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IX9915 데이터시트(PDF) 8 Page - IXYS Corporation

부품명 IX9915
상세설명  Low Voltage Error Amplifier with a 350V Darlington Transistor
PDF  10 Pages
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제조업체  IXYS [IXYS Corporation]
홈페이지  http://www.ixys.com
Logo IXYS - IXYS Corporation

IX9915 데이터시트(HTML) 8 Page - IXYS Corporation

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INTEGRATED CIRCUITS DIVISION
IX9915
8
www.ixysic.com
R01
always into pin FB). This error causes the regulated
output voltage to increase which increases the current
through R1 by an amount equal to IIB, thereby
restoring the current through R2 to its original value.
Reducing the VREG error created by the input bias
current to less than 1% is accomplished by setting the
value of R1 using the following formula:
Where:
2.2 Compensation
The dominate pole of the error amplifier is around
13kHz. In a typical system with a low-bandwidth
requirement, it doesn't need any external
compensation. Frequency response of the system can
be optimized for the specific application by placing a
compensation network between the OC and FB pins
of the IX9915. For designs with more critical
bandwidth requirements, measurement of the loop
response must be made and compensation adjusted
as necessary.
2.3 Design Example
A design example for the bleeder circuitry in LED lamp
exhibits the detailed steps. In this example, it will target
the predetermined voltage VLINE-TH=25V and
maximum bleeding current IH-MAX=25mA.
In order to flow the maximum bleeding current IH-MAX
through the Darlington transistor:
If taking RE=100:
In fact, the components in the dashed rectangle
function as a comparator, its gain:
If taking R0=40k, the gain of the comparator is
around 82dB. That is to say, once the error amplifier
starts to regulate, the Darlington transistor will be shut
off by this comparator. So, IOC can be ignored for
affecting the predetermined voltage:
Almost full power supply voltage will cross over R3,
taking R3=100k to minimize its power consumption:
Substituting:
• IQ=75A,
• I1=VREF / R2,
• VLINE-TH=25V
into formula (2):
R
1
V
REG
50
A
-------------
50
A
100
I
IB MAX

=
V
REG
I
HMAX
–
R
E
V
BE
+
=
(1)
V
REG
I
HMAX
–
R
E
V
BE
+
=
25mA
100
 1.5V
+
=
4V
=
A
R
0
R
2
g
m
R
1
R
2
+
------------------------------
=
R
2
R
1
R
2
+
------------------
V
REF
V
REG
-------------
1.299V
4V
-----------------
0.325
==
=
g
m
1S (typical)
=
V
LINE-TH
V
REG
I
1
I
Q
+
 R
3
+
(2)
P
V
rms

2
100k
-------------------
=
R
2
9.6k
R
1
R
2
V
REG
V
REF
-------------
1
–


=
R
1
20k



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