APU3037AM A-POWER [Advanced Power Electronics Corp.], APU3037AM Datasheet - Page 8

• Manufacturer Part Number: APU3037AM
• Description: 8-PIN SYNCHRONOUS PWM CONTROLLER
• Manufacturer: A-POWER [Advanced Power Electronics Corp.]

APU3037 / APU3037A
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
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
2p3R
V
(
OSC
ESR
2p 3 ESR 3 Co
V
IN
OUT
H(s) dB
m
m
3
3
and F
R
R
1
3
6
4
5
Fo3F
3C
R
V
R
REF
6
F
1
Fb
Gain(dB)
R
3R
6
O
LC
9
R
5
+ R
[ (1/5 ~ 1/10)3 f
2
5
ESR
5
3 R
5
)
3
E/A
OUT
F
3
R
4
Z
---(11)
5
) is given by:
R
+ R
1 + sR
Frequency
5
Comp
sC
6
---(8)
3
R
9
---(10)
C
4
4
C
S
g
9
1
9
4
m
V e
:
---(12)
---(9)
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
Where:
V
V
Fo = Crossover Frequency
F
F
R
g
For:
V
V
Fo = 30KHz
F
F
R
R
g
F
F
For:
Lo = 10mH
Co = 300mF
F
R
C
Choose C
F
C
for F
m
m
ESR
LC
ESR
LC
Z
Z
Z
P
IN
OSC
5
IN
OSC
5
6
4
9
POLE
and R
= Error Amplifier Transconductance
= 1K
= 1.65K
= 600mmho
= 2.17KHz
= 86.6KV
= 698pF
=
= Maximum Input Voltage
= 5V
= Resonant Frequency of the Output Filter
= 2.9KHz
4
= Zero Frequency of the Output Capacitor
= 26.52KHz
75%F
0.75 3
. This introduces one more pole which is mainly
= Oscillator Ramp Voltage
= 1.25V
2p 3 R
P
=
<<
p3R
6
9
= Resistor Dividers for Output Voltage
LC
f
= 680pF
2
S
4
Programming
=104.4KV. Choose R
2p
4
4
3f
3
1
1
S
C
C
L
-
9
1
9
O
3 C
C
+ C
POLE:
1
3 C
9
POLE
POLE
O
p3R
1
4
3f
4
=105KV
---(13)
S
9
, we get:
8/18