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:
1
F
ESR
=
---(8)
2p
3
ESR
3
Co
V
OUT
R
6
Fb
Where:
V
IN
= Maximum Input Voltage
V
OSC
= Oscillator Ramp Voltage
Fo = Crossover Frequency
F
ESR
= Zero Frequency of the Output Capacitor
F
LC
= Resonant Frequency of the Output Filter
R
5
and R
6
= Resistor Dividers for Output Voltage
Programming
g
m
= Error Amplifier Transconductance
For:
V
IN
= 5V
V
OSC
= 1.25V
Fo = 30KHz
F
ESR
= 26.52KHz
F
LC
= 2.9KHz
R
5
= 1K
R
6
= 1.65K
g
m
= 600mmho
This results to R
4
=104.4KV. Choose R
4
=105KV
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F
Z
≅
75%F
LC
R
5
V
REF
Gain(dB)
E/A
Comp
V
e
C
9
R
4
H(s) dB
F
Z
Frequency
Figure 6 - Compensation network without local
feedback and its asymptotic gain plot.
The transfer function (Ve / V
OUT
) is given by:
H(s) = g
m
3
F
Z
≅
0.75
3
1
2p
L
O
3
C
O
---(13)
(
R
5
R
6
+ R
5
sR
)
3
1 +sC C
4
9
9
---(9)
For:
Lo = 10mH
Co = 300mF
F
Z
= 2.17KHz
R
4
= 86.6KV
Using equations (11) and (13) to calculate C
9
, we get:
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
R
5
|H(s)| = g
m
3
3
R
4
---(10)
R
6
3R
5
F
Z
=
1
2p3R
4
3C
9
---(11)
C
9
= 698pF
Choose C
9
= 680pF
One more capacitor is sometimes added in parallel with
C
9
and R
4
. This introduces one more pole which is mainly
used to supress the switching noise. The additional pole
is given by:
1
F
P
=
C
9
3
C
POLE
2p
3
R
4
3
C
9
+ C
POLE
The pole sets to one half of switching frequency which
results in the capacitor C
POLE:
C
POLE
=
1
p3R
4
3f
S
-
1
C
9
≅
1
p3R
4
3f
S
The gain is determined by the voltage divider and E/A's
transconductance gain.
First select the desired zero-crossover frequency (Fo):
Fo > F
ESR
and F
O
[
(1/5 ~ 1/10)3 f
S
Use the following equation to calculate R
4
:
R
4
=
1
V
OSC
Fo3F
ESR
R
5
+ R
6
3
3
3
g
m
V
IN
F
LC2
R
5
---(12)
for F
P
<<
f
S
2
8/18
A-POWER [ ADVANCED POWER ELECTRONICS CORP. ]
