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APU3048

DUAL SYNCHRONOUS PWM CONTROLLER CIRCUITRY AND LDO CONTROLLER

Advanced Power Electronics Corp 

APU3048

For:

VIN1 = 12V

VOSC = 1.25V

FO1 = 30KHz

FESR1 = 26.5KHz

FLC1 = 2.8KHz

R8 = 1K

R6 = 1.64K

gm = 600mmho

ZIN

VOUT

C10

R8

R6

C12

R7

C11

Zf

This results to R9=46.4KV; Choose R9=46.4KV

To cancel one of the LC filter poles, place the zero be-

fore the LC filter resonant frequency pole:

FZ ≅ 75%FLC1

FZ ≅ 0.75 3

2p

For:

L3 = 10.2mH

Co = 300mF

Fz = 2.1KHz

R9 = 46.4KV

1

L3 3 CO

---(13)

Using equations (11) and (13) to calculate C9, we get:

C9 = 1630pF; Choose C9 = 1800pF

Using equations (11),(12) and (13) for Ch2, where:

VIN2 = 5V

VOSC = 1.25V

FO2 = 30KHz

FESR2 = 26.5KHz

FLC2 = 3.5KHz

R15 = 1K

R14 = 442V

gm = 600mhmo

Gain(dB)

H(s) dB

Fb

R5

VREF

E/A

Ve

Comp

FZ1

FZ2

FP2

FP3 Frequency

Figure 7 - Compensation network with local

feedback and its asymptotic gain plot.

In such configuration, the transfer function is given by:

Ve

1 - gmZf

VOUT = 1 + gmZIN

The error amplifier gain is independent of the transcon-

ductance under the following condition:

gmZf >> 1 and gmZIN >>1

---(14)

We get:

R11 = 38.9KV; Choose R11 = 39.2KV

C19 = 1554pF; Choose C19 = 1800pF

One more capacitor is sometimes added in parallel with

C9 and R4. This introduces one more pole which is mainly

used to supress the switching noise. The additional pole

is given by:

1

FP = 2p 3 R9 3 C18 3 CPOLE

C18 + CPOLE

The pole sets to one half of switching frequency which

results in the capacitor CPOLE:

1

CPOLE =

p 3 R9 3 fS - 1

C18

For FP <<

fS

2

1

≅ p 3 R9 3 fS

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.

By replacing ZIN and Zf according to figure 7, the trans-

former function can be expressed as:

1

(1+sR7C11)3[1+sC10(R6+R8)]

[ ( )] H(s)=

3

sR6(C12+C11) 1+sR7 C12C11

C12+C11

3(1+sR8C10)

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:

FP1 = 0

FP2

=

1

2p3R83C10

( ) FP3 =

1

2p3R73

C123C11

C12+C11

≅

1

2p3R73C12

FZ1

=

1

2p3R73C11

1

1

FZ2 = 2p3C103(R6 + R8) ≅ 2p3C103R6

9

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