electronics circuits and devices lab manual

electronics circuits and devices lab manual

Manual Authors: Eyad Al-kouz Princess Sumaya University for Technology Download file PDF Read file Download file PDF Read file Download citation Copy link Link copied Read file Download citation Copy link Link copied Abstract Electronics Lab.Eyad Al-Kouz Oct., 2010 I present it here as a very practical statement on how the human mind approaches the learning process: I h e ar, an d I fo rg et. I s e e, a nd I re me m b er. I d o, an d I un de r st an d. One of the best ways to understand something is to get your hands on it and actually experiment with it. In electronics, this means putting small circuits together, powering them up, and seeing first hand what they do. Table of Contents: Experiment (1): The voltage across the diode terminals determines whether or not the diode will conduct. If the anode is more positive than the cathode, the diode will conduct current and is said to be forward-biased. If the cathode is more positive than the anode, the diode will conduct only an extremely small leakage current and is said to be reverse-biased. When forward-biased, the voltage drop across a typical silicon diode is about 0.7V (germanium diodes drop about 0.3V). At forward voltages below this threshold, the diode only permits a small current to flow. This threshold is known as the knee of the diode characteristic curve. Since the relationship between voltage across and current through the diode changes in this region, the diode's resistance changes.Plot I on the vertical axis and V D on the horizontal axis. 3. Determine the static resistance of the diode at 0.1V, 0.5V, and 0.6V using the values obtained for I and V D from Table 1.1. 4. Graphically determine the dynamic resistance of the diode at 0.1 V, 0.5V, and 0.6V using the I-V characteristic curve obtained in question 2. 5. Using the values obtained for I and V D from Table 1.2, calculate the static resistance of the diode at -10V. Table 1.http://almondzwealth.com/administrator/imagetemp/dell-xps-8600-manual.xml

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1 In small-signal diode circuits, the changes in current and voltage occur over a very small portion of the characteristic curve of the diode. The diode is operated in its forward-biased region, which is only linear above the knee or the characteristic. (In the linear region, a small change in diode voltage yields a large change in diode current). In order to use the linear portion of the diode's characteristic, the small-signal is added to a dc voltage which shifts the operating region past the knee of the characteristic. Every possible combination or current I and voltage V D is a point which lies on this line. The quiescent current and voltage correspond to the point of intersection of the characteristic curve and the load line. The name load line comes from the fact that the slope of the line is inversely proportional to the series resistance, or load, in the circuit. Figure 2.I.1 shows how the load line is altered when the input voltage has a small ac component. When the ac component is zero, the diode current and voltage have their quiescent values, I Q and V Q. Using an oscilloscope set for ac input coupling, observe and accurately sketch the ac component of the voltage V D (t) across the diode. 4. To measure the ac voltage V R (t) across the resistor, it is necessary to interchange the diode and the resistor as shown in figure 2.I.3: Figure 2.I.3 5. Using an oscilloscope set for ac input coupling, observe and accurately sketch the ac component or the voltage V R (t) across the resistor. Having measured V R (t), the ac Compare this graphically obtained Q point with the measured Q point from procedure step 2. 2. Using the measurements of procedure steps 2 and 5, write an expression for the total diode current i D (t). 3. Using figure 2.I.1 as a guideline, draw the peak (maximum and minimum) ac load lines on the diode characteristic. Now draw the ac component of the diode voltage V D (t) and determine the resulting current i D (t).http://concretosmecanicos.com/nbloom/fckuploads/dell-xps-desktop-user-manual.xml

Compare the graphically obtained i D (t) with the results obtained analytically in question 2. Experiment (2): Diode Circuits Thus, in large-signal circuits, diode operation is not confined to the linear region. This means that the diode resistance changes from a very low value to a very high one. Consequently, the diode acts very much like a switch. One or the most important applications of the diode in large-signal circuits is to perform rectification, by which an alternating current is changed to a direct current. Thus, load current will flow in one direction only. It is important to note, however, that the diode will drop about 0.7V of the voltage applied to the circuit containing it. Therefore, the applied voltage must be greater than 0.7V in order for any current to flow through the load. A half-wave rectifier with graphs of applied voltage and resulting current is shown in figure 2.II.1. Figure 2.II.1 Another important application of large signal diode circuits is that of performing digital logic functions. Digital logic gates are circuits which perform logical functions such as AND or OR. In the AND function, the output is true (a high voltage) only if both In a typical logic circuit, an AND gate produces a 5V output only if inputs 1 and 2 are both 5V. The OR gate produces a 5V output if either input 1 or input 2 is 5V. 2. II.4 Procedure 1. To investigate the use of a diode in half-wave rectifier circuits, connect the circuit shown in fig. 2.II.2: Figure 2.II.2 2. With a dual-trace oscilloscope set for dc input coupling, measure the peak-to-peak values of the input voltage e(t) and the output voltage V R (t). Sketch both waveforms. 3. Reverse the diode terminals in figure 2.II.2 and repeat procedure steps 2. 4. Change the input signal e(t) in figure 2.II.2 to a 5Vp-p square wave and repeat procedure step 2. 5. To investigate the use of diodes in simple digital logic gates, connect the following circuit shown in figure 2.II.3: Figure 2.II.3 6.

Measure and record the values of V O at each of the combinations of values of V 1 and V 2 in Table 2.II.1. Calculate V O, assuming that each diode has a forward-biased voltage drop of 0.7V. Compare the calculated values of V O with the measured values. 5. What digital logic function is performed by the circuit in figure 2.II.3, and figure 2.II.4? V 1 (Volts ( V 2 (Volts ( V O (Volts ( 0 0 0 5 5 0 5 5 As will be demonstrated in a later experiment, the common base amplifier is also used in conjunction with FET amplifiers for high frequency amplification. When used as a small-signal amplifier; the input and output voltages and currents vary over a small range of the transistor's characteristic curves. In this situation, the amplifier is said to be operating in its Linear region, i.e. the gain of the amplifier is the same for all amplitude variations at the input and output. Small-signal amplifiers are often analyzed using ac equivalent circuits. Figure 3.I.1 shows the small-signal ac equivalent circuit of the common base amplifier in figure 3.I.3. Notice that no capacitors or dc voltage sources appear in the equivalent circuit, because they are assumed to be short-circuits to the ac signal. R 1 and R 2 in figure 3.I.3 are similarly shorted to ac ground.Adjust this potentiometer until V O is (one-half of the previous output). Remove the potentiometer and measure its resistance.Compare these with the measured values. 4. Explain the distortion observed in procedure step 6. Why does the output waveform distort when the amplitude of the input is increased above a certain value? It is extremely useful because it has very high input resistance, high current gain, very small output resistance, and approximately unity voltage gain. The high input resistance and low output resistance make the emitter follower an ideal buffer between a high impedance source and a low impedance load. A buffer is any circuit that keeps the source from being affected by a load.

For example, a common emitter amplifier with a 10k.However, the load in this case is in parallel with the emitter resistor, so the output voltage is in phase with the input voltage.It is extremely useful because it has high voltage gain, high current gain, moderate input resistance and moderate output resistance. The common emitter amplifier will be used as the example in most general amplifier experiments in this book. In many common emitter amplifiers, the emitter resistor is bypassed, by connecting a capacitor in parallel with it. At high frequencies, the capacitor effectively shorts the emitter resistor to ground, but at dc the capacitor is large impedance that does not affect the dc biasing of the circuit. The purpose of the emitter bypass capacitor is to increase the gain of the amplifier by eliminating ac degeneration. AC degeneration occurs when there is a voltage present across the emitter resistor that is out of phase with the output voltage. The small-signal ac equivalent circuit in figure 4.1 can be used to calculate the gain, input resistance, and output resistance of the common emitter amplifier of figure 4.3. The equivalent circuit does not show R E because it is assumed to be completely bypassed (shorted to ground) by C E at the frequency of operation. Note that the current-controlled current source in figure 4.1 is pointing down, unlike that of the common base amplifier. For this reason the common emitter amplifier is referred to as an inverting amplifier. The open-circuit voltage gain A V of the common emitter amplifier can be calculated using the appropriate one of the following equations (the minus sign means that the common emitter amplifier is an inverting amplifier): Resistors R l and R 2 are feedback resistors which generally improve the amplifier's characteristics at the expensive of voltage gain. At the same time, the voltage gain is stabilized to a particular value, which is also a desirable characteristic. Figure 5.I.

1 The op amp with feedback will have characteristics determined mostly by the two external resistors.These included summation, subtraction, multiplication, division, integration, and differentiation. Figure 5.II.2 shows an example of how an operational amplifier is connected to perform voltage summation. (In this figure, an ac and a dc voltage are summed). To perform integration, a capacitor is connected in the feedback path of the amplifier. However, any dc voltage appearing at the input of an integrator will cause the output voltage to rise (or fall) until it reaches its maximum possible value. To prevent this undesirable occurrence, a resistor, To perform differentiation, a capacitor is connected in series with the input. For this reason, a resistor is placed in series with the capacitor in the input. This establishes another break frequency that can be calculated as in the integrator. 5. II.4 Procedure 1. To demonstrate the use of an operational amplifier as a summing amplifier, connect the following circuit shown in figure 5.II.2: Sketch the output waveform. Be sure to note the dc level in the output. 3. Interchange the 5V dc power supply and the 2Vp-p signal generator. Repeat procedure step 2. 4. To investigate the use of an operational amplifier to perform mathematical integration, connect the following circuit shown if figure 5.II.3: 5. With V S adjusted to produce a 10Vp-p sine wave at 20Hz, and using a dual-trace oscilloscope set to ac input coupling, measure and record the peak value of the output voltage V O in Table 5.II.1. Note any phase shift of the output voltage V O with respect to the input voltage V S. Repeat this procedure for the remaining frequencies in Table 5.II.1. Using a dual-trace oscilloscope, sketch the output voltage V O with respect to the input voltage V S. 7. To demonstrate the use of an operational amplifier as a differentiator, connect the following circuit shown in figure 5.II.4: 8.

With V S adjusted to produce a 2Vp-p sine wave at 500Hz, and using a dual-trace oscilloscope set to ac input coupling, measure and record the peak value of the output voltage V O in Table 5.II.2. Note any phase shift of the output voltage V O with respect to the input voltage V S. Repeat this procedure for the remaining frequencies in Table 5.II.2. Figure5.II.4 9. Set V S to a 2Vp-p square wave at 200Hz. Using a dual-trace oscilloscope, sketch the output voltage V O with respect to the input voltage V S. Table 5.II.1 Frequency (Hz) Vo (Volts) Phase Shift 20 50 100 What differences, If any, were there between the sketches obtained in procedure step 6 and 9 and figure 5.II.1? 6. Calculate the theoretical break frequencies for both the integrator of figure 5.II.3, and the differentiator of figure 5.II.4. At very low frequencies, the capacitive reactance of the coupling capacitors may become large enough to drop some of the input voltage or output voltage. Also, the emitter bypass capacitor may become large enough that it no longer shorts the emitter resistor to ground. The following equation can be used to determine the lower cutoff frequency, where the voltage gain drops 3dB from its mid-band value (0.707 times the mid-band A V ): This fact can lead to problems when an amplifier is used for high-frequency amplification. A transistor has inherent shunt capacitances between each pair of its terminals. At high frequencies, these capacitances effectively short (shunt) the ac signal voltage. Therefore, in high-frequency amplifiers, shunt capacitance must be extremely small. In this experiment, artificial shunt capacitors will be installed in the amplifier circuit because it is extremely difficult to measure the actual interelectrode capacitances of the transistor. It is equally difficult to measure stray shunt-capacitance due to the wiring of the circuit.

Since the artificial capacitors are much larger than the real capacitance already present, the parallel combination of real capacitance and artificial capacitors is approximately equal to the value of the artificial capacitors. The objective is to investigate the high frequency response of the amplifier to gain insight into the problems associated with shunt capacitance, and to obtain practice measuring the upper cutoff frequency of an amplifier. Make sure the output voltage from the signal generator is constant. These values will be used to plot the low frequency response of the amplifier. 5. By making two capacitors very large, the effects of those capacitors on the lower cutoff frequency can be made negligible. The cutoff frequency due to the third capacitor can then be measured.From this graph of frequency response, obtain the lower cutoff frequency (-3dB point) and label the graph accordingly. 3. Ca1culate f 1 (C 1 ), f 1 (C 2 ), and f 1 (C E ) for the common emitter amplifier. Compare these with the 1ower cutoff frequencies measured in procedure steps 5-8. 4. Which capacitor had the most effect on the lower cutoff frequency. Compare the lower cutoff frequency due to this capacitor with the overall cutoff frequency determined in question 2. Comment and explain. It is easy to see that the resulting gain of two cascaded ideal amplifiers would be the product of the individual gains. However, since voltage sources and amplifiers have output resistance, and amplifiers and loads have input resistance, there are voltage divisions taking place between these stages or amplification. Figure 7.I.1 shows a two-stage amplifier. A O1 and A O2 are the open-circuit (un- loaded) voltage gains of each stage: Figure 7.I.1 To calculate the overall gain of the multistage amplifiers, the product of the individual open-circuit gains of each stage is reduced by the voltage divider at the input and the output of each stage.

The following equation can be used to calculate the overall gain of a two-stage amplifier. Compare the calculated gain with that measured in procedure step 5. 3. Explain the phase relationship of the output voltage V L with respect to the input voltage V S as observed in procedure step 5. FREQUENCY RESPONSE OF MULTISTAGE AMPLIFIERS 7.II.1 Reference Sections 11.1-11.3, Electronic Devices and Circuits: Gain Relations in Multistage Amplifiers, Methods of Coupling, RC Coupled BJT Amplifiers. 7.II.2 Objectives 1. To measure the low frequency response of a two-stage common emitter amplifier. 2. To measure the lower cutoff frequency due to each coupling capacitor of the two-stage common emitter amplifier. 7.II.3 Discussion Whenever several amplifiers are RC coupled in order to furnish adequate gain, there will be additional coupling capacitors affecting the lower frequency response. As was the case with the single-stage amplifier, there is a lower cutoff frequency associated with each coupling and bypass capacitor. The calculations for lower cutoff frequency are the same as for the single-stage amplifier, except there are additional capacitors that must be taken into account. In this experiment, no emitter-bypass capacitors will be used. A O1 and A O2 are the open-circuit (un-loaded) voltage gains of each stage. The capacitors C 1, C 2, and C 3 are coupling capacitors. The following equations are used to calcu1ate the lower cutoff frequency due to each coupling capacitor C 1, C 2, and C 3.The cutoff frequency due to the third capacitor can then be measured.Add asymptotic lines to this graph and show the break frequency due to each coupling capacitor.For example, a measurement of outside temperature would show a nearly sinusoidal variation with a positive peak in the early afternoon and a negative peak in the early morning. A voltage waveform representing this variation would have a frequency of about 11.6?Hz.

At a frequency this low, coupling capacitors would prevent any of the waveform from passing through the amplifier. Therefore, direct-coupled amplifiers are used. The circuit shown in figure 8.1 is an example of a direct-coupled amplifier. Notice that the dc collector voltage V C1 of the first stage is also the dc base voltage V B2 of the second stage. Since the collector-to-base junction of an NPN transistor must be reverse-biased to keep the collector voltage in the linear region of the output characteristics, the collector voltage of Q 2 must be more positive than the collector voltage of Q 1, after several stages of amplification, the collector voltage approaches the supply voltage and the voltage swing of the output is limited. To correct the problem of increasing collector voltages from one stage to the next, the second stage can be changed to a complementary (PNP) device, as shown in figure 8.2. Recall that the collector voltage of a PNP transistor must be more negative than its base voltage. Therefore, the collector voltage of Q 2 is less than the collector voltage of Q 1. Typically, direct-coupled amplifiers will consist of an NPN stage followed by a PNP stage, etc. For either the circuit in figure 8.1 or the circuit in figure 8.2, the no-load voltage gain is calculated just as it is with RC coupled amplifiers. In this experiment, the large value of R E2 in both circuits makes the loading effect of the second stage on the first Decrease the variable dc power supply until the input voltage V B1 is 1.40V dc. For example, a 2N5457 N-channel JFET will have an I DSS which can vary from 1.0 mA to 5.0 mA, and an V GS(off) which can vary from -0.5V to -6.0V. This wide variation makes it a challenge to create a circuit which will work with any 2N5457 JFET. It is often desired to know the I DSS and V GS(off) for a specific JFET. They can be measured using the circuit of figure 9.1. I DSS is measured in figure 9.1 (a) by measuring the voltage directly across 1k.

For example, if R S were to decrease, the bias line would be more vertical, thus intersecting the transfer curve higher up. As an example of its use to predict circuit operation, consider what would happen in the circuit if R 1 were to increase in value. The gate voltage would become smaller, but since R S is unchanged, the whole bias line would shift to the left. This would cause the bias line to intersect the transfer curve at a lower value of drain current. So, if R 1 increases, the drain current will decrease. This, in turn, would cause the drain to source voltage to increase. 9.3.2: JFET AMPLIFIERS A JFET parameter which controls voltage gain in FET amplifiers is called r m. In fact, most FET amplifier equations are almost the same as those used in BJTs, except that.The parameter r m, is slightly more difficult to obtain than was ?. Input resistance for FETs is generally regarded as infinite, consequently Ri is controlled by the biasing resistors only.The drain voltage should be between 13 and 14V, while the source voltage should be between 6.1 and 9.6V. 7. Complete the table 9.2 with your measurements. Hint: Voc means output voltage without load, and V L means the output load voltage. Using dual channel oscilloscope, measure and sketch the both signals, and observe the phase shift between input and output signals. 8. Remove the bypass capacitor on source resistor and repeat step7. The source voltage should be between 6.1 and 9.6V. Hint: Voc means output voltage without load, and V L means the output load voltage.The input stage to an op-amp is a differential amplifier. Most differential amplifiers are constructed as integrated circuit, but to facilitate experimentation, we will investigate a discrete version of the same circuit. Differential amplifiers can be operated in either of two manners: the input signals can be different, or the input signals can be identical.

If the input signals are different, the amplifier is said to be operating in its difference mode. This means that the output voltage will be proportional to the difference in the two input signals. If the input signals are the same, or the inputs are tied together, the amplifier is said to be operating in its common mode. Figure 10.1 shows a differential amplifier with small external emitter resistors designed to compensate for any differences in the values of r e of the two transistors. Where: r id: is the total ac resistance between the input terminals. The common mode rejection ratio (CMRR) is a ratio of signal gain to noise gain that is, how well the amplifier amplifies the wanted signal and cancels the unwanted noise. The single ended. CMRR is a ratio of the single-ended difference mode voltage gain to the single-ended common-mode voltage gain. The double-ended CMRR is a ratio of the double-ended difference mode voltage gain to the double-ended common- mode voltage gain. Typically, the CMRR is extremely high (75 to 100dB is not uncommon).Also measure the maximum resistance of the potentiometer (across its outer terminals). Figure 10.2 3. With V i1 and V i2 set to 0V (grounded), connect a digital voltmeter between the outputs V O1 and V O2 so it will read dc volts. Now adjust the 200? potentiometer until the voltmeter reads 0Vdc. Measure the ac voltage from the center tap of the potentiometer to ground. Connect a dual-trace oscilloscope to observe and measure the single-ended output voltages V O1 and V O2 with respect to ground. If the oscilloscope has a difference mode setting, also measure the output difference voltage (V O1 -V O2 ). In each case note the phase relation with the input. Also calculate the differential voltage gain (or double-ended voltage gain) using the same inputs. Compare these with the measured values obtained in procedure step 5. 3. Calculate the value of R id using the results of question 1 and the value for ?. 4.

using the results of procedure step 7, calculate the differential common mode gain and calculate the CMRR in dB. 5. What is the purpose of the common emitter stage which has the differential amplifier in its collector circuit? ResearchGate has not been able to resolve any references for this publication. Advertisement Recommendations Discover more Project Electronics Lab. This document compares the results obtained after two questionnaires were applied to two different groups of students at the College of Electrical Engineering (FIE) at the Universidad Tecnologica de Panama. The second questionnaire was applied to FIE faculty members. The aim of the program is to produce high quality electrical and electronic engineering graduates in Malaysia. There were no specific methods used in this program. The approach used more toward students' motivation. The results show that most of the students who joined this program had improved their academic performance. 23.1% of students able to get their GPA above 3.0 while 63.7% students improved their GPA as compared with previous semester. This has proved that MyUiTM Engineer program offered by the Faculty of Electrical Engineering, Universiti Teknologi MARA is of high quality especially to produce competent electrical and electronic engineers. Called SUPERB (Summer Undergraduate Program in Engineering Research at Berkeley), the program brings six to eight underrepresented minority students to Berkeley for a research experience. The goals of the program are to affirm the motivation of the students for graduate study while strengthening their qualifications. Based on evaluation of the program and annual tracking of the student participants, these goals have been met. About 75% of SUPERB alumni have attended or are attending graduate programs in engineering. In addition, other unanticipated positive effects have resulted.

The Berkeley REU model differs from some others in ways which present interesting challenges Read more Download citation What type of file do you want. RIS BibTeX Plain Text What do you want to download. Citation only Citation and abstract Download ResearchGate iOS App Get it from the App Store now. Install Keep up with your stats and more Access scientific knowledge from anywhere or Discover by subject area Recruit researchers Join for free Login Email Tip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password. Keep me logged in Log in or Continue with LinkedIn Continue with Google Welcome back. Keep me logged in Log in or Continue with LinkedIn Continue with Google No account. All rights reserved. Terms Privacy Copyright Imprint. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. You can download the paper by clicking the button above. Related Papers electronic devices and circuits By Joshua Refe Ece 311 By Nezar Alhazeem READ PAPER Download file. Download link for ECE 2nd SEM EC6211 CIRCUITS DEVICES Lab Manua l is listed down for students to make perfect utilization and score maximum marks with our study materials. This is a zip file. After downloading it to your computer, unzip its contents to any folder on your machine. For unzipping, you can can use any zip utility. Most likely, some zip utility is already available on your computer. If not, you can download any zip utility available on the internet (e.g., 7-zip ). The unzipped individual circuit files will have the “.lab” file extension. ElectricVLab provides a visualization of the operation of the circuit. You can interact with the circuit by changing the voltages, resistance values etc. You can also build and simulate your own circuits.