Datasheet

=
1
(R
Load
C
OUT
)2S
F
RCP-
10 kHz5 kHz
o
=
=
1
( )
R
2
xC
f
2S
F
CFZERO-
REF
V
¨
¨
©
§
OUT
V
=
2
R
1
¸
¸
¹
·
- x
1
R
10
100 1k 10k 100k 1M
FREQUENCY
-80
-60
-40
-20
0
20
40
60
80
dB
-180
-90
0
90
180
RHP-Zero
Ext (Cf)
gm-Pole
RC-Pole
Vi = 5V
Vo = 12V
Io = 500 mA
Co = 10 mF
Lo = 5 mH
Ext (Cf)-Pole
gm-zero
-Zero
D = 0.625
Cf = 220 pF
Fz-cf = 8 kHz
RHP-Zero = 107 kHz
Fp-rc = 660 Hz
Fp-cf = 77 kHz
LM2735
SNVS485F –JUNE 2007–REVISED APRIL 2013 www.ti.com
Figure 24. LM2735 With External Compensation
The simplest method to determine the compensation component value is as follows.
Set the output voltage with the following equation.
where
• R1 is the bottom resistor and R2 is the resistor tied to the output voltage. (10)
The next step is to calculate the value of C3. The internal compensation has been designed so that when a zero
is added between 5 kHz & 10 kHz the converter will have good transient response with plenty of phase margin
for all input & output voltage combinations.
(11)
Lower output voltages will have the zero set closer to 10 kHz, and higher output voltages will usually have the
zero set closer to 5 kHz. It is always recommended to obtain a Gain/Phase plot for your actual application. One
could refer to the Typical applications section to obtain examples of working applications and the associated
component values.
Pole @ origin due to internal gm amplifier:
F
P-ORIGIN
(12)
Pole due to output load and capacitor:
(13)
This equation only determines the frequency of the pole for perfect current mode control (CMC). I.e, it doesn’t
take into account the additional internal artificial ramp that is added to the current signal for stability reasons. By
adding artificial ramp, you begin to move away from CMC to voltage mode control (VMC). The artifact is that the
pole due to the output load and output capacitor will actually be slightly higher in frequency than calculated. In
this example it is calculated at 650 Hz, but in reality it is around 1 kHz.
The zero created with capacitor C3 & resistor R2:
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