Datasheet

1
Vdc max
325V
naux
R 17.0k
10.9
IQR
1.75mA
= = = W
( )
OFFSET EXTERNAL
IQR
VCS 6.6k R
100
= · W +
Vaux
IQR
R1
-
=
Vdc
Vaux
Naux
- =
CS
VCS _ OFFSET : V VCS _ CL 0.5V 0.382V 0.118V= - = - =
CS
VCS _ OFFSET : V VCS _ CL= -
325V
VCS _ CL 0.15 2.679Apk 160ns 0.382V
400 H
é ù
æ ö
= W · - · =
ê ú
ç ÷
m
è ø
ë û
Vdc max
VCS _ CL Rsense IL max_ CL tprop
Lp
é ù
æ ö
= · - ·
ê ú
ç ÷
è ø
ë û
2 94.9W
IL max_ LL 2.679Apk
0.86 400 H 76.8kHz
·
= =
· m ·
2 Pout _ LL
IL max_ LL
Lp Freq _ Comp
·
=
h · ·
( )
2
2
2
2
4
Freq _ Comp 76.8kHz
94.9W
2 400 H (19V 0.7V 0.167 325V)
0.86 400 H
2 400 H 94.9 (19 0.7 0.167 325V)
4 580ns
325V (19V 0.7V)
0.86 325V 19V 0.7V
= =
é ù
· m · + + · ·
ê ú
æ ö
· m
· m · · + + ·
ê ú
ç ÷
· + +
ê ú
ç ÷
· +
· · +
è ø
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LM5023
www.ti.com SNVS961D –APRIL 2013–REVISED JANUARY 2014
Step four is to calculate the peak current.
For the power supply to go into pulse-by-pulse current limit the voltage across the current sense resistor must be
0.5 V, so:
VCS_OFFSET is the required voltage offset that must be injected across the current sense resistor, R
SENSE
.
After calculating the required offset voltage, use the following equations to calculate the required current feed
forward:
While the main Flyback switch is on, Q1, the voltage on the Auxiliary winding will be negative and proportional to
the rectified line.
IQR should be chosen in the range of 1 ma to 4 ma. The demagnetization circuit impedance should be
calculated to limit the maximum current flowing through Pin 1 to less than 4 mA.
R
OFFSET
= 6.6 kΩ + R
EXTERNAL
(the 6.6 kΩ resistance is internal to the LM5023).
Where: Naux is the number of turns on the Flyback primary (Np) divided by the number of turns on the
transformer Auxiliary (Naux) winding. The current mirror in the QR pin input has a gain of 100; this will offset the
voltage on the current sense pin by:
Set IQR= 1.75 mA
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