AD826AR-REEL Analog Devices Inc, AD826AR-REEL Datasheet - Page 12

• Manufacturer Part Number: AD826AR-REEL
• Description: IC,Operational Amplifier,DUAL,BIPOLAR,SOP,8PIN,PLASTIC
• Manufacturer: Analog Devices Inc

Specifications of AD826AR-REEL

• Rohs Status: RoHS non-compliant
• Amplifier Type: Voltage Feedback
• Number Of Circuits: 2
• Slew Rate: 350 V/µs
• Gain Bandwidth Product: 50MHz
• -3db Bandwidth: 50MHz
• Current - Input Bias: 3.3µA
• Voltage - Input Offset: 500µV
• Current - Supply: 6.6mA
• Current - Output / Channel: 50mA
• Voltage - Supply, Single/dual (±): 5 V ~ 36 V, ±2.5 V ~ 18 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (3.9mm Width)
• Output Type: -

SINGLE-ENDED TO DIFFERENTIAL LINE DRIVER
Outstanding CMRR (> 80 dB @ 5 MHz), high bandwidth, wide
supply voltage range, and the ability to drive heavy loads, make
the AD826 an ideal choice for many line driving applications.
In this application, the AD830 high speed video difference
amp serves as the differential line receiver on the end of a back
terminated, 50 ft., twisted-pair transmission line (see Figure 40).
The overall system is configured in a gain of +1 and has a –3 dB
bandwidth of 14 MHz. Figure 39 is the pulse response with a
2 V p-p, 1 MHz signal input.
AD826
LOW DISTORTION LINE DRIVER
The AD826 can quickly be turned into a powerful, low distor-
tion line driver (see Figure 41). In this arrangement the AD826
can comfortably drive a 75 Ω back-terminated cable, with a
5 MHz, 2 V p-p input; all of this while achieving the harmonic
distortion performance outlined in the following table.
Configuration
1. No Load
2. 150 Ω R
3. 150 Ω R
In this application one half of the AD826 operates at a gain of
2.1 and supplies the current to the load, while the other pro-
vides the overall system gain of 2. This is important for two
reasons: the first is to keep the bandwidth of both amplifiers the
same, and the second is to preserve the AD826’s ability to oper-
ate from low supply voltages. R
be chosen to satisfy the following equation:
where M is defined by [(M+ 1) G
G
S
= System Gain.
L
L
Only
7.5 Ω R
I
N
BNC
1.05k
1.05k
5pF
5pF
C
R
C
–15V
= MR
15V
1/2
AD826
1/2
AD826
C
1.05k
1.05k
S
varies with the load and must
= G
0.01 F
0.01 F
0.1 F
0.1 F
L
D
] and G
2.2 F
2.2 F
2nd Harmonic
–78.5 dBm
–63.8 dBm
–70.4 dBm
36
36
D
= Driver’s Gain,
50 FEET TWISTED PAIR
Z = 72
75
36
36
100
90
10
0%
1k
1k
2V
2V
AD830
1/2
AD826
1/2
AD826
V
–15V
1.1k
S
15V
1k
0.1 F
0.1 F
1 F
1 F
0.01 F
0.1 F
0.01 F
0.1 F
R
7.5
C
75
200ns
V
OUT
R
L
REV. C
75