TPS54062 Datasheet, PDF (18/30 Pages) Texas Instruments – 4.7V to 60V Input, 50mA Synchronous Step-Down Converter with Low IQ

TPS54062
SLVSAV1 – MAY 2011
www.ti.com
Closing the Loop
There are several methods used to compensate DC/DC regulators. The method presented here is easy to
calculate and ignores the effects of the slope compensation that is internal to the device. Since the slope
compensation is ignored, the actual cross over frequency will usually be lower than the cross over frequency
used in the calculations. This method assume the crossover frequency is between the modulator pole and the
ESR zero and the ESR zero is at least 10 times greater the modulator pole. Use SwitcherPro™ software for a
more accurate design.
To get started, the modulator pole, fpole, and the ESR zero, fzero must be calculated using Equation 18 and
Equation 19. For Cout, use a derated value of 8.9 µF. Use Equation 21 and Equation 22, to estimate a starting
point for the crossover frequency, fco, to design the compensation. For the example design, fpole is 271 Hz and
fzero is 5960 kHz.
Equation 21 is the geometric mean of the modulator pole and the esr zero and Equation 22 is the mean of
modulator pole and the switching frequency. Equation 21 yields 40.29 kHz and Equation 22 gives 7.36 kHz. Use
a frequency near the lower value of Equation 21 or Equation 22 for an initial crossover frequency.
For this example, fco is 7.8 kHz. Next, the compensation components are calculated. A resistor in series with a
capacitor is used to create a compensating zero. A capacitor in parallel to these two components forms the
compensating pole.
To determine the compensation resistor, R4, use Equation 22. Assume the power stage transconductance,
gmps, is 0.65 A/V. The output voltage, Vo, reference voltage, V_REF, and amplifier transconductance, gmea, are
3.3 V, 0.8 V and 102 µA/V, respectively.
R4 is calculated to be 27.1 kΩ, use the nearest standard value of 27.4 kΩ. Use Equation 23 to set the
compensation zero to the modulator pole frequency. Equation 23 yields 0.021.4 µF for compensating capacitor
C5, a 0.022 µF is used on the board. Use the larger value of Equation 24 and Equation 25 to calculate the C6
value, to set the compensation pole. Equation 25yields 29 pF so the nearest standard of 27 pF is used.
f pole(Hz) =
V_O
I_O
1
´ C_O ´ 2 ´ p
(18)
f zero(Hz) =
R_C
´
1
C_O ´
2
´
p
(19)
f co1(Hz) = ( f zero ´ f pole)^0.5
(20)
f co2(Hz)
=
æ
çè
f sw
2
´
ö^0.5
f pole _÷ø
(21)
R_COMP
=
2
´ p ´ f_CO
gmps
´
V_O
V_REF ´ gmea
(22)
1
C_COMP = _{2 ´ p ´ RCOMP ´ fPOLE}
(23)
C_POLE1
=
R_C ´ C_O
R_COMP
(24)
1
C_POLE2 = _{RCOMP ´ fSW ´ p}
(25)
18
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