參數(shù)資料
型號: TPS63700DRC
廠商: Texas Instruments, Inc.
英文描述: DC-DC INVERTER
中文描述: 直流逆變器
文件頁數(shù): 14/22頁
文件大?。?/td> 1029K
代理商: TPS63700DRC
www.ti.com
R3
R2
VREF
VOUT
VREF
1
(1)
Inductor Selection
ILavg
VIN
VIN
VOUT
0.8
IOUT
(2)
L
VIN
VOUT
IL
VOUT
VIN
f
(3)
ILmax
VIN
VIN
VOUT
0.8
IOUT
IL
2
(4)
IL
VIN
VOUT
L
VOUT
VIN
f
(5)
TPS63700
SLVS530–SEPTEMBER 2005
APPLICATION INFORMATION (continued)
For example, if an output voltage of –5 V is needed and a resistor of 150 k
has been chosen for R2, a 680-k
resistor is needed to program the desired output voltage.
An inductive converter normally requires two main passive components for storing energy during the conversion.
An inductor and a storage capacitor at the output are required.
The average inductor current depends on the output load, the input voltage (VIN), and the output voltage VOUT.
It can be estimated with
Equation 2
, which shows the formula for the inverting converter.
with:
I
Lavg
= average inductor current
An important parameter for choosing the inductor is the desired current ripple in the inductor.
A ripple current value between 20% and 80% of the average inductor current can be considered as reasonable,
depending on the application requirements. A smaller ripple reduces the losses in the inductor, as well as output
voltage ripple and EMI. But in the same way, the inductor becomes larger and more expensive.
Keeping those parameters in mind, the possible inductor value can be calculated using
Equation 3
.
with:
I
L
= peak-to-peak ripple current
f = switching frequency
L = inductor value
With the known inductor current ripple, the peak inductor value can be approximated with
Equation 4
. The peak
current through the switch and the inductor depends also on the output load, the input voltage (VIN), and the
output voltage (VOUT). To select the right inductor, it is recommended to keep the possible peak inductor current
below the current-limit threshold of the power switch. For example, the current-limit threshold of the TPS63700
switch for the inverting converter is nominally 1000 mA.
with:
I
LMAX
= peak inductor current
With
Equation 5
, the inductor current ripple at a given inductor can be approximated.
Care has to be taken for the possibility that load transients and losses in the circuit can lead to higher currents as
estimated in
Equation 4
. Also, the losses caused by magnetic hysteresis losses and copper losses are a major
parameter for total circuit efficiency.
14
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