Datasheet

f
p max
=
2S 300k 10P 100 PF
x
x
x
0.5
1
2S 5: 100 PF
x
x
+
= 584 Hz
f
p min
=
2S 300k 10P 100 PF
x
x
x
0.5
1
2S 50: 100 PF
x
x
+
= 297 Hz
f
z
=
1
2S 20 m: 100 PF
x
x
= 80 kHz
10 100 1k 10k 100k 1M
FREQUENCY (Hz)
-60
-40
-20
0
20
GAIN (d
B)
0
-45
-90
-135
-180
PHA
SE (°)
Asymptotic
Gain
Phase
-20dB/dec
(f
p1
is at zero frequency)
-20
dB
/
dec
FREQUENCY
GAIN (dB)
B
f
z1
f
z2
f
p2
LM2717-ADJ
SNVS407C –DECEMBER 2005–REVISED MARCH 2013 www.ti.com
Figure 17. Control-Output Transfer Function Figure 18. Output-Control Transfer Function
The control-output corner frequencies, and thus the desired compensation corner frequencies, can be
determined approximately by the following equations:
where
• C
o
is the output capacitance
• R
e
is the output capacitance ESR
• f is the switching frequency (8)
where
• C
o
is the output capacitance
• R
o
is the load resistance
• f is the switching frequency (9)
Since fp is determined by the output network, it will shift with loading (Ro) and duty cycle. First determine the
range of frequencies (fpmin/max) of the pole across the expected load range, then place the first compensation
zero within that range.
Example: V
o
= 5V, R
e
= 20mΩ, C
o
= 100µF, R
omax
= 5V/100mA = 50Ω, R
omin
= 5V/1A = 5Ω, L = 10µH, f =
300kHz:
(10)
(11)
(12)
Once the fp range is determined, R
c1
should be calculated using:
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