AD8002 Datasheet, PDF (6/18 Pages) Analog Devices – Dual 600 MHz, 50 mW Current Feedback Amplifier

AD8002
+2
V_IN = 50mV
G = +1
+1 _{RF = 953⍀}
R_L = 100⍀
0
–1
SIDE 1
SIDE 2
–2
–3
75⍀
50⍀
50⍀
–4
953⍀
–5
–6
1M
10M
100M
1G
FREQUENCY – Hz
Figure 16. Frequency Response, G = +1
–40
G = +1
–50
R_L = 100⍀
V_OUT = 2V p-p
–60
–70
2ND HARMONIC
–80
3RD HARMONIC
–90
–100
10k
100k
1M
10M
FREQUENCY – Hz
100M
Figure 17. Distortion vs. Frequency, G = +1, R_L = 100 Ω
–40
G = +1
–50 ^{RL = 1k⍀}
–60
–70
2ND HARMONIC
3RD HARMONIC
–80
–90
–100
–110
10k
100k
1M
10M
FREQUENCY – Hz
100M
Figure 18. Distortion vs. Frequency, G = +1, R_L = 1 kΩ
0
+6
–3
+3
–6
0
–9
–3
–12
–6
–15
–9
–18
G = +2
–12
R_F = 681⍀
–21
V_S = ؎5V
–15
R_L = 100⍀
–24
–18
–27
1M
10M
100M
FREQUENCY – Hz
–21
500M
Figure 19. Large Signal Frequency Response, G = +2
+9
+6
R_L = 100⍀
G = +1
+3
R_F = 1.21k⍀
0
–3
–6
75⍀
–9
50⍀
50⍀
–12
–15
1.21k⍀
–18
–27
1M
10M
100M
FREQUENCY – Hz
500M
Figure 20. Large Signal Frequency Response, G = +1
+45
+40
V_S = ؎5V
+35
R_L = 100⍀
+30
G = +100
+25
R_F = 1000⍀
+20
G = +10
+15
R_F = 499⍀
+10
+5
0
–5
1M
10M
100M
1G
FREQUENCY – Hz
Figure 21. Frequency Response, G = +10, G = +100
–6–
REV. C