米6体育平台手机版

ZHCSQ75C June 2022 – March 2023 UCC28C50-Q1 , UCC28C51-Q1 , UCC28C52-Q1 , UCC28C53-Q1 , UCC28C54-Q1 , UCC28C55-Q1 , UCC28C56H-Q1 , UCC28C56L-Q1 , UCC28C57H-Q1 , UCC28C57L-Q1 , UCC28C58-Q1 , UCC28C59-Q1

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D|8
- MSOI002K

散热焊盘机械数据 (封装 | 引脚)

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9.2.2.14.2 Compensation Components

To compensate a peak-current-mode controller it’s very common to use a Type-II compensator. The Type-II compensator introduces a pole at DC, a relatively low frequency zero (f_Z,COMP), and a higher frequency pole (f_P,COMP). The pole at DC forces the system to have high gain at very low frequency and zero stead-state error.

Figure 9-4 Type-II Compensation with an Error Amplifier

The low frequency zero is formed by R18 and C19

Equation 34.

f_{Z, C O M P} = \frac{1}{2 \times π \times C 19 \times R 18}

The higher frequency pole is formed by R18 and C20

Equation 35.

f_{P, C O M P} = \frac{1}{2 \times π \times C 20 \times R 18}

The mid-frequency gain of the compensator is given by

Equation 36.

G_{C O M P} = \frac{R 18}{(R 17 + R 19)}

First, select a crossover frequency, 625 Hz. Then, use the frequency response of the plant (control-to-output) at low input voltage to determine how much gain the compensator must add to increase the crossover to the desired bandwidth. In Figure 9-5 the plant is measured to be -23.3 dB at 625 Hz. Therefore, the error amplifier must have a mid-frequency gain of 14.6 V/V.

Figure 9-5 Measuring the plant gain at the desired crossover frequency: -23 dB at 625 Hz

Choose (R17+R19) = 22.5 kΩ, and solve for R18

Equation 37.

R 18 = G_{C O M P} \times (R 17 + R 19) = 14.6 V / V \times 22.5 k Ω = 328 k Ω

Select a standard value for R18, like 324 kΩ.

Now that we know R18, it’s fairly straightforward to set f_Z,COMP = f_POLE and solve for C19

Equation 38.

f_{Z, C O M P} = f_{P O L E} = \frac{1}{2 \times π \times C 19 \times 324 k Ω} = 17 H z

Equation 39.

C 19 = 28 n F

Likewise, set f_P,COMP = f_ZERO and solve the following for C20

Equation 40.

f_{P, C O M P} = f_{Z E R O} = \frac{1}{2 \times π \times C 20 \times 324 k Ω} = 4.8 k H z

Equation 41.

C 20 = 102 p F

Finally, choose standard capacitor values, C19 = 22 nF and C20 = 100 pF. Notice that a slightly lower value was used for C19 than calculated. This was done to have a faster “reset” time during after a load transient response. If the complete loop is found to have too little phase margin then C19 can be increased at the cost of slower reset time.