AD8014ART-EBZ View Datasheet(PDF) - Analog Devices
Part Name
Description
Manufacturer
AD8014ART-EBZ Datasheet PDF : 11 Pages
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Typical Performance Characteristics– AD8014
15
G = +1
12 VO = 200mV p-p
9 RF = 1k⍀
RL = 1k⍀
6
VS = ؎5V
3
VS = +5V
0
–3
–6
–9
–12
–15
1
10
100
FREQUENCY – MHz
1000
Figure 1. Frequency Response, G = +1, VS = ±5 V and +5 V
2.0
1.0
0
–1.0
VO = 0.2V
VO = 0.5V
–2.0
–3.0
–4.0
–5.0
–6.0
–7.0
1
VS = ؎5V
G = –1
RF = 1k⍀
RL = 1k⍀
VO = 1V
VO = 2V
VO = 4V
10
100
FREQUENCY – MHz
1000
Figure 4. Bandwidth vs. Output Level—Gain of –1, Dual
Supply
12
9
6
3
VS = ؎5V
0 G = +2
RF = 500⍀
–3
VO = 2V p-p
–6
RL = 75⍀
RL = 50⍀
–9
–12
–15
1
10
100
FREQUENCY – MHz
1000
Figure 2. Frequency Response, G = +2, VO = 2 V p-p
12
9
VO = 0.5V p-p
VO = 1V p-p
6
3
0
–3
VO = 2V p-p
–6
–9
–12
1
VS = +5V
G = +2
RF = 1k⍀
RL = 1k⍀
VO = 3V p-p
10
100
FREQUENCY – MHz
1000
Figure 5. Bandwidth vs. Output Level—Single Supply,
G = +2
12
9
VO = 0.5V p-p
6
VO = 1V p-p
3
0
VO = 4V p-p
–3
–6 VS = ؎5V
G = +2
–9 RF = 1k⍀
RL = 1k⍀
–12
10
VO = 2V p-p
100
FREQUENCY – MHz
1000
Figure 3. Bandwidth vs. Output Voltage Level—
Dual Supply, G = +2
2
1
VO = 0.5V p-p
0
–1
–2
–3
–4
VO = 4V p-p
VO = 2V p-p
–5
–6
–7
–8
1
VS = +5V
G = –1
RF = 1k⍀
RL = 1k⍀
VO = 0.2V p-p
10
100
FREQUENCY – MHz
1000
Figure 6. Bandwidth vs. Output Level—Single Supply,
Gain of –1
Rev. C
–5–
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