IE1206 Embedded Electronics Le1 Le2 PIC-block Documentation, Seriecom Pulse sensors I , U , R , P , serial and parallel Le3 Ex1 KC1 LAB1 Pulse sensors, Menu program • 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 • • Written exam Le13 Trafo, Ethernet contact William Sandqvist william@kth.se
R L C CAP An impedance which contain inductors and capacitors have, depending on the frequency, either inductive character IND , or capacitive character CAP . An important special case occurs at the frequency where capacitances and inductances are equally strong, and their effects cancel each other out. The impedance becomes purely resisistiv. The phenomenon is called the resonance and the frequency on which this occurs is the resonant frequency . William Sandqvist william@kth.se
Reactance frequency dependency X X L L [ Ω [ Ω ] ] f f [ Hz ] [ Hz ] 1 = ω ⋅ = X L X L C ω ⋅ C ω = π f 2 William Sandqvist william@kth.se
R L C impedances X [ Ω ] f [ Hz ] • At a certain frequence X L and X C has the same amount. 1 = ω ⋅ = X L X L C ω ⋅ C ω = π f 2 William Sandqvist william@kth.se
William Sandqvist william@kth.se
How big is U ? (13.1) The three volt meters show the same, 1V, how much is the alternating supply voltage U ? ( Warning, teaser ) William Sandqvist william@kth.se
How big is U ? (13.1) The three volt meters show the same, 1V, how much is the alternating supply voltage U ? ( Warning, teaser ) William Sandqvist william@kth.se
How big is U ? (13.1) The three volt meters show the same, 1V, how much is the alternating supply voltage U ? ( Warning, teaser ) 1 Since volt meters show the = = = ω = R X X R L "same" and the current I is ω L C C common: William Sandqvist william@kth.se
If | X L |=| X C |=2 R ? Suppose the AC voltage U still 1 V, but the reactances are twice as big. What will the voltmeters show? 1 ω = = ⋅ L R 2 ω C William Sandqvist william@kth.se
If | X L |=| X C |=2 R ? Suppose the AC voltage U still 1 V, but the reactances are twice as big. What will the voltmeters show? 1 ω = = ⋅ L R 2 ω C William Sandqvist william@kth.se
If | X L |=| X C |=2 R ? Suppose the AC voltage U still 1 V, but the reactances are twice as big. What will the voltmeters show? 1 ω = = ⋅ L R 2 ω C At resonance, the voltage over the reactances can be many times higher than the AC supply voltage. William Sandqvist william@kth.se
Tesla coil Many builds "Tesla" coils to gain some excitement in life… William Sandqvist william@kth.se
Inductor quality factor Q Usually it is the internal resistance of the coil which is the resistor in the RLC circuit. The higher the coil AC resistance ω L is in relation to the DC resistance r , the larger the voltage across the coil at a resonance get. This ratio is called the coil quality factor Q . ( or Q-factor ). ω X L = = � ≈ ⋅ Q U Q U L r r UT IN William Sandqvist william@kth.se
Series resonance � � � � 1 1 = ⋅ � + ω + � = ⋅ + ω − U I r L I � r L � j j ( ) � � r ω ω C C � � � j � William Sandqvist william@kth.se
Series resonance =0 � � � � 1 1 = ⋅ � + ω + � = ⋅ + ω − U I r L I � r L � j j ( ) � � r ω ω C C � � � j � The Impedance is real when the imaginary part is ”0”. This will happen at angular frequency ω 0 ( frequency f 0 ). William Sandqvist william@kth.se
Series resonance =0 � � � � 1 1 = ⋅ � + ω + � = ⋅ + ω − U I r L I � r L � j j ( ) � � r ω ω C C � � � j � The Impedance is real when the imaginary part is ”0”. This will happen at angular frequency ω 0 (frequency f 0 ). [ ] 1 1 1 = ω − = � ω = = Z L f Im 0 ω C 0 0 π LC LC 2 William Sandqvist william@kth.se
Series resonance phasor diagram � � 1 = ⋅ + ω − U I r L � � j ( ) ω C � � r William Sandqvist william@kth.se
Series resonance phasor diagram � � 1 = ⋅ + ω − U I r L � � j ( ) ω C � � r William Sandqvist william@kth.se
Series resonance phasor diagram � � 1 = ⋅ + ω − U I r L � � j ( ) ω C � � r William Sandqvist william@kth.se
Series resonance circuit Q It is the resistance of the resonant circuit, usually coil r internal resistance, which determines how pronounced resonance phenomenon becomes. It is customary to "normalize" the relationship between the different variables by introducing the resonance angular frequency ω ω ω ω 0 together with the peak current I max in the function I ( ω ) with parameter Q : Normalized ω L 1 ω = = Q 0 chart of the 0 r LC series resonant I = I circuit. A high Q max � � ω ω � � corresponds to a + − Q 0 1 j ( ) � � ω ω � � narrow 0 resonance peak. William Sandqvist william@kth.se
Bandwidth BW At two different angular frequencies becomes imaginary Im and real part Re in the denominator equal. I is then I max / √ 2 ( ≈ 71%). The Bandwidth BW = ∆ω is the distans between those two angular frequencies. I = I max � � ω ω � � + − Q 0 1 j ( ) � � The equations give : ω ω � � 0 Re = Im � � ω [ ] 1 1 � � = ∆ ω = ω − ω = ω = ω ⋅ ω ω ω = ω ± + + BW 2 0 rad/s , 1 � ( ) � 2 1 Q 0 2 1 2 1 0 Q Q 2 2 2 � � William Sandqvist william@kth.se
• More convenient formulas π ∆ ω ∆ f L f 1 2 1 1 = = = � = f Q 0 ω 0 r Q f Q π LC 2 0 0 ∆ f If Q is high, no significant error is done if the ≈ ± f f f , 2 1 0 bandwidth is divided equally on both sides of f 0 . 2 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Example, series resonance circuit C = 25 nF f 0 = 100 kHz 71 % BW = ∆ f = 12,5 kHz Q = ? L = ? r = ? r =? f William Sandqvist william@kth.se
Example, series resonance circuit C = 25 nF f 0 = 100 kHz 71 % BW = ∆ f = 12,5 kHz Q = ? L = ? r = ? r =? f 100 = = = Q 0 8 ∆ f f 12 , 5 William Sandqvist william@kth.se
Example, series resonance circuit C = 25 nF f 0 = 100 kHz 71 % BW = ∆ f = 12,5 kHz Q = ? L = ? r = ? r =? f 100 = = = Q 0 8 ∆ f f 12 , 5 1 1 1 = � = = = f L 0 , 1 mH π π ⋅ ⋅ ⋅ ⋅ − 0 π f C LC 2 3 2 9 ( 2 ) ( 2 100 10 ) 25 10 2 0 William Sandqvist william@kth.se
Example, series resonance circuit C = 25 nF f 0 = 100 kHz 71 % BW = ∆ f = 12,5 kHz Q = ? L = ? r = ? r =? f 100 = = = Q 0 8 ∆ f f 12 , 5 1 1 1 = � = = = f L 0 , 1 mH π π ⋅ ⋅ ⋅ ⋅ − 0 π f C LC 2 3 2 9 ( 2 ) ( 2 100 10 ) 25 10 2 0 π ⋅ π ⋅ π ⋅ ⋅ ⋅ ⋅ − f L f L X 3 3 2 2 2 100 10 0 , 1 10 = = � = = ≈ Ω Q L r 0 0 8 r r Q 8 William Sandqvist william@kth.se
William Sandqvist william@kth.se
How big is I ? (13.2) The three ammeters show the same, 1A, how much is the AC supply current I ? ( Warning, teaser ) William Sandqvist william@kth.se
How big is I ? (13.2) The three ammeters show the same, 1A, how much is the AC supply current I ? ( Warning, teaser ) William Sandqvist william@kth.se
How big is I ? (13.2) The three ammeters show the same, 1A, how much is the AC supply current I ? ( Warning, teaser ) I L and I C becomes a circulating current decoupled from I R . I L , I C can be many times bigger than the supply current I = I R . This is parallel resonance. William Sandqvist william@kth.se
Ideal parallel resonance circuit 1 1 = = = Z R L C || || 1 1 1 1 + + ω + ω − C C j j ( ) ω ω R L R L j =0 The resonance frequency has exactly the same expression as for the series resonant circuit, but otherwise the circuit has reverse character , IND at low frequencies and CAP at high. At resonance, the impedance is real = R . 1 0 = f π LC 2 William Sandqvist william@kth.se
Ideal parallel resonance circuit 1 1 = = = Z R L C || || 1 1 1 1 + + ω + ω − C C j j ( ) ω ω R L R L j =0 The resonance frequency has exactly the same expression as for the series resonant circuit, but otherwise the circuit has reverse character , IND at low frequencies and CAP at high. At resonance, the impedance is real = R . 1 0 = f π LC 2 Actual parallel resonant circuit Actual parallel resonant circuits has a series resistance inside the coil. The calculations become more complecated and the resonance frequency will also differ slightly from our formula. William Sandqvist william@kth.se
William Sandqvist william@kth.se
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