AD8132ARZ Analog Devices Inc, AD8132ARZ Datasheet - Page 22

• Manufacturer Part Number: AD8132ARZ
• Description: IC AMP DIFF LDIST LP 70MA 8SOIC
• Manufacturer: Analog Devices Inc
• Type: Differential Ampr

Specifications of AD8132ARZ

• Amplifier Type: Differential
• Number Of Circuits: 1
• Output Type: Differential
• Slew Rate: 1200 V/µs
• -3db Bandwidth: 360MHz
• Current - Input Bias: 3µA
• Voltage - Input Offset: 1000µV
• Current - Supply: 12mA
• Current - Output / Channel: 70mA
• Voltage - Supply, Single/dual (±): 2.7 V ~ 11 V, ±1.35 V ~ 5.5 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (3.9mm Width)
• Number Of Channels: 1
• Number Of Elements: 1
• Power Supply Requirement: Single/Dual
• Common Mode Rejection Ratio: 70dB
• Voltage Gain Db: 0.129dB
• Input Resistance: 3@5VMohm
• Input Offset Voltage: 3.5@5VmV
• Input Bias Current: 7@5VnA
• Single Supply Voltage (typ): 3/5/9V
• Dual Supply Voltage (typ): ±3/±5V
• Power Supply Rejection Ratio: 70dB
• Power Dissipation: 250mW
• Rail/rail I/o Type: No
• Single Supply Voltage (min): 2.7V
• Single Supply Voltage (max): 11V
• Dual Supply Voltage (min): ±1.35V
• Dual Supply Voltage (max): ±5.5V
• Operating Temp Range: -40C to 125C
• Operating Temperature Classification: Automotive
• Mounting: Surface Mount
• Pin Count: 8
• Package Type: SOIC N
• No. Of Amplifiers: 1
• Gain Db Max: 1.015dB
• Bandwidth: 350MHz
• Supply Voltage Range: ± 1.35V To ± 5.5V
• Supply Current: 10.7mA
• Rohs Compliant: Yes
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Gain Bandwidth Product: -
• Lead Free Status / RoHS Status: Compliant, Lead free / RoHS Compliant

AD8132
THEORY OF OPERATION
The AD8132 differs from conventional op amps by the external
presence of an additional input and output. The additional
input, V
additional output is the analog complement of the single output
of a conventional op amp. For its operation, the AD8132 uses two
feedback loops as compared to the single loop of conventional
op amps. Although this provides significant freedom to create
various novel circuits, basic op amp theory can still be used to
analyze the operation.
One of the feedback loops controls the output common-mode
voltage, V
common mode, or average voltage, of the two differential outputs
(+OUT and −OUT). The gain of this circuit is internally set to
unity. When the AD8132 is operating in its linear region, this
establishes one of the operational constraints: V
The second feedback loop controls the differential operation.
Similar to an op amp, the gain and gain shaping of the transfer
function can be controlled by adding passive feedback networks.
However, only one feedback network is required to close the
loop and fully constrain the operation, but depending on the
function desired, two feedback networks can be used. This is
possible because there are two outputs that are each inverted
with respect to the differential inputs.
GENERAL USAGE OF THE AD8132
Several assumptions are made here for a first-order analysis; they
are the typical assumptions used for the analysis of op amps.
•
•
•
•
Though it is possible to operate the AD8132 with a purely
differential input, many of its applications call for a circuit
that has a single-ended input with a differential output.
For a single-ended-to-differential circuit, the R
is not driven is tied to a reference voltage or to ground. Additional
conditions are discussed in the following sections. In addition,
the voltage at V
Figure 67 shows a generalized schematic of such a circuit using an
AD8132 with two feedback paths.
The input bias currents are sufficiently small so they can be
neglected.
The output impedances are arbitrarily low.
The open-loop gain is arbitrarily large and drives the
amplifier to a state where the input differential voltage is
effectively 0.
Offset voltages are assumed to be 0.
OCM
OUT, cm
, controls the output common-mode voltage. The
OCM
. Its input is V
, and therefore V
OCM
(Pin 2) and the output is the
OUT, cm
, is assumed to be ground.
G
OUT, cm
of the input that
= V
OCM
Rev. I | Page 22 of 32
.
For each feedback network, a feedback factor can be defined as
the fraction of the output signal that is fed back to the opposite
sign input. These terms are
The feedback factor, β1, is for the side that is driven, and the
feedback factor, β2, is for the side that is tied to a reference
voltage (ground). Note that each feedback factor can vary
anywhere between 0 and 1.
A single-ended-to-differential gain equation can be derived
(this is true for all values of β1 and β2) from
This expression is not very intuitive, but some further examples
can provide better understanding of its implications. One
observation that can be made immediately is that a tolerance
error in β1 does not have the same effect on gain as the same
tolerance error in β2.
DIFFERENTIAL AMPLIFIER WITHOUT RESISTORS
(HIGH INPUT IMPEDANCE INVERTING AMPLIFIER)
The simplest closed-loop circuit that can be made does not
require any resistors and is shown in Figure 70. In this circuit,
β1 is equal to 0, and β2 is equal to 1. The gain is equal to 2.
A more intuitive method to figure the gain is by simple inspection.
+OUT is connected to −IN, whose voltage is equal to the voltage at
+IN under equilibrium conditions. Therefore, +V
V
swing in the opposite direction from +OUT due to the common-
mode constraint, its effect doubles the output signal and
produces a gain of 2.
One useful function that this circuit provides is a high input
impedance inverter. If +OUT is ignored, there is a unity-gain,
high input impedance amplifier formed from +IN to −OUT.
Most traditional op amp inverters have relatively low input
impedances, unless they are buffered with another amplifier.
V
constraint that +V
does not change +V
changing V
For example, if V
by 2 V. This makes V
as the average of the two differential output voltages. This means
that the gain from V
IN
OCM
, and there is unity gain in this path. Because −OUT has to
β1 = R
β2 = R
G
is assumed to be at midsupply. Because there is still the
=
2
(
β1
OCM
G1
G2
(
1
/(R
/(R
−
+
must show up at −OUT.
β1
β2
G1
G2
OCM
OUT
)
)
+ R
+ R
OUT
OUT, cm
OCM
must equal V
is raised by 1 V, then −V
F1
F2
(equal to V
)
)
to the differential output is 2.
also increase by 1 V because it is defined
IN
IN
). Therefore, the effect of
, changing the V
OUT
must increase
OUT
OCM
is equal to
voltage