Engineering Communications DV9N-34 LO-2. Simulation Exercises. MjD

Transcription

1 Engineering Communications DV9N-34 LO-2 Exercises MjD September 2010

2 Mike Doyle September 2010 Page - 2 of 19

3 Part-1 d.c. simulation Mike Doyle September 2010 Page - 3 of 19

4 Engineering Communications Learning Outcome 2 : For this part of engineering communications we have to examine simulation software for both electronic and mechanical systems. of Electronic Circuits Here we will use a freely available simulation package; Type the following into a web browser [Internet Explorer]:- After a short wait an applet should start, see Image 1 below Image 1: Start up applet. In this applet you can see the electronic components and also moving dots. These moving dots show the direction of current flow. At the bottom of the screen you can see the various voltage levels, more about this later. Let s draw up a simple circuit = remember the first circuit I did on the board last week? Mike Doyle September 2010 Page - 4 of 19

5 From the menu select circuits and then blank circuit You should now see Image 3. Image 2: Blank Circuit. Image 3: Select - Add Resistor. Image 4: Circuits menu. Right click anywhere on the black area and select Add Resistor (r) see Image 4. Nothing appears to happen, but if you click and drag anywhere on the black area a resistor should appear. See Image 5. Image 5: Resistor. Now right click on the black area again and select Add voltage source (2-terminal). Mike Doyle September 2010 Page - 5 of 19

6 Image 6: Insert Voltage source. This should allow us to draw a battery/cell. It is best to draw this from bottom to top. This makes sure the positive end [green] of the battery is at the top. In order to see the items settings move the cursor over the top of the battery and you will see the following information: It is a Voltage source, there is 0 Amps of current flowing out and it is 5 Volts. To change the voltage value right click and select Edit from the drop down. In the Voltage box change the value to 10 and click OK Check that the value has changed move the cursor over the battery Mike Doyle September 2010 Page - 6 of 19

7 Now, in the same way, change the resistor value to 10 (ohms) Right click the work area and select add wire (w) Click on the dot on the end of the battery and drag up to crate a wire which reaches the same level as the resistor. Now add a second wire from the new dot on the wire to the dot on the resistor. It should look similar to Image 7: Partial circuit. Image 7: Partial circuit. Now complete your circuit as shown in Image 8. Image 8: Complete circuit. Mike Doyle September 2010 Page - 7 of 19

8 To check the total circuit current move the mouse over the wire and check the text area. Image 9: Checking the current. We can see that at the point under the cursor the current value is 1 Amp [what we expected] and the voltage level is 10Volts. OK. This is exactly what we expected, remember Ohms Law last week? Current (I) = Voltage (V) divided by Resistance (R) I = 10/10 I = 1 Amp OK Lets draw up the Parallel circuit from last week. From the menu select circuits and then blank circuit Now build the following circuit: Mike Doyle September 2010 Page - 8 of 19

9 Wire 1 Wire Resistor 2 1 Wire 3 Image 10: Parallel resistive circuit. Resistor 2 Remember to change the battery voltage to 10 Volts and the resistor values to 10 Ohms. Moving the mouse over the components and wires should allow you to complete the following table. Voltage (V) Volts (V) Current (I) Amps (A) Resistance (R) Ohms(Ω) Power (P) Watts (W) Wire 1 Wire 2 Wire 3 Resistor 1 Resistor 2 Mike Doyle September 2010 Page - 9 of 19

10 Change the resistance of Resistor 2 to 30 Ohms and complete the next Table Voltage (V) Volts (V) Current (I) Amps (A) Resistance (R) Ohms(Ω) Power (P) Watts (W) Wire 1 Wire 2 Wire 3 Resistor 1 Resistor 2 Now draw up the circuit on the next page - Image 11: Full Circuit. NOW COMPLETE THE TABLE BELOW Voltage (V) Volts (V) Current (I) Amps (A) Resistance (R) Ohms(Ω) Power (P) Watts (W) I-total I-2 I-3 R1 R2 R3 Mike Doyle September 2010 Page - 10 of 19

11 R4 R5 R6 R7 R8 R9 Continue over the page Mike Doyle September 2010 Page - 11 of 19

12 As you have seen previously, Resistance (R), Current (I), and Voltage (V) are all inter-related and we can use Ohms Law to calculate one of the entities if we know the other two. This relationship (Ohms Law) can be stated simply using a triangle. Another relationship can be identified in a similar manner that of Power (P) measured in Watts. The Power dissipated in a circuit can be calculated in it s simplest way by using Voltage (V) and Current (I) as P = I x V this too can be expressed in the triangular form. How can you prove that the information provided by the simulation program is correct? Mike Doyle September 2010 Page - 12 of 19

13 Part-2 a.c. simulation Mike Doyle September 2010 Page - 13 of 19

14 As before we will use the Java circuit simulator applet but this time examine the characteristics of an a.c. circuit. The a.c. circuit we will examine is the Bridge Rectifier circuit. But first an introduction to the semiconductor diode - Circuit symbol > The semiconductor diode, or just diode, acts like a 1 way valve for the current in an electrical circuit. It allows current to flow only in the direction of the arrow of the symbol. For the circuit symbol shown above current would only flow from left to right any current trying to flow from right to left would be blocked. The diode must have a particular set of conditions before it allows any current to flow at all. First it must be part of a valid electrical circuit with both a source and a load and must be connected in such a way that ~0.7V is dropped across it. Draw up the following circuit: Make sure that the battery/cell is set to 5 volts Now move the cursor and check the voltage at the left of the diode note the voltage in the table Voltage at left of diode Voltage at right of diode Volt drop Now move the cursor to the right of the diode and again note the voltage in the table. Calculate the voltage across the diode. Now change the battery/cell voltage to 0.71volts. Now move the cursor and check the voltage at the left of the diode note the voltage in the table Voltage at left of diode Voltage at right of diode Volt drop Mike Doyle September 2010 Page - 14 of 19

15 Now move the cursor to the right of the diode and again note the voltage in the table. Calculate the voltage across the diode. Also note that there is virtually no current flow now. This is all ok, but the source is d.c. and we want to examine a.c. sources. Remove the battery/cell from you circuit and add an A/C Source (2- terminal) Mike Doyle September 2010 Page - 15 of 19

16 What is happening? To see things a bit clearer we can use a tool called an oscilloscope. An oscilloscope allows us to view the shape of the voltage/current wave form. Right click the wire underneath the ac source and click What you should see is wave forms appearing as shown below. There are actually two traces shown the green is the voltage and the yellow is the current. We are not really interested in the current just now and we can switch it of by right clicking on the waveform and deselecting by clicking on it. You should now see a single trace. Note the voltage value shown, top left, this is the peak or maximum voltage the waveform reaches. Lets look at the waveform at the load now. Right click the load resistor and select, right click the new waveform and deselect the. What are you seeing? Well the diode allows current to flow during the positive ½ cycle of the waveform and blocks the current in the negative ½ cycle virtually removing the negative ½ cycle. What about the difference in the peak voltages shown on the waveforms? Can you explain the difference between them? Peak Peak Voltage Voltage Voltage difference input load Mike Doyle September 2010 Page - 16 of 19

17 This type of circuit is known as a half-wave rectifier as it removes half the wave. This type of circuit can be used for producing d.c. as it removes the negative ½ of the wave. This is a bit wasteful as ½ of the available power is wasted. A better and more widely used circuit is the full-wave rectifier. Mike Doyle September 2010 Page - 17 of 19

18 Build the circuit shown here on the right. Change the ac source voltage to 20 volts What can you say about this circuit? Hint look at the current through the load. Now use the oscilloscope to view both the input and output voltage waveforms Explain what you are seeing. Mike Doyle September 2010 Page - 18 of 19

19 Again can you explain the difference in the peak voltages shown. Peak Voltage input Peak Voltage load Voltage difference Mike Doyle September 2010 Page - 19 of 19