Week 5 8-14 February 2016

During this week's lab we work with an Oscilloscope and Function Generator along with a transistor on an integrated circuit.

      1. Functional check: Oscilloscope manual page 5. Perform the functional check (photo).

First we performed an a Functional Check on the Oscilloscope as outlined on page 5 of the Oscilloscope Manual.

Testing a 10x Oscilloscope Probe as part of the Functional Check. 

      2. Perform manual probe compensation (Oscilloscope manual page 8) (Photo of overcompensation and proper compensation).

During the Functional check we discovered we found one of our probes was out of calibration, so we performed manual probe compensation in accordance with the Oscilloscope Manual Page 8. 

Manual Probe Compensation Photos

Probe in a state of overcompensation.

Probe reading after proper calibration.

      3. What does probe attenuation (1x vs 10x) do (Oscilloscope manual page 9)?

Some probes have multiple attenuation settings that can be selected using a mechanical switch on handle of the probe.  It is important to insure the setting you select matches the setting on the Oscilloscope.  The attenuation settings change the impedance through the probes you are using, allowing a more accurate reading.  If your probes settings do not match those of your Oscilloscope you will be unable to utilize the machine's full bandwidth potential.   

      4. How do vertical and horizontal controls work? Why would you need it (Oscilloscope manual pages 34-35)?

Prior to starting this week's experiment we took some time to familiarize ourselves with the oscilloscope's controls.  Our particular oscilloscope has two separate input channels and the control panel allows the user to choose whether to display data from either channel individually or both channels simultaneously.  

The Horizontal and Vertical controls operate similarly to the controls on an old tube style television and allow you to control where the data is displayed on the screen.  This is especially useful when displaying data from both channels as you can choose to either separate the data on different parts of the screen or directly overlay both channels data atop one another for easy comparison.

The Scale controls are also important as they allow the user to either expand or condense the data being displayed as needed to take proper reading.

Another important control is the Auto Set button which displays the scales the data on the display in an "optimized" manner with the press of a single button.  While it is not always exactly what is called for it provides a good starting point for finer adjustment.  

Also in conjunction with the Measure button the five buttons to the immediate right of the O-scope's screen allow the user to customize the information on the display.    

     5. Generate a 1 kHz, 1 Vpp around a DC 2 V from the function generator (use the output connector). DO NOT USE oscilloscope probes for the function generator. There is a separate BNC cable for the function generator.

a.       Connect this to the oscilloscope and verify the input signal using the horizontal and vertical readings (photo).

b.      Figure out how to measure the signal properties using menu buttons on the scope.

Our first task was to connect our function generator to our O-Scope and attempt to generate a 2VDC/RMS signal.  Our instructions indicated we should have been able to accomplish this by generating 1 kHz / 1Vpp signal but even after increasing our output to amplitude to 2V (4Vpp) we were only able to achieve 1.50 V RMS 

Function Generator putting out a 1 kHz 4Vpp Signal

Despite increasing our input voltage up to 4Vpp we were only able to achieve 1.50 V RMS

Note the RMS data displayed on the right hand side of the screen.  The data displayed is customizable using the Menu/Measure buttons.  In this case we selected Min/Max and Cycle/Cursor RMS Voltages.

  

      6. Connect function generator and oscilloscope probes switched (red to black, black to red). What happens? Why?

In accordance with the lab instructions we reversed the the black and red leads on the O-scope probe, while we expected to see a the waveform reverse, but instead the oscilloscope only displayed a flat zero volt signal.  

We believe the reason for this that according to the O-scope manual (page v) the ground lead is at ground potential.  Because of this the signal goes directly to ground and bypasses the oscilloscope's measurement circuitry.

7.  After calibrating the second probe, implement the voltage divider circuit below. Measure the following voltages using the Oscilloscope and comment on your results:

&

8. For the same circuit above, measure V_a and V_b using the handheld DMM both in AC and DC mode. What are your findings? Explain.

Circuit 7 & 8 Schematic

When attempting to implement the circuit below we were unable to achieve an input of .5 VAC/5 VDC utilizing our function generator.
We ended up using an input amplitude of 1.5 Volts with an offset of +3.72 Volts.  When we attempted to either increase the offset or input amplitude the function generator gave a warning message.

The data we collected for Steps 7&8 is below 

When taking our readings voltage division between the three resistors worked as expected with all three resistors dropping between 102-106 mV.  The measured RMS voltage as measured by the O-Scope was almost exactly one -tenth of the DMM's VDC reading.  While the DMM's AC reading was approximately equal to the O-Scope's V-RMS reading multiplied by sqrt(2).

      9.       For the circuit below

a.       Calculate R so given voltage values are satisfied. Explain your work (video)

b.      Construct the circuit and measure the values with the DMM and oscilloscope (video). Hint: 1kΩ cannot be probed directly by the scope. But R6 and R7 are in series and it does not matter which one is connected to the function generator. 

Step 9 Schematic Diagram

We Calculated the Resistance using the following Steps

    1.) Multiply 5V (RMS) by Sqrt 2 to Get an input Voltage of of 7.07 Vpp

    2.) Calculate VR7 by subtracting VR6 from the 7.07V input.  VR7=5.07 Vpp

    3.) Calculate R7 using the equation R7 = R6(VR7/VR6)

    4.) R7 = 2535 Ohms or 2.5 Kilo-Ohms 

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This is a short video describing how we calculated our R7 value for the given circuit.

This video chronicles the readings taken of circuit as requesting in step 9B, which confirmed the calculations we made in step 9A

      10.   Operational amplifier basics: Construct the following circuits using the pin diagram of the Op-Amp. The half circle on top of the pin diagram corresponds to the notch on the integrated circuit (IC). Explanations of the pin numbers are below:

a.       Inverting amplifier: R_in = 1kΩ, R_f = 5kΩ (do not forget -10 V and +10 V). Apply 1 V_pp @ 1kHz. Observe input and output at the same time. What happens if you slowly increase the input voltage up to 5 V? Explain your findings. (Video)

b.      Non-inverting amplifier: R₁ = 1kΩ, R₂ = 5kΩ (do not forget -10 V and +10 V). Apply 1 V_pp @ 1kHz. Observe input and output at the same time. What happens if you slowly increase the input voltage up to 5 V? Explain your findings. (Video)

Circuit 10 A (Inverting Amplifier) Schematic

Chris Explains the data we collected from the Inverting Amplifier.
A complete breakdown of this data can be found in the Table Below.

Circuit 10 B Schematic

Chris Explains the data we collected from the Non-Inverting Amplifier.
A complete breakdown of this data can be found in the Table Below.

Week 4: 1-5, February 2016

  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. Use 10K potentiometer to generate the voltage. Use 5 V but 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). The circuit should look like below:

Circuit 1 Schematics

With the circuit setup in accordance with the schematic above, we were able to use the 10k potentiometer to apply various voltages to the base of the transmitter.  As the voltage increases from 0-0.6V there is only a slight increase in current at the Collector Node.  When the voltage increases to 0.65V the current begins to increase exponentially reaching a peak of 4.18 mA when we stopped the experiment at 0.66V to prevent potential damage to the transistor.

This table shows the values we recorded for VBE and IC

Chart showing the Exponential increase in Current at the Transistor's Collector Node as Voltage at the Base increases past its trigger point.

  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. The circuit should look like below:

Circuit 2 Schematic

This table above displays the values we found for this particular experiment.

With 0 Volts present at the Transistor Base only 1 mA of Current was present at the collector base

With 1 Volt present at the Transistor Collector Current increased gradually from 0 to 14.8 mA as the Base voltage increased from 0 to 0.7v between 0.7 to 0.8v the current began increase exponentially peaking at 79.2.

With 2 Volt present at the Transistor Collector Current increased gradually from 0 to 29.3 mA as the Base voltage increased from 0 to 0.7v between 0.7 to 0.8v the current began increase exponentially peaking at 142.

With 4 Volt present at the Transistor Collector Current increased gradually from 0 to 23.8 mA as the Base voltage increased from 0 to 0.7v between 0.7 to 0.8v the current began increase exponentially peaking at 272.

  
  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?

I_C and I_B are related by beta, which is equal to I_C divided by I_B . For example, if beta was equal to 100, that would mean that whatever the current for I_B was, the the current for I_C would be 100 times greater. Our data did not support this. We believe that our transistor was damaged in the previous circuit, yielding us inaccurate results. 

  4.     (Table) Explain photocell outputs with different light settings. Create a table for the light conditions and photocell resistance.
   
    A photocell provides different resistance values depending on the amount of light that it receives. The more light it receives, the less resistance it supplies. Our data we collected supports this.

This table illustrates the resistance value measured across the Photo Sensor under various lighting conditions. 

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

This table illustrates the current drawn by the motor as the voltage increases from 0-5V in 1 volt increments. 

  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.

    With 2 volts applied increasing pressure was applied to the shaft of the motor.  The amount of current being drawn increased from 36.1 mA with no load applied, to 97.3 mA at the point sufficient pressure was applied to stop the motor. 

The table above shows the current being drawn by the electric motor as various amounts of pressure are applied to the spinning shaft.

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

Rube Goldberg Circuit Schematic

The Circuit above operated by using a photo sensor in conjunction with a 1k W resistor to control the voltage present at the base of the transistor.  A second power supply provides power to the transistor's collector node.  As the light intensity increase the photo sensor's resistance decreases, this increases the voltage present at the transistor base, allowing voltage to flow from the Collector to the Emitter providing power to the motor.

Video of the rebuilt Rube Goldberg Circuit in action.

 
  8.     Explain the 47 Ohm Resistor'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.

  The 47 Ohm resistor in the Rube Goldberg circuit serves as a buffer that prevents excessive current from the power supply from damaging the transistor.  Ohm's Law dictates the if the buffer's resistance is doubled the current value is halved, and if the resistance is halved the current will double.  This prediction is confirmed by the data we collected which can be seen on the table below.

Chart of Current and Voltage present at the Transistor's Collector Node.  

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

  Our team's approach to the Rube Goldberg circuit is to utilize the a paddle attached to the motor to knock over a Formation of Dominoes that can then be used to trigger the next portion of the circuit. 

A successful test of our Rube Goldberg Circuit knocking over a Formation of Dominoes

* Video of our previous "failed" tests are posted on our Lab Team's You Tube Channel at

Lab Team 4 on YouTube  

Week 3: 25-31 January, 2016

  1.)  Compare the calculated and measured equivalent resistance values between the nodes A and B  for three circuit configurations given below.

Experiment 1 Schematics



The Table above shows the measured resistance values
from Point A to Point B for circuits a b and c. 

All measurements were within 1 Ohm of the values
calculated using the formulas above.  



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

     
    If you put the Multimeter in series like you are supposed to, you will get the current reading that is going through the resistor. If you put the Multimeter in parallel, the current will just go through the Multimeter, skipping the resistor, essentially shorting the circuit.

  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.

       

    By applying 5 Volts to both the a 47 and 120 Ohm resistor connected in series, we were able to collect the following voltage and current Data. 



Measured and Calculated Voltage and Current across
a 47 and 120 Ohm resistor connected in Series 

All Measured Voltages were within 1% of the calculated values

The Measured current was within 0.7 mA of the calculated value.



  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.

Experiment 4 Schematic

The Voltage Measurement across the circuit above were as follows

Voltage across the 120 Ω resistor:  4.98 V 

Current through the 120 Ω resistor:  39.8 mA

Voltage across the 120 Ω resistor:  4.97 V 

Current through the 120 Ω resistor:  94.8 mA

These readings are consistant with the expected values since voltage is constant in a parallel circuit. The voltage measured at any point in the circuit should be equal to the 5 V applied by the power supply.

According to Ohm's Law Current = Voltage/Resistance, as such

5 V divided 120 Ohms = 41.7 mA  which is very close to the measured 39.8 mA 

and

5V divided by 47 Ohms = 106.4 mA  which is also close the measured 94.8 mA 

The variations between the calculated and measured values can most likely be attributed to a higher than expected resistances caused by using multiple sets of alligator clips and meter leads to construct and measure the circuit. 

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

a.       Current on 2 kΩ resistor, 

b.      Voltage across both 1.2 kΩ resistors. 

Photo of the circuit as laid out on our breadboard

Alternate view of the circuit as laid out on our breadboard



Experiment 5 Schematic with labelled Mesh Loops



Theoretical Voltages and Current Readings calculated using Mesh Analysis on the TI-Nspire CAS student software.

A.)  The current measured across the 2 kΩ resistor was 1.83 mA  Which closely matches the calculated I1 current of 2.05 mA.

B.) The Voltages were as follows 

(R1 refers to the 1.2 kΩ resistor drawn horizontally in the schematic above)

(R2 refers to the 1.2 kΩ resistor drawn vertically in the schematic above)

Measured Voltage Across R1:  1.149 V      

Measured Voltage Across R2:  1.045 V

Voltage across both 1.2 kΩ resistor was somewhat higher than the 0.88 V and 0.75 V calculated above.

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

    The Equivalent Resistance for the circuit is equal to the applied voltage over the calculated resistance.  Since we are reading at the power supply we can use 5V as the Test voltage and 2.056 mA as calculated above for the I1 current. Using these values the equivalent resistance is calculated as 2.43 kΩ by dividing 5V by .002056A.

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

    Without Power resistance was measured at 2.73 kΩ which is close to the 2.43 kΩ calculated in question 6.

    With power applied the meter read "OL", this is because the current flowing through the circuit was overloading the Multimeter rendering it unable to take an accurate reading.

  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)

This video shows us using the potentiometer, and demonstrating how you can use it to put out a

constant 10KΩ, or using the first and second, or second and third pin to put out a variable resistance. 

     The resistance of the Potentiometer is measured as follow

    

    - T-1  to T-3 will always read the Potentiometer's maximum

       resistance (in this case 10 kΩ)

    - T-1 to T2 will vary from 0 to the Potentiometer's max value

       based on the knob's position
    - T-2 to T-3 will also range from 0 to the Potentiometer's max value
       again based on the knob's position.

-   * Note:  (T1 to T2) + (T2 to T3) = Potentiometer's Max value

-

  9.)  What would be the minimum and maximum voltage that can be obtained at V1 by changing the knob position of the 10 KΩ pot? Explain.

    If you have the potentiometer set to 0 Ω , there will be no voltage drop across it, because there would be no resistance. However, as long as the potentiometer is at anything other than zero, the voltage drop will always be 5 volts, with a varying amount current.

Experiment 9 Schematic

  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)

Experiment 10 Schematic

A short Video showing how the potentiometer setting

affects voltage at V1 and V2

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



Experiment 11 Schematic

      If you turn the potentiometer all the way down to zero, turning it's resistance to zero, you would short the circuit, and all of the current would then go through the potentiometer since there is no resistance. We unfortunately did exactly this our first time doing this circuit, and fried the potentiometer.


    *  Question 11 was not added to this week's blog requirements until after we discovered the our potentiometer's inability to handle the circuit's maximum current.  

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).

Short Video showing the effects of the potentiometer setting on both the 1 kΩ resistor and 5 KΩ potentiometer's current readings. 

    As the video above demonstrates the potentiometer has no effect on the current running through the 1 kΩ resistor, this is because current is a function of Voltage and Resistance, and since voltage is constant throughout a parallel circuit and the resistor's value is fixed its current remains the same regardless of what value the potentiometer is set at.  The only exception to this rule would be if the potentiometer was set to 0 resistance in which case all current would bypass the 1 kΩ resistor and it would draw zero current.  This would be a bad idea however as this short circuit condition would damage the potentiometer as we discussed in question 11.  What the potentiometer setting does effect is the amount of current going through the potentiometer itself.  As resistance increases the amount of current decreases  and as resistance decreases the amount of current passing through the potentiometer will increase until a value of 0 Ω is reached at which point maximum current will pass through the potentiometer.



  13.)   Explain what a voltage divider is and how it works based on your experiments.

Voltage divider is the process of how voltage splits across different resistors in a series circuit as demonstrated above (See # 10).  For that circuit a 5 volt drop was observed across the two resistors. However, when the potentiometer was set to zero, the 1 kΩ resistor would drop the entire 5 volts, and when the potentiometer was to 10 kΩ, the 1 kΩ resistor about only about 0.45 volts  and the remaining 4.55 volts would drop across the potentiometer. 

You can think that the total voltage as being divided into 11 parts (since there was 11 kΩ of  total resistance), since 10 of the 11kΩ of resistance was provided by the potentiometer it receives 10 parts of the voltage,leaving only 1 part for the 1 kΩ resistor. 

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



    Current divider is the similar as voltage divider, but deals with current in a parallel circuit. This process can be seen in the circuit from #11. When two resistors are in parallel, the current wants to take the path of least resistance. In the circuit from #11, the current has two possible paths, the left loop through the potentiometer, or the right loop through the  1 kΩ resistor.  When the potentiometer was set to 10K, most of the current (about 90%)bypassed the potentiometer and travelled through the 1 kΩ resistor.  As the potentiometer's resistance decreased more current travelled through the potentiometer and less travelled through the 1 kΩ resistor 
    
    Similarly to Voltage Divider you can think that the current  as being divided into 11 parts (since there was 11 kΩ of  total resistance), however; when using the current divider rule as resistance increases current decreases so when the potentiometer is providing 10 parts of the total resistance it only receives 1 part of the total current, leaving 10 parts for the 1 kΩ resistor.

 
    Post contributors (C Ruby & K Pawlak)