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| 型號(hào)： | SG3842M |
| 廠商： | MICROSEMI CORP-ANALOG MIXED SIGNAL GROUP |
| 元件分類： | 穩(wěn)壓器 |
| 英文描述： | CURRENT MODE PWM CONTROLLER |
| 中文描述： | 1 A SWITCHING CONTROLLER, 500 kHz SWITCHING FREQ-MAX, PDIP8 |
| 封裝： | ROHS COMPLIANT, PLASTIC, DIP-8 |
| 文件頁(yè)數(shù)： | 10/14頁(yè) |
| 文件大?。?/td> | 224K |
| 代理商： | SG3842M |
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C
U R R E N T
-M
O D E
PWM C
O N T R O L L E R
SG1842/SG1843 Series
P R O D U C T D A T A B O O K 1 9 9 6 / 1 9 9 7
Copyright 2000
Rev. 1.6 4/00
10
P
R O D U C T I O N
D
A T A
S
H E E T
A P P L I C AT I O N I N F O R M AT I O N
OSCILLATOR
The oscillator of the 1842/43 family of PWM's is designed such
that many values of R
and C
will give the same oscillator
frequency, but only one combination will yield a specific duty
cycle at a given frequency.
FIGURE 13
— OSCILLATOR TIMING CIRCUIT
A set of formulas are given to determine the values of R
and
C
for a given frequency and maximum duty cycle. (Note: These
formulas are less accurate for smaller duty cycles or higher
frequencies. This will require trimming of R
T
or C
T
to correct for
this error.)
Given: Frequency
≡
f
Maximum Duty Cycle
≡
D
m
Calculate: R
T
= 267
(
)
where .3 < D
m
< .95
C
T
=
(μF)
For Duty-Cycles above 95% use:
Example:
A Flyback power supply requires a maximum of 45% duty
cycle at a switching frequency of 50kHz. What are the values
of R
T
and C
T
Given:
f
=
50kHz
D
m
=
0.45
Calculate: R
T
= 267
= 674
C
T
=
= .025μF
(1.76)
1/.045
-1
(1.76)
.55/.45
- 1
1.86 * 0.45
50000 * 674
(1.76)
1/D
m
-1
(1.76)
(1-D
m
)/D
m
- 1
1.86 * D
m
f * R
T
V
REF
R
T
/C
T
GND
R
T
C
T
.001
1
C
T
Value - (μF)
1000
f
.002
.02
100
10
.005
.01
.05
0.1
R
T
= 680
R
T
= 2k
R
T
= 5k
R
T
= 10k
R
T
= 20k
R
T
= 30k
R
T
= 50k
R
T
= 70k
R
T
= 100k
FIGURE 14
— OSCILLATOR FREQUENCY vs. R
T
FOR VARIOUS C
T
F
≈
where R
T
≥
5k
1.86
R
T
C
T
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