IR3621 datasheet(15/29 Pages) IRF

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IR3621&(PbF)

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Where:

VIN = Maximum Input Voltage

VOSC = Oscillator Ramp Voltage

O1 = Crossover Frequency

FESR = Zero Frequency of the Output Capacitor

FLC = Resonant Frequency of the Output Filter

R5 and R9 = Resistor Dividers for Output Voltage

Programming

gm = Error Amplifier Transconductance

This results to R4=4.8K

Choose R4=5K

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

fore the LC filter resonant frequency pole:

Using equations (13) and (15) to calculate C9, we get:

Same calcuation For V1.8V will result to: R3 = 4.2K and

C8 = 10nF

One more capacitor is sometimes added in parallel with

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

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 CPOLE:

≅ 8.3nF; Choose C9 =8.2nF

FLC = 5.06kHz

R5 = 1K

R9 = 2.14K

gm = 1400

µmho

For V2.5V:

VIN = 12V

VOSC = 1.25V

O1 = 40KHz

FESR = 13.3kHz

R4 =

---(14)

O1×F

ESR

FLC2

VOSC

VIN

R5 + R9

For:

Lo = 1.1

µH

Co = 990

µF

≅ 75%FLC

≅ 0.75×

× CO

---(15)

FZ = 3.61kHz

R4 = 5K

FP =

π×R4×

×CPOLE

C9 + CPOLE

CPOLE =

≅

for FP <<

π×R4×fS

π×R4×fS -

For a general solution for unconditional stability for ce-

ramic output capacitor with very low ESR or any type of

output capacitors, in a wide range of ESR values we

should implement local feedback with a compensation

network. The typically used compensation network for a

voltage-mode controller is shown in Figure 16.

Figure 16- Compensation network with local

feedback and its asymptotic gain plot.

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 ZIN and Zf according to Figure 16, the trans-

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:

1 - gmZf

1 + gmZIN

VOUT

gmZf >> 1

and

gmZIN >>1

---(16)

H(s) =

1+sR7

×(1+sR8C10)

(1+sR7C11)

×[1+sC10(R6+R8)]

[

(

)]

sR6(C12+C11)

C12C11

C12+C11

VOUT

VP2=VREF

C10

C12

C11

FZ1

FZ2

FP2

FP3

E/A

ZIN

Frequency

Gain(dB)

H(s) dB

Comp

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