TPS40422_16 Datasheet, PDF (55/71 Pages) Texas Instruments – Dual-Output or Two-Phase Synchronous Buck Controller with PMBus Interface

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TPS40422
SLUSAQ4E – OCTOBER 2011 – REVISED SEPTEMBER 2016
9.2.1.2.7.2 FB1 (Pin 2) and FB2 (Pin 8) Output Voltage Setting
A feedback divider between the DIFFO pin and AGND sets the output voltage. The components in Figure 9 that
determine the nominal output voltage are R₁ and R₂.
Calculation for 1.2-V output: Select the high-side resistor (R₁) to be 47.5 kΩ and use Equation 31 to calculate
low-side resistor R₂.
R₂
=
V_FB
×
R₁
V_OUT F V_FB
=
47.5 kÀ
0.6 V ×
1.2 V F 0.6 V
=
47.5 kÀ
where
• V_FB is the feedback voltage of 600 mV
• V_OUT is the desired output voltage of 1.2 V
• R₁ and R₂ are resistors for the voltage divider from the output to the feedback
(31)
9.2.1.2.7.3 Compensation Network Using COMP1 (Pin 3) , COMP2 (Pin 7), FB1 (Pin 2) FB2 DIFFO1 (Pin 8) (Pin 39)
The device uses voltage mode control topology in a single phase dual-output configuration. In this example, a
Type III compensation network is implemented to compensate for the double pole, close the loop and stabilize
the system. TI provides a compensation calculator tool to streamline the compensation design process.
The Loop Compensation Tool (SLUC263) provides the recommended compensation components as a starting
point and approximate bode plots. It is always recommended to measure the real system bode plot after the
design and adjust the compensation values accordingly. The chosen compensation values are derived from the
tool calculation along with the Venable K-factor method and optimization based on the measured data.
• R₁ = R₂ = 47.5 kΩ
• R₃ = 4.75 kΩ
• R₄ = 20 kΩ
• C₁ = 470 pF
• C₂ = 1.2 nF
• C₃ = 120 pF
In this design example, the desired crossover frequency chosen is approximately 4 × f_DP (20 kHz). Use
Equation 32 to calculate the double pole frequency formed by the output inductor and capacitor bank.
1
1
f_DP = _{tN × ¾LC} = _{tN × ¥820 nH × 1004 µF} = 5547 Hz
(32)
Because the Venable K-factor method was used to derive these compensation values, it is important to ensure
that the poles and zeroes are coincident. A way to confirm that the poles and zeroes are coincident is to
calculate the poles and zeroes from the gain of the Type-III compensation network.
GAIN_{TYPE F3}
= ^{R1 + R3}
R₁ × R₃ × C₃
×
@s
+
R₄
1
×
C₂^A
×
ls
+
:R₁
+
1
R₃;
×
C₁^p
s
×
@s
+
C₂
R₄ ×
+ C₃
C₃ ×
C₂^A
×
@s
+
R₃
1
×
C₁^A
(33)
The first and second zeroes should yield approximately equal frequency values of 6480 Hz and 6631 Hz.
1
1
f_Z1 = _{tN × :R1 + R3; × C1} = _{tN × :47.5 kÀ + 4.75 kÀ; × 470 pF} = 6480 Hz
(34)
1
1
f_Z2 = _{tN × R4 × C2} = _{tN × 20 kÀ × 1.2 nF} = 6631 Hz
(35)
Analogous to the zeroes being coincident, the first and second poles are coincident as well and are shown in
Equation 36 and Equation 37.
1
1
f_P1 = _{tN × R3 × C1} = _{tN × 4.75 kÀ × 470 pF} = 71290 Hz
(36)
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