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ISL78206AVEZ Datasheet(PDF) 16 Page - Intersil Corporation

Part No. ISL78206AVEZ
Description  40V 2.5A Buck Controller with Integrated High-side 40V 2.5A Buck Controller with Integrated High-side
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Maker  INTERSIL [Intersil Corporation]
Homepage  http://www.intersil.com/cda/home
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ISL78206AVEZ Datasheet(HTML) 16 Page - Intersil Corporation

 
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ISL78206
16
FN8618.2
March 25, 2015
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Power Stage Transfer Functions
Transfer function F1(S) from control to output voltage is:
Where
Transfer function F2(S) from control to inductor current is given
by Equation 13:
Where
.
Current loop gain Ti(S) is expressed as Equation 14:
The voltage loop gain with open current loop is Equation 15:
The Voltage loop gain with current loop closed is given by
Equation 16:
If Ti(S)>>1, then Equation 16 can be simplified as Equation 17:
Equation 17 shows that the system is a single order system.
Therefore, a simple type II compensator can be easily used to
stabilize the system. While type III compensator is needed to
expand the bandwidth for current mode control in some cases.
A compensator with 2 zeros and 1 pole is recommended for this
part as shown in Figure 23. Its transfer function is expressed as
Equation 18:
Where
,
Compensator design goal:
Loop bandwidth fc:
Gain margin: >10dB
Phase margin: 45°
The compensator design procedure is as follows:
1. Position
CZ2 and CP to derive R
3 and C3.
Put the compensator zero
CZ2 at (1 to 3)/(R
oCo)
Put the compensator pole
CP at ESR zero or 0.35 to 0.5 times
of switching frequency, whichever is lower. In all-ceramic-cap
design, the ESR zero is normally higher than half of the switching
frequency. R3 and C3 can be derived as following:
Case A: ESR zero
less than (0.35 to 0.5) fs
Case B: ESR zero
larger than (0.35 to 0.5) fs
2. Derive R2 and C1.
The loop gain Lv(S) at cross over frequency of fc has unity gain.
Therefore, C1 is determined by Equation 24.
The compensator zero
CZ1 can boost the phase margin and
bandwidth. To put
CZ1 at 2 times of cross cover frequency f
c is a
good start point. It can be adjusted according to specific design.
R1 can be derived from Equation 25.
Example: VIN = 12V, VO = 5V, IO = 2A, fs = 500kHz,
Co = 60µF/3mΩ, L = 10µH, Rt = 0.20V/A, fc = 50kHz,
R1 = 105k, RBIAS =20kΩ.
F1 S

vˆo
------
Vin
1
S
esr
------------
+
S
2
o
2
-------
S
oQp
--------------- 1
++
---------------------------------------
==
(EQ. 12)
esr
1
RcCo
--------------- Qp Ro
Co
LP
-------
o
1
LPCo
-------------------
=
,
,
=
F2 S

o
----
Vin
Ro RLP
+
-------------------------
1
S
z
------
+
S
2
o
2
-------
S
oQp
--------------- 1
++
---------------------------------------
==
(EQ. 13)
z
1
RoCo
---------------
=
Ti S

RtFmF2 S
H
e S

=
(EQ. 14)
Tv S

KFmF1 S
A
v S

=
(EQ. 15)
Lv S

Tv S

1Ti S

+
------------------------
=
(EQ. 16)
Lv S

Ro RLP
+
Rt
-------------------------
1
S
esr
------------
+
1
S
p
-------
+
----------------------
Av S

He S

----------------
p
1
RoCo
---------------
,
=
(EQ. 17)
FIGURE 23. TYPE III COMPENSATOR
R2
C1
R1
RBIAS
R3
C3
VO
VCOMP
VREF
Av S

comp
O
-----------------
1
SR1C1
-------------------
1
S
cz1
-------------
+

 1
S
cz2
-------------
+


1
S
cp
----------
+


---------------------------------------------------------
=
=
(EQ. 18)
cz1
1
R2C1
---------------
cz2
1
R1 R3
+
C
3
----------------------------------
=
cp
1
R3C3
---------------
=
,
=
1
4
---to 1
10
------

 f
s
cz2
3
RoCo
---------------
=
(EQ. 19)
1
2
R
cCo
----------------------
C3
RoCo 3RcCo
3R1
----------------------------------------
=
(EQ. 20)
R3
3RcR1
Ro 3Rc
------------------------
=
(EQ. 21)
1
2
R
cCo
----------------------
C3
0.33RoCofs 0.46
fsR1
--------------------------------------------------
=
(EQ. 22)
R3
R1
0.73RoCofs 1
-----------------------------------------
=
(EQ. 23)
C1
R1 R3
+
C
3
2
f
cRtR1Co
----------------------------------
=
(EQ. 24)
R2
1
4
f
cC1
--------------------
=
(EQ. 25)


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