Experiment 8: Op Amp Design Project Accelerometer Sensor
for Automotive Safety and Efficiency
Partner: Keegan Tawa
TA: Alex Cocking
Group 2
Oct 30, 2013
INTRODUCTION
Define the problem (given inputs and desired outputs):
Our design application involves the measurements of an automobile’s
acceleration using IC accelerometer. We will need to design and build the interface
circuitry of the accelerometer that will preform the following functions: 1) Provide a
linear display of the acceleration measurements. 2) Indicate when the automobile’s
acceleration/deceleration exceeds 0.1 G in magnitude. 3) Signal when a significant
deceleration (< -6 G) exists to warrant airbag deployment.
Design specifications and constraints:
Since we are not using an actual Analog Accelerometer, we are going to build a
circuit to give us the desired single ended output voltage that has a voltage range of 0.5V
to 4.5V. This signal is called AOP. We are going to use a differential amplifier with an
input of ±10V Ramp Signal.
For the Signal Conditioning Circuitry, we need to present two outputs: 1) An
output presenting the range of values ±10 G, which corresponds to ±5V. 2) An output
presenting the range of values ±0.2 G, which corresponds to 0-5 V. As an input, we are
going to use the AOP we obtained from the previous step.
For the Acceleration Indicator /Monitor, we are going to use LEDs as indicators
for our circuitry. As defined in the problem, we will need a circuit that shows when the
driver accelerates or decelerates too rapidly (Magnitude exceeds ±0.1 G), and indicator
for air bag deployment or possibly some collision avoidance measure. The indicating
signal would be a a deceleration of 6 Gs in the shape of a half sine wave for a period of
150 ms(6.67Hz).
The supply voltages are going to be ±15 V for our design.
Also, we need to make sure that our output current preferably DOESN’T exceed
1mA to not load down the op amp’s output.
CIRCUIT DESIGN AND SUPPORTING ANALYSIS
Schematics:
Difference Amplifier Schematic:
Voltage Polarity Monitor Schematic:
Signal Conditioning Circuitry Schematic
Circuit Analysis:
Supporting analysis equations:
For the difference amplifier circuit that outputs AOP, resistors ratios were given.
By setting R=1kΩ we obtain the following:
Rf
Ri1
Ri2
Ri3
Nominal
1 KΩ
29 KΩ
6 KΩ
6 KΩ
Actual
0.99 KΩ
28.7 KΩ
5.98 KΩ
5.97 KΩ
For the Signal Conditioning Circuitry:
1) The one outputting Vo1:
Vo = -Rf/R1*15 + [1+Rf/R1*(1+R1/R2)]*AOP…(1)
Vo = -6.25 + 2.5*AOP… (2)
By solving (1)&(2) we get the following:
Rf/R1 = 6.25/15 = 5/12 …(3)
1+Rf/R1(1+R1/R2) = 2.5…(4)
Setting Rf = 5 KΩ, R1 = 12 KΩ ====== R2 = 4.6 KΩ
Nominal
Actual
Rf
5 KΩ
4.98 KΩ
R1
12 KΩ
11.99 KΩ
2) The one outputting Vo2:
Vo = -Rf/R1*15 + [1+Rf/R1*(1+R1/R2)]*AOP…(1)
Vo = -2.5/4 + 1.25*AOP… (2)
By solving (1)&(2) we get the following:
Rf/R1 = 2.5/(4*15) = 5/120…(3)
1+Rf/R1(1+R1/R2) = 1.25…(4)
Setting Rf = 5 KΩ, R1 = 120 KΩ ====== R2 = 24 KΩ
R2
4.6 KΩ
4.56 KΩ
Nominal
Actual
Rf
5 KΩ
4.99 KΩ
R1
120 KΩ
119.6 KΩ
R2
24 KΩ
23.5 KΩ
For the Voltage Polarity Monitor’s comparators:
1) The one outputting Vo3:
We need a positive voltage at the output when:
0<Vo2<1.25V, 3.75V<Vo2<5V
V- = +3.75V. Vin = +15V
V+ = +1.25V. Vin = +15V
Using voltage division:
R2/(R1+R2)*15 = 3.75
R4/(R3+R4)*15 = 1.25
Setting R2 = R4 = 5 KΩ ====== R1 = 15 KΩ. R3 = 55 KΩ
2) The one outputting Vo4:
We need a positive voltage at the output when:
Vo1< -3V
V+ = -3V. Vin = -15V
Using voltage division:
R6/(R5+R6)*(-15)=(-3)
Setting R6 = 5 KΩ ======= R5 = 20 KΩ
Nominal
Actual
R1
15 KΩ
14.8 KΩ
R2
5 KΩ
4.99 KΩ
R3
55 KΩ
54.7 KΩ
R4
5 KΩ
5 KΩ
R5
20 KΩ
19.8 KΩ
R6
5 KΩ
4.98 KΩ
Discussion of design reasoning:
To check voltage polarity of two sources, comparators work best in this situation.
We can put our input in one terminal of the op amp, and fix the other terminal to some
other value. When one exceeds the other, we will have the polarity of the higher voltage
input’s terminal on our output.
For signal conditioning, we can use a non-inverting amplifier to condition our
outputs. Simply by inputting AOP to the positive terminal and a DC source to the
negative terminal with couple resistors.
Multism
Multisim schematic of all circuitry:
Simulated scope plots for VO1:
Simulated scope plots for VO2:
Simulated scope plots for VO3:
Simulated scope plots for VO4:
DATA
Table comparing measured and specified design specifications:
Max (V)
(Specified)
Min (V)
(Specified)
Max (V)
(ACTUAL)
Min (V)
(ACTUAL)
AOP
4.5
Vo1
5.0
Vo2
5.0
Vo3
VLED(~1.7)
Vo4
VLED(~1.7)
0.5
-5.0
0.0
0.0
VEE(-15)
4.43
5.20
5.20
1.80
1.70
0.3
-5.10
-0.2
0.08
-12.9
Scope plot of VO2&VO3:
Scope plot of VO1&VO4:
DISCUSSION
Percent error between measured values and theoretical values:
AOP
Vo1
Vo2
Vo3
%Err
1.56%
4%
4%
5.88%
(MAX)
%Err
40%
2%
2%
UNDEF
(MIN)
Vo4
0%
14%
Percent error between measured values and simulated values:
Vo1
Vo2
Vo3
Vo4
%Err
3.86%
4%
0.58%
1.16%
(MAX)
%Err
2.2%
7%
88%
7.86%
(MIN)
Discussion of the significance of your results:
All of our calculation and analysis was based on the ideal characteristics of the
OP AMP. From the start, we would know that we would have some slight errors in our
calculations. Also, our actual resistor values were a bit off what we actually wanted them
to be. The input output linear relationship in the OP AMP depends on those resistor
values. However, our results were almost as expected.
When we compare our measurements with the simulation and the theoretical
values, our simulation shows closer results to our measurements. It is the fact that
simulations account for non-ideality.
SUMMARY, CONCLUSIONS, AND ATTACHMENTS
Summary and Conclusions (What you observed and learned):
In order to make the interface circuitry for the accelerometer, we don’t need to
have the actual device. Instead, we created a test signal that emulates the AOP signal out
of the accelerometer.
Op amps in reality are non-ideal devices. However, we can use the ideality
characteristics in analyzing the circuitry and designing it. The results may be off a little,
but it is easier to analyze this way.