NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Instrumentation Amplifier
An instrumentation amplifier is a circuit which has an output of
V o = k 1 (V a − V b )
where the gain, k1, can be adjusted. Several variations follow:
Single Op-Amp Design:
R1a
R2a
Va
Vp
+
Vo
Vm
-
R1b
R2b
Vb
Single Op-Amp Instrumentation Amplifier
Again, we have three voltage nodes so we need to write three equations:
Vp = Vm
⎛ V m −V b ⎞ + ⎛ V m −V o ⎞ = 0
⎝ R 2b ⎠ ⎝ R 1b ⎠
⎛ V p −V a ⎞ + ⎛ V p ⎞ = 0
⎝ R 2a ⎠ ⎝ R 1a ⎠
Solving
R +R
R
R
V o = ⎛⎝ R 1a2b ⎞⎠ ⎛⎝ R 1b1a +R 2b2a ⎞⎠ V a − ⎛⎝ R 1b2b ⎞⎠ V b
If R1a = R1b and R2a = R2b, then
R
V o = ⎛⎝ R 12 ⎞⎠ (V a − V b )
The problems with this amplifier are
If you want to adjust the gain, you need to adjust two resistors (R1a and R1b)
If the resistor pairs are not exactly the same, you get a common-mode gain (there is a term which
includes Va + Vb).
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Two Op-Amp Instrumentation Amplifier
R2
R1
Vm1
Vb
Vp1
Vx
-
R4
R3
Vm2
Vp2
+
Vo
+
Va
Two op-amp instrumentation amplifier
The node equations are:
V b = V p1 = V m1
V a = V p2 = V m2
⎛ V m1 ⎞ + ⎛ V m1 −V x ⎞ = 0
⎝ R1 ⎠ ⎝ R2 ⎠
⎛ V m2 −V x ⎞ + ⎛ V m2 −V o ⎞ = 0
⎝ R3 ⎠ ⎝ R4 ⎠
Simplifying:
R
R
R
V o = ⎛⎝ 1 + R 43 ⎞⎠ V a − ⎛⎝ R 43 ⎞⎠ ⎛⎝ 1 + R 21 ⎞⎠ V b
This circuit has a high input impedance (good) but still requires you to adjust two resistors to adjust the
gain.
Example: Find R1, R2, R3, and R4 so that the gain is
V o = 10(V a − V b )
Solution:
⎛ 1 + R 4 ⎞ = 10
R3 ⎠
⎝
let R3 = 1k, R4 = 9k
⎛ R 4 ⎞ ⎛ 1 + R 2 ⎞ = 10
R1 ⎠
⎝ R3 ⎠ ⎝
R2
R1
= 0.1111
Let R1 = 10k, R2 = 1.111k
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Two Op-Amp Instrumentation Amplifier with a DC Offset: (AMP04)
R1
Rg
Vb
Vm2
-
Va
Vp2
Vo
+
R2
+
Vp1
Vm1
R1
R2
Vref
Two Op-Amp Instrumentation Amplifier with a DC Offset:
The equations for this circuit are:
V m1 = V p1 = V a
V m2 = V p2 = V a
⎛ V m2 −V b ⎞ + ⎛ V m2 −V o ⎞ + ⎛ V m2 −V x ⎞ = 0
⎝ Rg ⎠ ⎝ R1 ⎠ ⎝ R2 ⎠
⎛ V m1 −V ref ⎞ + ⎛ V m1 −V x ⎞ = 0
⎝ R1 ⎠ ⎝ R2 ⎠
Substituting:
R
V o = ⎛⎝ R 1g ⎞⎠ (V a − V b ) + V ref
Note here:
By adjusting a single resistor, Rg, you can adjust the gain.
You can also provide a DC offset to the output with Vref. This is useful when the output needs a
DC offset, such as 2.5V, with a signal riding on top of this offset.
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Three Op-Amp Instrumentation Amplifier:
Three Op-Amp Instrumentation Amplifier
The voltage node equations are:
⎛ V 1 −V 3 ⎞ + ⎛ V 1 −V 2 ⎞ = 0
⎝ R ⎠ ⎝ Rg ⎠
⎛ V 2 −V 4 ⎞ + ⎛ V 2 −V 1 ⎞ = 0
⎝ R ⎠ ⎝ Rg ⎠
From the single op-amp instrumentation amplifier:
V out = V 4 − V 3
Solving:
V out = ⎛⎝ 1 + 2 RRg ⎞⎠ (V 2 − V 1 )
This amplifier has
High input impedance (good)
A single resistor to adjust the gain (also good)
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Design Example in MATLAB
Design a circuit which outputs -10V to +10V as the temperature goes from -30C to +30C. Assume the
following thermistor
The specifications for this thermistor are given in a table in the data sheets:
-30C
-20C
-10C
0C
10C
20C
30C
17.04k
9.486k
5.447k
3.225k
1.976k
1.248k
0.809k
Curve-fitting this,
R ≈ 1000 ⋅ e −0.0516(T−25C) Ω
First, choose the instrumentation amplifier you want to use. Let's use the single op-amp version along
with a voltage divider
R2
+10V
R1
Vp
Y
Va
R
Vm
R2
R1
R
Y = ⎛⎝ R 12 ⎞⎠ (V p − V m )
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Second, determine the voltage at Va vs. temperature. Using the thermistor characteristics in MATLAB:
-->T = [-30:30]';
-->R = 1000 * exp(-0.0516*(T-25));
-->plot(T,R);
-->xlabel('Temperature (C)');
-->ylabel('Resistance (Ohms)');
The voltage divider converts resistance to voltage. In middle of the range, R = 3000, so use a 3k resistor
for the voltage divider:
-->Va = (R ./ (3000 + R)) * 10;
-->plot(T,Va);
-->xlabel('Temperature (C)');
-->ylabel('Va (Volts)');
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Third, compute the required gain. Since the output is to go from -10V to +10V (20V swing), the gain
required is 3.09
-->max(Va)
8.5060928
-->min(Va)
2.0479145
-->gain = 20 / ( max(Va) - min(Va) )
3.0968485
Since the output voltage increases as the input voltage drops, connect Va to the - input.
Fourth, find the offset voltage. The offset you need is from
Y = gain ⋅ (V p − V m )
At +30C
10V = 3.0968 ⋅ (V p − 2.0479V)
V p = 5.2770
-->Offset = 10/gain + min(Va)
5.2770036
-->Y = gain*(Offset - Va);
-->plot(T,Y);
-->xlabel('Temperature (C)');
-->ylabel('Output Voltage (V)');
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
ECE 321 - JSG
Voltage - Temperature Relationship for Instrumentation Amplifier
Note:
The endpoints are (-30C, -10V) and (+30C, +10V) as was the requirement
The relationship isn't linear - but it's closed
It's not surprising that the resulting relationship isn't linear
The thermistor has a highly nonlinear temperature vs resistance relationship
The voltage divider has a nonlinear resistance vs. voltage relationship
The resulting temperature - voltage relationship isn't that bad, however, considering how nonlinear the
circuit it.
Also note, you can do the same with light, magnetic field, dust, tilt, acceleration, etc. Just replace the
thermistor with a different sensor and redo the calculations for R.
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November 4, 2015
NDSU
52: Instrumentation Amplifiers
100k
+10V
ECE 321 - JSG
309k
5.277V
3k
Y
R
100k
309k
Instrumentation Amplifier: Y goes from -10V at -30V to +10V at +30V
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November 4, 2015