tÔ phẠm hÀ triỀu lÊ nguyÊn trƯỜng sƠn nguyỄn ngỌc sƠn
DESCRIPTION
TÔ PHẠM HÀ TRIỀU LÊ NGUYÊN TRƯỜNG SƠN NGUYỄN NGỌC SƠN. OP-AMP TL072CN. Sơ đồ chân. Transistor tầng lái sc2383. Transistor tầng khuếch đại công suất. D468 npn. B562-pnp. Diode FR207. Sơ đồ tương đương ac bán kỳ dương. Sơ đồ tương đương ac bán kỳ âm. Dòng ic ở 2 bán kỳ đối xứng. Xét Q1 - PowerPoint PPT PresentationTRANSCRIPT
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TÔ PHẠM HÀ TRIỀULÊ NGUYÊN TRƯỜNG SƠNNGUYỄN NGỌC SƠN
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0
F R E Q = 5 0V A M P L = 1V O F F = 0
Q 3
d 4 6 8
R b
3 3 k
C 2
4 7 0 u
R 2
5 0 K
C 3
4 7 0 u
R 1
5 0 K
R 3
2 7 8
V 21 2 v d c
R f
4 4 KR K
1 0 K
0
U 1 A
TL 0 7 2 C N
3
2
84
1
+
-
V+
V-
O U T
D 1
F R 2 0 7
21
V
D 2
F R 2 0 7
21
C i
4 7 0 u
Q 2
b 5 6 2
0
00
R E 1
1
0
R E 2
1
R i
5 0
R L
1 5
0
Q 1
2 3 8 3
0
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OP-AMP TL072CN
Sơ đồ chân
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Transistor tầng lái sc2383
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Transistor tầng khuếch đại công suất
D468 npn
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B562-pnp
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Diode FR207
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ohmicq
hfehiehie 100
2
2.2523
mAohm
ICQICQ 501.2
2.62.5,632
mAhfe
icmibm
ARL
PiLmicm
825.1200
365.0
22
365.02
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Sơ đồ tương đương ac bán kỳ dương
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Sơ đồ tương đương ac bán kỳ âm
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3'33
11.33
hfeRlhieR
IbhfeRib
2'223
11)2.3(2
hfeRlhieRDR
IbhfeRDRib
23 ibib Dòng ic ở 2 bán kỳ đối xứng
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Xét Q1Bỏ qua ib2,ib3Rdc= R3Rac=R3//(hie2+Rl'hfe2) bỏ qua RdPhương trình đường tải 12-1.3=Rdc. ICQ1 +VCEQ1
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ICQTU
33.3//3
7.10
RkR
Dòng Id=ICQ1,để diode đủ dẫn ta chọn dòng id=0.02A
02.033.3//3
7.10
RkR
R3=278 ohm
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)0___()(6 IvV
áp hồi tiếp trên ngỏ ra tầng công suất là 6 V,áp tại ngõ vao V- của opam là
ta sẽ chọn RK,Rf sao cho V+-V->0.24
thì áp ra OPAMP sẽ là 6V
kRb 33125.02.0
65.06
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vi=1V ,ta chọn R1=R2 sao cho R1//R2>>richọn R1=R2=50k =>vi
V+
V+
4.5
Rk
RfRk
v
vA
i
lmv
47.5. RLiLmvLm
v+ = v- = vi
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chọn Rk=10k =>Rf=44k
các tụ bypasss được chọn sao cho trong khoãng âm tầng thì bị bypass, chon Ci=Ck=470u
)(2
1
RLREfCL
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Chọn tầng số cắt thấp là 20 hz => CL=470u
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VLm
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ÁP NGÕ RA Q1
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V
R 2
5 0 K
0
1 . 5 2 3 V
Q 3
d 4 6 8
Q 2
b 5 6 2
1 0 . 6 7 V
R 3
2 7 8
R E 2
1
R K
1 0 K
00
R b
3 3 k
C i
4 7 0 u
R 1
5 0 K
Q 1
2 3 8 3
7 0 6 . 4 m V
0
R f
4 4 K
0
C 3
4 7 0 u
1 0 . 2 0 V
1 2 . 0 0 V
F R E Q = 5 0V A M P L = 1V O F F = 0
R E 1
1D 2
F R 2 0 7
21
1 1 . 0 4 V
0
R i
5 0
V 21 2 v d c
0
C 2
4 7 0 u
0
1 0 . 6 7 V
R L
1 51 0 . 6 7 V
U 1 A
TL 0 7 2 C N
3
2
84
1
+
-
V+
V-
O U T
6 . 0 0 0 V
D 1
F R 2 0 7
21
1 0 . 6 7 V
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ÁP NGÕ RA OPAMP
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