AD8014
–5–rev. b
典型 效能 characteristics–
频率 – mhz
12
–15
1 100010010
–12
–9
–6
–3
3
6
9
0
g = +1
V
O
= 200mv p-p
R
F
= 1k
R
L
= 1k
V
S
=
5V
V
S
= +5v
15
normalized 增益 – db
图示 1. 频率 回馈, g = +1, v
S
=
±
5 v 和 +5 v
V
S
=
5V
g = +2
R
F
= 500
V
O
= 2v p-p
频率 – mhz
12
–15
1 100010010
–12
–9
–6
–3
3
6
9
0
R
L
= 50
R
L
= 75
normalized 增益 – db
图示 2. 频率 回馈, g = +2, v
O
= 2 v p-p
V
S
=
5V
g = +2
R
F
= 1k
R
L
= 1k
频率 – mhz
12
–12
10
1000
100
0
–9
–6
–3
3
6
9
V
O
= 0.5v p-p
V
O
= 1v p-p
V
O
= 4v p-p
V
O
= 2v p-p
normalized 增益 – db
图示 3. 带宽 vs. 输出 电压 level—
双 供应, g = +2
频率 – mhz
2.0
–7.0
1 100010010
–6.0
–5.0
–4.0
–3.0
–1.0
0
1.0
–2.0
normalized 增益 – db
V
S
=
5V
g = –1
R
F
= 1k
R
L
= 1k
V
O
= 2v
V
O
= 4v
V
O
= 0.2v
V
O
= 0.5v
V
O
= 1v
图示 4. 带宽 vs. 输出 level—gain 的 –1, 双
供应
频率 – mhz
12
1 100010010
–12
–9
–6
–3
3
6
9
0
normalized 增益 – db
V
S
= +5v
g = +2
R
F
= 1k
R
L
= 1k
V
O
= 1v p-p
V
O
= 3v p-p
V
O
= 2v p-p
V
O
= 0.5v p-p
图示 5. 带宽 vs. 输出 level—single 供应,
g = +2
频率 – mhz
2
1 100010010
–8
–7
–5
–4
–2
0
1
–3
–6
–1
normalized 增益 – db
V
S
= +5v
g = –1
R
F
= 1k
R
L
= 1k
V
O
= 2v p-p
V
O
= 0.2v p-p
V
O
= 4v p-p
V
O
= 0.5v p-p
图示 6. 带宽 vs. 输出 level—single 供应,
增益 的 –1
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