iru3037acftr International Rectifier Corp., iru3037acftr Datasheet - Page 8

• Manufacturer Part Number: iru3037acftr
• Description: 8 Pwm 200 Khz Sync Contr In A 8-pin Tssop Package
• Manufacturer: International Rectifier Corp.

IRU3037/IRU3037A & (PbF)
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)| = g
F
Fo > F
R
Figure 6 - Compensation network without local
ESR
Z
4
=
=
feedback and its asymptotic gain plot.
=
V
2π×R
V
(
ESR
OSC
2π × ESR × Co
V
IN
OUT
H(s) dB
m
m
×
and F
R
R
×
1
×
6
4
5
Fo×F
×C
R
V
R
REF
Gain(dB)
6
F
1
Fb
R
×R
O
6
9
LC
R
5
+ R
≤ (1/5 ~ 1/10)× f
2
5
ESR
5
× R
5
)
×
E/A
OUT
×
F
R
4
Z
---(11)
5
) is given by:
R
+ R
1 + sR
5
Frequency
Comp
sC
6
---(8)
×
R
---(10)
9
C
4
4
C
9
S
g
1
Ve
9
4
m
:
---(12)
---(9)
www.irf.com
This results to R
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 supress 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:
V
V
Fo = 30KHz
F
F
R
R
g
F
F
For:
Lo = 10µH
Co = 300µF
F
R
C
Choose C
F
C
Where:
V
V
Fo = Crossover Frequency
F
F
R
g
for F
m
ESR
LC
m
Z
Z
Z
P
IN
OSC
ESR
LC
IN
OSC
5
5
6
4
9
POLE
= 2.17KHz
and R
= Error Amplifier Transconductance
= 86.6KΩ
= 698pF
= 1K
= 1.65K
= 600µmho
=
= Maximum Input Voltage
= 5V
= Resonant Frequency of the Output Filter
= 2.9KHz
4
75%F
0.75 ×
= Zero Frequency of the Output Capacitor
= 26.52KHz
= Oscillator Ramp Voltage
= 1.25V
. This introduces one more pole which is mainly
2π × R
P
=
<<
π×R
6
9
= Resistor Dividers for Output Voltage
LC
f
2
= 680pF
S
4
Programming
=104.4KΩ. Choose R
2π
4
4
×f
×
1
1
S
C
C
-
L
9
1
9
O
× C
POLE:
+ C
C
1
× C
9
POLE
POLE
O
π×R
1
4
×f
4
---(13)
=105KΩ
S
9
, we get: