iru3138 International Rectifier Corp., iru3138 Datasheet - Page 10

• Manufacturer Part Number: iru3138
• Description: Synchronous Pwm Controller For Termination Power Supply Applications -
• Manufacturer: International Rectifier Corp.

IRU3138
The IRU3138’s error amplifier is a differential-input
transconductance amplifier. The output is available for
DC gain control or AC phase compensation.
The E/A can be compensated with or without the use of
local feedback. When operated without local feedback,
the transconductance properties of the E/A become evi-
dent and can be used to cancel one of the output filter
poles. This will be accomplished with a series RC circuit
from Comp pin to ground as shown in Figure 9.
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:
|H(s)| is the gain at zero cross frequency.
10
F
H(s) = g
|H(s=j32p3F
F
Figure 9 - Compensation network without local
ESR
Z
=
V
feedback and its asymptotic gain plot.
OUT
=
H(s) dB
2p3R
R
R
(
2p3ESR3Co
6
5
Vp=V
m
1
3
Fb
Gain(dB)
4
3C
R
REF
1
O
6
)| = g
R
9
+ R
5
E/A
5
F
m
)
Z
3
3
OUT
Frequency
R
1 + sR
Comp
6
) is given by:
R
3R
---(17)
sC
5
R
C
4
5
9
9
4
3R
V e
C
---(14)
Optional
9
4
---(15)
---(16)
www.irf.com
First select the desired zero-crossover frequency (Fo):
Use the following equation to calculate R
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
9
and R
For F
R
Choose C
Fo > F
R
Where:
V
V
Fo = Crossover Frequency
F
F
R
g
This results to R
Choose R
F
F
For:
Lo = 1.1mH
Co = 990mF
F
C
For:
V
V
Fo = 40KHz
F
C
4
m
=17.8K and F
ESR
LC
Z
Z
P
4
IN
OSC
5
ESR
POLE
IN
OSC
9
=
and R
= Error Amplifier Transconductance
=
= Maximum Input Voltage
= Resonant Frequency of the Output Filter
= 5V
P
4
= Zero Frequency of the Output Capacitor
75%F
0.753
V
= 12KHz
= Oscillator Ramp Voltage
2.4nF; Choose C
. This introduces one more pole which is mainly
= 1.25V
2p3R
=
V
<< f
OSC
ESR
IN
p3R
6
POLE
4
S
and F
= Resistor Dividers for Output Voltage
3
=17.8K
LC
/2
2p
4
=47pF.
Programming
Fo3F
3
4
1
3f
S
F
C
1
C
=400KHz will result to C
O
4
LC
9
=17.32K
9
S
L
[ (1/5 ~ 1/10)3f
3C
1
+ C
2
O
-
ESR
3 C
C
POLE:
1
POLE
POLE
3
9
9
=2.2nF
O
R
F
R
R
g
F
R
5
m
R
LC
Z
5
6
4
p3R
+ R
5
= 1K
= 1K
= 600mmho
= 3.6KHz
= 17.8K
= 4.82KHz
6
3
1
4
---(19)
3f
S
g
1
S
m
4
:
9
POLE
, we get:
=44pF.
01/29/04
---(18)
Rev. 1.0