Analog Circuits Utilizing Electronic Workbench - Case Study Example

Running Header: Analogue Electronics Analogue Electronics Name & ID Course Name & Code Instructor’s Name 10 December 2009 Table of Contents Table of Contents 2 Task 1: Regulated Power Supply 3 Task 2 – Oscillator Circuits 8 Task 3: Wien Bridge Oscillator 10 Conclusions 11 Task 1: Regulated Power Supply Introduction The aim of this report is to be able to enter and simulate analogue circuits utilising electronic workbench. The most popular and common use of the diode (diodes are usually semiconductors in nature) is in the rectification of current. Rectification means converting alternating current to direct current. To effectively rectify AC power, four diodes are used in full wave rectifiers that employs the two half cycles of a sine wave. The formula that is used to calculate DC is easy to use and apply and to smoothen the signal effectively, capacitors and resistors are mostly employed. This was evident in the laboratory research as shown: Part 1 Step 0: After initialising the windows session, the electronic workbench is located as shown in figure 1a. Step 1 From the start menu section, we locate the Electronic Workbench and then we run the program as shown in figure 1b Step 2 With the help of the mouse cursor, we select and click “diode” library, and then we select the Full-Wave Bridge Rectifier. Then, the Full-Wave Bridge Rectifier is dragged and dropped into the worksheet window as shown in figure 2. Through the provision Save as, we saved the session with the name hamed.ewb. Step 3 In the basic section, numerous components that is applicable in electronic analysis, which includes resistor, nodes, inductors and capacitors to name some. As shown in figure 3, we selected capacitor and resistor, which are paramount in fulfilling the requirements of this project. This program is easily to use and like the movement of the full wave, these two components are dragged and dropped into the electronic worksheet, and then with the help of lines, the nodes are connected. Step 4 From the source panel, we select the sine wave generator and we place it at the bridge input. The same process is repeated as illustrated in figure 4; we drag and drop the requirement. Generally, the following illustrates the design parameters: 20V RMS supply (Peak of 20*s^0.5 = 28.28 V) at frequency = 50 Hz Load resistor = 1k Peak to peak ripple should be less than 1.5 V Thus, from this information, the capacitor value can be calculated From the notes, the formula is Vr = Iave/ (2fc) Where Iave represents (Vpeak - Vr/2-1.4)/ Reload Assumptions is that two diode drops give 1.4 V Iave is 26.13 mA C=174.23F Step 5 To obtain the required vales, the schematic components should be edited. The program used makes this requirement easier because double-clicking on the resistor brings up editor box (5a) while double clicking on the sine wave generator brings up editor box (figure 5b) Figure 5b: Alternating Current Voltage Source Properties Double clicking on the capacitor component also produces editor box as seen in figure 5c Step 6 As shown in figure 6, to start the simulation process, the switch ON mode is clicked with value ‘1’ into the right top side of the window. Like the other processes, the simulation is probed through dragging and dropping an oscilloscope. Double clicking the oscilloscope opens up a new screen and the colour of the wires can be changed. Moreover, the time base is adjusted to 0.01 s/div; Channel B: 20 V0Div and Channel A: also 20V/Div; these features can be seen in figure 7. At this time, the simulation has been paused. Step 8 Measuring the rectified voltage as shown as red in figure 8a and 8c is achieved through expanding the Oscilloscope widow using the option “Expand” on the bottom of the window. Through these process figures 8a, 8b, and 8c is seen. The red line, which refers to the peak voltage is 27.10V as shown in figure 8a while at the bottom of the ripple is about 25.45 as shown in figure 8c. Thus, how do these values obtained compare with the expected values The calculated value of peak voltage is 28.28V while the peak-to-peak ripple should not be more than 1.5. Through analysis, it can be assumed that the desire voltage has an error of about 4%. Part 2 Like part 1, the same procedures are repeated by these new design parameters are included: 25V rms supply Ripple of 2V peak to peak 820 Ohms load Thus, for capacitance the formula and calculation is: C = ( Vpeak – Vripple/2 – 1.4diode ) / (2 f Vripple R) C = ( 25 V * 20.5 – 2 V/2 – 1.4 V ) / (2 * 50 Hz * 2 V 820 Ohms ) C = 200 uF The simulation is then ran and compared with the values that were expected Same diagram as in Part 1 is used by some values are changed. The new values are: Capacitance = 200uF Resistor = 820 Ohms VAC = 25rms at 50Hz (Vpeak = 35.35 V) Utilising these values we obtain figure 9. Comparison of the values can be obtained through oscilloscope and the voltage rectified with those expected values are compared. The information obtained is recorder in that the peak voltage simulated is 34.20V as seen in figure 10a, and after the first half cycle the voltage remains the same has seen in figure 10b, the same value is obtained at the end of the cycle as shown by figure 10c; this means that the rectified voltage is constant. However, the value that was expected is 35.35 V, which shows that there is an error of 3%. Part 3 The aim of this part is to investigate an Operational Amplicator that is controlled by linear regulator circuit with the current limit show in figure 11. After ensuring that the circuit is effectively operating, some questions that should be answered include about its functionality Relationship between the output and zener voltage Amount of load required for the current limit to start operating What is the value of the short circuit output current Within the conditions of short circuit, what is the power dissipation in Q1 Steps that are important in limiting power dissipation in Q1 to its rated value of 1W Modify the circuit so that with the used of a potentiometer, the output voltage moves from 0 to its maximum value with the given zener. Task 2 – Oscillator Circuits Schmitt Trigger Such types of amplifies usually have feedback network connected to the non-inverting input. For example, the output amplifiers is saturated with high output voltage of Vh, this will result in the non-inverting input to have a voltage of Vh*R2/ (R1+R2). However, if this signal on the inverting input exceeds this amount of voltage it will result in output swing to its lowest saturation value of –VL that in turn results in a new voltage on the non-inverting input provided by the formula: -VL*R2/ (R1+R2) Thus, the output voltage will remain low until the signal reaches or becomes lower than inverting-input but when the amount of output retains normalcy, the output then swing back high again. Hence, two systems that involve threshold levels are evident and provided by the following formulae: Va = Vh*R2/ (R1+R2) Vb = -VL*R2/ (R1+R2) Thus, this result in a situation that can be referred to as hysteresis in which the signal is supposed to cover a distance of Va-Vb, to change the position of the output. Schmitt trigger is the phenomenal that is commonly referred to hysteresis. Those devices that possess this feature offers strategic provisions that eliminate sound or noise effect on a signal. Sometimes, noise degradation can result because of unwanted swings that are up to Va above zero and Vb below zero without it affecting switching functionality. For example, simulating the circuit provided in figure 12, when R1 = 10 k and R2 = 4.7 k and bringing into consideration the sine generator to input frequency of 1 kHz and amplitude of 8V. Analysing Va out high and Vb out low from the Op Amp and evaluation of switching points V and V. These values are compared with the expected values: Va = 8 * 4.7 k / (10 k + 4.7 k) = 2.55 V Phase Shift RC oscillator In this situation, simulation of the oscillator will not start on its own. Thus, an input that is sine generator is applied via R6 (1M). The simulation was performed and the output was examined. The high resonant frequency of the oscillator is 1/ (2 6^0.5 C R) Where the values R = 1 k and C = 10nF, and the resonance that is obtained occurs at the correct frequency. Thus, Fresonant1 = 1 / (2 6^0.5 10 x 10-9 F 1000) = 6.497 kHz Re-run the simulation with R = 1 k and C = 1nF. Fresonant2 = 1 / (2 6^0.5 1 x 10-9 F 1000) = 64.974 kHz From this calculation, it is easy to calculate the resonance. Thus, a question may be asked in that why does the frequency plot roll off when the frequency supersedes resonance? Because the characteristic of the term resonance is that, there is a peak around it. Task 3: Wien Bridge Oscillator Simulation Circuit The diagram in figure 19 shows a Wien Bridge oscillator that uses LM741 operational amplifier that runs from +/-9V. R1 = R2 = R = 10 k C1 = C2 = C = 10nF R3 = 20 k, R4 = 10 k, R5 = 1M. When the frequency of oscillation is given by 1/(2  R C) the value of frequency shown will be: Frequency = 1/ (2 *  * 10 x 103 * 10 x 10-9 F) = 1.591 kHz To ensure that oscillation takes place, the amplifier gain is supposed to be 3, and so it is given by R3 = 2R4 The nature of simulation states that the circuit finds a solution where the output is zero. This means that it is paramount to introduce disturbance into the circuit and also to increase the amplifier gain slightly. Thus, the values that will be used are R3 = 20 k, and R4 = 10 k To cause disturbance, the sine generator circuit is connected via 1M resistor to the amplifier input. This results in the generator been set at 1Mv, and the frequency does not play a major role and can be set, for example, to 1 kHz. Therefore, R is set to its half value (R1 = R2 = 5 k), and the entire simulation is re-run Freq = 1/ (2*  * 5 x 103 * 10 x 10-9 F) = 3.183 kHz. The simulation was run using different values and the start up transient that was expected is near the expected value. Building a Wine Bridge Oscillator With the help of the figure 22, it is required to built and test the circuit The circuit, to some extent, is similar to earlier R3=R4 design but have new values. R4=100k while R3 = 200k. This means that R3 is in series with a pot of 10k that can likely increase the feedback gain. Conclusions In this laboratory research work, electronic workbench has become the fundamental means to prove response of some studied analogue circuits such as the rectifier circuits, operational amplifiers and oscillators. Rectifier Bridge is inherent when conversing the current from AC to DC. It is easy to calculate the values of resistor and capacitor, and the values obtained from the experiment are within the values of analytical approximation. On the other hand, operational amplifier is the most common circuit that is used to rectify, amplify or even regulate any alternate current source. Numerous circuits that employ operational amplifiers exists that may include general oscillator diagrams and Schmitt Trigger. All the simulations values obtained from the experimented are similar to calculated ones. Read More