AD829
Because the closed-loop bandwidth is a function of RF and
CCOMP (see Figure 39), it is independent of the amplifier closed-
loop gain, as shown in Figure 41. To preserve stability, the time
constant of RF and CCOMP needs to provide a bandwidth of
<65 MHz. For example, with CCOMP = 15 pF and RF = 1 kΩ, the
small signal bandwidth of the AD829 is 10 MHz. Figure 40
shows that the slew rate is in excess of 60 V/µs. As shown in
Figure 41, the closed-loop bandwidth is constant for gains of
−1 to −4; this is a property of the current feedback amplifiers.
RF
CCOMP
50Ω
COAX
CABLE R1
VIN
C1*
50Ω
0.1µF
+VS
IN4148
7
2–
5
AD829 6
3+ 4
VOUT
RL
1kΩ
0.1µF
–VS
*RECOMMENDED VALUE
OF CCOMP FOR C1
<7pF
≥7pF
0pF
15pF
CCOMP SHOULD NEVER EXCEED
15pF FOR THIS CONNECTION
Figure 39. Inverting Amplifier Connection Using Current Feedback
Compensation
5V
100%
90
200ns
Data Sheet
Figure 42 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;
RF = 3 kΩ and CCOMP = 1 pF. Figure 43 shows the excellent pulse
response as a unity-gain inverter, this using component values
of RF = 1 kΩ and CCOMP = 4 pF.
5V
200ns
100%
90
10
0%
Figure 42. Large Signal Pulse Response of the Inverting Amplifier Using
Current Feedback Compensation, CCOMP = 1 pF, RF = 3 kΩ, R1 = 3 kΩ
100%
90
10ns
10
0%
Figure 40. Large Signal Pulse Response of Inverting Amplifier Using Current
Feedback Compensation, CCOMP = 15 pF, C1 = 15 pF RF = 1 kΩ, R1 = 1 kΩ
15
12 GAIN = –4
–3dB @ 8.2MHz
9
GAIN = –2
6
–3dB @ 9.6MHz
3
GAIN = –1
0
–3dB @ 10.2MHz
–3
–6 VIN = –30dBm
–9
VS = ±15V
RL = 1kΩ
RF = 1kΩ
–12 CCOMP = 15pF
C1 = 15pF
–15
100k
1M
10M
FREQUENCY (Hz)
100M
Figure 41. Closed-Loop Gain vs. Frequency for the Circuit of Figure 38
10
0%
20mV
Figure 43. Small Signal Pulse Response of Inverting Amplified Using Current
Feedback Compensation, CCOMP = 4 pF, RF = 1 kΩ, R1 = 1 kΩ
Rev. I | Page 14 of 20