12
FN6442.1
September 29, 2008
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, and over-damped for
Q > 1, and under-damped for Q < 1. An under-damped
condition can result in current loop instability.
where D is the percent of on time during a half cycle. Setting
Q = 1 and solving for Se yields:
Since S
n
and S
e
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.
Vn can be solved for in terms of input voltage, current
transducer components, and output inductance yielding:
where R
CS
is the current sense burden resistor, N
CT
is the
current transformer turns ratio, L
O
is the output inductance,
V
O
is the output voltage, and Ns and Np are the secondary
and primary turns, respectively.
The inductor current, when reflected through the isolation
transformer and the current sense transformer to obtain the
current feedback signal at the sense resistor yields:
where V
CS
is the voltage across the current sense resistor
and I
O
is the output current at current limit.
Since the peak current limit threshold is 1.00V, the total
current feedback signal plus the external ramp voltage must
sum to this value.
Substituting Equations 16 and 17 into Equation 18 and
solving for R
CS
yields:
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 to the current feedback signal 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
IN
is the input voltage that corresponds to the duty
cycle D and Lm is the primary magnetizing inductance. The
effect of the magnetizing current at the current sense
resistor, R
CS
, is:
If V
CS
is greater than or equal to Ve, then no additional
slope compensation is needed and R
CS
becomes:
If V
CS
is less than Ve, then Equation 19 is still valid for the
value of R
CS
, but the amount of slope compensation added
by the external ramp must be reduced by V
CS
.
Adding slope compensation is accomplished in the ISL6755
using the CTBUF signal. CTBUF is an amplified
representation of the sawtooth signal that appears on the CT
pin. It is offset from ground by 0.4V and is 2x the peak-to-
peak amplitude of CT (0.4V to 4.4V). A typical application
Fm
1
SnTsw
--------------------
=
(EQ. 10)
Fm
1
Sn Se+Tsw
---------------------------------------
1
m
c
SnTsw
----------------------------
==
(EQ. 11)
m
c
1
Se
Sn
-------
+=
(EQ. 12)
Q
1
m
c
1D–0.5–
-------------------------------------------------
=
(EQ. 13)
S
e
S
n
1
---
0.5+
1
1D–
-------------
1–
=
(EQ. 14)
V
e
V
n
1
---
0.5+
1
1D–
-------------
1–
=
(EQ. 15)
V
e
T
SW
V
O
R
CS
N
CT
L
O
------------------------------------------
N
S
N
P
--------
1
---
D0.5–+
= V
(EQ. 16)
V
CS
N
S
R
CS
N
P
N
CT
------------------------
I
O
DT
SW
2L
O
---------------------
V
IN
N
S
N
P
--------
V
O
–
+
= V
(EQ. 17)
R
CS
N
P
N
CT
N
S
------------------------
1
I
O
V
O
L
O
--------
T
SW
1
---
D
2
----
+
+
------------------------------------------------------
=
(EQ. 19)
I
P
V
IN
DT
SW
L
m
-------------------------------
= A
(EQ. 20)
V
CS
I
P
R
CS
N
CT
--------------------------
= V
(EQ. 21)
R
CS
N
CT
N
S
N
P
--------
I
O
DT
SW
2L
O
-----------------
V
IN
N
S
N
P
--------
V
O
–
+
V
IN
DT
SW
L
m
-------------------------------
+
--------------------------------------------------------------------------------------------------------------------------------------
=
(EQ. 22)
ISL6755
