Datasheet
f
VI
2
+
P
IN
(
2 p
)
2
DV
VAOUT
V
OUT
R
IN
C
OUT
C
f
R
f
+
1
2 p f
VI
C
f
C
Z
+
1
2 p
f
VI
10
R
f
Current Loop
G
ID
(s) +
V
OUT
R
SENSE
s L
BOOST
V
P
G
EA
+
1
G
ID
+
1
0.383
+ 2.611
C
Z
+
1
2 p R
f
f
C
C
P
+
1
2 p R
f
f
s
2
UCC2817-EP
UCC2818-EP
SLUS716 – DECEMBER 2008 ...........................................................................................................................................................................................
www.ti.com
In this example, C
f
= 150 nF. Resistor R
f
sets the dc gain of the error amplifier and, thus, determines the
frequency of the pole of the error amplifier. The location of the pole can be found by setting the gain of the loop
equation to one and solving for the crossover frequency. The frequency, expressed in terms of input power, can
be calculated by the equation:
f
VI
for this converter is 10 Hz. A derivation of this equation can be found in the Unitrode Power Supply Design
Seminar SEM1000, Topic 1 ( A 250-kHz, 500-W Power Factor Correction Circuit Employing Zero Voltage
Transitions).
Solving for R
f
becomes:
or R
f
= 100 k Ω .
Due to the low output impedance of the voltage amplifier, capacitor C
Z
was added in series with R
F
to reduce
loading on the voltage divider. To ensure the voltage loop crossed over at f
VI
, C
Z
was selected to add a zero at
1/10th of f
VI
. For this design, a 2.2- µ F capacitor was chosen for C
Z
. The following equation can be used to
calculate C
Z
:
The gain of the power stage is:
R
SENSE
has been chosen to give the desired differential voltage for the current sense amplifier at the desired
current limit point. In this example, a current limit of 4 A and a reasonable differential voltage to the current
amplifier of 1 V gives a R
SENSE
value of 0.25 Ω . V
P
in this equation is the voltage swing of the oscillator ramp, 4 V
for the UCC2817. Setting the crossover frequency of the system to 1/10th of the switching frequency, or 10 kHz,
requires a power-stage gain at that frequency of 0.383. In order for the system to have a gain of 1 at the
crossover frequency, the current amplifier must have a gain of 1/G
ID
at that frequency. G
EA
, the current amplifier
gain is then:
R
I
is the R
MOUT
resistor, previously calculated to be 3.9 k Ω (see Figure 3 ). The gain of the current amplifier is
R
f
/R
I
, so multiplying R
I
by G
EA
gives the value of R
f
, in this case approximately 12 k Ω . Setting a zero at the
crossover frequency and a pole at one-half the switching frequency completes the current loop compensation.
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