参数资料
| 型号： | SC470ITSTRT |
| 厂商： | Semtech |
| 文件页数： | 12/31页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM 14-TSSOP |
| 标准包装： | 1 |
| PWM 型： | 控制器 |
| 输出数： | 1 |
| 电源电压： | 5 V ~ 5.5 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 ~ 125°C |
| 封装/外壳： | 14-TSSOP（0.173"，4.40mm 宽） |
| 包装： | 标准包装 |
| 产品目录页面： | 1358 (CN2011-ZH PDF) |
| 其它名称： | SC470ITSDKR |
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POWER MANAGEMENT
NOT RECOMMENDED FOR NEW DESIGN
SC470
Application Information (Cont.)
For our example:
( V BAT ( MIN ) ON _ VBAT ( MIN ) ) Hz
f SW _ VBAT ( MIN )
=
V OUT
? t
I RIPPLE_VBAT(MIN) = 1.74A P-P and I RIPPLE_VBAT(MAX) = 2.18A P-P
( V BAT ( MAX ) ON _ VBAT ( MAX ) ) Hz
I INDUCTOR ( MIN ) OUT ( MAX ) +
= I
A ( MIN )
and,
f SW _ VBAT ( MAX ) =
V OUT
? t
From this we can calculate the minimum inductor
current rating for normal operation:
I RIPPLE _ VBAT ( MAX )
2
t ON is generated by a one-shot comparator that samples
V BAT via R tON , converting this to a current. This current is
used to charge an internal 3.3pF capacitor to V OUT . The
equations on page 11 reflect this along with any internal
components or delays that influence t ON .
For our example we select R tON = 1M ? :
t ON_VBAT(MIN) = 563ns and t ON_VBAT(MAX) = 255ns
For our example:
I INDUCTOR(MIN) = 7.1A (MIN)
Next we will calculate the maximum output capacitor
equivalent series resistance (ESR). This is determined by
calculating the remaining static and transient tolerance
allowances. Then the maximum ESR is the smaller of the
calculated static ESR (R ESR_ST(MAX) ) and transient ESR
(R ESR_TR(MAX) ):
f SW_VBAT(MIN) = 266kHz and f SW_VBAT(MAX) = 235kHz
Now that we know t ON we can calculate suitable values
for the inductor. To do this we select an acceptable
R ESR _ ST ( MAX ) =
( ERR ST ? ERR DC ) ? 2
I RIPPLE _ VBAT ( MAX )
Ohms
inductor ripple current. The calculations below assume
50% of I OUT which will give us a starting place.
Where ERR ST is the static output tolerance and ERR DC is
the DC error. The DC error will be 0.85% plus the
L VBAT ( MIN ) = ( V BAT ( MIN ) ? V OUT ) ?
t ON _ VBAT ( MIN )
( 0 . 5 ? I OUT )
H
tolerance of the feedback resistors, thus 1.85% total
for 1% feedback resistors.
For our example:
and,
L VBAT ( MAX ) = ( V BAT ( MAX ) ? V OUT ) ?
For our example:
t ON _ VBAT ( MAX )
( 0 . 5 ? I OUT )
H
ERR ST = 48mV and ERR DC = 22mV, therefore,
R ESR_ST(MAX) = 24m ?
? ? I OUT + RIPPLE _ VBAT ( MAX ) ? ?
?
?
L VBAT(MIN) = 1.3μH and L VBAT(MAX) = 1.6μH
We will select an inductor value of 2.2μH to reduce the
ripple current, which can be calculated as follows:
R ESR _ TR ( MAX ) =
? I ?
( ERR TR ? ERR DC )
2
Ohms
and,
I RIPPLE _ VBAT ( MIN ) = ( V BAT ( MIN ) ? V OUT ) ?
t ON _ VBAT ( MIN )
L
A P ? P
Where ERR TR is the transient output tolerance. Note that
this calculation assumes that the worst case load
transient is full load. For half of full load, divide the I OUT
term by 2.
I RIPPLE _ VBAT ( MAX ) = ( V BAT ( MAX ) ? V OUT ) ?
? 2005 Semtech Corp.
t ON _ VBAT ( MAX )
L
A P ? P
12
For our example:
ERR TR = 96mV and ERR DC = 22mV, therefore,
www.semtech.com
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