . 82
( ) 16 . ? 0 + ( ) 18 . ? 0
4 . 82
– 5 . 18
from C IN (time point A). 50% of the way through, TG2
turns on and the total current is 13A (time point B).
Shortly thereafter, TG1 turns off and the current drops to
10A (time point C). Finally, TG2 turns off and the current
I RMS =
2 2
2 2
spends a short time at 0 before TG1 turns on again (time
point D).
I AVG = ( 3 A ? 0 . 5 ) + ( 13 A ? 0 . 16 ) +
( 10 A ? 0 . 16 ) + ( 0 A ? 0 . 18 ) = 5 . 18 A
Now we can calculate the RMS current. Using the same
waveform we used to calculate the average DC current,
subtract the average current from each of the DC values.
= 4 . 55 A RMS
If the circuit is likely to spend time with one side operating
and the other side shut down, the RMS current will need
to be calculated for each possible case (side 1 on, side 2
off; side 1 off, side 2 on; both sides on). The capacitor
must be sized to withstand the largest RMS current of the
three—sometimes this occurs with one side shut down!
( ) 67 . ? 1 0 + ( ) 33 . ? 0 = 1 . 42 A
I RMS 1 =
– 2
( ) 32 . ? 0 + ( ) 68 . ? 0
I RMS 2 =
6 . 8
– 3 . 2
Squareeachcurrenttermandmultiplythesquaresbythe
same period percentages we used to calculate the aver-
age DC current. Sum the results and take the square root.
The result is the approximate RMS current as seen by the
input capacitor with both sides of the LTC1702 at full load.
Actual RMS current will differ due to inductor ripple cur-
rent and resistive losses, but this approximate value is
adequate for input capacitor calculation purposes.
50% 16% 16% 18%
7.8
Side 1 only :
I AVE 1 = ( 3 A ? 0 . 67 ) + ( 0 A ? 0 . 33 ) = 2 . 01 A
2 2
Side 2 only :
I AVE 2 = ( 10 A ? 0 . 32 ) + ( 0 A ? 0 . 68 ) = 3 . 2 A
2 2
= 4 . 66 A RMS > 4 . 55 A RMS
RMS
4.8
Consider the case where both sides are operating at the
same load, with a 50% duty cycle at each side. The RMS
0
–2.2
current with both sides running is near zero, while the
RMS current with one side active is 1/2 the total load
current of that side. The 2-phase, 5V to 2.5V circuit in the
– 5.2
0
A
B
TIME
C
D
1702 SB2
applications section takes advantage of this phenom-
enon, allowing it to supply 40A of output current with only
120 μ F of input capacitance (and only 40 μ F of output
Figure SB2. AC Current Calculation
the single case (Figure 7)! The peak current deltas are still
only 10A, requiring the same 0.01 ? ESR rating. As long as
the capacitor we chose for the single side application can
support the slightly higher 4.8A RMS current, we can add
the second channel without changing the input capacitor
at all. As a general rule, an input bypass capacitor capable
of supporting the larger output current channel can sup-
port both channels running simultaneously (see the
capacitance!).
Tantalum capacitors are a popular choice as input capaci-
tors for LTC1702 applications, but they deserve a special
caution here. Generic tantalum capacitors have a destruc-
tive failure mechanism when they are subjected to large
RMS currents (like those seen at the input of a LTC1702).
At some random time after they are turned on, they can
blow up for no apparent reason. The capacitor manufac-
turers are aware of this and sell special “surge tested”
2-Phase Operation section for more details).
1702fa
19
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