Daniel Bruce and Moaad Alzahrani

Blog sheet week 4
(January 30 - February 3)

1.       (Table and graph) Use the transistor by itself. The goal is to create the graph for I_C (y axis) versus V_BE (x axis). Connect base and collector. DO NOT EXCEED 1 V for V_BE. Make sure you have the required voltage value set before applying it to the base. Transistor might get really hot. Do not TOUCH THE TRANSISTOR! Make sure to get enough data points to graph. (Suggestion: measure for V_BE = 0V, 0.5V, and 1V and fill the gaps if necessary by taking extra measurements). 

Table 1.1 Voltage of Base Emitter and the current flow through the collector

Graph 1.1 Shows relationship Vbe vs Ic

The graph above shows that the current of the collector is dependent on the voltage of the base emitter; also note that there is almost no current flow until the voltage of the base emitter reaches about .7 volts. This point of current flow is known as saturation.

2.       (Table and graph) Create the graph for I_C (y axis) versus V_CE (x axis). Vary VCE from 0 V to 5 V. Do this measurement for 3 different VBE values: 0V, 0.7V, and 0.8V.

Table 2.1 Voltage base emitter and collector and corresponding current of collector

Graph 2.1 Shows the Vce vs Ic relationship in a transistor from our experiment

The graph above shows that the current in the collector increases rapidly with an increase in voltage of the collector. But the graph also shows that the base emitter requires around .7 or .8 Volts to achieve this increase in the collector. It should also be noted that when we applied .8 volts to the base emitter the transistor was behaving abnormally and was giving us unusually low readings of current through the collector. We later found that the transistor was faulty in our experiments; and we did not have enough time to repeat the experiment to get new readings, but it is still useful in showing the relationship of increase in Vc results in increase of Ic.

3.       (Table) Apply the following bias voltages and fill out the table. How is I_C and I_B related? Does your data support your theory?

VBE (V)	VCE (V)	IC (mA)	IB (mA)
0.7 V	2 V	138	.152
0.75 V	2 V	276	.281
0.8 V	2 V	385	1.12

Table 3.1 Changing base emitter voltage and measuring corresponding collector and base current

From our measurements we see a trend of the current through the base is much smaller than the current through the collector. Around .7 Volts we see a significant amount of current flow. As the voltage is increased in the base emitter the amount of current flow through the collector begins to become excessive causing heat build up in the transistor.

4.       (Table) Explain photocell outputs with different light settings. Create a table for the light conditions and photocell resistance.

Table 4.1 Shows resistance values of photo cell as varying degrees of light

When there is no light on the photocell the resistance increases to the maximum amount. As we increased the amount of light on the photocell, the resistance seemed to dropped quite a bit giving a range of about .3K ohms to 12K ohms.

5.       (Table) Apply voltage (0 to 5 V with 1 V steps) to DC motor directly and measure the current using the DMM. 

Table 5.1 Shows voltage values and corresponding current values through the dc motor

We saw that increasing the voltage on the dc motor caused the current flow to increase as well  as increase the rotational speed of the motor.

6.       Apply 2 V to the DC motor and measure the current. Repeat this by increasing the load on the DC motor. Slightly pinching the shaft would do the trick. 

When we applied 2V to the motor and let it spin freely it consumed about 32.9 mA, when we pinched the shaft on the motor the current rose to 161 mA.

7.       (Video) Create the circuit below (same circuit from week 1). Explain the operation in detail


.

Video 7.1 Explains the operation of the circuit below we constructed

We showed how the transistor, motor and photocell are able to work together and be activated by light to provide current to the transistor; which then allows current to flow to the motor.

  Circuit Diagram 7.1 Electrical circuit with transistor,motor and photocell

8.       Explain R₄’s role by changing its value to a smaller and bigger resistors and observing the voltage and the current at the collector of the transistor. 

Resistor 4's role is to limit the amount of current that flows through the motor. When we put a resistor of higher value  the motor would barely spin even with light shining on it. We also put a wire in place of the resistor (very low resitance value); this caused the motor to spin faster.

9.       (Video) Create your own Rube Goldberg setup. 

Video 9.1 Demonstration of our Rube Goldberg circuit

The video shows the 3 major components (transistor, motor, and photocell) we learned about this week working in unison to create a Rube Goldberg.

Daniel Bruce and Moaad Alzahrani

Week 3

      1.     Compare the calculated and measured equivalent resistance values between the nodes A and B for three circuit configurations given below. Choose your own resistors. (Table)

Part A	R1	R2	R3	R4	REQ Calculated		REQ Measured
	121 ohm	48 ohm	100 ohm		25.6 ohm		27.1 ohm
Part B	R1	R2	R3	R4	REQ Calculated		REQ Measured
	121 ohm	48 ohm	100 ohm		152.2 ohm		153.4 ohm
Part C	R1	R2	R3	R4	REQ Calculated		REQ Measured
	121 ohm	48 ohm	100 ohm	2K ohm	167.9 ohm		142 ohm

Table 1.1 Shows the caculated and measured resistance values of our circuits.

Most of our calculated readings were very close to the measured readings, except for part C. I assume we may have had some wires touching unknowingly decreasing our resistance values.

      2.     Apply 5V on a 120 Ω resistor. Measure the current by putting the multimeter in series and parallel. Why are they different?

      When measuring the current in series we are measuring the current flowing through the resistor.   When measuring the current in parallel we are shorting the circuit causing almost no current to flow through the resistor, and almost all of the available current to flow through the meter resulting in a wrong measurement.

3    3.     Apply 5 V to two resistors (47 Ω and 120 Ω) that are in series. Compare the measured and calculated values of voltage and current values on each resistor.

   
      Table 3.1 Below is a table of our measurements of each resistor's current values and voltage values.
    

		120 ohm		48 ohm
Calculated voltage		3.57 V		1.42 V
Measured voltage		3.56 V		1.39 V
Calculated current		29 mA		28 mA
Measured current		28 mA		28 mA

      4.     Apply 5 V to two resistors (47 Ω and 120 Ω) that are in parallel. Compare the measured and calculated values of voltage and current values on each resistor.

		120 ohm		48 ohm
Calculated voltage		5V		5V
Measured voltage		5V		5V
Calculated current		41 mA		88 mA
Measured current		37 mA		85 mA

      5.     Compare the calculated and measured values of the following current and voltage for the circuit below: (breadboard photo)

Figure 5.1 Shows the connections on our breadboard for measuring current and voltage

a.     Current on 2 kΩ resistor, 

Calculated
I=V/R ,  5.0V/2.5K=2.0mA
      Measured
I=V/R , 5.3V/2.5K=2.5mA

b.     Voltage across both 1.2 kΩ resistors. 

Calculated
1.2K Resistor A ;V=I*R , .15mA*1.2K ohm= .18V
1.2K Resistor B ;V=I*R , .88mA*1.2K ohm= 1.056V
      Measured
1.2K Resistor A ;V=I*R , .148mA*1.2K ohm=.1776V
1.2K Resistor B ;V=I*R , .885mA*1.2K ohm=.1.062V

      6.     What would be the equivalent resistance value of the circuit above (between the power supply nodes)?
        

       We measured 2.06K ohms of the circuit between the power supply nodes, this is the equivalent resistance value between the power supply nodes.

      7.     Measure the equivalent resistance with and without the 5 V power supply. Are they different? Why?

       Yes, because the power introduced voltage to the circuit. The meter is supposed to produce a small current and read the voltage; when the power supply added voltage to that reading the meter is assuming the resistance is very high in the circuit; this gave us a reading of "over limit" on the meter.

      8.     Explain the operation of a potentiometer by measuring the resistance values between the terminals (there are 3 terminals, so there would be 3 combinations). (video)

       

Video 8.1 Showing the operation of the potentiometer

   The video above shows the variable resistance between the middle terminal and any one of the other terminals. Also we wanted to point out that maximum resistance is always obtained by connecting the two outside terminals regardless of the position of the knob on the potentiometer.

      9.     What would be the minimum and maximum voltage that can be obtained at V1 by changing the knob position of the 5 KΩ pot? Explain.

We used a 10K ohm potentiometer and the minimum and maximum voltage that can be obtained at V1 is the same at any position of the 10 K ohm potentiometer. Changing the resistance value does not change the voltage drop across the potentiometer. The full 5V will always be dropped in this circuit because it is the only resistor in the circuit.

     10.  How are V1 and V2 (voltages are defined with respect to ground) related and how do they change with the position of the knob of the pot? (video)

Video 10.1 Shows the voltage drop vs resistance relationship of a series circuit using a potentiometer and resistor

     11.  For the circuit below, YOU SHOULD NOT turn down the potentiometer all the way down to reach 0 Ω. Why?

We should not turn down the potentiometer all the way because the resistance of the potentiometer will be almost zero. This is a parallel circuit, and if the potentiometer has too low of resistance, this will in effect become a short to ground; causing maximum current to flow through it and may burn up the potentiometer.

     12.  For the circuit above, how are current values of 1 kΩ resistor and 5 KΩ pot related and how do they change with the position of the knob of the pot? (video).

      

Video 12.1 Shows the current flow vs resistance of a parallel circuit using a potentiometer and resistor

     13.  Explain what a voltage divider is and how it works based on your experiments.
   
      A voltage divider effectively turns a higher voltage into one that is smaller by placing a resistor is series with the target device or component, causing the output voltage after the resistor to be lower to the desired level for the component. When we placed a potentiometer in front of a resistor we were able to effectively change the voltage on the resistor by changing the position of the knob on the potentiometer. When the resistance was increased on the potentiometer the voltage on the resistor was lowered.

     14.  Explain what a current divider is and how it works based on your experiments.

      A current divider allows current be adjusted to a desired level for a component by placing a resistor in parallel to the component. We used a potentiometer placed parallel of a resistor. As we increased the resistance of the potentiometer the current flow through the resistor was increased. When the resistance of the potentiometer was decreased this would direct less current flow through the resistor and more through the potentiometer.