ISL78840ASRH, ISL78841ASRH, ISL78843ASRH, ISL78845ASRH
10
FN6991.1
August 19, 2010
where D is the percent of on time during a switching
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
(ΔI) during the on time and Ve is the voltage that must
be added by the external ramp.
For a flyback converter, Vn can be solved for in terms of
input voltage, current transducer components, and
primary inductance, yielding Equation
12:where RCS is the current sense resistor, fsw is the
switching frequency, Lp is the primary inductance, VIN is
the minimum input voltage, and D is the maximum duty
cycle.
The current sense signal at the end of the ON time for
CCM operation is Equation
13:where VCS is the voltage across the current sense
resistor, Ls is the secondary winding inductance, and IO is
the output current at current limit. Equation
13 assumes
the voltage drop across the output rectifier is negligible.
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 as shown in Equation
14.Substituting Equations
12 and
13 into Equation
14 and
solving for RCS yields Equation 15: Adding slope compensation is accomplished in the
ISL7884xASRH using an external buffer transistor and
the RTCT signal. A typical application sums the buffered
RTCT signal with the current sense feedback and applies
the result to the CS pin as shown in Figure
6.Assuming the designer has selected values for the RC
filter (R6 and C4) placed on the CS pin, the value of R9
required to add the appropriate external ramp can be
found by superposition.
The factor of 2.05 in Equation
16 arises from the peak
amplitude of the sawtooth waveform on RTCT minus a
base-emitter junction drop. That voltage multiplied by
the maximum duty cycle is the voltage source for the
slope compensation. Rearranging to solve for R9 yields
The value of RCS determined in Equation 15 must be rescaled so that the current sense signal presented at the
CS pin is that predicted by Equation
13. The divider
created by R6 and R9 makes this necessary.
Example:
VIN = 12V
VO = 48V
Ls = 800H
Ns/Np = 10
Lp = 8.0H
IO = 200mA
Switching Frequency, fsw = 200kHz
Duty Cycle, D = 28.6%
R6 = 499Ω
Solve for the current sense resistor, RCS, using
S
e
S
n
1
π
---
0.5
+
1
1D
–
-------------
1
–
=
(EQ. 10)
V
e
V
n
1
π
---
0.5
+
1
1D
–
-------------
1
–
=
(EQ. 11)
V
e
DT
SW VIN RCS
L
p
----------------------------------------------------
1
π
---
0.5
+
1
1D
–
-------------
1
–
=
V
(EQ. 12)
V
CS
N
S RCS
N
P
------------------------ I
O
1D
–
() V
O f
sw
2L
s
--------------------------------------------
+
=
V
(EQ. 13)
V
e
V
CS
+
1
=
(EQ. 14)
R
CS
1
Df
sw VIN
L
p
-------------------------------
1
π
---
0.5
+
1D
–
------------------ 1
–
N
s
N
p
-------
I
O
1D
–
() V
O fsw
2L
s
--------------------------------------------
+
+
-----------------------------------------------------------------------------------------------------------------------------------------------------
=
(EQ. 15)
CS
RTCT
R6
C4
R9
ISL788
43A
SRH
VREF
FIGURE 6. SLOPE COMPENSATION
V
e
2.05D R
6
R
6
R
9
+
----------------------------
=
V
(EQ. 16)
R
9
2.05D
V
e
–
() R
6
V
e
----------------------------------------------
=
Ω
(EQ. 17)
R
′
CS
R
6
R
9
+
R
9
--------------------- R
CS
=
(EQ. 18)
ISL78840ASRH, ISL78841ASRH, ISL78843ASRH, ISL78845ASRH