Lab 4: Transistors

           A transistor can be think of as a switch or a current amplifier. It's purpose is to limit the flow of electric current going through it. The current through the transistor may be increased by applying a controlling current at the base pin of the transistor.

In this lab, we will be using 2N 3904 transistor.

 

 

Step 1)  Testing the Transistor 

When current is allowed to go into the base pin of the transistor, the LED turns on as the transistor opens up, and allows a higher output current from the emitter pin. This means......the transistor is FUNCTIONAL!

 

Step 2)   Examining the sensitivity of the transistor



 We will limit the control current to very small by using the hugh resistance in human's skin.

Human's skin resistance has an average resistance of 400 kilo ohm.

 

No current is going through the LED if the controlling circuit is opened.

 

When the controlling circuit is closed, the LED is lit with a very dim light. The transistor is opened even with a very small current.

 

Transistor are extemely sensitive!!

 

 

For the last Experiment, we tests the beta value of our transister as well as it saturation limit:

 

We first connect a ammeter to the in series to the base, and another ammerter in series with the emitter.

 

Transistorsare used to limit the current to protect the transistor from over heating or burning any of the circuit's components.

 

Here are the results:

 

 

From the graph,

     the slope gives our beta value of approximately 80;

    

     the transistor begins to saturate when the base current reaches 0.75mA as the slope of the graph

     begins to reaches 0.

Lab 3: Voltage Divider

          The voltage divider is often used for controling the voltage through an electronic devices especially when the voltage source is an unregulated power source. Otherwise, the device could be damaged if the voltage aross it is too high.

          In this experiment, a circuit is designed to power 3 loads using an unregulated power supply and voltage divider at a given acceptable range of voltage for the loads.

Each load will be represent by a resistor (R1, R2, R3).

 

All 3 loads can be combined into a single resistar with equivilant resistance.

 

Things that we are trying to solve for this lab given that the load resistance (R1= R2= R3 = 1k ohm) and the Vbus must remain between 5.75V and 6.25V:

1. Calculated for Vs and Rs, and construct the designed circuit including calculated data.
2. Record the maximum and the minimum BUS current.
3. Record the normal resistance and the actual resistance in the resistors and the resistance box.
4. Measure and adjust the voltage in the unregulated power supply to 6V; check that  the max current and power in within tolerance range of the power supply and resistor box.
5. Collect any data for the circuit with different numbers of load connected, and determine the power dilivered by the loads(s).
6. Complete the post lab analysis.

Vmax happens when Req is at max, and vice versa. Req is max when only 1 load is turned on, and at min when all 3 loads are turned on.

By solving the system of equations, we calculated Vs to be 6.53V and Rs is 45.5ohm

Max BUS current = 6.25mA;

Min BUS currernt = 17.24mA.

 

All resistance values and the voltage of the power supply calculated on above picktures will be used in the circuit.

 

A schematic drawing of the circuit on paper.

The overall view of the circuit with DMM measuring the instantanous voltage, and resistance box connected in series with breadboard and power supply.

 

A closer view to the breadboard with each resistor being a load in this lab.

 

 

Using another DMM as ammeter to measure the total currernt to the loads from the power supply.

 

All the measurements for the parameter of the circuit, as well as the calculated Power by the loads.

 

Power calculation for 2 loads:  P = VI = (10.97 mA)*(5.08 V) = 55.7 mW

Load	Voltage (V)	Variation (%)
1 Load	5.48	-8.67
2 Load	5.08	-15.33
3 Load	4.55	-24.17

The voltage across the parallel loads are shown in the above table.
Clearly, the voltage across the parallel load is dropping as more loads are added into the circuit. This is the result of using an unregulated power supply as it does not maintain a constant voltage output. The unregulated power supply is the main cause for the enlarging variation in the supplied voltage.


V_Bus,min = R_eq*V_s/(R_s + R_eq).
New R_eq = 1/4 kΩ.
V_s = 6.53 V,
R_s = 45.5 Ω
V_Bus,min = 5.52 V
The new variation using V_s = 6V would be -7.92 %.

Variation by 1% ==> 5.94 V < V_s < 6.06 V
V_s = 6.12 V
R_s = 10.2 Ω

Using FreeMat

What is FreeMat?

        FreeMat is a free version of the popular MATLAB technical desktop software application which allow the user to solve mathematical equations as well as graphing them. FreeMat works alot like a graphic calculator but the user interface in a programming interface.

 

Graphing y=sin(x).

Graphing y=cos(x)

 

By typing "hold on", two graphs can be drawn on the same page. If "hold off", further equatins will be open on a new page.

          FreeMat also allows the users to wrtie Script Files to perform particular functions or operations. In fact, Script Files are like modules or functions in C++ language.

A Script File to perform square, and square root calculation for a predefined variable. 

 

 

Another main function for FreeMat is to solve for solution for matrix equation. This function is particularly useful for solving complex Kirchoff's law which may form a 6x6 matrix or even bigger.

The martix can be solved by using the inverse function from the liberary of FreeMat. As the as the marix is set up correctly, FreeMat will solve it with ease.

 

There is an easier way to generate point plots for a specific range by using the "linspace" function. This function becomes very handly when compare or adding multiple equations within the same range of data.

3 different exponential graphs shown on the same window.

 

Adding graphs together can be done by simply defining a new variable as the sum of two equations and plot it using the plot function. This will come in handy when adding sinusoid waves together instead of solving it by hands with trig. subsitution.

 

 

Additional Assignments:

 

 

1) Circuit A has a time constant of 100ms, while circuit B has a time constant of 200ms. Because of the design of the circuit, the output is 2e^(-t/tau), tau as in time constant. Plot the two outputs and identify which circuit will have the lower output sooner.

 

The graph shows that circuit A will have a lower output sooner than B.

 

If the output is changed into 2(1-e^(-7/tau)), replot the graph.

 





Instead of decreasing the output, the output increase.

 

 

2) Determine what the output would be when adding the following two sinusoids: 3sin(2t+10degree) and 5cos(2t-30degree).

The graphical result of the sum of the two sinusoid equations.

 

   Replot the graph with the frequency change to 10Hz.

 

The period changes as frequency changes.





 



Lab 2 : Introduction to Biasing.

3/4/2013

          Introduction: Applying "Biasing" technique to connect electronic components using lower voltage than the voltage supplied by the power source without burning the components.

step 1) Calculate the required baising resistors for each LED.
step 2) Build the circuit with meansuring devices connected to the circuit.
step 3) Measure the current and the voltage across different elements.
step 4) Calculate the effiency and errors in the circuit.

STEP 1

The biasing circuit consist of a 9V power supply , 2 biasing resistors, and 2 LED lights. (2V green LED, and 5V yellow LED)

 

STEP 2

The biasing resistors are approximate at 176 ohm (in series with the 5V LED), and 350 ohm (in series with the 2V LED).

The resistors available to us with the closest values are 220 ohm and 350 ohm.

 

STEP 3

Both LEDs are lit to the approximate same brightness after the voltages across them are "fixed" to their optimal voltages (2 and 5 volt).

 

Voltage across the 5V LED. Error occurs in every circuit. In this case, we are reading a higher voltage than it should be across the 5V LED.

Measuring the current in the branch with the 5V LED. Note tha the DMM is connected in opposite polarity.

 

All measured values for the elements in the circuit.

 

STEP  4

1) Assume that a 9V Alkaline battery has a useful life of 0.2A-hr, with both LEDs in the experiment circuit, how long can the circuit operate before the battery dies?

                                    

                                       0.2 A-hr = 33.64mA * t

                                                   t = 5.95 hr                   The circuit can operate for about 6 hrs.

 

2) Find the percent error between the achieved LED current, and the desired current.

 

                      5V LED:     (22.75-13.65)mA / 22.75mA * 100% = 40%

                      2V LED:     (20.5-20.0)mA / 20.0mA * 100% =2.5%

                

               The achieved LED current was so far off because the actual resistance in the resistors were different from the desired resistance values calculated from the desired current. Moreove, the inexpansive resistors used in the experiment were not within its designed tolerance of 1%.

 

3) Determine the circuit efficiency with both LED in the circuit.

 

            Pout / Pbattery = 0.1452/0.303 * 100% = 48%                Low efficiency.

 

4) Determine whether the efficiency would increase or not if a 6V battery is used instead of 9V. Also, determine what battery would be the most efficient in the circuit.

 

The efficiency increases to 59.9% by replacing the 9V battery with a 6V.

 

The best theoretical value of battery voltage for the circuit would be 5V, such that no resistor is need for the 5V LED but the only for the 2V LED. With fewer resistor and less resistance, the circuit wastes less power to the resistor.

 

 

Bouns!!!!!

 

Why must both LED be in this circuit for it to always work properly??

 

 

The circuit is acutally an example of voltage divider, the current supplied by the battery, Isupply , is increased by adding the parallel LED branch such that the potential drop in R3  will remain at 4V. By doing so, the voltage between the 5V LED and the parallel branch will be at 5V. If the parallel branch is removed, the potential drop in R3 will decrease, and the voltage throughthe 5V resistor will exceed its limited voltage.

Lab1: Intrduction to DC Circuits

2/27/2013

          When learning about electric circuit, lots of elements are assumed to be ideal to simplify calculations. However, ideal case rarely occurs in real conditions. The cables in any circuit have internal resistance that consume the power supplied by the batteries. Pouillet's Law stated that electrical resistance of any substance is proportional to its length and cross-section-area, R=pL/A, where p is the electrical resistivity (a constant).

          In designing a circuit with the load seperated far away from the power source, the maximum permissible stretch or the cable (or wire) must be determined so that there will always be a substantial voltage across the load after consumption by the cable.

          The goal for this lab was to determine the maximum stretch of AWG#30 cable between the battery and  load by determining the permissible cable resistance. Meanwhile, the team would also determine the distribution efficiency and the approximate discharging time of the battery by taking these assumptions:
                     1)  The Load is rated to consume 0.144W when supplied 12V.
                     2)  The Load will operate properly as long as voltage across it is greater than 11V.
                     3)  Battery voltage will remain constant at around 12V and it has a capacity of 0.8Ahr.
    

By using P = V^2 /R, the resistance of the load was calculated to be 1000ohm. A resistance box  of 1W power rating would be used instead of the cables to determine the permissible resistance.

 

A model of the circuit was first constrcuted to test the equipemts as well as the connections.

 

The 1000ohm resistor serving as the load had a measured value of 980ohm and wattage at 0.125W.

 

Color Code	Nominal Value	Measured Value	Within Tolerance?	Wattage
Black-Black-Black-Brown	1000 ohm	980 ohm	No, off by more than 1%	1/8 Watt
(Resistor Box)	94 ohm	90 ohm	Yes, within 1%	1 Watt.

 

 The power supply supplied 12.10V instead of the 12V, and the maximum supply current is 2A.

 

After starting the experiment with the actual circuit with ammeter connected, the resistance in the resistance box was 84ohm as the voltage across the load is dropped to 11V. The current was measured to be 11.44mA.

 

 

 

Data Calculation:

 

a)  With the Amp-hour capacity rating of the battery (0.8Ahr) and the current in the circuit (11.44mA) known, the discharge time could be calculated as:

 

 

0.8Ahr = 11.44mA * t  

t = 69.93hr

 

for which it would take the battery close to 70hrs to discharge.

 

b) The power to the load and the power to the cable could be calculated by P = I^2 * R where I was same for both load and cable, and R was individual resistance.

 

Power to the Load = (11.44mA)^2 * 1000ohm = 0.131W

Power to the Cable =  (11.44mA)^2 * 84ohm = 0.011W

 

   Distrubution efficiency = Pload / (Pload + Pcable) *100 = 92.25

 

c) The power capability of the resistor box was not exceed because the box's rating was at 1W while the power it adsorbed was only at 0.011W.

 

d)  Given that the resistance of AWG#30 is 0.35ohm/m, the maximum distane of between the battery and the load can be calcuated as:

84ohm = 0.3451ohm/m * 2L

L = 121.70m        

 

e)  If instead of using AWG#30, AWG#28. The signal through this tether were 20mA 5V TTL. The longest length of #28 wire that could be used while keeping the signals within sspec would be:

 

 

5.0V - 2.6V = 20mA * R         

R = 2.4 / 20mA

L(212.9 ohm/km) = 2.4 / 20mA                     

L = 563.64m

distance/2 = 563.64m           

distance = 281.82m        

 

   Which mean the load can be at max a distance of 281.82m from the power source and it will take around 564m of AWG#28 wire. 

 

 f) If, in the robo-sub project, 48 volts at 10A was sent down to the sub. 36 volt must reach the sub so that the voltage regulator will stay within regulation. The minimum cable guage that could be used for the project was:

48V = IR + 36V

12V = 10R        

R = 1.2ohm

 

let the tether be 60ft:

r = 1.2ohm / 60ft = 0.02ohm/ft

 

AWG#22 would be the minimum cable that could be used.

 

 

End of Lab 1