Datasheet

AD8139
Rev. B | Page 20 of 24
For proper operation, the voltages at V
AN
and V
AP
must stay
within their respective linear ranges.
Calculating Input Impedance
The input impedance of the circuit in Figure 59 depends on
whether the amplifier is being driven by a single-ended or a
differential signal source. For balanced differential input
signals, the differential input impedance (R
IN, dm
) is simply
R
IN, dm
= 2R
G
(18)
For a single-ended signal (for example, when V
IN
is grounded
and the input signal drives V
IP
), the input impedance becomes
)(2
1
F
G
F
G
IN
RR
R
R
R
+
−
=
(19)
The input impedance of a conventional inverting op amp
configuration is simply R
G
, but it is higher in Equation 19
because a fraction of the differential output voltage appears at
the summing junctions, V
AN
and V
AP
. This voltage partially
bootstraps the voltage across the input resistor R
G
, leading to
the increased input resistance.
Input Common-Mode Swing Considerations
In some single-ended-to-differential applications, when using a
single-supply voltage, attention must be paid to the swing of the
input common-mode voltage, V
ACM
.
Consider the case in
Figure 61, where V
IN
is 5 V p-p swinging
about a baseline at ground, and V
REF
is connected to ground.
The circuit has a differential gain of 1.6 and β = 0.38. V
ICM
has
an amplitude of 2.5 V p-p and is swinging about ground. Using
the results in Equation 16, the common-mode voltage at the
inputs of the AD8139, V
ACM
, is a 1.5 V p-p signal swinging
about a baseline of 0.95 V. The maximum negative excursion
of V
ACM
in this case is 0.2 V, which exceeds the lower input
common-mode voltage limit.
DGND AGND REFGND REF REFBUFIN PDBUF
ADR431
2.5V
REFERENCE
47µF
AD7674
IN+
IN–
AVDD
DVDD
AD8139
+
–
8
V
REF
2.5V
2
1
6
3
4
5
V
OCM
200Ω
200Ω 324Ω
324Ω
5
V
20Ω
15Ω
15Ω
V
IN
2.7nF
2.7nF
0.1µF
0.1µF 0.1µF 0.1µF
+1.7V
+0.95V
V
ACM
WITH V
REF
= 0
+0.2V
+2.5V
GND
–2.5V
04679-052
Figure 61. AD8139 Driving AD7674, 18-Bit, 800 kSPS ADC