Op Amp Applications

In this experiment, we will use AS35 , which is an electronic temperature sensor that will  produce output voltage that is proportional to the temperature around it to a scale factor of 10mV/degree C. The device only operate at 4-20V and an external resistor to provide an appropriate load.






As shown above, we chose R1 to be 5000 ohms while R2 was chosen to be 4000 ohms . This was so that the ratio R2/R1 was 4/5 and Vo was multiplied by a factor of 1.8. To provide the correct level-shifting, Vref was solved for so that (4/5)Vref was equal to the amount of shift required, 320mV. It turned out to be -400mV, so a voltage divider was used to obtain -400mV from a 4.5V source.

Vref has to be negative if the overall shift is to be positive. The design also has constraint which the output current from the op-amp is much less than the maximum allowable output current specified in the data sheep which is 25 mA for LM741.

The initial circuit output unusually large values and the sensor smelled burnt. When the simple circuit on the first page was built, it still output over 2V. We realized that the manufacturer's diagram was wrong after we disassembled the circuit, so we had to use new values for some resistors. The right column still lists the actual values


Data Collect:



Vc=0.228V
Vf=0.711V
Theoretical Value of Tf = (1.8*0.228+0.32)*100=73.04 degree Fahrenheit
% error = (73.04-71.1)/73.04*100% = 2.66%
The % error between the two values is 2.66%. The error may be come from the measurement was taken not too long after it was overheated due to the incorrect diagram

Impedance an d AC analysis I

The purpose of the first part is to investigate a real-world inductor which is equivalent to connect with RL and L in series, so the impedance Z can be determined as ZL = RL + jwL and determine the inductance by looking for the ratio of magnitude of the terminal voltage and current phasors by following circuit.

Next, set the RMS voltage of the function generator with 20KHz, then  device the circuit and  record the reading from the multimeter and ammeter

V in,rms =4.53V

I in,rms =88mA

The reading is different from the FB display because there is some internal resistance due to the function generator.

Next, make computations:

|Z|=|V|/|I| = 4.52/88mA=51.36ohm

Z=R_ext+R_L+jwL

|Z|=sqrt((R_ext+R_L)^2+(wL)^2))

w=2πf==125664 rad/s

Since we know the value of Z, R_ext, R_L and w

we will need to find L, which is about 0.4mH

Investigating a Series RLC circuit:

Data

Scope measurement at 20.97 kHz.

Vp-p,ch1 = 23.08 V

Vp-p,ch2 = 19.49 V

Δ t = 19.46 us

The phase difference = Δ t * f *360 degree = 19.46 us* 20.97kHz * 360 degree = 146 degrees

Then, use DDM to measure the voltage and current at different frequency

Question:

1. Why is the input current that largest at 20.0kHz(12.7K)?

The input current is largest because the impedance of the element is the smallest

2. Calculate the theoretical voltage phasor across the real inductor at 20.97KHz (use the DMM measurement value as the source voltage phasor magnitude). Compare this with the scope measurements. Convert the scope measurement to RMS for comparison purposes.

Theoretical: (23.08/(2sqrt(2))*2pi*20.97*1000*0.4*10^-3/(68.5+3.4)=5.98V

Experimental: 19.49/(2sqrt(2))=6.89

% Error: (6.89-5.98)/5.98=15.2%

3. Does the circuit look more capacitive or inductive at frequencies below 20kHZ?

capacitive

4. Does the circuit look more capacitive or inductive at frequencies above 20kHZ?

inductive

AC Signal #1

In the lab, the purpose is to explore the phase difference between AC sine signals at the same frequency in a simple RC circuit; besides, by adjust the frequency of the signal and resistance in the circuit, the phases will be different. Also, we can learn how to read the value of current or voltage on the scope display.

The measurement of phase difference is time difference between the positive peaks multi by the frequency and 360. In this scope display, since CH1 takes its peak before the CH2 signal, we say that the CH1 signal leads the CH2 signal.

Set the FG to produce 1 V peak-peak sine wave at 1k Hz. The RMS value for DMM is 0.318 V. The complex impedance of the 100 nF capacitor can be calculated as 1/(2*pi*C) = 1592 ohms. Next, we connect CH1 to FG and CH2 to the top of capacitor, and then Resistor box and C connected as shown above.

DATA:

CH2 Vcap, pk-pk  = 0.852 V

CH2 Vcap, rms  = 0.23 V

tx =105.41 us

According to the method mentioned before

phase change = tx * f * 360 = 37.9 degrees

CH1 is leading CH2 by 37.9 degrees

Next adjust the frequency of input signal to 10 kHz rather than 1k Hz.

CH2 Vcap, pk-pk =0.154V

CH2 Vcap, rms = 0.033 V

tx=23.78 us

phase change = tx * f * 360 = 85.6 degrees

Adjust the resistor box to 10 Kohms.

CH2 Vcap, pk-pk = 0.178 V

CH2 Vcap,rms = 0.049 V

tx = 221.62 us

Phase change= tx * f * 360 = 79.8 degrees

LOW frequency range is the capacitor voltage amplitude greatest.

HIGH frequency range is the capacitor voltage amplitude smallest.

If a filter is a device that selectively passes through a certain range of f, the circuit is being LOW PASS 

FreeMat Assignment

 Q1:

 

Q2:

 

Q3：

 

Q4：

integrator op amp

The purpose of this lab was to learn how to use an oscilloscope to measure time varying signals. For this lab we used a function generator to produce the signals. We used the function generator to create both periodic sine waves, and square waves to display on the oscilloscope. We then used the divisions on the screen, and the increments per division to calculate the amplitudes, frequencies, and voltages when crossed with a battery. To verify our scope calculations, we measured the output of the function generator with a DMM. Then we used what we learned to calculate two mystery signals.The Syscomp CircuitGear was used to generate the waves and record their integrals as shown in sample shots below.

 The figures below shows the input signals (red) and their integrals (blue):

Using the function generator, we can add an offset to the function, which will be displayed by the oscilloscope. By observing the difference, we can come to understand how step functions apply.

Question Answer:

If there was a DC component in the input waveform and the 10M ohm resistor wasn't present, then charges will eventually build up on the capacitor, and the entire feedback loop would become an open circuit. The 10M ohm resistor prevents this from happening. Integrating a constant would continue increasing the output voltage until the circuit reaches saturation.

Capacitor Charging/Discharging

In the lab, we considered charging and discharging a capacitor. TO avoid the need for sophisticated electronics in a pluse-forming network, we will consider the ideal capacitor. Like the figure below, in the charge position, as long as Vs>Vcap, energy is transferred from the source to the capacitor. IN the discharge position, the energy is absorbed by the resistance Rdischarge. In either case, C,Rcharge, and Rdischage control the rate at which energy is transferd.

In the experiment, the two modes of operation are done and the circuit is shown in the figure below. We can compute the Thevenin voltage and resistance as seen by the capacitor in each case. We substituted this voltage for the source voltage in the charging mode and we replaced the charging or discharging resistance by the appropriate Thevenin resistance.

In the RC charging circuit, the voltage and current across capacitor can be express as V(t)=Vs(1-e^(-t/tau)) and I(t)=Vs/R*e^(-t/tau), where tau is time constant which is Rcharge*C. In addition, in the discharging circuit the voltage and current across the capacitor are V(t)=Vic*e^(-t/taud) and I(t)=-Vic/Rdiscahrge*e^(-t/taud), where taud is time constant in discharging circuit. However, in real capacitor model the capacitor connected to resistor Rleak in parallel, so we must apply Thevenin's theorem to find out the real value of resistance.

 Then the circuit was built using a 65k charge resistor, a 6.47k discharge resistor, three21 uF capacitors. The three capacitors were placed parallel to simulate a 66 uF capacitor.

Vs = 9.01 V

Vfinal = 8.562 V = Vs*Rleak/(Rleak+65k), so R(leak)=1.24 Mohms

The Thevenin equivalent voltage and resistance seen by the capacitor during charging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 65k in parallel which is 61.8k ohms, and the Thevenin equivalent voltage and resistance seen by the capacitor during discharging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 6.47k in parallel which is 6.44k ohms

Questions:

1. The capacitance can be found using the energy equation w = 0.5Cv^2= 1.42F.

2. The capacitors in series add up to equivalents of C/2. There are 4 parallel branches, so the total capacitance is 2C. 2C has to equal 1.42F, C=0.71F.

Operational Amplifiers

The idea behind the op amps lab is to apply operational amplifiers in a real world scenario. There may be many instances in the real world where a circuit may be required to have a different set of voltages at different parts of it. For example, a sensor which captures certain information may not be outputting an ideal voltage for it to work with the attached processing unit. Enter the Op Amp which can make it work! The following circuit was built and all the values were recorded.

Once the circuit was constructed, it was apparent that the circuit contains a feedback loop. By putting a resistor in between the output voltage and the input voltage, the operation amplifier can be made to output the desired voltage gain.

Given

  V1 = 12.13V

  V2 =  12.06V                                                                          

 Result

   Ri = 1K Ω     Rf = 10KΩ

     Rx = 1152 Ω       Ry= 104.7 Ω

Vin (V)	Vout (V)	GAIN	VRi (V)	IRi (uA)	VRf (V)
0	-0.05	0	0.007	7.00	0.06
0.25	-2.54	-10.16	0.269	269	2.57
0.50	-5.08	-10.17	0.536	536	5.12
0.75	-7.56	-10.08	0.797	797	7.62
1.00	-9.91	-9.91	0.946	946	9.91

The power supply current

Iv1=2.32mA

Iv2=1.49mA

Pv1= 12.13*2.32m=28.14mw; Pv2= 12.06*1.49m=17.97mw which satisfy the power supply constraint to supply no more than 30mW each. It came out that the current being drawn from sources were about 7.5 and 1.8 milliamps. These were not expected but that may be in part due to the fact that the power supplies (voltage supplies) were also being used to power the voltage dividers.