APU3048 Advanced Power Electronics Corp., APU3048 Datasheet - Page 9

• Manufacturer Part Number: APU3048
• Description: The APU3048 IC combines a Dual synchronous Buck controller and a linear regulator controller, providing a cost-effective, high performance and flexible solution for multi-output applications
• Manufacturer: Advanced Power Electronics Corp.

Specifications of APU3048

• Vin(min): 1.25
• Vin(max): 25
• Vout(min): 1.25
• Vout(max): 5
• Iout(max): 10
• Frequency: 200KHz
• Scp: ?
• Otp: ?
• Package: 16-pin TSSOP, SOP-16

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
Using equations (11),(12) and (13) for Ch2, where:
We get:
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
For a general solution for unconditionally stability for any
type of output capacitors, in a wide range of ESR values
we should implement local feedback with a compensa-
tion network. The typically used compensation network
for voltage-mode controller is shown in Figure 7.
9
and R
For:
V
V
F
F
This results to R
F
F
For:
L
Co = 300mF
Fz = 2.1KHz
R
C
V
V
F
F
R
C
F
C
For F
3
O1
ESR1
Z
Z
O2
ESR2
P
IN1
OSC
9
9
IN2
OSC
11
19
POLE
= 10.2mH
= 1630pF; Choose C
= 46.4KV
=
= 38.9KV; Choose R
= 1554pF; Choose C
= 30KHz
= 30KHz
= 12V
= 5V
4
75%F
0.75 3
. This introduces one more pole which is mainly
= 1.25V
= 1.25V
2p 3 R
P
= 26.5KHz
= 26.5KHz
=
<<
p 3 R
LC1
f
2
S
9
2p
3
9
3 f
9
1
=46.4KV; Choose R
C
1
C
18
L
18
1
S
3
3 C
POLE:
+ C
- 1
3 C
C
9
11
19
F
R
R
g
18
= 1800pF
POLE
F
R
R
g
POLE
O
LC2
15
14
m = 600mhmo
LC1
= 39.2KV
8
6
m = 600mmho
= 1800pF
= 1K
= 1.64K
= 1K
= 442V
= 3.5KHz
= 2.8KHz
p 3 R
---(13)
1
9
9
=46.4KV
3 f
9
, we get:
S
In such configuration, the transfer function is given by:
The error amplifier gain is independent of the transcon-
ductance under the following condition:
By replacing Z
former function can be expressed as:
As known, transconductance amplifier has high imped-
ance (current source) output, therefore, consider should
be taken when loading the E/A output. It may exceed its
source/sink output current capability, so that the ampli-
fier will not be able to swing its output voltage over the
necessary range.
The compensation network has three poles and two ze-
ros and they are expressed as follows:
H(s)=
H(s) dB
g
F
F
F
F
V
V
P1
P3
Z1
Z2
m
Figure 7 - Compensation network with local
OUT
Z
e
sR
Z
= 0
=
=
=
IN
feedback and its asymptotic gain plot.
f
Gain(dB)
>> 1
6
=
2p3C
2p3R
R
(C
2p3R
C
8
1 +
1
12
10
1 -
+C
F
IN
V
Z
1
g
g
1
10
OUT
and Z
7
11
7
3C
and
m
m
3
3(R
)
R
1
R
3
Z
Z
6
(
5
1
f
IN
11
F
[
C
f
C
6
F
1+sR
Z
V
according to figure 7, the trans-
12
+ R
12
2
P2
Fb
REF
g
3C
(1+sR
+C
m
=
8
Z
)
7
11
2p3R
11
IN
(
)
R
7
>>1
C
C
C
7
F
2p3C
12
E/A
11
12
P
+C
)3[1+sC
1
2
C
8
2p3R
3C
11
C
APU3048
11
)]
12
1
10
10
3R
F
1
C
3(1+sR
7
P
---(14)
Comp
3C
3
11
10
6
(R
Frequency
12
6
+R
Z
8
V e
f
8
C
)]
10
9
)