AD829
Figure 11. Large Signal Pulse Response of Inverting
Amplifier Using Current Feedback Compensation,
CCOMP = 15 pF, C1 = 15 pF, RF = 1 kΩ, R1 = 1 kΩ
Figure 13. Large Signal Pulse Response of the Inverting
Amplifier Using Current Feedback Compensation,
C
COMP = 1 pF, RF = 3 kΩ, R1 = 3 kΩ
15
GAIN = –4
12
–3dB @ 8.2MHz
9
GAIN = –2
6
–3dB @ 9.6MHz
3
GAIN = –1
0
–3dB @ 10.2MHz
–3
V
V
R
R
C
C
= –30dBM
= ؎15V
= 1k⍀
IN
–6
–9
S
L
F
= 1k⍀
= 15pF
–12
–15
COMP
= 15pF
1
100k
1M
10M
FREQUENCY (Hz)
100M
Figure 14. Small Signal Pulse Response of Inverting
Amplifier Using Current Feedback Compensation,
CCOMP = 4 pF, RF = 1 kΩ, R1 = 1 kΩ
Figure 12. Closed-Loop Gain vs. Frequency for the Circuit
of Figure 9
Figure 13 is an oscilloscope photo of the pulse response of a
unity gain inverter that has been configured to provide a small
signal bandwidth of 53 MHz and a subsequent slew rate of
180 V/µs; resistor RF = 3 kΩ and capacitor CCOMP = 1 pF. Figure 14
shows the excellent pulse response as a unity gain inverter, this
using component values of RF = 1 kΩ and CCOMP = 4 pF.
15
C
= 2pF
= 3pF
= 4pF
GAIN = –4
GAIN = –2
COMP
12
9
C
COMP
6
3
Figures 15 and 16 show the closed-loop frequency response of the
AD829 for different closed-loop gains and different supply voltages.
GAIN = –1
C
COMP
0
–3
–6
–9
–12
–15
If a noninverting amplifier configuration using current feedback
compensation is needed, the circuit of Figure 17 is recommended.
This circuit provides a slew rate twice that of the shunt com-
pensated noninverting amplifier of Figure 18 at the expense of
gain flatness. Nonetheless, this circuit delivers 95 MHz bandwidth
with 1 dB flatness into a back terminated cable, with a differ-
ential gain error of only 0.01% and a differential phase error of
only 0.015° at 4.43 MHz.
V
= ؎15V
= 1k⍀
= 1k⍀
S
R
R
V
L
F
= –30dBM
IN
1
10
FREQUENCY (MHz)
100
Figure 15. Closed-Loop Frequency Response for the
Inverting Amplifier Using Current Feedback Compensation
–12–
REV. G
ADI [ ADI ]
