iru3038 International Rectifier Corp., iru3038 Datasheet - Page 8

• Manufacturer Part Number: iru3038
• Description: Synchronous Pwm Controller For Termination Power Supply Applications
• Manufacturer: International Rectifier Corp.

IRU3038
Note that this method requires that the output capacitor
should have enough ESR to satisfy stability requirements.
In general, the output capacitor’s ESR generates a zero
typically at 5KHz to 50KHz which is essential for an
acceptable phase margin.
The ESR zero of the output capacitor expressed as fol-
lows:
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:
The gain is determined by the voltage divider and E/A's
transconductance gain.
First select the desired zero-crossover frequency (Fo):
Use the following equation to calculate R
8
F
H(s) = g
|H(s)| =
F
Fo > F
R
Figure 6 - Compensation network without local
ESR
Z
4
=
=
feedback and its asymptotic gain plot.
=
V
2p 3 R
V
V
(
OSC
ESR
2p3ESR3Co
OUT
IN
H(s) dB
g
R
R
m
m
and F
3
3
6
5
Vp=V
3
1
Fo3F
R
4
1
Fb
Gain(dB)
3 C
6
F
R
R
O
+ R
REF
LC
6
5
[ (1/5 ~ 1/10)3 f
R
2
3 R
ESR
9
5
5
)
E/A
3
3
F
5
OUT
Z
3 R
R
1 + sR
5
) is given by:
---(11)
R
Frequency
+ R
sC
Comp
5
4
6
9
R
3
4
C
C
---(8)
4
9
9
g
S
V e
1
---(10)
m
4
:
---(9)
---(12)
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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 (11) and (13) 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
9
and R
F
Where:
V
V
Fo = Crossover Frequency
F
F
R
g
For:
V
V
Fo = 30KHz
F
F
R
R
g
This results to R
F
F
For:
Lo = 10mH
Co = 300mF
F
R
C
C
for F
P
ESR
LC
ESR
LC
Z
Z
Z
IN
OSC
5
m = Error Amplifier Transconductance
IN
OSC
5
6
m = 600mmho
4
9
POLE
=
and R
= 1K
= 1K
= 3.8KHz
= 26.1KV
= Maximum Input Voltage
= 5V
= Resonant Frequency of the Output Filter
= 5KHz
4
= Zero Frequency of the Output Capacitor
= 26.5KHz
75%F
0.75 3
1800pF
. This introduces one more pole which is mainly
= Oscillator Ramp Voltage
= 1.25V
2p 3 R
P
=
<<
p 3 R
6
= Resistor Dividers for Output Voltage
LC
f
2
S
Programming
2p
4
3
4
3 f
4
1
1
=26.52KV. Choose R
C
C
L
9
1
9
S
O
3 C
+ C
POLE:
- 1
3 C
C
POLE
POLE
9
O
p 3 R
---(13)
1
4
4
3 f
=26.1KV
9
, we get:
S
09/12/02
Rev. 2.0