LMV716 Datasheet, PDF (11/25 Pages) National Semiconductor (TI) – 5 MHz, Low Noise, RRO, Dual Operational Amplifier with CMOS Input

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Three-Op-Amp Instrumentation Amplifier
A typical instrumentation amplifier is shown in Figure 30.
V₁
+
V₀₁
R₂
-
R₁
R₁
R₁₁ = _a
-
V₂
+
R₁
V₀₂
R₂
LMV716
SNOSAT9B – APRIL 2006 – REVISED MARCH 2013
KR₂
-
+
V_OUT
KR₂
Figure 30. Three-Op-Amp Instrumentation Amplifier
There are two stages in this configuration. The last stage, the output stage, is a differential amplifier. In an ideal
case the two amplifiers of the first stage, the input stage, would be set up as buffers to isolate the inputs.
However they cannot be connected as followers due to the mismatch of real amplifiers. The circuit in Figure 30
utilizes a balancing resistor between the two amplifiers to compensate for this mismatch. The product of the two
stages of gain will be the gain of the instrumentation amplifier circuit. Ideally, the CMRR should be infinite.
However the output stage has a small non-zero common mode gain which results from resistor mismatch.
In the input stage of the circuit, current is the same across all resistors. This is due to the high input impedance
and low input bias current of the LMV716. With the node equations we have:
GIVEN: I _R1 = I_R11
(3)
By Ohm’s Law:
V_O1 - V_O2 = (2R₁ + R₁₁) IR₁₁
= (2a + 1) R₁₁ x I_R11
= (2a + 1) V _R11
(4)
However:
V _{R11 = V1 - V2}
(5)
So we have:
(6)
Now looking at the output of the instrumentation amplifier:
KR₂
V_O = _R2 (V_O2 - V_O1)
= -K (V_O1 - V_O2)
(7)
Substituting from Equation 6:
V_O = -K (2a + 1) (V₁ - V₂)
(8)
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