Laboratory #6 Thevenin Equivalents

This lab we did on this Monday that used thevenin method to make the complicated circuits look easier.
When the circuit have many sources and multiple loads, then we could calculate the thevenin to make the problem easier.

Before the lab began, we had the set up number for resistors and the voltage of the sources. They were Rc1= 100 ohms, Rc2=Rc3=39 ohms, RL1=680 ohms and Vs1=Vs2= 9V. Then, we could calculate theoretical value for the Vx by using the nodal analysis. (9-Vx)/39+(9-Vx)/100=Vx/680, so the Vx=8.6433V. The second part was to calculate the Vy by connecting the two points. (Vs2-Vy)/Rc2+(Vs1-Vy)/Rc1=Vy/RL1+Vy/Rc3, the Vy=5.11V. Furthermore, we got the Isc=Vy/Rc3=0.131A  and Rth= 65.97ohms. If the ab point carried 8v; then, the resistor should be (8.64-8)/66=8/X, so the resistor X=825 ohms. I would be the same with Isc and V was the same with Vth=8.64 V. After we calculated all the theoretical values, we did set up the circuit like the image, and we almost got perfect numbers. We were used Rth=66.6 ohms and RL2=825 ohms and Vth=8.64V, all the numbers were real close to the nominal value. The value we measured were 7.98V for Vload 2 when RL2=RL2, min, the percentage error was 0.25%, and when RL2= infinity, the Vload 2=8.64 the measured value was 8.63V so the percentage error was 0.11%. I thought this experiment was perfect.

Laboratory #5PSpice

This lab that we basically studied how to use the PSpice software to solve problems.

The first image which i did in the class is just what i followed by the handout of the pspice tutorial. The thing that what i learned is the software need to run as administrator, otherwise people would not find the running system.

The second image is the last problem from the homework. As the practice goes on, i noticed we always need to set up the ground first, and put the parts step by step. In addition, we need put more attention on the positive and negative sign. In a word, it is a great software to solving electronic circuit problems.

Laboratory #4 Nodal Analysis

Today, we just used the method of nodal analysis to figure out the voltage of each nodal and test it out the error. In another way, we also could use the voltage of each nodal to figure out the power supply.

First of this lab, we established two equation which were 0-V2/Rc1= V2-V1/Rc1+V2-V3/Rc2 and
V2-V3/Rc2=V3/RL2+V3-V4/Rc3. So we had V2=10.25V and V3=8.67V.
Furthermore, we got Ibat1=V1-V2/Rc1=0.0175A, and Ibat2=V4-V3/Rc3=0.0015A which these are the current of the battery. Because the power equal to the current times the voltage, so Pbat1=0.179375W, Pbat2=0.013005W.Then, we did the set up and use the actual equipment to do this experiment that we had Rc1=99.4ohms, Rc2=219ohms, Rc3=223ohms, RL1=995ohms, RL2=999ohms, Vbat1=12.11V and Vbat2=9.08V. We had been through a hard time to measure the value, and we got them all finally that they were Ibat1=17.8mA, Ibat2=1.53mA, V2=10.3V and V3=9.41V. Luckily, these numbers percent error were really small. They are 1.7%, 2%, 0.4% and 8.5%.
For doing the more calculation, the Pbat1= Vb1*Ib1= 12.11V*0.0178A=0.215558W and Pbat2=9.08V*0.00153A=0.013892W.
If the V2=V3=9V, the new set up will be -9V/1000=9-V1/100, and 9/1000+9-V4/220=0. The result will be V1=9.9V which is battery 1 and V4=10.98 which is battery 2.
After finishing this experiment, I learned a couple easy ways for measuring stuff. But i also get touch with a lot of mistake which could occur that makes the lab harder for doing the wrong way.

Laboratory #3 Voltage Dividers

This experiment is mainly about making a electrical circuit which let the load part work in a certain range of voltage, so it could work properly.

 Before we set up the circuit, we did all the calculations that we need to get the sources. Reqmax=1k ohms, Reqmin=333.333 ohms. Because we want Vbus be the max so the resistance should be the max, so Vbus min need to use the Reqmin. The equation which i set up for Vbus max is [(Vs-5.25V)/Rs]*1000ohms=5.25V, Vbus min=[(Vs-4.75V)/Rs]*333.333ohms=4.75V. According to the two equation, we got Vs=5.54V, Rs=55.55 ohms. After we knew the resistance and voltage, we could know the Ibus, max=Vs/(Rs+Reqmin)=0.0142A, Ibus,min=Vs/(Rs+Reqmax)0.005248A. It's the time to set up the circuit which as the image that R1, R2, R3 were parallel and serious with Rs. We measured the resistors that were all 988ohms, the power supply was 6.07v, and the Resistor box was 56ohms. As what I read from the power supply and the resistor box, the maximum current were both within the capability.
The calculation of the power with two loads is P= Ubus*Ibus=5.35*10.89*10^-3=0.0583W. To figure out the actual percentage in load voltage variation, we need to first find out the Vaverage=(5.65V+5.05V)/2=5.35, so the %Diff=[(5.65-5.05)/5.35]*100%= 3.21%.
If we add the forth 1kohms in the load, we should get [5.54/(55.55+250)]*250=4.53, and we tested which i got closed number that was 4.95V.
To make the load voltage variation to 1%, we could just change the Rs to make it happen. We actually tested in the circuit. We just lower 23ohms from the Rs, so the Rs=33ohm, and the voltage variation is zero percent. The voltage was just 5.35V which is the Vaverage.

Laboratory #2 Biasing

    This lab that we worked on last night was using two LEDs and trying light them up appropriately and efficiently.

Before we did the experiment, we calculate all the data that we need in the lab. First were the two resistance of  tow LED's. Rled1=219.78 ohms   Rled2 =100 ohms. According to the equivalent circuit in the lab book which the LEDs in parallel, we calculated the two resistance R1 and R2 which would share some voltage from the LEDs. R1=175.842 ohms R2=350. Because we could not get the exact resistor, we just get the close ones which were 148 ohms and 353 ohms. Then, we set the circuit as the image and recorded all the data for doing the further calculation. We had the current and the voltage of the led one were 15.5 mA and 6.69V. The current and the voltage of the LED two were 21mA and 1.64V.
Question: a.(Fig 2) 0.2 A-hr/(0.02275A+0.02A)= 4.678 hours
                b. LED 1 %Error = (15.5mA -22.75mA) /22.75mA *100%= 31.86%
                    LED 2%Error=(21mA-20mA) *100% = 5%\
                c.  P1=0.0155A*6.69V=0.103695 W
                    P2=0.021A*1.64V = 0.03444W      
                    so Ptotal=0.138135 and the Psupply= 9V*0.0365A=0.3285W
                      The efficiency= (Ptotal/Psupply) *100%= (0.138135/0.3285) *100%= 42.05%
                d. if the battery change to 6V, the efficiency would go up. It because the Psupply go smaller and                                            
                    also the lost energy would go smaller too.
After this experiment, we should know if we want the equipment work efficiency, we better let it work in the appropriate voltage range.

Laboratory #1 Power Loss in Cable

We did the lab to measure how many power loss in the cable. Before we did set this up, we measured the voltage of the power supply, the resistant of the load and recorded the data for the calculation at the later part which were 12.07V and 975 ohms. Then, we built the circuit as the image and read the voltage and the current of the circuit which the number were 11.96V and 12.5mA. According to the requirement, we changed the resistant of the cable to make the voltage of the road resistant drop 1 volt; then, we recorded the data again which were 88ohms, 10.96V and 11.45 mA.
After them all, we could do the calculation. Time of discharge: 0.8 Amp-hr= 0.01145A*t so the t=69.869 hours.  Power of the load: P=V^2/R=10.96^2/975=0.123202W
Power of the Cable: P=I^2R= (0.01145^2)*88=0.011537W
The Efficiency= (Pout/Pout+P lost)*100%=0.123202/0.123202+0.11577*100%=91.43%
If the AWG #30 wire is 0.3451 ohms/m, so the maximum distance should be 88ohms/(2*0.3451)=127.499 meters. The power lost which could lost from the heat of the wires or the room temperature may consider.