A5975DTR STMicroelectronics, A5975DTR Datasheet - Page 21

• Manufacturer Part Number: A5975DTR
• Description: IC REG SW STEP-DOWN 3A 8SOIC
• Manufacturer: STMicroelectronics
• Series: -r
• Type: Step-Down (Buck), PWMr

Specifications of A5975DTR

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.235 V ~ 35 V
• Current - Output: 3A
• Frequency - Switching: 250kHz
• Voltage - Input: 4 V ~ 36 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (0.154", 3.90mm Width) Exposed Pad
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant
• Other names: 497-11426-2

A5975D
7.2
LC filter
The transfer function of the LC filter is given by:
Equation 6
where R
If R
Equation 7
The zero of this transfer function is given by:
Equation 8
F
increase the phase margin of the loop.
The poles of the transfer function can be calculated through the following expression:
Equation 9
In the denominator of A
Equation 10
If the damping coefficient δ is very close to zero, the roots of the equation become a double
root whose value is ω
Similarly for A
Equation 11
0
is the zero introduced by the ESR of the output capacitor and it is very important to
LOAD
LOAD
>>ESR, the previous expression of A
A
LC
s ( )
is defined as the ratio between V
LC
=
, the poles can usually be defined as a double pole whose value is:
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- -
s
2
F PLC1 2
⋅
n
L C
.
⋅
LC
,
A
OUT
the typical second order system equation can be recognized:
LC
=
s ( )
Doc ID 018760 Rev 1
–
--------------------------------------------------------------------------------------------------------------------------------------------- -
⋅
ESR C
(
ESR
=
F
F
s
O
---------------------------------------------------------------------------------------- -
L C
⋅
2
PLC
=
⋅
+
+
R
OUT
R
2 δ ω
LOAD
----------------------------------------------- -
2 π ESR C
OUT
=
LOAD
⋅ ⋅
⋅ ⋅
1
------------------------------------------- -
2 π ⋅
±
⋅
+
⋅
s
(
)
(
ESR C
n
LC
ESR C
2
2 L C OUT
1
+
•
1
⋅
OUT
+
⋅ ⋅
+
s
s
can be simplified and becomes:
ESR C
1
ESR C
⋅
+
L C
⋅
⋅
⋅
(
ω
⋅
ESR C
and I
OUT
OUT
2
OUT
n
OUT
⋅
⋅
⋅
OUT
OUT
)
OUT
⋅
2
s
–
OUT
.
(
⋅
4 L C OUT
⋅
s
s
⋅ ⋅
⋅
+
)
R
1
LOAD
+
)
L
)
Closing the loop
+
R
LOAD
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