ISL6742 Intersil Corporation, ISL6742 Datasheet - Page 14

• Manufacturer Part Number: ISL6742
• Description: Advanced Double-Ended PWM Controller
• Manufacturer: Intersil Corporation

Slope Compensation
Peak current-mode control requires slope compensation to
improve noise immunity, particularly at lighter loads, and to
prevent current loop instability, particularly for duty cycles
greater than 50%. Slope compensation may be
accomplished by summing an external ramp with the current
feedback signal or by subtracting the external ramp from the
voltage feedback error signal. Adding the external ramp to
the current feedback signal is the more popular method.
From the small signal current-mode model [1] it can be
shown that the naturally-sampled modulator gain, Fm,
without slope compensation, is
where Sn is the slope of the sawtooth signal and Tsw is the
duration of the half-cycle. When an external ramp is added,
the modulator gain becomes
where Se is slope of the external ramp and
The criteria for determining the correct amount of external
ramp can be determined by appropriately setting the
damping factor of the double-pole located at half the
oscillator frequency. The double-pole will be critically
damped if the Q-factor is set to 1, over-damped for Q > 1,
and under-damped for Q < 1. An under-damped condition
may result in current loop instability.
where D is the percent of on time during a half cycle (half-
period duty cycle). Setting Q = 1 and solving for Se yields
Since Sn and Se are the on time slopes of the current ramp
and the external ramp, respectively, they can be multiplied
by Ton to obtain the voltage change that occurs during Ton.
where Vn is the change in the current feedback signal during
the on time and Ve is the voltage that must be added by the
external ramp.
Fm
Fm
m
Q
S
V
e
e
c
=
=
=
=
=
=
-------------------------------------------------
π m
S
V
1
(
------------------- -
SnTsw
-------------------------------------- -
(
n
n
Sn
+




c
1
Se
-------
Sn




(
1
-- -
π
1
-- -
π
+
1 D
+
+
Se
–
1
1
0.5
0.5
) Tsw
) 0.5




–
------------ - 1
1 D
------------ - 1
1 D
1
–
1
–
=
)
–
–
--------------------------- -
m
c




SnTsw
1
14
(EQ. 13)
(EQ. 15)
(EQ. 10)
(EQ. 12)
(EQ. 14)
(EQ. 11)
ISL6742
Vn can be solved for in terms of input voltage, current
transducer components, and output inductance yielding
where R
current transformer turns ratio, L
V
and primary turns, respectively.
The current sense signal, which represents the inductor
current after it has been reflected through the isolation and
current sense transformers, and passed through the current
sense burden resistor, is:
where V
and I
Since the peak current limit threshold is 1.00V, the total
current feedback signal plus the external ramp voltage must
sum to this value when the output load is at the current limit
threshold.
Substituting Equations 16 and 17 into Equation 18 and
solving for R
For simplicity, idealized components have been used for this
discussion, but the effect of magnetizing inductance must be
considered when determining the amount of external ramp
to add. Magnetizing inductance provides a degree of slope
compensation and reduces the amount of external ramp
required. The magnetizing inductance adds primary current
in excess of what is reflected from the inductor current in the
secondary.
where V
cycle D and Lm is the primary magnetizing inductance. The
effect of the magnetizing current at the current sense
resistor, R
V
V
V
R
∆
∆ V
O
I
e
CS
e
CS
P
CS
+
=
is the output voltage, and Ns and Np are the secondary
=
V
O
=
=
T
----------------------------------------- -
CS
V
-------------------------------
=
SW
is the output current at current limit.
N
------------------------ I
N
N
----------------------- -
IN
CS
IN
CS
N
∆ I
------------------------- -
S
P
P
=
CS
CT
N
⋅
L
⋅
⋅
⋅
⋅
P
is the input voltage that corresponds to the duty
DT
N
V
R
N
is the voltage across the current sense resistor
N
m
is the current sense burden resistor, N
S
1
CS
⋅
CT
, is
O
CS
CT
⋅
CT
R
SW
L
⋅
CS
O
R



yields
⋅
----------------------------------------------------- -
I
CS
O
O
+
+
⋅
D T
-------------------- - V
V
------- - T
N
------- -
N
L
A
2L
V
O
⋅
O
S
P


SW
O
1
-- -
π
SW
1
+



D 0.5


1
-- -
π
–
IN
+
O
⋅
D
--- -
2
N
------- -
N
is the output inductance,
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



S
P
–
V
O
Ω
V






V
CT
July 25, 2005
(EQ. 16)
(EQ. 17)
(EQ. 21)
(EQ. 19)
(EQ. 20)
(EQ. 18)
is the
FN9183.1