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參數(shù)資料
| 型號： | TPS5103IDBR |
| 英文描述： | SMPS Controller |
| 中文描述： | 開關(guān)電源控制器 |
| 文件頁數(shù)： | 15/31頁 |
| 文件大?。?/td> | 1083K |
| 代理商： | TPS5103IDBR |
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TPS5102
DUAL, HIGH-EFFICIENCY CONTROLLER FOR NOTEBOOK PC POWER
SLVS239 - SEPTEMBER 1999
15
POST OFFICE BOX 655303
DALLAS, TEXAS 75265
APPLICATION INFORMATION
output voltage setpoint calculation (continued)
If a higher precision resistor is used, the voltage setup can be more accurate.
In some applications, the output voltage is required to be lower than the reference voltage. With a few extra
components, the lower voltage can be easily achieved. The drawing below shows the method.
TPS5102
VCC
VO
R(top)
R(bottom)
Rz1
Zener
INV
Rz2
In the schematic, the Rz1, the Rz2, and the zener are the extra components. Rz1 is used to give the zener
enough current to build up the zener voltage. The zener voltage is added to INV through Rz2. Therefore, the
voltage on the INV is still equal to the IC internal voltage (1.185 V) even if the output voltage is regulated at a
lower setpoint. The equation for setting up the output voltage is shown below:
Rbtm
Vr
Rtop
)
Vo
–
Vr
(
)
Vr
Vz
(
2
Rz
+
–
=
When Rz2 is the adjusting resistor for low output voltage; Vz is the zener voltage; Vr is the internal reference
voltage; Rtop is the resistor of the voltage sensing network; Rbtm is the bottom resistor of the sensing
network;V
O
is the required output voltage setpoint.
Example: Assuming the required output voltage setpoint is V
O
= 0.8 V, V
Z
= 5 V; Rtop = 1 k
; Rbottom = 1 k
,
Then the Rz2 = 2.43 k
.
output inductor ripple current
The output inductor current ripple can affect not only the efficiency, but also the output voltage ripple. The
equation is exhibited below:
Iripple
Vin
Vout
Iout
Lout
(Rdson
R
L
)
D
Ts
Where Irippleis the peak-to-peak ripple current (A) through the inductor; Vinis the input voltage (V); Voutis the
output voltage (V); Ioutis the output current; Rdsonis the on-time resistance of MOSFET (
); Dis the duty cycle;
and Tsis the switching cycle (S). From the equation, it can be seen that the current ripple can be adjusted by
changing the output inductor value.
Example: Vin = 5 V; Vout = 1.8 V; Iout = 5 A; Rdson = 10 m
; RL = 5 m
; D = 0.36; Ts = 10
μ
S; Lout = 6
μ
H
Then, the ripple Iripple = 2 A.
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