IE1206 Embedded Electronics
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1 IE1206 Embedded Electronics Le1 Le3 Le4 Le2 Ex1 Ex2 PIC-block Documentation, Seriecom Pulse sensors I, U, R, P, serial and parallel KC1 LAB1 Pulsesensors, Menu program Start of programing task Kirchhoffs laws Node analysis Two-terminal R2R AD Le5 Ex3 KC2 LAB2 Two ports, AD, Comparator/Schmitt Le6 Le8 Ex6 Le13 Ex4 Ex5 Le10 Le7 Le9 Le11 Le12 Ex7 Display Written exam KC3 LAB3 Transients PWM Step-up, RC-oscillator Phasor j PWM CCP CAP/IND-sensor KC4 LAB4 LP-filter Trafo LC-osc, DC-motor, CCP PWM Display of programing task Trafo, Ethernet contact
2 Two terminal circuits Black box? =!
3 The power supply VOLTAGE knob to set the constant voltage. Coarse and fine adjustments. Buttons to select the display of voltage or current. Voltage / Amps C.V. Continuous Voltage. Led indicating that the unit operates as a voltage generator. + and poles - + ( GND is to connect the metal casing to +/- to suppress interference ).
4 The power supply CURRENT knob to set the current limit. Coarse and fine adjustments. C.C. Continuous Current. Led indicating that the unit operates as a current generator. To set the current limit you show Amps and then short voltage poles. The set current then becomes the maximum current that can occur.
5 Voltage and Current generator (Ex. 8.1) What value will the U get in these idealized and usually unrealistic circuits?
6 Voltage and Current generator (Ex. 8.1) What value will the U get in these idealized and usually unrealistic circuits?
7 Voltage and Current generator (Ex. 8.1) What value will the U get in these idealized and usually unrealistic circuits?
8
9 Simplify (8.2)
10 Simplify (8.2)
11 Simplify (8.2)
12
13 Equivalents step by step (8.4) Electronics prefix [V] [k] [ma]
14 Equivalents step by step (8.4) Electronics prefix [V] [k] [ma] 105=50
15 Equivalents step by step (8.4) Electronics prefix [V] [k] [ma] 105=50
16 Equivalents step by step (8.4) Electronics prefix [V] [k] [ma] 105=50 2(5 8) 1, ,
17 At last Voltage divider: 0,5 U 6,67 1,49 V 0,5 1,73 Step by step the circuit gets simpler while the numerical values becomes more complicated! You will need a calculator. Even with adapted numbers in the exercises you can come to select a computation path that generates unwieldy decimal numbers towards the intended simple answer.
18
19 ( Wheatstone bridge equivalent ) Determine the Wheatstone bridge Thevenin equivalent.
20 ( Determine R I ) Voltage turned down to zero R I
21 ( Determine E 0 ) U U E V
22 ( Determine R I E 0 ) U U E V Done!
23
24 Equivalent circuits (instead of mesh analysis)!
25 Equivalent circuits (instead of mesh analysis)!
26 Equivalent circuits (instead of mesh analysis)! 1 2 0, , , ,75 5 3
27 Equivalent circuits (instead of mesh analysis)! 1 2 0, , , I 3,33 3,75 0,67 4 1,88 0,064 A 5 6 3,75 5 3
28
29 Example (8.10) What value has the voltage U AB? Use Thevenine equivalent to at same time get value of U AB!
30 Example (8.10) a) Derive a Thevenin s equivalent, E 0 R I, to the circuit with the two voltage sources and the three resistors. b) How big is the voltage drop U AB over 1 k resistor in the original circuit?
31 Example (8.10) Let s calculate the voltage drop U AB over the 1 k resistor in the circuit, from the Thevenin s equivalent, as then U AB will be the same as the E 0! RI RI R I is the equivalent resistance when the both voltage sources are turned down to zero: 1 1 R I k k 1k 1k
32 Example (8.10) I K short circuit current. I K E R 0 I K I Suppose A and B are directly connected to each other. The third 1 k resistor will be short circuit and get no current and can therefore be ignored. The short circuit current will come from the two voltage sources through their 1k resistors: 12V 6V I K 18 ma 1k 1k
33 Example (8.10) The Thevenin equivalent will have the same short circuit current I K = 18 ma. This makes it easy to calculate E 0 : I K E R E 0 I K RI I 6 V And the voltage drop U AB is the same E 0. U AB = 6 V.
34 Example (8.10) No current! U AB = 6 V What would happen if one removed the 6V battery?
35 Example (8.10) U AB = 6 V What would happen if one removed the 6V battery? This is now another two terminal circuit. U AB is unchanged U AB = 6V, but R I increases to R I = 0,5 kω. ( I K = 6/0,5 = 12 ma ).
36
37 Tips & Tricks U, I Series connected Transform to voltage source! Add resistors Add voltages
38 Tips & Tricks U, I Parallel connected Transform to current source! Parallel resistors Add currents
39
40 Example (8.9) a) Derive a Thevenin s equivalent, E 0 R I, to the circuit with the two current sources. b) Calculate how big the current I would be if you connected a resistor R 4 = 2 k to the circuit (or it s equivalent).
41 Example (8.9)
42
43 Example (8.11) a) Derive a Thevenin s equivalent, E 0 R I, to the circuit with the voltage source and the current source and the three resistors. (The 6 k resistor is not includes in the circuit). b) Calculate how big current I would flow in a resistor R = 6 k connected to A-B? What direction will the current have?
44 Example (8.11) The current source with the 1 k resistor can be transformed to a voltage source. The circuit then becomes a 1 V voltage source with a voltage divider. E ,4 V R 0 I ,2 k The open circuit voltage is 0,4 V, and the internal resistance 3k 2k = 1,2 k. Note. The voltage source 0,4V is opposite to the definition of the figure (-0,4V).
45
46 ( Dependent sources ) Electronics semiconductor components must be described by dependent sources. Such source has an entity E or I that is decided by some other current or voltage in the circuit.
47 ( Eg. Transistor ). I C I B Ethis example could be an transistor, and the calculation of its operating point...
48 ( Eg. Transistor ). Derive the value of resistor R B so the voltage drop over resistor R C will be the half of E? The current source I C is depending on current I B by the equation: I C = I B. =? E 2 We do not introduce any new special symbols dependent sources.
49 ( Eg. Transistor ). Derive the value of R B so the voltage drop over R C will be the half of E? E 10 V U I I B RC RC E RC I RC 2 E I B 2R E U R B U BE BE 0,5 V O R B 40 I I B ,5 10 RC 0, R 10 k 3 0,5 10 0, k Calculations with depending sources can thus take place in a similar way as with independent sources, but beware C
50 Avoid Do not use the superposition principle whith dependent generators. To reset a source can break the dependence with the the rest of the circuit. Do not reset dependent sources to find the internal resistance of a two terminal circuit. To reset a source can break the dependence with the the rest of the circuit. However, it will always work to use calculations on open and shorted two terminal circuits.
51 Eg. current depending voltage source Suppose we got an emf E that in some way is dependent of its own current I, eg. E = 5I. It will then act as an resistor with the value 5! If you reset all sources in such a circuit, you will no longer se all resistors that exists in the circuit.
52 A Spice-simulation It is possible to simulate circuits with dependent generators.
53
54 7.4 Depending source
55 7.4 Depending source Kirchhoff current law: I 1 + I 2 +I 3 = 0 Kirchhoffs voltage law (the mesh with the not depending emf): 2I I 1 = 0 2I 1 + 0I 2 + 1I 3 = 3 Kirchhoffs voltage law (the mesh with the depending emf): 1I 3 ( 10I 3 ) + 3I 2 = 0 0I 1 + 3I 2 + 9I 3 = 0
56 7.4 Depending source The values are the same as used throughout the course example. It is possible to calculate circuits with dependent generators.
57
58 Node analysis U AB U A U B A B potential U A potential U B 0-potential I U R AB U A U R B
59 Node analysis OHM s law
60 (Current source at node analysis) I?
61 (Current source at node analysis) I? I 1 A
62 (Current source at node analysis) I1 I2 1 0 I1 I2 1
63 (Current source at node analysis) I I 2 1 I U 0 U R 12 2 I 1 I 2 1
64 (Current source at node analysis) I I I I U 0 U R 12 2 U E R 1 I 1 U 24 6 I 2 1
65 (Current source at node analysis) 20 V U U U U U U U R E U I U R U I I I I I
66 (Node analyses currents) Check: I I I , ,67 6 I 1 0,67 1,67 2 1
67
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