APU3137 View Datasheet(PDF) - Advanced Power Electronics Corp
Part Name
Description
Manufacturer
APU3137 Datasheet PDF : 17 Pages
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APU3137
The APU3137’s error amplifier is a differential-input
transconductance amplifier. The output is available for
DC gain control or AC phase compensation.
The E/A can be compensated with or without the use of
local feedback. When operated without local feedback,
the transconductance properties of the E/A become evi-
dent and can be used to cancel one of the output filter
poles. This will be accomplished with a series RC circuit
from Comp pin to ground as shown in Figure 9.
Note that this method requires that the output capacitor
should have enough ESR to satisfy stability requirements.
In general, the output capacitor’s ESR generates a zero
typically at 5KHz to 50KHz which is essential for an
acceptable phase margin.
The ESR zero of the output capacitor expressed as fol-
lows:
FESR
=
1
2π×ESR×Co
---(14)
VOUT
R6 Fb
R5
Vp=VREF
E/A
Comp
Ve
C9
R4 Optional
Gain(dB)
H(s) dB
First select the desired zero-crossover frequency (Fo):
Fo > FESR and FO ≤ (1/5 ~ 1/10)×fS
Use the following equation to calculate R4:
R4
=
VOSC
VIN
×
Fo×FESR
FLC2
×
R5 + R6
R5
×
1
gm
---(18)
Where:
VIN = Maximum Input Voltage
VOSC = Oscillator Ramp Voltage
Fo = Crossover Frequency
FESR = Zero Frequency of the Output Capacitor
FLC = Resonant Frequency of the Output Filter
R5 and R6 = Resistor Dividers for Output Voltage
Programming
gm = Error Amplifier Transconductance
For:
VIN = 5V
VOSC = 2.5V
Fo = 20KHz
FESR = 12KHz
FLC = 3.43KHz
R5 = 1K
R6 = 2.15K
gm = 600µmho
This results to R4=26.7K
Choose R4=30K
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
FZ ≅ 75%FLC
FZ ≅ 0.75×
2π
For:
Lo = 2.17µH
Co = 990µF
1
LO × CO
---(19)
FZ = 2.57KHz
R4 = 20K
FZ Frequency
Figure 9 - Compensation network without local
feedback and its asymptotic gain plot.
The transfer function (Ve / VOUT) is given by:
( ) H(s) =
gm×
R5
R6 + R5
×
1
+ sR4C9
sC9
---(15)
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
|H(s=j×2π×FO)|
=
gm×
R5
R6×R5
×R4
---(16)
FZ
=
1
2π×R4×C9
---(17)
|H(s)| is the gain at zero cross frequency.
Using equations (17) and (19) to calculate C9, we get:
C9 ≅ 2006pF; Choose C9 =3300pF
One more capacitor is sometimes added in parallel with
C9 and R4. This introduces one more pole which is mainly
used to suppress the switching noise. The additional
pole is given by:
1
FP
=
2π×R4×
C9×CPOLE
C9 + CPOLE
The pole sets to one half of switching frequency which
results in the capacitor CPOLE:
1
CPOLE =
π×R4×fS -
1
≅
1
π×R4×fS
C9
For FP << fS/2
R4=30K and FS=200KHz will result to CPOLE=53pF.
Choose CPOLE=47pF.
10/17
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