THEVENIN EQUIVALENTS

-----INTRO-----

Given a minimum acceptable voltage across Load #2, determine the smallest equivalent Load #2 resistance that can be successfully used.

Use the Thevenin process to reduce part of the network down to an equivalent source and series resistance.

The image below is a closer look at the circuit that we will be working with:

It consists of 2 power sources (volts) and 4 resistors of various resistance.

-----CALCULATIONS-----



 

NOTE: Vx=Vth, knowing this and using nodal analysis allowed us to come up with an answer for the thevenin voltage, 8.64V! Then we removed the power sources to calculate the thevenin resistance, 65.94Ohms!

The completed thevenin circuit is pictured above.

The calculations include an 8V because the experiment is asking for the load required when 8V are applied. We were able to conclude our Rload of 824.40V by using Voltage Division.

-----EXPERIMENT-----

 

 The Set Up

Resistors used: 100Ohms, 2 x 39Ohms, 680Ohms

Voltage (power supply): 2 x 9Volts

Measurements:

(Notice there is only minimal error making us comfortable to continue)

The next step asked for us to disassemble our thevenin circuit to test the original circuit so we could compare them two.

First, verify max power:

R12 was set to equal 0.5RTh

This experiment demonstrated the resemblance that a thevenin circuit, calculated properly, can work as an equivalent to an original circuit. Sometimes it is recommended since the circuit becomes a little simpler to work with.

TRANSISTOR SWITCHING

The circuit below is representing the circuit on paper:

 

If everything is placed in the correct position then the green diode will turn on!

 After we made sure that the circuit was operating properly we then experimented by holding the tip of the wires with our fingertips and allowing the current to flow through our hand!

 In order to get a more precise understanding on how a transistor works we experimented further...

Notice that R1+R2 = R3+R4, therefore the potential on the base of the transistor should be halfway between the two extremes. We tested this theory by using a potentiometer to test amps passing through A1 and A2. The data chart below shows our results:

Graph showing the linear relationship with a positive slope:

 

From this graph we can say that the current emerging from the emitter of the transistor through A2 is fairly constant with A1. We know that the ratio of the current coming out of the emitter is called the beta value; therefore the transistor's amplifying power is 5.93.

This transistor is acting both as a resistor and an amplifier!

NODAL ANALYSIS

Analyze this circuit using Nodal Analysis!

V2 & V3 are unknown. V1=Vbat1 & V4=Vbat2

Calculating Unknown Nodes

Results:

Our actual values resulted in:

EXPERIMENT

2 power sources of 12V each

5 resistors

Calculations:

Compare voltages above with results from below:

This experiment successfully demonstrated  how nodal analysis worked. Through a series of calculations we were able to find the missing variables required to find the power supplied by each battery.

INTRODUCTION TO BIASING

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This problem is one of BIASING.
Figure out the correct voltage and current needed for the component to operate properly.

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Circuit representation of the experiment:

Circuit representation of the experiment with measurement equipment:

Actual experiment:

(Includes - 1 battery, 2 voltmeters, 2 LEDs, 1 breadboard)

 Available Resistors of 350 Ohms and 220 Ohms were used in this experiment

We measured current and voltage drop 

 Data gathered from experiment:

Conclusion: The experiment was a success, our theoretical values were very close to our experimental values. The difference may be accounted for by the change of resistors. The resistance that the experiment was asking for was not available in class. The experiment resulted in a 40% power out of the power supply meaning that 60% was consumed by the resistors in place. The specific voltage and current values accepted before the diodes burn out are listed in the above chart. After 12 minutes the battery voltage will drop too low for the circuit to operate properly.

INTRODUCTION TO DC CIRCUITS

Circuit Set Up

1 Power Supply = Battery

1 fixed Resistor = Load

1 Resistor Substitution Box = Cable

2 DMM

Calculation the Theoretical Value for Load Resistance

Theoretical Resistance Value using Color Code

Assembly Components Ratings:

Resistor Box Rating = 1W

Power Supply Ratings:

Max Supply Voltage = 12V

Max Supply Current = 2A

EXPERIMENT

Total Cable Resistance= 40+30+20+4+3+2+1=100 Ohms

The load is receiving 11V from a battery providing 11.14mA

CALCULATIONS

What's the amount of time the battery can supply this load? 71.81hrs

What's the distribution efficiency of this circuit? 90.3%

CONCLUSION

The load receives its minimum of 11V from a battery providing 11.14mA through a cable resistance of 100Ohms. It takes about 72 hours before the load completely discharges. This circuit is about 91% efficient. The load's power is only 1/4W while the cable has 1W, we did not exceed the power capability of the resistor box. It was also observed that a longer cable would create a greater resistance.