Semiconductor Devices and Circuits

ETE 204

Course Description:

Characteristics and applications of solid-state diodes. Characteristics and biasing of BJT devices in CB, CE, CC amplifier configurations – load lines, input/output impedance and mid-band gain calculations. Characteristics and biasing of JFET devices and amplifiers, including load lines, input/output impedances and mid-band gain calculation.

Lab 1 Diodes and LEDs        2

OBJECTIVES        2

EQUIPMENT  REQUIRED        2

BACKGROUND        2

LABORATORY        4

Part A)        4

Part B)        5

Lab 2 Series/Parallel Diode Circuits        9

OBJECTIVES        9

EQUIPMENT  REQUIRED        9

BACKGROUND        9

LABORATORY        11

Part 1: Simple series diode circuit        11

Part 2: Simple parallel diode circuit:        12

Part 3: Diode and resistor in parallel        12

Part 4: Series diode with parallel diode/resistor:        13

Lab 3 Clippers and Clampers        15

OBJECTIVES        15

EQUIPMENT  REQUIRED        15

BACKGROUND        15

Clippers        15

Clampers        16

LABORATORY        17

Part 1: Clipper Circuits        17

Part 2: Clamper Circuits        21

Lab 4 Power Supplies and Rectification        24

Introduction:        24

OBJECTIVES        24

EQUIPMENT  REQUIRED        24

BACKGROUND        25

LABORATORY        27

Part 1: Full-Wave Rectifier, Unregulated        27

Part 2: Complete DC Power Supply        33

Lab 5 BJT Transistor Biasing        36

INTRODUCTION        36

OBJECTIVES        36

EQUIPMENT  REQUIRED        36

LABORATORY        37

Part 1: Fixed or Base Bias        37

Part 2: Voltage Divider Bias        39

Lab 1 Diodes and LEDs

OBJECTIVES

EQUIPMENT  REQUIRED

BACKGROUND

A diode is a semiconductor device consisting of a pn junction. Diodes have numerous applications in electronics and circuit design in general. Diodes can be used to convert voltage from AC to DC (rectification), reverse-polarity protection, emission of light (LED or Light Emitting Diode), voltage limiting, voltage reference, and many more applications. A diode is a non-linear device, meaning that the plot of current over voltage for it is non-linear. When the voltage across the diode is below the “knee” or threshold voltage, then the diode does not conduct and essentially no current flows (besides a negligible leakage current). As the voltage is increased beyond the knee voltage, then the diode begins to conduct, and the current can change dramatically in this region without the voltage changing significantly. For this reason, the threshold voltage is often approximated and used to solve circuit calculations involving diodes.

Diodes are labeled with a band to denote the cathode (K). The side of the diode without the band is the anode (A). A positive voltage must be applied to the anode for current to flow through it. See Figure 1 for how the schematic symbol of a diode relates to the physical package:

https://upload.wikimedia.org/wikipedia/commons/thumb/8/83/Diode_pinout_en_fr.svg/1431px-Diode_pinout_en_fr.svg.png

Figure 1: Typical Diode Package

https://en.wikipedia.org/wiki/Diode#/media/File:Diode_pinout_en_fr.svg

For a silicon diode like the one used in the lab today, the threshold voltage is approximately 0.7 V. To analyze most diode circuits, simply the approximate threshold voltage can be used. Consider the following circuit in Figure 2:

Figure 2: Simple Diode Circuit

Assuming that the supply voltage exceeds the threshold voltage of the diode, the diode is “on” and the circuit can be analyzed with Equation 1:

where  is the supply voltage,  is the voltage drop of the diode, and  is the resistance. Note that when V1 in Figure 2 is a positive voltage, the diode is said to be forward biased and when V1 is negative the diode is reverse biased. Current does not flow through the diode when it is reverse biased, except for a negligible leakage current. In summary, Equation 1 can only be used when the diode is on.

LABORATORY

Part A)

Creating a plot of voltage versus current for a diode:

 Data Table for Figure 2        

(V)

 (V)

  (V)

   (mA)

 (mW)

0.00

0V

0

0

0

0.25

0V

0.25

0

0

0.50

0.03V

0.47

64.5E-6

30.3E-6

0.75

0.19V

0.56

408.6E-6

228.8E-6

1.00

0.4V

0.6

860.2E-6

516.1E-6

2.00

1.34V

0.66

2.9E-3

1.9E-3

3.00

2.32V

0.68

5.0E-3

3.4E-3

4.00

3.30V

0.7

7.1E-3

5.0E-3

5.00

4.28V

0.72

9.2E-3

6.6E-3

6.00

5.27V

0.73

11.3E-3

8.3E-3

7.00

6.26V

0.74

13.5E-3

10.0E-3

8.00

7.26V

0.74

15.6E-3

11.6E-3

9.00

8.25V

0.75

17.7E-3

13.3E-3

10.0

9.25V

0.75

19.9E-3

14.9E-3

      Figure A (Graph of Data Table Figure 2)

Analysis

A silicon made diode has a threshold voltage of 0.7V so when the source voltage starts getting to 0.7 volts the diode would start to activate which means it would start to conduct and allow current to flow. Our data confirms that once the circuit passes the knee voltage current rapidly increases.

What was the largest value for  that was observed, and the smallest value, while  was non-zero?

Largest value for  

Smallest value for

Is 0.7 V a good approximation to the diode knee voltage?

        Yes 0.7 V is a good approximation because our data as in Figure A agrees with the statement.

Part B)

Diode Polarity

Forward Biased

9mA

Reverse Biased

0mA

Does the diode conduct equally in both directions?

No it does not equal the same current because in reverse biased the diode becomes an open therefore no current will flow through. While in forward biased the diode will conduct as long as it meets the knee voltage.

Calculate  using Equation 1 with a diode threshold voltage of 0.7 V. How does this current compare with the measured value?

    7.10mA(Calculated)

Measured= 7.1mA.     The calculated and measured current actually ended up being the same. This could be because of rounding or because the quality of the resistor used was very great.

LEDs

This part of the laboratory will explore LEDs. Unlike conventional silicon diodes like the 1N4148, LEDs cannot be approximated by a 0.7 V drop when conducting. LED voltage drop is color dependent. Most LEDs will drop around 1.5 to 2.0 V. Equation 1 can still be used when analyzing LED circuits, but just keep in mind that the diode voltage drop will not be 0.7 V when the diode is on.

Design problem:

It is desired to create an LED circuit which provides an LED current of 10 mA from a 5 V supply. Calculate the necessary resistance value to provide this assuming the “on” voltage for the LED is 1.6 V.

 

Choose the nearest standard resistor value to your calculated value above and build the circuit of Figure 2 using your resistor and an LED, with a 5V supply. Use Figure 3 to connect the LED with the correct orientation.

http://www.societyofrobots.com/images/electronics_led_diagram.pngfigure 3: LED (http://www.societyofrobots.com)

+5V  

http://ledcalculator.net/Images/Schematic/terminal_single_start.gif

http://ledcalculator.net/Images/Schematic/led.gif

220Ω

http://ledcalculator.net/Images/Schematic/resistor.gif

http://ledcalculator.net/Images/Schematic/terminal_single_end.gif

  GND

Figure A

(mA)

(mW)

Red

1.93

13.39

25.84

Green

3.07

8.4

25.79

Infrared

1.23

16.45

20.23

Which color LED has the largest voltage drop?

The green LED had the largest voltage drop.

Which color LED has the smallest voltage drop?

Infrared had the smallest.

How close are you to the design goal of 10 mA with each LED?

Red= Not that close compared to other LED since we ended up having 13.39mA more than the design goal.

Green= The closest LED to match the design goal of 10 with a current of 8.4mA.

Infrared= Exceeded the design goal by 6.45mA.

What is a possible advantage to a LED with a lower threshold voltage?

It will create a higher current with very little voltage. This means it would extend how long the LED can be activated.

Bi-color LEDs:

A bi-color LED is actually two LEDs connected backwards to each other, enclosed in a single package. A schematic representation for a two-lead bi-color LED will look like Figure 4:

     Figure 4: Bi-color LED (2-leads)                        Figure 5: Bi-color LED (3-leads), common cathode

The other possible arrangement of a bi-color LED is shown in Figure 5. Notice how both cathodes (K) are connected together. This configuration is called “common cathode” because the two LEDs share a common cathode connection. It is also possible for the LEDs to have their anodes connected in common, and this configuration is called “common anode.”

Obtain a bi-color LED from the instructor. Using the circuit of Figure 2, with the same supply voltage and resistor from before, measure . Then reverse the polarity of the diode and measure again.

LED Color

(mA)

Red

13.3

Green

12.6

Is the current equal through both colors of the LED? Why or why not?

Yes because the LED draws the same current, what changes is the voltage, a RED LED typically uses 1.63 ~ 2.3 Volts and green uses 1.9 ~ 4.0 Volts.

Lab 2 Series/Parallel Diode Circuits

 OBJECTIVES

EQUIPMENT  REQUIRED

BACKGROUND

This laboratory will examine diodes in series and parallel.

Series diodes can be analyzed in much the same way as before in Laboratory 1. The main difference is that instead of a single  drop, there will be many voltage drops in a series circuit with more than one diode. The current can be found using the same equation used in Laboratory 1, except that the  term will be the sum of all diode drops in the series circuit.

where  is the supply voltage,  is the sum of all diode voltage drops, and  is the resistance. Like before, this equation assumes that the supply voltage  exceeds the combined voltage drops of the diodes. Having many diodes in series is useful for the application of LED lighting. Using a series circuit with many LEDs on a single branch, allows many LEDs to be lit using a single current limiting resistor, as opposed to each LED being in parallel and having its own current limiting resistor. An example of a series diode circuit can be seen in Figure 3.

Parallel diode circuits (where two diodes are connected directly in parallel) are not particularly useful, but they should still be examined to better understand how diodes work. On paper, two diodes connected in parallel will share the current equally if they have the same threshold voltage. This configuration can be seen in Figure 1. In reality, two diodes rarely have the exact same threshold voltage and thus they do not share current equally. The diode with the lower threshold voltage will “win out” and will conduct nearly all of the current, while the diode with the higher voltage drop will conduct hardly any current. A more practical parallel diode circuit can be seen in Figure 2, where each diode has its current limiting resistor. Each branch of this circuit can be analyzed like in Laboratory 1.

Figure 1: Impractical Parallel Diode Circuit

Figure 2: Practical Parallel Diode Circuit

LABORATORY

Part 1: Simple series diode circuit

R1=463Ω

                                                 R2=976Ω

Figure 3: Series Diode Circuit

Construct the circuit shown in Figure 3. Measure the current through the circuit, then measure the voltages across diode D1, D2, and then the combination of D1 and D2 together.

7.50mA

.70V

.70V

1.42V

For a theoretical comparison, assume the diode threshold voltage is 0.7 V and solve for the circuit current:

Compare the theoretical and measured currents and find the percent error.

Percent Error

7.50mA

7.78mA

3.6%

= ==7.78mA

% Error = x 100=x 100=3.6%

Part 2: Simple parallel diode circuit:

Figure 4: Parallel Diode Circuit

Measure the voltage across either diode, then measure the current from the supply, and the current through each diode individually. LED=Green

8.99mA

0mA

9.00mA

.72V

Which diode is conducting the majority of the current?

The D1 diode (1N4148)  is conducting the majority of the current.

 Is the LED on? Why or why not?

The LED is off because D1 has a lower active voltage so nothing would go through the LED.

Part 3: Diode and resistor in parallel

Figure 5: Parallel Diode and Resistor

Construct circuit of Figure 5. Measure the current through diode D1 and the supply current. Next, measure the voltage across the D1/R2. The current through R2 can be calculated from the voltage across it.

5 V Supply

0.72V

9.04mA

8.25mA

Now change the supply voltage to 10 V and repeat the previous measurements.

10 V Supply

.76V

19.55mA

18.60mA

Answer the following questions comparing when the supply was 5V versus when it was 10V.

Did the voltage across D1/R2 change significantly? Why or why not?

The voltage across D1/R2 did not change significantly because the diode is active at 0.7V.

Did the current through resistor R2 change significantly? Why or why not?

The current of R2 did not change because the voltage of the diode did not change.

Did the current through diode D1 change significantly? Why or why not?

Yes it changed. The current by a factor of 2.25 once the power supply was increased to 10V.

What did the diode have to do in order to maintain the voltage at the anode?

The diode would have to absorb the current difference of the voltage difference.

Part 4: Series diode with parallel diode/resistor:

Figure 6: Series/Parallel Configuration

Measure the voltage at the anode of D1 (with respect to ground), then the voltage at the anode of D2. Measure the total supply current. Calculate the theoretical supply current by using the voltage on both sides of R1, assuming the diode threshold voltage is 0.7 V.

Percent Error

7.56mA

7.78mA

2.8%

0.71

0.70V

= ==7.78mA

% Error = x 100=x 100= 2.8%


Lab 3 Clippers and Clampers

 OBJECTIVES

EQUIPMENT  REQUIRED

BACKGROUND

Clippers

A clipper circuit is a diode circuit designed to “clip” or remove a portion of a waveform that is above or below a certain voltage. A simple clipper circuit can be seen in Figure 1. This circuit clips negative values due to the orientation of the diode. During the positive swing of the input voltage, the diode is reverse biased and the input waveform can be seen at the output. On the negative portion of the waveform the diode is forward biased and will not allow the voltage to increase beyond the diode threshold voltage of approximately 0.7 V. In this particular configuration, the negative portion of the waveform is clipped off because of the orientation of the diode. If the diode is reversed, then the positive portion of the waveform is clipped. It is also possible to shift the voltage at which the diode clips by inserting a DC voltage in series with the diode. Clipper circuits are used frequently in ICs to prevent an input or output voltage from exceeding the power supply positive or negative voltages, which could damage a device. ESD (Electro Static Discharge) can cause damage to devices and these limiter circuits are used to help mitigate this problem.

Figure 1: Simple Clipper Circuit

Clampers

Clamping circuits get their name because they “clamp” an AC voltage around a DC bias. Clampers cause the varying voltage to be shifted up or down, instead of being centered on zero volts. The peak to peak voltage is unchanged in a clamped circuit, the only thing that changes is the point at which the waveform is centered. To understand the operation of a clamper, let us walk through a couple of cycles of the AC input voltage and examine the output. See Figure 2 for a simple clamping circuit. During the first positive cycle of the input, the diode is reverse biased and the series capacitor (coupling capacitor) passes the AC waveform through to the output. The output will resemble the input during this first positive cycle. On the first negative cycle, the diode becomes forward biased when the input voltage goes below 0.7 V. This charges the capacitor up to . As the voltage is reduced in magnitude from its peak negative value, the capacitor can only discharge through the resistor, because the diode is blocking. Then, on the next positive cycle the capacitor still has its charge from the previous negative cycle, so the voltage of the source and capacitor add because they are in series. This causes the voltage to rise up to . Therefore the whole waveform is shifted up by the amount of voltage the capacitor maintains:  Like in the case of the clipper circuit, this level can be shifted by adding a DC voltage in series with the diode. An important design guideline is ensure that the RC time constant of the circuit is at least 100 times the period of the input waveform. This ensures that the capacitor does not discharge significantly in between refresh cycles.

Figure 2: Diode Clamper Circuit

LABORATORY

Part 1: Clipper Circuits

Figure 4: Clipper with DC Bias

  1. Predict which portion of the waveform will be clipped (positive/negative/both) and at what voltage(s).

Negative, at -2.7V

  1. Construct the circuit of Figure 4. Measure  on an oscilloscope.

R1 = 2.15k

The peak is 4.6V and it clips at -3.2V

  1. Does your prediction agree with what you obtained from the oscilloscope?

Yes

  1. Perform a screen capture from the oscilloscope. Include capture in report.

C:\Users\Mauricio\AppData\Local\Microsoft\Windows\INetCache\Content.Word\TEK0000.jpg

  1. Change the voltage to 1 V and reverse the polarity of the DC supply.
  2. Predict which portion of the waveform will be clipped (positive/negative/both) and at what voltage(s).

Negative, .3V

  1. Perform a screen capture from the oscilloscope. Include capture in report.

  1. Construct the circuit using PSPICE or Multisim. Take a screen capture of  for 3 periods of the input waveform. Include this in your report.

                        

Figure 5: Parallel Clipper

  1. Predict which portion of the waveform will be clipped (positive/negative/both) and at what voltage(s).

the waveform will be clipped at Both positive and negative at .7V and -.7V

  1. Construct the circuit of Figure 5. Measure  on an oscilloscope.

The circuit clips at .656 and -.648 volts

  1. Does your prediction agree with what you obtained from the oscilloscope?

Yes our prediction agrees with our data.

  1. Perform a screen capture from the oscilloscope. Include capture in report.

  1. Add a DC voltage between the cathode of D2 and ground. Choose a voltage of either 1 or 2 volts and choose a polarity.

D2 to positive 1v to gnd

  1. Predict which portion of the waveform will be clipped (positive/negative/both) and at what voltage(s).

Both, 1.7V and -.7V

  1. Does your prediction agree with what you obtained from the oscilloscope?
  1. Yes our prediction agrees although it differs ever so slightly.
  1. Perform a screen capture from the oscilloscope. Include capture in report.

C:\Users\Mauricio\AppData\Local\Microsoft\Windows\INetCache\Content.Word\F0005TEK.JPG

Figure 6: Series Clipper

  1. Predict which portion of the waveform will be clipped (positive/negative/both) and at what voltage(s). For the conducting portion of the waveform, what will be the maximum voltage?

The waveform will be clipped at negative, because the diode will be reverse bias, the max voltage would be 8V.

  1. Construct the circuit of Figure 6. Measure  on an oscilloscope.

 V out was measured to be 8.88V.

  1. Does your prediction agree with what you obtained from the oscilloscope?

Yes our prediction agrees with that of what we obtained from the oscilloscope with a slight variation of 0.88V.

  1. Perform a screen capture from the oscilloscope. Include capture in report.

Part 2: Clamper Circuits

R1 = 2.2k

Figure 7: Positive Clamper

  1. Predict the maximum, minimum, peak to peak, and average value of the output waveform.

Maximum voltage would be 9.3V. Minimum voltage would be 0.7V. Peak to Peak would be 10V. Average value would be 4.65V.

  1. Calculate the RC time constant necessary to provide effective clamping, so that the output waveform will not be significantly distorted.

        [f=1khz,T-.001s,RCtc=>100*T]

Tau=RC=1/[2pi(freq)]

tau=RC=1/[2(3.14)(1khz)]=159us

  1. Choose a value for R1 based on the given value for C1 and the RC time constant.

159us = R(1uF)

R=159

  1. Construct the circuit of Figure 7, using your calculated value for R1. Measure  on an oscilloscope.

        Our mistake was using a 2.2k ohm resistor. Our measured Vout is 8.8V

  1. Does your prediction agree with what you obtained from the oscilloscope?

No our prediction does not agree due to the issue of us using the incorrect resistor.

  1. Perform a screen capture from the oscilloscope. Include capture in report.

  1. Change R1 to a value 1/10 of the previously calculated value. Predict what effect this will have on the operation of the circuit.

This will affect the circuit in that the voltage will drop by half and the negative will double.

  1. Re-construct the circuit using the new value for R1. Does your prediction agree with your results?

        No, it does not I fear it is due to our first mistake of using the wrong resistor.

  1. Perform a screen capture from the oscilloscope. Include capture in report.

Figure 8: Clamper with DC Bias

  1. Predict the maximum, minimum, peak to peak, and average value of the output waveform.

maximum: 21.3V minimum:1.3V

peak-to-peak: 20V avg value: 10V

  1. Calculate the RC time constant necessary to provide effective clamping, so that the output waveform will not be significantly distorted.

Tau=RC=1/[2(3.14)(1khz)]=159us

  1. Use the value for R1 that was calculated in Part 3 of the previous section.
  2. Construct the circuit of Figure 8. Measure  on an oscilloscope.

Vout: 10V and -7V

  1. Does your prediction agree with what you obtained from the oscilloscope?

No this does not agree with our prediction I fear either we misread the oscilloscope or used the wrong resistor value.

6.        Perform a screen capture from the oscilloscope. Include capture in report.

Lab 4 Power Supplies and Rectification

Introduction:

In this lab we will use an oscilloscope in order to visualize how clippers and full rectifiers work. By building each component of a full rectifier separately and in conjunction with each other we see how each part: transformer, rectifier, filter, and regulator operate.

OBJECTIVES

EQUIPMENT  REQUIRED

BACKGROUND

This laboratory will explore DC power supplies. A power supply is an important and often overlooked part of most any design. The job of the power supply is to convert a changing AC voltage to a steady DC voltage. For most power supplies, the AC mains (120 V wall outlet) is rectified and “smoothed” to provide a steady output voltage. There are a few basic subsystems to a power supply:

        Transformer

        Rectifier

        Filter or smoothing circuit

        Voltage Regulator

http://electrical4u.com/electrical/wp-content/uploads/2014/01/components-of-typical-linear-power-supply.gif

Figure 1: Complete Linear Regulated Power Supply

http://electrical4u.com/electrical/wp-content/uploads/2014/01/components-of-typical-linear-power-supply.gif

The transformer is used to step-down the 120 VAC mains to a lower AC voltage before rectification. A transformer consists of a primary and secondary side, which are completely isolated (no common connection). The AC mains voltage connects to the primary side of the transformer and the output is taken at the secondary side. The output of a transformer can in general be step-up or step-down, but for power supplies it is usually step-down. The secondary voltage can be calculated from the turns ratio of the primary and secondary windings, like in Equation 1. Remember that a transformer simply changes the amplitude of the AC waveform, the waveform shape remains intact.

(1)

The rectifier has the job of converting an AC voltage into a pseudo-DC voltage. The rectifier can either block the negative portions of the AC waveform (half-wave rectifier) or invert the negative portions to make them positive (full-wave rectifier). A half-wave rectifier is not efficient because the entire negative cycle of the waveform is unused. Full-wave rectifiers are preferred because they make more efficient use of the input voltage. Half-wave rectifiers are seldom used, and will not be examined in this laboratory.

C:\Users\Jason\Dropbox\Teaching\ETT 201\Lab 1\rect2.gif

Figure 2: Half-Wave Rectifier

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/ietron/rectbr2.gifhttp://hyperphysics.phy-astr.gsu.edu/hbase/electronic/ietron/rectbr2.gif

Figure 3: Full-Wave Bridge Rectifier

Adding a capacitor in parallel with the output provides “smoothing” or filtering of the unrectified waveform. This provides a “DC” voltage with a slight ripple depending on the values of R and C.

LABORATORY

Part 1: Full-Wave Rectifier, Unregulated

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/ietron/rectbr2.gif

Figure 4: Full-Wave Bridge Rectifier, Unregulated

  1. Construct the circuit of Figure 4 using a bridge rectifier package, not individual diodes. DO NOT include the resistor and capacitor yet.

Use an oscilloscope to measure the output waveform from the transformer. Take a scope screen capture and show a measurement for Vpp on the capture.

  1. Use an oscilloscope to examine the output of the rectifier. Record the peak value of the waveform and the frequency of the waveform. Perform a scope screen capture of the waveform.

  1. Now add a resistor and capacitor as shown in Figure 4. Use a 100 µF capacitor and a 2.2 (2.18) kΩ resistor.Record the DC voltage from the output with a multi-meter.

  1. Measure the output ripple using an oscilloscope with AC coupling. Include a scope screen capture showing the measurement.

  1. Calculate an approximate output peak-peak ripple voltage using the following formula. Don’t forget that the frequency of a full-wave rectified sine wave is twice the input sine wave:

where  is the peak voltage of the AC input to the rectifier,  is the threshold voltage of the diodes in the bridge rectifier,  is the frequency of the output waveform,  is the load resistor, and  is the filter capacitor.

= 295.45mV

  1. Change  to 463 Ω, and then calculate and measure the output ripple voltage again.
  2. Was the output ripple increased, decreased, or did it remain the same?

        The ripple was increased.

  1. Does this trend agree with the calculation?

        Yes this trend agree with the calculation

  1. Now, increase the capacitor value to something in the range of 330 µF to 470 µF.What effect should this change make to the ripple voltage?

        The ripple voltage should decrease.

  1. Take another scope capture after the capacitor was changed. Measure the ripple voltage and compare from before.

        

  1. What type of filter does a capacitor in parallel with the output provide, HPF (High Pass Filter) or LPF (Low Pass Filter)? Think of the reactance of the capacitor as frequency increases, and the frequency of the output we are trying to “pass.”

        The type of filter that the  capacitor in parallel with the output provides a Low pass filter.


  1. Use PSPICE or Multisim to simulate the circuit of Figure 4, using R = 470 (463) Ω and C removed. Include a screen capture of output waveform showing at least 2 periods of the ripple signal.

Add the capacitor at 100 µF and perform and then simulate and capture again.

Part 2: Complete DC Power Supply

http://www.eleinmec.com/figures/016_04.gif

Figure 6: Complete Linear Regulated Power Supply

  1. Construct the circuit of Figure 6. Use a value of 330 µF to 470 µF for the “smoothing” capacitor. Use a load resistance of 2.15 kΩ. Measure the output of the regulator with a voltmeter.

        4.98V

  1. Measure the output ripple of the regulator (not the rectifier). Is the ripple greater, smaller, or unchanged compared to the unregulated circuit using the same R and C values currently used?

        

  1. If standard 0.25 W resistors were used as the load, what is lowest value of resistance that could be chosen without exceeding the maximum power dissipation of the resistor?

        7805 supports 1.5A

        If our output voltage is 4.98V

        P=IV = .25W=1.5A*4.98V = 50mA

        R=V/I = 4.98V/50mA = 99.6Ω

  1. Estimate the maximum power dissipation of the regulator under full load, at the regulator input voltage of your circuit. This can be done by knowing the voltage across the regulator, and the current through it. P = IV

P=1.5A*(9.20V-4.98V) = 6.33W

  1. Suppose that your input voltage to the regulator increased to 12 V. What would the new power dissipation of the regulator be?
  2. P=1.5A*(9.20V-4.98V) = 6.33W
  3. Remove the bridge rectifier circuit and connect a DC power supply directly to the input of the voltage regulator. Also, remove the load resistor. Set the power supply to 8.5 VDC.Connect a multi-meter to measure the output voltage of the regulator. Slowly decrease power supply voltage until a significant change in output voltage occurs. Measure and record the regulator input voltage at this point (DC supply).

        Regulator input 8.5V output = 4.98V

Regulator input 5.61V output =4.75V

  1. Set the power supply back to 8.5 VDC. Connect a load resistor of 462 Ω. Slowly decrease the power supply voltage again and then record the input value when the output wchanges significantly.

        Regulator input 8.5V output = 4.98V

Regulator input 6.07V output =4.75V

  1. When the output starts to fall away from the regulated 5 V, this is known as regulator “dropout.” In order for a (linear) voltage regulator to maintain regulation, its input voltage must be larger than its output by a certain amount (often 2 to 3 V) called the “dropout” voltage. Based on the two dropout measurements performed above, what can be concluded regarding regulator dropout voltage and load resistance? Does the load affect what input voltage the regulator “drops out” at?

As the load resistance increases the dropout voltage increases.

  1. Based on both power dissipation and dropout voltage, approximately what input voltage is ideal to minimize power consumption, yet maintain voltage regulation?


Lab 5 BJT Transistor Biasing

INTRODUCTION

        In this lab we are to learn the difference between the different type of biasing circuits used with BJT transistor. When transistors are manufactured the their is a large tolerance between the variance of β, which is why we will test two of the same model of transistor to learn the difference of their β variance on Q-pt. stability.

OBJECTIVES

EQUIPMENT  REQUIRED

LABORATORY

Part 1: Fixed or Base Bias

Figure 1: Fixed or Base Bias Configuration

Measure R1 and R2 with a multimeter. Record the values and use them in the following calculations: R2= 463k R1= 1.81k

Assume β=100. By hand calculation, find , , and .

19.78uA

6.42V

1.978mA

What mode of operation is this transistor in? Why? 

The mode of operation of this transistor configuration is linear because Vce is in between 0V and 10V

Construct the circuit of Figure 1. 

Measure , , and  with a multi-meter.

Calculate  for this transistor. This will be transistor #1.

Transistor #1 (measured)

β

52.4uA

7.5V

21.4mA

408.97

Record data for this transistor. Now swap the current measured transistor for a different one, the new transistor will be transistor #2.

Measure , , and  with a multi-meter again for transistor #2. Recalculate β.

Transistor #2 (measured)

β

51.8uA

7.6V

21.4mA

413.12

Compare calculated  and  with both measured values.

Calculated

Measured Transistor #1

Measured Transistor #2

6.42V

1.978mA

7.5V

21.4mA

7.6

21.4mA

Is there a large variation between calculated and measured or between the two measured transistors? Why/Why not?

There is  a large variation between our calculated and measured transistors due to our theoretical β being 100 and our actual β being approximately 400.

Change the value of R2 to be 10kΩ. Measure the resistance with a multimeter and record the value. Recalculate the circuit by hand with the new value. Then assemble the circuit and measure the same variables.

R2=9.91k

Calculated

939.4uA

-160.03

93.94mA

Measured

924uA

9.3mV

5.73mA

What mode of operation is the transistor operating in? Why?

The mode of operation of this transistor configuration is saturated because Vce is very close to 0V.

Part 2: Voltage Divider Bias

Figure 2: Voltage Divider Bias


Based on the circuit of Figure 2, measure , , , and  with a multi-meter. Record the values and use them in the following calculations: R1=9.91k R2=1.8k Rc=1.81k Re=470

Assume β=100.

Based on the values of  and  , is this a “stiff” voltage divider? Recall that for a stiff voltage divider: Yes it is a stiff voltage based off our values of Re and R2. 48k > 18k

What if β changed to a value of 50, is it still a stiff voltage divider? It would still remain as a stiff voltage because Re would still be greater than R2. 23.5k > 18k

By hand calculation, find , and  using the appropriate method and β=100.

9.6

176uA

Construct the circuit of Figure 2.

Measure , and  with a multimeter.

Now swap the transistor for a different one. Re-measure.

Transistor #1 (measured)

Transistor #2 (measured)

1.589mA

5.37

1.609mA

5.43

Compare calculated  and  with both measured values.

Calculated

Measured Transistor #1

Measured Transistor #2

9.6V

179uA

5.37

1.589mA

5.43

1.609mA

Is there a large variation between calculated and measured or between the two measured transistors? Why/Why not?  

There is a large variation between the calculated and measured. This is because the calculated values was with the assumption that β would be 100 but our β wasn’t 100. Between the two measured transistors there was a very small variation so it would be safe to assume their  β were the same.

Compare fixed-bias with voltage divider bias. Which one provided a more stable Q-pt?

Voltage divider bias provides a more stable.

What mode of operation is this transistor in? Why?

The mode of operation of this transistor configuration is linear because Vce is in between 0V and 10V.

Change the value of R2 to be 10kΩ. Measure the resistance with a multi-meter and record the value. Recalculate the circuit by hand with the new value. Then assemble the circuit and measure the same variables. R2=9.91k

Measured

25.6uA

9.90V

What mode of operation is the transistor operating in? Why?

The mode of operation of this transistor configuration is linear because Vce is in between 0V and 10V.

Change the value of R2 to be 470Ω. Measure the resistance with a multi-meter and record the value. Recalculate the circuit by hand with the new value. Then assemble the circuit and measure the same variables.

Measured

.5uA

10.01

What mode of operation is the transistor operating in? Why?

The mode of operation of this transistor configuration is saturated because Vce is 10V.