ISL85033IRTZ-T Intersil, ISL85033IRTZ-T Datasheet - Page 20

• Manufacturer Part Number: ISL85033IRTZ-T
• Description: IC REG BUCK ADJ DUAL 28TQFN
• Manufacturer: Intersil
• Type: Step-Down (Buck)r

Specifications of ISL85033IRTZ-T

• Internal Switch(s): Yes
• Synchronous Rectifier: Yes
• Number Of Outputs: 2
• Voltage - Output: 0.8 ~ 28 V
• Current - Output: 3A
• Frequency - Switching: 300kHz ~ 2MHz
• Voltage - Input: 4.5 ~ 28 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 28-WFQFN exposed Pad
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Power - Output: -

C
where Io [A] is the output load current, R
(Ω) is the ESR of the output capacitor Co.
Example: V
C
R1 = 96kΩ.
The compensation capacitors are:
C
parasitic capacitance from V
is optional).
Theory of Compensation
The sensed current signal is injected into the voltage loop
to achieve current mode control to simplify the loop
compensation design. The inductor is not considered as a
state variable for current mode control and the system
becomes a single order system. It is much easier to
design a compensator to stabilize the voltage loop than
voltage mode control. Figure 46 shows the small signal
model of the synchronous buck regulator.
FIGURE 46. SMALL SIGNAL MODEL OF SYNCHRONOUS
PWM Comparator Gain F
The PWM comparator gain Fm for peak current mode
control is given by Equation 13:
Where S
S
Where:
R
sensing resistance and gain of the current amplifier in
current loop.
F
S
T
1
o
1
m
n
n
+
+
^
V
is trans-resistance, and is the product of the current
=
= 47µF/R
= 815pF, C
i
is given by Equation 14.
^
=
=
IN
IN
C
----------------------------------------- C
R
---------------- -
v ˆ
o
comp
t
V
---------------------
d ˆ
×
e
in
I
o
V
L
is the slew rate of the slope compensation and
–
o
×
I L
I
o
V
=
×
d
R
^
BUCK REGULATOR
c
Fm
Fm
= 5V, I
o
+
(
1
------------------------------- -
(
10
2
= 5mΩ, then the compensation resistance
S
1:D
1:D
d
^
e
= 2.5pF (There is approximately 3pF
)
3
+
,
1
S
V
2
o
IN
n
=
)T
d ^
He(S)
He(S)
= 3A, f
s
C
-----------------------------------------
o
^
L i
×
20
T
T
R
i
(S)
COMP
R
c
s
1
L
L
V
×
^
= 500kHz, f
COMP
(
m
10
R
to GND; therefore, C
)
T
6
-Av(S)
-Av(S)
1
Rc
Rc
Co
Co
c
Ro
Ro
(Ω) and Rc
T (S)
T
= 50kHz,
v
(S)
V
^
O
(EQ. 12)
(EQ. 13)
(EQ. 14)
K
ISL85033
ISL85033
2
CURRENT SAMPLING TRANSFER FUNCTION H
In current loop, the current signal is sampled every
switching cycle. Equation 15 shows the transfer function:
Where Q
Power Stage Transfer Functions
Transfer function F
calculated in Equation 16:
Where
Transfer function F
given by Equation 17:
Where
Current loop gain T
The voltage loop gain with open current loop is calculated
in Equation 19:
T
The voltage loop gain with current loop closed is given by
Equation 14:
Where
voltage error amplifier. If T
can be simplified as shown in Equation 21:
From Equation 21, it is shown that the system is a single
order system, which has a single pole located at
the half switching frequency. Therefore, a simple type II
compensator can be easily used to stabilize the system.
F
F
T
H
L
L
v
2
i
1
v
e
v
( )
( )
( )
( )
( )
S
( )
( )
S
S
S
S
S
S
=
=
=
=
=
=
=
V
---------- -
ω
R
S
------ -
ω
KF
ω
V
I ˆ
----
v ˆ o
------
d ˆ
----------------------- -
1
K
o
FB
n
d ˆ
esr
T
2
2
n
z
o
T
+
F
+
m
=
and ω
=
v
=
R
-------------------- -
=
m
T
F
( )
-------------- -
ω
=
o
-------------------- -
R
S
i
V
---------- -
F
-------------- -
R
1
V
( )
R
n
V
S
S
+
o
2
FB
( )A
-------------- - Q
R
Q
o
in
V
T
o
1
S
( )H
+
R
C
c
in
S
n
-------------------------------------- -
1
S
------ -
ω
,
n
C
L
R
o
+
2
2
o
1
--------------------- -
o
are given by
L
v
V
2
1
1
1
1
+
,
.
( )
e
+
i
FB
(S) from control to inductor current is
(S) from control to output voltage is
(S) is expressed as Equation 18:
S
+
+
( )
-------------- -
ω
-------------------------------------- -
S
------ -
ω
----------- -
ω
p
S
------ -
ω
----------- -
ω
o
2
2
o
S
S
esr
S
≈
Q
S
esr
p
+
is the feedback voltage of the
R
1
p
-------------- -
ω
---------------- ω
H
A
o
+
+
o
v
e
S
------
ω
i
1
Q
( )
S
( )
(S)>>1, then Equation 20
C
------ - ω
S
S
L
z
p
o
+
,
,
1
Q
p
n
o
=
≈
=
-------------- -
R
-------------- -
–
o
2
-- -
π
LC
1
1
C
,
=
o
o
ω
n
=
December 8, 2010
πf
ω
s
p
.
(EQ. 15)
(EQ. 16)
(EQ. 17)
(EQ. 18)
(EQ. 19)
(EQ. 20)
(EQ. 21)
before
FN6676.2
e
(S)