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參數(shù)資料
| 型號： | TB62732FU |
| 廠商： | Toshiba Corporation |
| 元件分類： | LED驅(qū)動器 |
| 英文描述： | The step up type DC-DC converter only for white LED driver lighting |
| 中文描述： | 唯一的白光LED驅(qū)動器照明加強(qiáng)型DC - DC轉(zhuǎn)換 |
| 文件頁數(shù)： | 7/15頁 |
| 文件大小： | 177K |
| 代理商： | TB62732FU |
TOSHIBA
T B 6 2 7 3 2 F U
27, Feb 2002 (Ver.04)
7
THE SETUP ON THE OUTPUT SIDE CAPACITOR
The C2 is upper 0.47(uF) above is recommended from the consideration to the IF peak.
Capacitor C2 (uF)
Ripple current (mA)
0.01
15-25
0.1
5-8
0.47
2-4
1
1-3
THE SETUP ON THE INDUCTANCE
The minimum inductance with the outside is calculated with the next formula.
L(uH) = ((K*Po)-Vin min.*Io)*(1/fOSC min.)*2*(1/(Ip min.*Ip min.)) - - -
Form 2
Each clause is as mentioned in the following.
Po (W) : output power (the electric energy which should be necessary the LED load)
Po (W) = (VF LED*IF LED)+(Vf schottky*IF LED)+(R_sens*IF LED*IF LED)
( Forward current of LED is IF LED (mA) = Setup currnet Io (mA), Forward voltage of LED is VF LED (V),
Forward voltage of schottky diode is VF schottky (V), Setup resistance of output currnet is R_sens (ohms) )
Vin min (V) : Minimum input voltage(battely voltage)
When there is a resistance element on the input voltage side, that one for the voltage descent is taken into
consideration to the minimum input voltage.
The input Iin is estimated roughly in Form 3.
Iin (mA) = Vo / Vin * IF
--- Form 3
Example, the voltage drop of 1(V) occurs when it becomes Iin = 0.1(A) and has the line resistance of
1 (ohms).
At this time, Vin=3.1 (V) becomes minimum Vin value because the minimum Vcc specifications of spec
is Vcc=3.0 (V).
Io (A) : The average current value established with R_sens. Show the fig-5 on next page.
fOSC(Hz) : The switching frequency of the internal MOS transistor.
The specification of fOSC(MHz) = 0.7 min, 1.0 typ and 1.3 max.
Ip (A) : Peak current value to supply to the inductance.
The specification of Ip (mA) = 320 min, 350 typ, 380 max.
For example, the following condition is substituted for the formula.
It is supposed under condition.
Input voltage Vin : Vin=3.0-4.3(V),
Output side capacitance C2 : C2=0.47(uF) - - - C2 is ignored by the calculation.
VF LED = 16(V), schottly diode VF: Vf schottky = 0.3(V),
Setup resistence R_sens : R_sens = 2.7(ohms), Setup current Io : Io = 19.6(mA).
L (uH) = ((16*0.0196+0.3*0.0196+2.7*0.0196*0.0196)-
Vin
*0.0196)*(1/700e3)*2*(1/(0.32*0.32))
= 7.19(uH, Vin = 3.0V) and 6.59 (uH,Vin = 4.3V)
Therefore, 7.19(uH) in Vcc=3.0V whose input voltage is low is chosen.
It is sufficient by the above calculation on the standard condition.
If the worst case is taken into consideration, the coil of about 1.1 times of the calculation is chosen.
L(uH)=7.19(uH)*1.1 >= 7.90 (uH)
Note
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