Contax SW II - REV 4.01 Manual do Utilizador descarregar pdf (Página 15)

R2J20702NP Target Specification

REJ03G1782-0401 Rev.4.01 Page 15 of 27

Jun 17, 2010

Vout

Amplifier output: to current-sense comparato

50 kΩ

0.6 V reference

RfCf

Figure 3 Error Amplifier Compensation

Design example

Specification: L = 360 nH, Co = 600 F, Fsw = 500 kHz, Vin = 12 V, Vout = 1.8 V, R1 = 2 k, R2 = 1 k,

RCS = 750 

1. Flat-band gain of error amplifier

The flat-band gain is; Af = Rf / (R1 // R2) / 2  {R2 / (R1 + R2)}

Hence, Rf = 2  Af  R1 (1)

In the Bode plot, the total gain should be less than 1 (0 dB) at the switching frequency.

The total gain at Fsw (= Asw) depends on the flat-band gain, so Af should be expressed as follows.

Af = Asw  2   Fsw  Co  RCS / Nt (2)

Here, Nt = Idh / Ics = 22000

In the typical way, the value chosen for Asw is in the range from 0.1 to 0.5, since this produces a stable control loop.

The transient response will be faster if a larger Asw is adopted ,but the system might be unstable.

We choose 0.25 for Asw in the example below.

Af = 0.25  2   500 kHz  600 F  750  / 22000 = 16.06

Rf = 2  16.06  2 k = 64.240 k

Therefore, we select a value of 62 k for Rf.

2. Selecting the Cf value to determine the frequency of the zero

The frequency of the zero established by Cf and Rf is about ten times the frequency of the pole for the power stage

and modulator.

We must start with the dc gain of the power stage and modulator.

0 = ……(3)

2 × Nt / RCS × L × Vin × Fsw

SQRT {Vin

− 8 × L × Vin × Fsw × (VCS0 × Nt / RCS) }

Here VCS0 is the peak ac voltage on the CS pin when the load current is zero, thus

VCS0 = 0.5  RCS  (Vin – Vout)  Vout / (L  Vin  Fsw) / 22000 (4)

= 0.5  750   (12 V – 1.8 V)  1.8 V / (360 nH  12 V  500 kHz) / 22000

= 0.145 V

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