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AN2495 Datasheet(PDF) 20 Page - STMicroelectronics

Part # AN2495
Description  The purpose of this application note is to explain the design
PDF  36 Pages
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Manufacturer  STMICROELECTRONICS [STMicroelectronics]
Direct Link  http://www.st.com
Logo STMICROELECTRONICS - STMicroelectronics

AN2495 Datasheet(HTML) 20 Page - STMicroelectronics

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Frequency response and loop compensation
AN2495
20/36
Doc ID 13127 Rev 5
From the L6565 datasheet we know that Icomp = 5 mA (source current) and Rcomp = 15 kΩ.
and can obtain:
Equation 39
R24 = 1.5 kΩ is a good choice. It is good practice to put a 3.3 kΩ resistor (R23 = 3.3 kΩ) in
parallel to the photodiode.
The resistive partition must be set with high precision according to the following equation:
Equation 40
There is no next close commercial value so it is a good idea to put two 47 k
Ω resistors
(R19 = R20 = 47 kΩ) in series.
Now C11 (Ccomp), C13 (CF) and R21 (RF) must be set in order to satisfy frequency response
requirements.
A good choice is to set the pole of G2(s) so as to cancel the low frequency zero of G1(s):
Equation 41
The next close commercial value for C11 = 2.2 nF has been chosen.
Similarly C13 and R21 have been determined by setting the corresponding zero close to the
pole of G1(s) and imposing the open loop gain to cross the 0 dB axis only once at
f = fc = 10 kHz:
Equation 42
In such a way we ideally get a phase margin of 90 degrees and an adequate closed-loop
bandwidth.
R
B
R
24
V
out
V
ref
V
diode
–
()
–
I
co mp
-------------------------------------------------------
24V
3.5V
–
5mA
-----------------------------
4.1k
Ω
=
=
<
=
R
high
R
22
V
out
V
re f
–
V
ref
---------------------------
•
23.2k
Ω
=
=
1
2
π R
co mp
•
C
co mp
•
-----------------------------------------------------
5kHz
C
11
C
com p
2.12nF
≅
=
⇒
=
1
2
π
R
H
R
F
+
() C
F
•
•
-----------------------------------------------------
400 Hz
C
13
10nF
=
⇒
=
Gj
ω
()
ω 2π 10
4
– rad
•
=
1R
21
15k
≅
⇒
=
⎩
⎪
⎨
⎪
⎧
/sec



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