|
The Instrumentation Amplifier.
Analog Devices defines an instrumentation amplifier or I-amp as "a closed loop gain
block that has a differential input and an output terminal that is single-ended with
respect to a reference terminal." Also, unlike an op-amp which sets it closed loop gain
by a external resistor network connected between its input and output, an I-amp uses an
internal isolated gain network.

In Figure 4, a classic three op-amp instrumentation amplifier is shown. The three op-amp
is the most straight forward implementation of an Instrumentation Amplifier. It consists of
two non-inverting input buffer amplifiers, followed by a difference amplifier. It should
be noted that there are a number of other implementations using two amplifiers for example,
each with their own weaknesses and strong points, but still preserving the basic definition
of an Instrumentation Amplifier above.
There are three major advantages to this circuit arrangement. First, if the buffer
amplifiers are matched, the input impedances are identical, and very high.
Second, the feedback network is completely isolated from the signal inputs.
Finally, as we will see, the gain of the amplifier is set by a single resistor Rg.
Analysis of the circuit is straight forward.
Using superposition, when vin+ = 0:
va = vin- x (R1 + Rg) / Rg
vb = vin- x (R1' / Rg)
When vin- = 0,
vb = vin+ × (R1' + Rg) / Rg
va = vin+ × (R1 / Rg)
Combining terms and subtracting we get::
(vb - va) = (Vin+ × ((R1'+ Rg) / Rg ) - Vin- x (R1' / Rg))
- (Vin- x ((R1 + Rg) / Rg) - Vin+ x (R1 / Rg))
If R1 = R1'
= Vin+ x (2R1 / Rg + 1) - Vin- x (2R1 / Rg + 1)
= (Vin+ - Vin-) x (2R1 / Rg + 1)
We know from above that the output of the difference amplifier stage, if the
resistors are matched:
Vout = (vb - va) × (R3 / R2)
Substituting for vb and va, we get
Vout = (Vin+ - Vin-) x (2R1 / Rg + 1) × (R3 / R2)
We know that common mode rejection depends upon close matching of the resistors of the
difference amplifier, but what effect does a mis- match of R1 and R1' at the inputs
have on common mode voltage error? It turns out none.
Vcm = (vb - va) = (Vin+ × ((R1' + Rg) / Rg ) - Vin- × (R1' / Rg))
- (Vin- × ((R1 + Rg) / Rg) - Vin+ × (R1/Rg))
but Vcm = Vin+ = Vin- by definition
Voutc = Vcm ((R1'+ Rg)/Rg - R1'/Rg - R1'/Rg - ( R1+Rg)/Rg + R1/Rg) )
= Vcm (R1'/Rg + 1 - R1'/Rg - R1/Rg - 1 + R1'/Rg)
= Vcm (0)
The gain of the amplifier can be set by the single resistor Rg, with no increased common
mode voltage error.
|