ada4930-1 Analog Devices, Inc., ada4930-1 Datasheet - Page 22

• Manufacturer Part Number: ada4930-1
• Description: Ultralow Noise Drivers For Low Voltage Adcs Ada4930-1/ada4930-2
• Manufacturer: Analog Devices, Inc.

ADA4930-1/ADA4930-2
Input Common-Mode Adjustment with Resistors
The circuit shown in Figure 54 shows an alternate method to
bias the amplifier inputs, eliminating the dc source.
Define β1 = R
and R
Set R
Write a nodal equation at V
Determine V
voltage from the Specifications section. For a 3.3 V supply,
V
Determine the minimum input voltage, V
the source. Recognize that once properly terminated, the source
voltage is ½ of its open circuit value. Therefore, V
Rearrange the V
P min
V
SOURCE
2V p-p
F1
V
R
N
= 0.3 V.
P
CM
1
= R
= R
=
Figure 54. Single-Supply Biasing Scheme with Resistors
=
F2
β1
G2
β1β2
50Ω
2
R
||R
= R
P min
+
V
S
P
/R
1
S
β2
CM
R
F
P
. This is the minimum input common-mode
F1
equation for R
F
to maintain a balanced system, as shown.
||R
⎛
⎜
⎜
⎝
R
V
V
and β2 = R
⎛
⎜
⎜
⎝
T
3.3V
S
S
3.3V
R
R
F2
β1
G
F
V
β1β2
.
1
IN
V
+
R
R
V
301Ω
IN
CM
CM
β2
R
OCM
301Ω
R
G2
P
G1
+
V
and solve for V
2
P
N
V
/R
min
CM
OCM
F2
−
301Ω
301Ω
ADA4930
, where R
R
R
+
R
F1
F2
R
+V
–V
G
V
F
1
S
S
S
V
R
2
IN min
IN
R
CM
min
F
P
P
.
= R
⎞
⎟
⎟
⎠
at the output of
−
2
G1
IN min
V
R
||R
L
OCM
V
CM
= −0.5 V.
OUT, dm
⎞
⎟
⎟
⎠
||R
F1
Rev. A | Page 22 of 28
Calculate the following:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Figure 55. Single-Supply, Single-Ended Input System with Bias Resistors
2V p-p
β1 and β2. For the circuit shown in Figure 54, β1 = 0.5 and
β2 = 0.5.
R
The new values for β1 and β2. β1 = 0.4925 and β2 = 0.4925.
The input impedance using the following:
R
V
The new values for R
R
The new values for β1 and β2. β1 = 0.284 and β2 = 0.317.
The new value of R
Repeat Step 3 through Step 8 until the values of R
remain constant between iterations. After four iterations,
the final circuit is shown in Figure 55.
R
R
CM
T
G1
V
TH
IN
, R
S
IN-SE
−
= 133.88 Ω.
for V
= 1.067 V.
SE
TH
50Ω
= 399.35 Ω.
R
, and V
=
S
R
P min
G1
65.1Ω
⎛
⎜
⎜
⎜
⎜
⎝
= 0.3 V and V
1
TH
+V
R
+V
−
T
. R
S
S
1
V
1.87kΩ
1.87kΩ
V
R
R
INP
T
CM
142Ω
170Ω
P
V
CM
CM
R
R
= 57.16 Ω, R
OCM
G1
G1
G2
. R
⎞
⎟
⎟
⎟
⎟
⎠
and R
=
CM
R
= 4759.63 Ω.
G1
IN min
G2
301Ω
301Ω
⎛
⎜
⎜
⎜
⎜
⎝
ADA4930
R
R
. R
β1
F1
F2
+V
–V
= −0.5 V. R
TH
G2
S
S
+
β2
= 26.67 Ω, and
= 160.55 Ω and
β1
−
+
R
R
β2
G1
F1
CM
R
β1β2
L
= 9933 Ω.
V
OUT, dm
G1
⎞
⎟
⎟
⎟
⎟
⎠
and R
G2