Instrumentation Amplifier

Transcription

1 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 OpAmp Design: Va R2a a Vp Vo Vm Vb R2b b Single OpAmp Instrumentation Amplifier Again, we have three voltage nodes so we need to write three equations: Solving V p = V m Vm V b b V p V a a Vm Vo b V p a V o = a b b b a a V a b b V b If a = b and R2a = R2b, then V o = (V a V b ) The problems with this amplifier are If you want to adjust the gain, you need to adjust two resistors (a and b) If the resistor pairs are not exactly the same, you get a commonmode gain (there is a term which includes Va Vb). page 1 November 4, 2015

2 Two OpAmp Instrumentation Amplifier R2 Vx R3 R4 Vb Vm1 Vp1 Vm2 Vp2 Vo Va Two opamp instrumentation amplifier The node equations are: V b = V p1 = V m1 V a = V p2 = V m2 V m1 V m1 V x V m2 V x Simplifying: V m2 V o R 4 V o = 1 R 4 V a R 4 1 V b This circuit has a high input impedance (good) but still requires you to adjust two resistors to adjust the gain. Example: Find, R2, R3, and R4 so that the gain is Solution: V o = 10(V a V b ) 1 R 4 = 10 let R3 = 1k, R4 = 9k R 4 1 = Let = 10k, R2 = 1.111k page 2 November 4, 2015

3 Two OpAmp Instrumentation Amplifier with a DC Offset: (AMP04) Vb Rg Vm2 Vo Va Vp2 Vp1 Vm1 R2 Vref R2 Two OpAmp 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 R g Substituting: Note here: V m1 V ref V m2 V o V m2 V x V m1 V x V o = R g (V a V b ) V ref 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. page 3 November 4, 2015

4 Three OpAmp Instrumentation Amplifier: Three OpAmp Instrumentation Amplifier The voltage node equations are: V 1 V 3 R V 2 V 4 R V 1 V 2 R g V 2 V 1 R g From the single opamp instrumentation amplifier: V out = V 4 V 3 Solving: V out = 1 2 R R g (V 2 V 1 ) This amplifier has High input impedance (good) A single resistor to adjust the gain (also good) page 4 November 4, 2015

5 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 Curvefitting this, R 1000 e (T 25C) Ω First, choose the instrumentation amplifier you want to use. Let's use the single opamp version along with a voltage divider 10V R2 Vp Va Y R Vm R2 Y = (V p V m) page 5 November 4, 2015

6 Second, determine the voltage at Va vs. temperature. Using the thermistor characteristics in MATLAB: >T = [30:30]'; >R = 1000 * exp(0.0516*(t25)); >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)'); page 6 November 4, 2015

7 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) >min(va) >gain = 20 / ( max(va) min(va) ) 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 At 30C Y = gain (V p V m) 10V = (V p V) V p = >Offset = 10/gain min(va) >Y = gain*(offset Va); >plot(t,y); >xlabel('temperature (C)'); >ylabel('output Voltage (V)'); page 7 November 4, 2015

8 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. page 8 November 4, 2015

9 10V 5.277V 100k 309k 3k Y R 100k 309k Instrumentation Amplifier: Y goes from 10V at 30V to 10V at 30V page 9 November 4, 2015