Datasheet

A
CM
+
1
R1 C2
+ 3.978 10
5
Tv (j × 2 × π × fc) = 1
z
1
f = = 6kHz
2 × π × (R1+ R2) × C1
p
1
f = = 176.8kHz
2 × π × R2 × C1
Tv(s) = Gc(s) × Gvc(s)
1 s × (R1+ R2) × C1+ 1
Gc(s) = ×
R1× C2 s × (s × R2 × C1+ 1)
R2
R1
C2
C1
+
V
REF
TPS40140
SLUS660H –SEPTEMBER 2005–REVISED JUNE 2013 www.ti.com
Figure 6-1. Type II Compensator
The compensator transfer function is:
(51)
The loop gain transfer function is:
(52)
Assume the desired crossover frequency is 60 kHz, then set the compensator zero about 1/10 of the
crossover frequency and the compensator pole equal to the ESR zero. The compensator gain is then
calculated to achieve the desired bandwidth. In this design, the compensator gain, pole and zero are
selected as following:
(53)
(54)
(55)
Using Equation 54, the compensator gain is solved as 3.978 × 10
5
.
(56)
where
• C1 = 2.6 nF
• C2 = 250 pF
• R2 = 350 Ω
• Set R1 equal to 10-kΩ, and then calculate all the other components
In the real laboratory practice, the final components are selected as following to increase the phase
margin and reduce PWM jitter.
• R1 = 10 kΩ
• C2 = 330 pF
• R2 = 50 Ω
• C1 = 2.2 nF
48 DESIGN EXAMPLES Copyright © 2005–2013, Texas Instruments Incorporated
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