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IE1206 Embedded Electronics PIC-block Documentation, Seriecom Pulse - PowerPoint PPT Presentation

IE1206 Embedded Electronics PIC-block Documentation, Seriecom Pulse sensors Le1 Le2 I , U , R , P , serial and parallel KC1 LAB1 Pulse sensors, Menu program Le3 Ex1 Start of programing task Ex2 Le4 Kirchhoffs laws Node


  1. IE1206 Embedded Electronics PIC-block Documentation, Seriecom Pulse sensors Le1 Le2 I , U , R , P , serial and parallel KC1 LAB1 Pulse sensors, Menu program Le3 Ex1 • Start of programing task • • • Ex2 Le4 Kirchhoffs laws Node analysis Two-terminals R2R AD Two-terminals, AD, Comparator/Schmitt Ex3 Le5 KC2 LAB2 Transients PWM Le6 Ex4 Le7 KC3 LAB3 Step-up, RC-oscillator Phasor j ω PWM CCP CAP/IND-sensor Ex5 Le8 Le9 LC-osc, DC-motor, CCP PWM Le11 KC4 LAB4 Ex6 Le10 LP-filter Trafo Display Le12 Ex7 • • Display of programing task • • Le13 Trafo, Ethernet contact Written exam William Sandqvist william@kth.se

  2. Easy to generate a sinusoidal voltage Our entire power grid works with sinusoidal voltage. When the loop rotates with constant speed in a magnetic field a sine wave is generated. So much easier, it can not be … William Sandqvist william@kth.se

  3. The sine wave – what do you remember? y T period Y RMS ˆ Y top amplitude , t time ˆ Y 1 = ˆ ω ω = π = = y t Y t f f Y ( ) sin( ) 2 T 2 William Sandqvist william@kth.se

  4. (11.1) Phase ϕ y If a sine curve does not begin with 0 the function expression has a phase angle ϕ . = ˆ ω + ϕ y t Y t ( ) sin( ) Specify this function mathematically: = ⋅ π ⋅ ⋅ + ϕ u t t ( ) 6 sin( 2 1000 ) � � 3 = = ⋅ ϕ � ϕ = = = ° u � � ( 0 ) 3 6 sin( ) arcsin 0 , 52 rad ( 30 ) � � 6 = ⋅ ⋅ + u t t ( ) 6 sin( 6283 0 , 52 ) William Sandqvist william@kth.se

  5. Apples and pears? In circuit analyses it is common (eg. in textbooks) to expresses the angle of the sine function mixed in radians ω · t [rad] and in degrees ϕ [ ° ]. This is obviously improper, but practical (!). The user must "convert" phase angle to radians to calculate the sine function value for any given time t . ( You have now been warned …) Conversion: x [ ° ]= x [rad] ⋅ 57,3 = ⋅ ⋅ + ° u t t ( ) 6 sin( 6283 30 ) x [rad]= x [ ° ] ⋅ 0,017 ? ? William Sandqvist william@kth.se

  6. William Sandqvist william@kth.se

  7. Mean and effective value All pure AC voltages, has the mean value 0. • More interesting is the effective value – root mean square, rms. T � u t t 2 ( ) d T � = = = U u t t U 0 1 ( ) d 0 T med T 0 William Sandqvist william@kth.se

  8. (11.2) Example. RMS. The rms value is what is normally used for an alternating voltage U . 1,63 V effective value gives the same power in a resistor as a 1,63 V pure DC voltage would do. RMS, effective value T � u t t 2 ( ) d − − + − + ⋅ ⋅ ⋅ ⋅ 2 2 3 3 ( 2 ( 2 ) 0 ) 5 10 8 5 10 = = = = U 0 1 , 63 V − − ⋅ ⋅ T 3 3 15 10 15 10 William Sandqvist william@kth.se

  9. Sine wave effective value Ex. 11.3 sin 2 has the sin 2 x ( ) mean value ½ 1 Therefore: RMS, 2 effective value ˆ U U = 2 1 1 � = = x x sin 2 ( ) d 2 2 • Effective value is often called RMS ( R oot M ean S quare ). William Sandqvist william@kth.se

  10. William Sandqvist william@kth.se

  11. Addition of sinusoidal quantities ˆ = ω + ϕ y A t sin( ) 1 1 1 + y = y ? 1 2 = ˆ ω + ϕ y A t sin( ) 2 2 2 William Sandqvist william@kth.se

  12. Addition of sinusoidal quantities When we shall apply the circuit laws on AC circuits, we must add the sines. The sum of two sinusoidal quantities of the same frequency is always a new sine of this frequency, but with a new amplitude and a new phase angle. ( Ooops! The result of the rather laborious calculations are shown below). = ˆ ⋅ ω + ϕ = ˆ ⋅ ω + ϕ = + = y t t y t t y t y t y t ( ) A sin( ) ( ) A sin( ) ( ) ( ) ( ) 1 1 1 2 2 2 1 2 � � ˆ ˆ ϕ + ϕ A sin( ) A sin( ) � � = ˆ + ˆ + ˆ ˆ ϕ − ϕ ⋅ ω + t 2 2 1 1 2 2 A A 2 A A cos( ) sin arctan � � 1 2 1 2 1 2 ˆ ϕ + ˆ ϕ � A cos( ) A cos( ) � 1 1 2 2 William Sandqvist william@kth.se

  13. Sine wave as a pointer A sinusoidal voltage or current, ˆ = ⋅ ω ⋅ y t Y t ( ) sin( ) can be represented by a pointer that rotates (counterclockwise) with the angular velocity ω [rad/sec] . Wikipedia Phasors William Sandqvist william@kth.se

  14. Simpler with vectors If you ignore the "revolution" and adds the pointers with vector addition, as they stand at the time t = 0, it then becomes a whole lot easier! Wikipedia Phasors http://en.wikipedia.org/wiki/Phasors William Sandqvist william@kth.se

  15. Pointer with complex numbers A AC voltage 10 V that has the Imaginary axis phase 30° is usually written: 10 ∠ 30° ( Phasor ) Once the vector additions require more than the most common Real axis geometrical formulas, it is instead = + z a b j preferable to represent pointers with complex numbers. ° ∠ ° ⇔ ⋅ ⇔ ⋅ ° + ⋅ ⋅ ° j 30 10 30 10 e 10 cos 30 10 j sin 30 In electricity one uses j as imaginary unit, as i is already in use for current. William Sandqvist william@kth.se

  16. Phasor Sinusoidal alternating quantities can be represented as pointers, phasors. ”amount” ∠ ∠ ”phase” ∠ ∠ A pointer (phasor) can either be viewed as a vector expressed in polar coordinates, or as a complex number. It is important to be able to describe alternating current phenomena without necessarily having to require that the audience has a knowledge of complex numbers - hence the vector method. The complex numbers and j ω -method are powerful tools that facilitate the processing of AC problems. They can be generalized to the Fourier transform and Laplace transform, so the electro engineer’s use of complex numbers is extensive. William Sandqvist william@kth.se

  17. peak/effective value - phasor Imaginary axis Real axis = + z a b j The phasor lengths corresponds to sine peak values, but since the effective value only is the peak value scaled by 1/ √ 2 so it does not matter if you count with peak values or effective values - as long as you are consistent! William Sandqvist william@kth.se

  18. William Sandqvist william@kth.se

  19. The inductor and capacitor counteracts changes The inductor and capacitor counteracts changes, such as when connecting or disconnecting a source to a circuit. What if the source then is sinusoidal AC – which is then changing continuously? ? William Sandqvist william@kth.se

  20. Alternating current through resistor A sinusoidal current i R ( t ) through a resistor R provides a proportional sinusoidal voltage drop u R ( t ) according to Ohm's law. The current and voltage are in phase. No energy is stored in the resistor. Phasors U R and I R become parallel to each other. = ˆ ⋅ ω = ⋅ � = ⋅ ˆ ⋅ ω i t I t u t i t R u t R I t ( ) sin( ) ( ) ( ) ( ) sin( ) R R R R R R = ⋅ U R I • Vector phasor R R = ⋅ U R I • Complex phasor R R The phasor may be a peak pointer or effective value pointer as long as you do not mix different types. William Sandqvist william@kth.se

  21. Alternating current through inductor A sinusoidal current i L ( t ) through an inductor provides, due to self-induction, a votage drop u L ( t ) which is 90 ° before the current. Energy stored in the magnetic field is used to provide this voltage. The phasor U L will be ω L · I L and it is 90° before I L . The entity ω L is the ”amount” of the inductor’s AC resistance, reactance X L [ Ω ]. π i t d ( ) = ˆ ⋅ ω = � = ω ⋅ ˆ ⋅ ω = ω ⋅ ˆ ⋅ ω + i t I t u t L u t L I t L I t L ( ) sin( ) ( ) ( ) cos( ) sin( ) L L L L L t L d 2 = ω ⋅ U L I • Vector phasor L L When using complex pointers one multiplies ω L with ”j”, this rotates the voltage pointer +90 ° (in complex plane). The method automatically keeps track of the phase angles! = ω ⋅ = ⋅ • Complex phasor U L I X I j j L L L L William Sandqvist william@kth.se

  22. Alternating current through capacitor A sinusoidal current i C ( t ) throug a capacitor will charge it with the ”voltage drop” u C ( t ) that lags 90° behind the current. Energy is storered in the electric field. Phasor U C is I C /( ω C ) and it lags 90° after I C . 1 = ω ⋅ The entity 1/( ω C ) is the ”amount” of the U I C C C capacitor’s AC resistance, reactance X C [ Ω ]. • Vector phasor Q u t q d ( ) 1 d 1 1 � = � = ⋅ = ⋅ � = ⋅ U i t u t i t t C ( ) ( ) ( ) d C C C C t C t C C d d π 1 1 ˆ ˆ ˆ = ⋅ ω � = − ⋅ ⋅ ω = ⋅ ⋅ ω − i t I t u t I t I t ( ) sin( ) ( ) cos( ) sin( ) ω ω C C C C C C C 2 William Sandqvist william@kth.se

  23. Complex phasor and the sign of reactance If you use complex phasor you get the -90 ° phase by dividing (1/ ω C ) with ” j ”. The method with complex pointer automatically keeps track of the phase angles if we consider the capacitor reactance X C as negative , and hence the inductor reactance X L as positive . 1 - 1 1 = ⋅ = ⋅ � = − U I I X • Complex phasor j C C C C ω ω ω C C C j William Sandqvist william@kth.se

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