A5975ADTR STMicroelectronics, A5975ADTR Datasheet - Page 21

• Manufacturer Part Number: A5975ADTR
• Description: IC REG SW STEP-DOWN 2.5A 8SOIC
• Manufacturer: STMicroelectronics
• Series: -r
• Type: Step-Down (Buck), PWMr

Specifications of A5975ADTR

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.235 V ~ 35 V
• Current - Output: 2.5A
• Frequency - Switching: 500kHz
• 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: 497-11425-2

A5975AD
7.2
Equation 4
whereas the zero is defined as:
Equation 5
F
the frequency of the double pole of the LC filter (see below). F
frequency.
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
P1
0
is the zero introduced by the ESR of the output capacitor and it is very important to
LOAD
is the low frequency which sets the bandwidth, while the zero F
LOAD
>>ESR, the previous expression of A
A
LC
s ( )
is defined as the ratio between V
=
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- -
s
2
F PLC1 2
⋅
n
L C
.
⋅
LC
A
OUT
,
the typical second order system equation can be recognized:
LC
=
s ( )
Doc ID 018761 Rev 1
⋅
–
--------------------------------------------------------------------------------------------------------------------------------------- -
(
ESR C OUT
ESR
F
=
F
P2
s
O
F
---------------------------------------------------------------------------------------- -
L C
2
⋅
=
Z1
=
⋅
+
+
R
--------------------------------------------------- -
2 π R
R
2 δ ω
LOAD
----------------------------------------------- -
2 π ESR C
OUT
=
LOAD
⋅ ⋅
⋅ ⋅
⋅ ⋅
---------------------------------
2 π R
1
±
⋅
⋅
+
⋅
s
)
(
ESR C
n
LC
c
2
(
2 L C OUT
1
⋅
+
ESR C OUT
1
⋅
OUT
⋅
1
1
+
⋅ ⋅
+
s
(
s
can be simplified and becomes:
C
c
ESR C
ESR C
⋅
+
⋅
⋅
0
⋅
(
⋅
ω
C
ESR C
and I
+
OUT
2
c
OUT
C
n
⋅
⋅
p
)
⋅
OUT
OUT
OUT
⋅
)
s
2
OUT
–
.
⋅
4 L C OUT
⋅
s
s
P2
⋅ ⋅
⋅
+
)
R
1
is usually at a very high
LOAD
Z1
is usually put near to
+
L
)
Closing the loop
+
R
LOAD
21/50