iru3039 International Rectifier Corp., iru3039 Datasheet - Page 11

• Manufacturer Part Number: iru3039
• Description: Ic Ctrlr Pwm Sync Switch 20-mlpq
• Manufacturer: International Rectifier Corp.

The IRU3039’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 12.
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.
Figure 12 - Compensation network without local
F
H(s) = g
|H(s=j×2π×F
F
ESR
Z
=
feedback and its asymptotic gain plot.
=
2π×R
V
(
2π×ESR×Co
OUT
H(s) dB
R
R
m
1
×
6
5
Vp=V
4
×C
R
1
Gain(dB)
Fb
O
6
R
)| = g
9
+ R
REF
5
5
)
m
E/A
×
×
F
OUT
Z
R
1 + sR
6
---(17)
) is given by:
R
×R
Frequency
5
sC
Comp
5
9
R
4
×R
---(14)
C
C
4
9
9
Ve
4
---(15)
---(16)
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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:
V
V
Fo = 20KHz
F
Fo > F
R
F
F
For:
Lo = 4.7µH
Co = 660µF
F
C
Where:
V
V
Fo = Crossover Frequency
F
F
R
g
This results to R
Choose R
C
for F
ESR
IN
OSC
m
Z
Z
P
ESR
LC
4
IN
OSC
POLE
5
9
= 18V
=
and R
= Error Amplifier Transconductance
=
= Maximum Input Voltage
= Resonant Frequency of the Output Filter
= 12KHz
= 3.3V
75%F
0.75×
4
5300pF; Choose C
= Zero Frequency of the Output Capacitor
V
= Oscillator Ramp Voltage
2π×R
P
. This introduces one more pole which is mainly
V
=
ESR
OSC
<<
IN
π×R
6
×
and F
4
= Resistor Dividers for Output Voltage
=14K
LC
f
2
S
4
2π
Programming
×
Fo×F
4
1
×f
C
1
F
C
O
4
LC
9
9
=12.08K
L
S
≤ (1/5 ~ 1/10)×f
×C
1
+ C
2
O
-
ESR
× C
POLE:
C
POLE
1
POLE
×
9
9
IRU3039(PbF)
=5600pF
O
R
F
R
R
g
F
R
m
LC
5
5
6
Z
R
4
π×R
+ R
= 3.16K
= 700µmho
= 1K
5
= 2.1KHz
= 14K
= 2.8KHz
6
1
×
4
---(19)
×f
S
g
1
S
4
m
:
9
, we get:
---(18)
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