Thevenin Equivalent

According to the very interesting thevenin theorem, every single circuit could be reduced to a power supply and a resistor. To grasp the concept we did the following lab experiment. In the given circuit, we are interested in information about the R_L2 resistor, so we reduce everything else to an equivalent voltage source and resistor called V thevenin and R thevenin respectively.

 Assuming the following information are given we will try to solve the problem.

R_C1 = 100Ω ,R_C2=R_C3=39Ω , R_L1=680Ω and V_s1=V_s2=9V

  If we short the circuit where R_L2is then we can solve for the voltage at the main junction and find the I through R_C3.

 

At that junction:

 I_SC being the current through Rc3 and Vy the voltage at the main junction.

Now to find R thevenin we break the circuit at R_L2.

 

 Using node voltage analysis at the same junction as before gives: 

 now we can find the thevenin resistance:

Given that  V_{Load2, min}= 8V we can determine smallest acceptable RL2.

 

Now that we have all the values we do the same experiment through 2 different setup. One we will use all the initial elements given and the second time we use the thevenin voltage and resistance equivalent. 

First we construct the circuit with thevenin equivalents. But before we start we have to measure the actual value of the resistors and thevenin voltage.

Again for both parts we used the variable voltage sources in order to get as close as possible to the nominal value.

Component measurements for the second setup is as follow:

 

 This table also shows that the power increases at first but after reaching R_L2  = R_Th  starts decreasing again. Finally we can say with certainty that we can change any complicated circuit with its thevenin equivalent.

2N3904 Transistor

To get a better understanding of how transistor work, we were given a lab manual to put a NPN 2N3904 transistor in a circuit and measure current through its base and emitter. We made changes to values of the current by changing the resistance value in the resistor box.

the schematic for the circuit is as follow:

 

Without even graphing the linear relationship between the currents presents itself. As values of base currents increase the values for the emitter current change with almost the same amount.

Since I_E=(1+β)I_B then the equation for emitter current as a function of times becomes:

 

I_E=I_B+βI_B , when compared to y=mx+b we can deduce that the β is the slope of the graph and the I_B is the initial value. As a result β=0.968 and I_B = 3.73. this value of base emitter is where the linear portion of their relationship starts. Also if we use resistor values below 1.9K or above 10K ohm we will be out of the linear portion of the currents relationships.

Nodal Analysis

To construct a reliable power system we have to design it in a way that by damaging one element we wont destroy the whole system. To get a better understanding of this and the nodal analysis, we modeled a circuit by replacing cable resistances and loads by  equivalent resistors and batteries by voltage sources.
The modeled circuit is as follow:

 
to use the node voltage analysis we put the ground on the bottom and labeled the top two corners and the two junctions.

 

since we have 4 unknowns we have to write 4 equations.

 Given

R_C1=100Ω, R_C2=R_C1=220Ω, R_L1=R_L2=1KΩ

V_bat1=12V, V_bat2=9V

and the equations we should be able to solve the problem for  V₁, V₂,V₃, V₄. 
 In this case V₁ and V₄ are easy to find since they are just equal to the battery voltages.
 V₁ = 12V and V₄ = 9V.
Solving the equations gives,  V₂ = 10.2554V and V₃ = 8.6736V.

Next to be able to construct the circuit we have to know the exact values of all the elements.

The following table is the measured values for the resistors. 

After constructing the circuit these are the values obtained:

Here is a closer look at our circuit setup.

V_bat1 = V₂ + ( I_bat1×R_C1) = 10.39 + 0.01729×100 =12.119 ± 0.5 V  
V_bat2 = V₃ + ( I_bat2×R_C3) = 8.75 + 0.0015×220 =9.08 ± 0.5 V
P_bat1 = 209.5 ± 2 mW
P_bat2 = 13.62 ± 2 mW

Now if we want V₂ = V₂ = 9V then by carrying the equation wrote earlier we obtain these values:

In order to get the precise values of the battery voltages we used the variable voltage sources.

The small amount of error in the results  comes from the fact as always to make calculations easier we ignored the effects of wiring, which is minimal but still big enough to interfere.

Intro to Numerical Computation by FreeMat

In this lab we got introduced to a powerful freeware called MatLab. This program is used for calculations specially matrices. Matlab makes solvig the long equations with too many variables and equations easy.
To learn the basics of the program we went through a handout and the last problem we did on the handout was this:
Given the circuit below find the current through R3.
To solve I went through voltage drops in  2 loops, and I wrote a constrain equation for the bottom junction.

 

20I1+10R3=15

5I2-10R3=7

I1-I2-I3=0

 

 

so I3= -0.186A

Introduction to Biasing

In this lab experiment we powered two different LED's by a 9V source. The LED's were different in their rating one was rated at 2V and 20mA and the other 5V and 22.75mA. To meet the LED's specification we cannot simply put them in series or parallel since in both ways as explained by the manual they will burn out fast. One solution is to add resistors to the circuit to ensure these specifications are met.

From LED's ratings we can obtain their resistance.

R_LED1=219.8 Ω

R_LED2=100 Ω

so in the circuit if we replace the LED's with equivalent resistors we can determine R1 and R2 values easier.

 

since resistors only exist in discrete values we had to choose a combination of resistors in order to get as close as possible to desired values.
 For R1 we chose a 100, three 22 and a 10 Ω resistor, and for R2 a 220 and a 130 Ω resistor.
We measured the voltages across and currents through the LED's in 3 different configurations.
Configuration 1:Both LED's in the circuit
Configuration 2: Remove LED2, 9V supply
Configuration 3:  Remove LED1
Configuration 4:  Both LED's in the circuit, 6V supply
in all 3 configurations the current through the power supply was also measured. 

 In the table above the value of the current through LED2 is missing but it is 20.3 mA.

The reason for the relatively huge uncertainties is that I did the experiment twice and measured the values each time a few times.  

Assuming the useful life of a 9V battery to be 0.2A-hr, with both LED's in the circuit the battery will last 6.45 ± 0.85hr or 387 ± 51min.

Following is the error percentage between the theoretical and actual value:

 

 the huge error percentage for LED1 comes from several different places. One source of error could be due to loss of power to wires and that not every single element's value was close to its theoretical value. But this source of error applies to both LED's. The only thing uncommon between the two LED's is that the assumption for their ratings might be wrong. On different websites like ebay or amazon the forward voltage varies between 2.5-5V and current from 20-30 mA. This gives rise to a huge miscalculation in analysis of the circuit.

efficiency of the circuit is:

P_out=P_LED1+P_LED2=85.34±5 + 42.81±5 mW

P_in=P_Battery=279±5 mW

  

Efficiency of the circuit when the battery is 6V is:

P_out=P_LED1+P_LED2=25±5 + 41.2±5 mW = 66.2 mW

P_in=P_Battery=95.4±5 mW

As the voltage developed by the battery goes down the efficiency goes up since to deliver the needed specifications to the loads we need less resistance and eventually the most efficiency circuit is produced when the output voltage is 5V since in this case there is no need for any resistance in the circuit leg that contains the LED2. As a result all the power dissipated in that leg will only be due to the load.

In that case R1 is 0 and R2 is 150 ohm to meet the specifications and the efficiency becomes:

P_out=P_LED1+P_LED2=113.75 ±5 + 40±5 mW = 513 mW

P_in=P_Battery=213.75±5 mW

efficiency = 72%

Introduction to DC Circuits

In this lab we are given a circuit in which we have a load that is rated to consume a constant 0.144W. Given a 12V DC battery supply we have to find the maximum resistance and length of the cable we can use to have minimum of 11V delivered to the load. As we learned from the last lab wires can drop the voltage noticeably due to their resistance. so for example in a ROV that has to work through a tether the length of the wire quickly becomes a problem.
Even though it was possible to find the most efficient wire for the circuit analytically, we found the answers by constructing the following lab experiment.

 we modeled the system by replacing the  wires by two resistors and since the wires are in series we combined them and used one single resistor and to be able to change the value of the cable resistors we used a resistor box.

Since the value for the load resistor dos not change, as seen at the bottom right of the picture we used a 1K resistor. with 0.144W and 12V drop the resistor becomes 1Kohm resistor .

in the circuit while using two meters to measure current and voltage across the load we change the value of the resistor through resistor box to get 11 volts across the load.

We note that maximum voltage and current of the supply is 12V and 2A.

The table shows the information of the resistors after constructing the circuit.

Color Code	Nominal Value	Measured Value	Within Tolerance	Wattage
Brown-Black-Red-Gold	1KΩ	0.975KΩ	Yes	0.25W
NA	86Ω	86.1	Yes	0.6W

the first one is the 1K resistor with 5% tolerance and the second one is the value of the resistor box.
R-resistorbox=86.1±10

As seen on the above picture while the voltage drop across load is 11V the current through the circuit is 11.32 ±3 mA.
Since the battery's capacity is 0.8 Amp-Hour we can calculate the time that the battery can supply this much current. if we divide the  0.8 Amp-hour by the 11.32 mA we'll get 70.7±10 hours, which tell us that the battery will put out 11.32A for about 73 hours. However we are assuming that the battery supplies 12V constantly and the capacity does not change by other factors such as time or temperature.
since the actual value for the load is 975 ohm and voltage drop for it is 11, the power to it is 0.124 W.
And power dissipated in cables is 0.0132W. This amount of power is well below the 0.3W rating for each element of the resistor box so we are not exceeding the power capability of the resistor box.

As instructed if we were to try to use a AWG#30 wire which has 0.3451Ω/m, the maximum distance that we can have between battery and the load, considering the cable has both forward and backward conductor, is 125m or 410ft.
Next we use a 60ft AWG#28 wire as the tether for the robosub while considering that signals are 20mA 5V TTL with nominal rating 0-0.4 V for low and 2.6-5V for high. V=RI
For a AWG#28 wire we have 0.0764 Ω/ft.

Voltage  (V)	Resistance (Ω)	Max Length (ft)	Max Length (m)
5	250	1636±10	499±10
4.6	230	1505±10	456±10
2.4	120	785±10	239±10

the limiting factor here will be the 2.4 drop across the cables so if we use a 239 m cable we should be able to provide the necessary voltage for the signal transfer.

the next question is if we send 48V at 10A down the tether, and we want 36V to reach the sub, what is the minimum cable gauge. 

since we want to loose a maximum of 12V at 10A to cable the resistance of the cable is 1.2Ω. If we want the cable to be 785ft then the Ω/ft of the cable is 0.001529, which is closest to AWG#12 with 0.00187 Ω/ft.

At the end I have to mention that we made a critical mistake when we obtained the constant value for the load. we used 12V as voltage drop across the load while it should have been 11 volts across. So the resistance of the load should have been 840Ω not 1KΩ. As a result the final values differ from what we should have gotten, so to fix the problem the only thing that we can do now beside redoing the whole experiment is to increase the uncertainty. For that reason the current through the circuit should be 13.09mA not 11.32mA. Resistance of the resistor box then has to be 76.4Ω. Total working time of the battery becomes 61.1 hours.  This error also shows the reason behind the huge difference between the desired power rating of 0.144 for the load and obtained value of 0.124.

Using a Multimeter

In this short lab the goal was to demonstrate that we know how to use a multimeter properly. i.e. we wouldn't burn the fuse :(.
we were supposed to create the following simple circuit and measure everything. Everything includes currents resistance values and voltage drops across elements.

But first to refresh our memory on how to measure resistance values we chose 4 different resistors and determined their values by reading the colors then measuring them.

Then we measured the value for an unregulated 12V DC adapter which came out to be about 16.9V. The number makes sense since the 12V labeled on the power supply means the company guarantees a minimum of 12 volts.

Then we constructed the circuit shown at the beginning.

The obtained values from measuring the current in series and voltage in parallel are as follow:

The reason that the total voltage does not equal to 5V is the fact that the wires used to construct the circuit have some low resistance as well. In this case about 10 ohm. However wires contribute the most in error we cannot assume that they are the only source since the actual value for resistors are slightly different than the measured value from the multimeter.

Overal now we can say we are able to measure current, voltage and resistance in circuit elements which is a very important tool in this course :).