IE1206 Embedded Electronics Le1 Le2 PIC-block Documentation, Seriecom Pulse sensors I , U , R , P , serial and parallell Le3 Ex1 Pulsesensors, Menuprogram KC1 LAB1 • Start of programing task • • • Ex2 Le4 Kirchhoffs laws Node analysis Two ports R2R AD Two ports, AD, Comparator/Schmitt Le5 Ex3 KC2 LAB2 Transients PWM Le6 Ex4 Le7 KC3 LAB3 Step-up, RC-oscillator Phasor j ω PWM CCP KAP/IND-sensor Ex5 Le8 Le9 LC-osc, DC-motor, CCP PWM Le11 Ex6 Le10 KC4 LAB4 LP-filter Trafo Display Le12 Ex7 • • Display of programing task • • Written exam Le13 Trafo, Ethernetcontact William Sandqvist william@kth.se
Quick Formula for exponential • Rising process t − = 1 − x t e τ ( ) • Falling process t − = x t e τ ( ) The Quick Formula directly provides the equation for a rising/falling exponential process: t x 0 = process start value − = − − x t x x x e τ x ∞ = process end value ( ) ( ) ∞ ∞ 0 τ = process time constant William Sandqvist william@kth.se
Time constants τ = R ⋅ C L τ = R • More complex circuits one simplifies with equivalent circuits to one of these elementary shapes. (If this is not possible advanced courses will have a transform methood available). William Sandqvist william@kth.se
Continuity requirements Summary The Capacitor has voltage inertia In a capacitor, charging is always continuous The capacitor voltage is always continuous . The Inductor has current inertia In an inductor the magnetic flux is always continuous In an inductor current is always continuous . William Sandqvist william@kth.se
”All” by ”the rest” ”rest” ”all” t t − x � − x � t X x − − = − � = − � = − � = − τ ⋅ x X e e t τ τ � � ( 1 ) 1 ln 1 ln τ X X X � � X " all" = τ ⋅ = τ ⋅ t ln ln − X x " rest" William Sandqvist william@kth.se
William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? u C0 = 5 V u C ∞ = 15 V τ = 2000 ⋅ 1000 ⋅ 10 -6 = 2 s William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? t − = − − ⋅ u C0 = 5 V x t x x x e τ ( ) ( ) ∞ ∞ 0 u C ∞ = 15 V t − − ⋅ = − − ⋅ = − ⋅ t u t e e 0 , 5 τ = 2000 ⋅ 1000 ⋅ 10 -6 = 2 s ( ) 15 ( 15 5 ) 2 15 10 C William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? t − = − − ⋅ u C0 = 5 V x t x x x e τ ( ) ( ) ∞ ∞ 0 u C ∞ = 15 V t − − ⋅ = − − ⋅ = − ⋅ t u t e e 0 , 5 τ = 2000 ⋅ 1000 ⋅ 10 -6 = 2 s ( ) 15 ( 15 5 ) 2 15 10 C Note: Capacitor voltage is continuous – If you put a voltage across a capacitor it can not charge instantaneously (would require infinite current). The voltage will not change at once. William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? William Sandqvist william@kth.se
Capacitor charging (10.5) R = 2000 Ω and C = 1000 µ F Obtain an expression for u C ( t ) Draw function u C ( t ) Calculate how long it takes for u C to reach +10V? − " all" 15 5 = τ ⋅ = ⋅ = t ln 2 ln − " rest" 15 10 = ⋅ = 2 0 , 695 1 , 39 s William Sandqvist william@kth.se
William Sandqvist william@kth.se
Neon lamp (10.9) Flash-circuit with neon lamp. William Sandqvist william@kth.se
Neon lamp (10.9) a) When will the first flashing light be? The circuit’s Thevenin equivalent: R I = 600||400 = 240 k Ω E 0 = 200 ⋅ 400/1000 = 80V The capacitor is charged from 0V up to 80V at 65V the neon lamp lights up (and discharges the capacitor to 55V when it goes off). − τ = ⋅ = ⋅ ⋅ ⋅ = R I C 3 6 240 10 2 , 2 10 0 , 528 − all 80 0 = τ ⋅ = ⋅ = t s ln 0 , 528 ln 0 , 88 − rest 80 65 William Sandqvist william@kth.se
Neon lamp (10.9) b) How long will it take until the next blink? The capacitor is now charging from 55V up to 80V at 65V when the neon lamp lights up (and discharges the capacitor to 55V, then it goes off). τ = 0 , 528 Flash frequency: − all 80 55 1 1 = τ ⋅ = ⋅ = t s = T = = f ln 0 , 528 ln 0 , 27 3 , 7 Hz − rest 0 , 27 80 65 William Sandqvist william@kth.se
Neon lamp (10.9) c) If R 2 is removed, how long does it then between flashes? If R 2 is removed E will not be votage divded. E = 200. Timeconstant will be changed. The capacitor is charging now from 55V up to 200V at 65V when the neon lamp lights up (and discharges the capacitor to 55V when it goes off). − τ = ⋅ = ⋅ ⋅ ⋅ = R C 3 6 600 10 2 , 2 10 1 , 32 Flash frequency: 1 − all 1 1 200 55 = T = = f = τ ⋅ = ⋅ = 11 Hz t s ln 1 , 32 ln 0 , 094 0 , 094 − rest 200 65 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Schmitt-trigger (10.10) William Sandqvist william@kth.se
Trigger levels? (10.10) 0 , 5 1 = 5 5 + 1 0 , 5 3 0V ? Voltage ? divider. 0V 5V 5V Voltage ? divider. ? 1 2 = 5 5 + 1 0 , 5 3 William Sandqvist william@kth.se
RC-oscillator (10.10) 2 3 1 3 The comparator charges the capacitor to the upper trigger level, then it turns the output on and discharges the capacitor to the lower trigger level. The frequency of the output of the comparator depends on the product R · C . Since C is constant so will the R controls the frequency. William Sandqvist william@kth.se
RC-oscillator frequency (10.10) t = t 1 2 = Ω R 5 k − − τ = ⋅ = ⋅ ⋅ ⋅ = ⋅ R C 3 9 3 5 10 150 10 0 , 75 10 − ⋅ all 1 5 5 − − = τ ⋅ = ⋅ ⋅ = ⋅ ⋅ = t 3 3 3 ln 0 , 75 10 ln 0 , 75 10 ln 2 5 , 2 ms ⋅ 1 rest 1 5 3 1 1 = = ⋅ = ⋅ ⋅ − = = = = t t T t f 3 2 2 5 , 2 10 10 , 4 ms 962 Hz − 2 1 1 ⋅ T 3 10 , 4 10 The supply voltage 5V went shorten away. The frequency is thus independent of changes in the supply voltage! William Sandqvist william@kth.se
William Sandqvist william@kth.se
Inductor connection and disconnection (10.8) E is a DC source. At the time t 1 the switch is closed. William Sandqvist william@kth.se
Inductor connection and disconnection (10.8) E is a DC source. At the time t 1 the switch is closed. a) How large is the current through the coil in the first moment? William Sandqvist william@kth.se
Inductor connection and disconnection (10.8) E is a DC source. At the time t 1 the switch is closed. a) How large is the current through the coil in the first moment? Answer : The inductor has has ”current inertia”. The first moment ( t 1 ) the current will be the ”same” i = 0. William Sandqvist william@kth.se
Inductor connection and disconnection (10.8) b) How large is the current through the inductor after a long time interval? William Sandqvist william@kth.se
Inductor connection and disconnection (10.8) u L = 0 b) How large is the current through the inductor after a long time interval? William Sandqvist william@kth.se
Inductor connection and disconnection (10.8) u L = 0 b) How large is the current through the inductor after a long time interval? Answer : After a long time, the changes have faded away. The voltage across the inductor (is due to changes) then is 0, the inductor is "shorting" the 100 Ω parallel resistor. The 100 Ω series resistor limits the current from the voltage source. i = 10V/100 Ω = 0,1 A. William Sandqvist william@kth.se
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