Datasheet

R1 = R 2 = 10k
L
R1//R2 = = 5kΩ
DCR × C1
BIAS
OUT
R1
R = 0.7 × = 8.75kΩ
V - 0.7
( )
1.058
PH
3
R 1.33 39.2 10 ƒ 7 ,
-
= ´ ´ ´ -
P
SRtot
+ P
SRcond
) P
DIODE
+ 1.47 W
diode OUT d f sw
P = 2 × I × t × V × f = 0.77W
P
SRcond
+
ǒ
I
SRrms
Ǔ
2
R
DS(on)
(sr)
N
+ 0.7 W
I
SRrms
+
(
1 * D
)
ǒ
I
OUT
2
)
I
RIPPLE
2
12
Ǔ
Ǹ
+ 18.7 A
missing equation
P
SWtot
+ P
SWcond
) P
SWsw
+ 0.91 W
TPS40140
www.ti.com SLUS660H –SEPTEMBER 2005–REVISED JUNE 2013
The calculated total loss in the high-side MOSFET is:
(31)
The power losses in the low-side SR MOSFET is calculated Equation 32:
(32)
The RMS current in the low-side MOSFET is shown in Equation 33.
(33)
The R
DS(on)
is 4 mΩ when the MOSFET gate voltage is 4.5 V.
The total conduction loss in the two low-side MOSFETs is shown in .
where
• N is the number of MOSFETs. Here, it is equal to 2. (34)
The total power loss in the body diode is:
(35)
So the calculated total loss in the SR MOSFET is:
(36)
6.1.5 Step 5: Peripheral Component Design
6.1.5.1 Switching Frequency Setting (RT pin 5):
(37)
In the design, a 62-kΩ resistor is selected. The actual switching frequency is 510kHz.
6.1.5.2 Output Voltage Setting (FB1 pin 36):
Substitute R1 with 10 kΩ and then calculate R
BIAS
.
(38)
6.1.5.3 Current Sensing Network Design (CS1 pin 31 and CSRT1 pin32)
For small pulse width, to avoid the sub-harmonics brought by the loop delay, a resistor divider is usually
used to attenuate the current feedback information as described in the Inductor DCR Current Sense
section.
Choosing C1 a value for 0.1-μF, and let R1 and R2 be equal, calculating R1 and R2 with the following
equations:
(39)
(40)
A simplified equation to determine if the design produces sub-harmonics is shown in Equation 41.
Copyright © 2005–2013, Texas Instruments Incorporated DESIGN EXAMPLES 45
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