IRU3073CQ IRF [International Rectifier], IRU3073CQ Datasheet - Page 11

• Manufacturer Part Number: IRU3073CQ
• Description: SYNCHRONOUS PWM CONTROLLER WITH OVER-CURRENT PROTECTION / LDO CONTROLLER
• Manufacturer: IRF [International Rectifier]

The transfer function (Ve / V
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
|H(s)| is the gain at zero cross frequency.
First select the desired zero-crossover frequency (Fo):
Use the following equation to calculate R
Rev. 1.0
09/17/03
Fo > F
Figure 11 - Compensation network without local
H(s) = g
|H(s=j32p3F
F
R
Where:
V
V
Fo = Crossover Frequency
F
F
R
g
m
Z
ESR
LC
4
IN
OSC
5
=
=
and R
= Error Amplifier Transconductance
= Maximum Input Voltage
= Resonant Frequency of the Output Filter
feedback and its asymptotic gain plot.
V
= Zero Frequency of the Output Capacitor
V
2p3R
= Oscillator Ramp Voltage
OUT
V
ESR
H(s) dB
(
OSC
IN
R
R
6
6
5
m
and F
Vp=V
= Resistor Dividers for Output Voltage
1
3
3
4
Gain(dB)
Fb
3C
Programming
Fo3F
R
O
6
REF
F
O
)| = g
R
+ R
9
LC
[ (1/5 ~ 1/10)3f
5
2
ESR
E/A
5
m
)
F
3
3
3
Z
OUT
R
R
1 + sR
Frequency
6
5
) is given by:
R
3R
Comp
---(17)
R
+ R
sC
5
5
R
5
C
6
9
4
3
4
3R
9
C
Ve
S
9
C
g
1
4
POLE
4
m
:
---(15)
---(16)
---(18)
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To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
Using equations (17) and (19) to calculate C
One more capacitor is sometimes added in parallel with
C
used to suppress the switching noise. The additional
pole is given by:
The pole sets to one half of switching frequency which
results in the capacitor C
For a general solution for unconditionally stability for
ceramic capacitor with very low ESR and any type of
output capacitors, in a wide range of ESR values we
should implement local feedback with a compensation
network. The typically used compensation network for
voltage-mode controller is shown in Figure 12.
9
and R
For:
V
V
Fo = 20KHz
F
This results to R
Choose R
F
F
For:
Lo = 3.3mH
Co = 660mF
C
F
C
for F
ESR
Z
Z
P
IN
OSC
9
POLE
=
= 5V
4
= 12KHz
75%F
0.753
2590pF; Choose C
. This introduces one more pole which is mainly
= 1.25V
2p3R
P
=
<<
p3R
4
=24K
LC
f
2
S
2p
4
3
4
1
3f
C
1
C
4
9
9
S
=23.14K
L
3C
1
+ C
O
-
3 C
C
POLE:
1
POLE
POLE
9
9
=2200pF
O
F
R
R
g
F
R
m
LC
Z
5
6
4
p3R
= 1K
= 2.15K
= 700mmho
= 2.5KHz
= 24K
= 3.41KHz
1
4
---(19)
3f
IRU3073
S
9
, we get:
11