ISL6251, ISL6251A
Current loop gain Ti(S) is expressed as the following
equation:
Ti(S) = 0.25 RTF2(S)M
where RT is the trans-resistance in current loop. RT is
usually equal to the product of the charging current sensing
resistance and the gain of the current sense amplifier, CA2.
For ISL6251, ISL6251A, RT = 20R1.
The voltage gain with open current loop is:
Tv(S) = KM F1(S)AV(S)
Where
K
= VFB
Vo
, VFB is the
feedback voltage of the voltage
error amplifier. The Voltage loop gain with current loop
closed is given by:
Lv
(S
)
=
Tv (S)
1 + Ti (S)
If Ti(S)>>1, then it can be simplified as follows:
LV(S)
=
-4---V-----F---B--
VO
(---R-----O-----+------R----L----)--
RT
--1-----+------ω---------Se------s-------r
1 + ω--S---P--
AV
(S),
ωP
≈
--------1---------
ROCO
From the above equation, it is shown that the system is a
single order system, which has a single pole located at ωp
before the half switching frequency. Therefore, simple type II
compensator can be easily used to stabilize the system.
iˆin
iˆL L
vˆ o
vˆ in
+
ILdˆ 1:D Vindˆ
RT
Rc
Ro
VCA2
Co
Ti(S)
dˆ
K
11/Vin
0.25VCA2
+
-
Tv(S)
vˆcomp -Av(S)
FIGURE 14. SMALL SIGNAL MODEL OF SYNCHRONOUS
BUCK REGULATOR
Figure 15 shows the voltage loop compensator, and its
transfer function is expressed as follows:
Av
(S)
=
vˆ comp
vˆ FB
1+ S
= gm
ω cz
SC1
where
ω cz
=
1
R1C1
Vo
VFB -
VREF
gm
+
VCOMP
R1
C1
FIGURE 15. VOLTAGE LOOP COMPENSATOR
Compensator design goal:
• High DC gain
•
Loop bandwidth fc:
⎜⎛ 1
⎝5
−
1
20
⎟⎞
⎠
fs
• Gain margin: >10dB
• Phase margin: 40°
The compensator design procedure is as follows:
1. Put compensator zero at:
ωcz
=
(1 − 3) 1
RoCo
2. Put one compensator pole at zero frequency to achieve
high DC gain, and put another compensator pole at either
ESR zero frequency or half switching frequency,
whichever is lower.
The loop gain Tv(S) at cross over frequency of fc has unity
gain. Therefore, the compensator resistance R1 is
determined by:
R1
=
8----π----f--C----V----O-----C-----O----R-----T-
gmVFB
where gm is the trans-conductance of the voltage loop error
amplifier. Compensator capacitor C1 is then given by:
C1
=
1
R1 ωcz
Example: Vin = 19V, Vo = 16.8V, Io = 2.6A, fs = 300kHz,
Co = 10μF/10mΩ, L = 10μH, gm = 250μs, RT = 0.8Ω,
VFB = 2.1V, fc = 20kHz, then compensator resistance
R1 = 10kΩ. Choose R1 = 10kΩ. Put the compensator zero at
1.5kHz. The compensator capacitor is C1 = 6.5nF.
Therefore, choose voltage loop compensator: R1 = 10k,
C1 = 6.5nF.
17
FN9202.2
May 10, 2006