1

Q =

= 0.33

S

[(3.36)(0.44) - 0.5]

1

2S(100 x 10

-6)(0.01)

fESR =

= 159kHz

fp1 =

(500 x 10

3)(3.3 x 10-6)(100 x 10-6)

+

1

(100 x 10

-6)(0.83)

1

2S

1

(3.36)(0.44) - 0.5

>

@ = 2.86 kHz

ADC =

0.83

1.8 x 0.02

[(3.36(0.44) - 0.5]

(500 x 10

3)(3.3 x 10-6)

1+

1

0.83

= 15.5

H = feedback gain =

1.27

2.5

= 0.508

FC(s) =

(sCC1RC + 1)

s

2C

C1CC2RCRGM + s(CC2RGM + CC1(RGM + RC)) + 1

, CC2 used

FC(s) =

(sCC1RC + 1)

sCC1RGM (RGM + RC) + 1

, CC2 not used

Fh(s) =

+

1

s

2

1

S

fs

2

s

1

S

fsQ

+ 1

FP(S) =

1 +

s

S

fESR

1 +

s

S

fp1

RGM + RC

S

fESR RGMRC

CC2 =

(F)

LM3477

www.ti.com

SNVS141K – OCTOBER 2000 – REVISED MARCH 2013

(47)

PLOTTING THE OPEN LOOP RESPONSE

The open loop response is expressed as:

T = ADC x ACM x H x Fp(s) x Fc(s)

Where ADC and H are given above and

ACM = GM x RGM

(48)

(49)

(50)

(51)

One can plot the magnitude and phase of the open loop response to analyze the frequency response.

EXAMPLE: COMPENSATION DESIGN

4.5V

≤ VIN ≤ 5.5V

VOUT = 2.5V

IOUT = 3A (R = 0.83Ω)

RSN = 0.02Ω

L = 3.3µH

RSL = 0Ω

COUT = 100µF

RESR = 0.01Ω

First, calculate the power stage parameters using VIN(MIN) and R(MAX):

(52)

(53)

(54)

(55)

(56)

In this example, a crossover frequency of 20kHz is chosen, so: fC = 20000. RC is now calculated using the power

stage information and the target crossover frequency fC:

Copyright © 2000–2013, Texas Instruments Incorporated

Submit Documentation Feedback

25

Product Folder Links: LM3477