Datasheet

ADA4930-1/ADA4930-2
Rev. A | Page 21 of 28
Terminating a Single-Ended Input in a Single-Supply
Applications
When the application circuit of Figure 50 is powered by a single
supply, the common-mode voltage at the amplifier inputs, V
P
and V
N
, may have to be raised to comply with the specified input
common-mode range. Two methods are available: a dc bias on
the source, as shown in Figure 51, or by connecting resistors R
CM
between each input and the supply, as shown on Figure 54.
Input Common-Mode Adjustment with DC Biased Source
To drive a 1.8 V ADC with V
CM
= 1 V, a 3.3 V single supply
minimizes the power dissipation of the ADA4930-1/ADA4930-2.
The application circuit of Figure 50 on a 3.3 V single supply with a
dc bias added to the source is shown in Figure 51.
ADA4930
R
L
V
OUT, dm
1.990V p-p
3.3V
R
S
50
R
G1
142
V
P
V
N
R
G2
142
R
F2
301
R
F1
301
V
OCM
V
S
2V p-p
V
DC
R
T
64.2
64.2
50
09209-151
Figure 51. Single-Supply, Terminated Single-Ended-to-Differential System with G = 1
To determine the minimum required dc bias, the following steps
must be taken:
1.
Convert the terminated inputs to their Thevenin equivalents,
as shown in the Figure 52 circuit.
ADA4930
R
L
V
OUT, dm
1.99V p-p
3.3V
VON
VOP
R
TH
28.11
R
G1
142
V
P
V
N
R
G2
142
R
F2
301
R
F1
301
V
OCM
V
TH
1.124V p-p
V
DC-TH
0
9209-159
R
TH
28.11
Figure 52. Thevenin Equivalent of Single-Supply Application Circuit
2. Write a nodal equation for V
P
or V
N
.
()
THDCTH
ON
THDCTH
P
VVVVVV
−−
−−
++
++=
28.11142301
301
OP
THDC
N
VVV
28.11142301
301
++
+=
−
Recognize that while the ADA4930-1/ADA4930-2 is in its
linear operating region, V
P
and V
N
are equal. Therefore,
both equations in Step 2 give equal results.
3.
To comply with the minimum specified input common-mode
voltage of 0.3 V at V
S
= 3.3 V, set the minimum value of V
P
and V
N
to 0.3 V.
4.
Recognize that V
P
and V
N
are at their minimum values when
V
OP
and V
S
are at their minimum (and therefore V
ON
is at its
maximum).
Let
V
P min
= V
N min
= 0.3 V, V
OCM
= V
CM
= 1 V, V
TH min
= −V
TH
/2
V
ON max
= V
OCM
+ V
OUT, dm
/4 and V
OP min
= V
OCM
− V
OUT, dm
/4
Substitute conditions into the nodal equation for V
P
and solve
for V
DC-TH
.
0.3 = −1.124/2 + V
DC-TH
+ 0.361 × (1 + 1.99/4 = 1.124/2 – V
DC-TH
)
0.3 + 0.562 − 0.361 − 0.18 − 0.203 = 0.639 V
DC-TH
V
DC-TH
= 0.186 V
Or
Substitute conditions into the nodal equation for V
N
and
solve for V
DC-TH
.
0.3 = V
DC-TH
+ 0.361 × (1 − 1.99/4 − V
DC-TH
)
0.3 – 0.361 + 0.18 = 0.639 × V
DC-TH
V
DC-TH
= 0.186 V
5.
Converting V
DC-TH
from its Thevenin equivalent results in
V 0.330.186 =×
+
=
TH
TH
S
DC
R
RR
V
The final application circuit is shown in Figure 53. The
additional dc bias of 0.33 V at the inputs ensures that the
minimum input common-mode requirements are met when
the source signal is bipolar with a 2 V p-p amplitude and
V
OCM
is at 1 V.
3.3V
ADA4930
R
L
V
OUT, dm
1.990V p-p
R
S
50
R
G1
142
R
G2
142
R
F2
301
R
F1
301
V
OCM
V
S
2V p-p
R
T
64.2
64.2
09209-160
V
P
V
N
50
V
DC
0.33V
Figure 53. Single-Supply Application Circuit with DC Source Bias