IE1206 Embedded Electronics Le1 Le2 PIC-block Documentation, Serial com Pulse sensors I , U , R , P , serial and parallell Le3 Ex1 KC1 LAB1 Pulse sensors, Menu program • 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 KC4 LAB4 Le11 Ex6 Le10 LP-filter Trafo Display Le12 Ex7 • • Display of programing task • • Written exam Le13 Trafo, Ethernet contact William Sandqvist william@kth.se
Closed circuit? • discussion. • • • Current can only flow 220 V 220 V through a circuit on a) b) condition there is a closed circuit. 220 V Describe in words the 220 V c) action of circuits a) … d) d) when when you operate the two switches. 103 ( All the circuits are perhaps not as useful … ) William Sandqvist william@kth.se
William Sandqvist william@kth.se
Series resistors William Sandqvist william@kth.se
Series resistors no current = not included in circuit! William Sandqvist william@kth.se
Series resistors no current = not included in circuit! = + + + = R 4 2 3 8 17 ERS William Sandqvist william@kth.se
William Sandqvist william@kth.se
Two resistors in Parallell William Sandqvist william@kth.se
Equivalent resistance (1.2) R 1 = 1 Ω R 2 = 21 Ω R 3 = 42 Ω R 4 = 30 Ω R ERS = 30//(1+21//42) ⋅ 21 42 = = � + = 21 // 42 14 ( 1 21 // 42 ) 15 + 21 42 ⋅ 30 15 = = � = Ω R 30 // 15 10 10 ERS + 30 15 William Sandqvist william@kth.se
William Sandqvist william@kth.se
N same value in parallell = = = = R R R R � N 1 2 N 1 1 1 = + + = � R R R R ERS R = R N ( ) ERS N William Sandqvist william@kth.se
OK to move … Redrawn: William Sandqvist william@kth.se
William Sandqvist william@kth.se
Equivalent resistance (1.6) R TOT = 2+(12//12)//(24//24) // means parallell connection = = 12 // 12 6 24 // 24 12 ⋅ 6 12 = = ( 12 // 12 ) //( 24 // 24 ) 4 + 6 12 = + = Ω R 2 4 6 TOT William Sandqvist william@kth.se
Equivalent resistance (1.1) R TOT = 1//(0,5+0,5) +1//(0,5+0,5) =1//1+1//1= 0,5+0,5 = 1 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Equivalent resistance (1.8) R TOT = (2+20//5)//(20//5+2) ⋅ 20 5 + = + = + = = ( 2 20 // 5 ) 2 2 4 6 6 // 6 3 + 20 5 = Ω R 3 TOT William Sandqvist william@kth.se
William Sandqvist william@kth.se
Potentiometer Appearance at our labs. William Sandqvist william@kth.se
Equivalent resistance (1.10) William Sandqvist william@kth.se
Equivalent resistance (1.10) a ) R ERS = 10/2 = 5 k Ω William Sandqvist william@kth.se
Equivalent resistance (1.10) a ) R ERS = 10/2 = 5 k Ω b ) R ERS = 5/2 + 5/2 = 5 k Ω William Sandqvist william@kth.se
Equivalent resistance (1.10) a ) R ERS = 10/2 = 5 k Ω b ) R ERS = 5/2 + 5/2 = 5 k Ω c ) R ERS = 0 Ω ! William Sandqvist william@kth.se
William Sandqvist william@kth.se
Voltage divider Voltage division Divided Total factor Voltage Voltage According to the voltage divider formula tyou get a divided voltage, for example U 1 across the resistor R 1, by multiplying the total voltage U with a voltage division factor. This voltage division factor is the resistance R 1 divided by the sum of all the resistorss that are in the series connection. William Sandqvist william@kth.se
Resistive sensors, rotate and slide resistances R TOT R TOT R = R TOT ⋅ x x x R R x relative movement/rotation 0 < x <1 William Sandqvist william@kth.se
Potentiometer with load (1.11) = ⋅ U E x x Without R B { 0 ... ... 1 } William Sandqvist william@kth.se
Potentiometer with load (1.11) 10 U [V] 9 8 7 6 5 4 3 2 x 1 0,2 0,4 0,6 0,8 1,0 At x = 0 and x = 1 then U = 0 and U = 5V. At x = 0,5 the load R B draws current from the voltage divider and this ” reduce” U . William Sandqvist william@kth.se
Potentiometer with load ? Would you happen to wish for any of the non-linear relationship that exists in the figure, it costs apparently just an extra resistor R 2 ! William Sandqvist william@kth.se
William Sandqvist william@kth.se
Seriel circuit (3.1) Determine the current I , its magnitude and direction. + − = + + = 8 6 12 2 3 , 6 2 , 4 4 , 8 10 , 8 2 = = I 0 , 19 A 10 , 8 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Serial – parallel circuits (3.4) 4 Ω Calculate current I = ? And voltage I =? U = ? for the serial-parallel circuit in + 0,5 Ω U =? - the figure. E 4 Ω 4 Ω Calculate the equivalent resistance: 10 V Ω 1,5 R ERS = 2//(4//4) = 2//2 = 1 Ω 149 Calculate voltage over the equivalent resistor U RERS 1 = = U 10 2 RERS + 4 1 Current I = U RERS /4 = 2/4 = 0,5 A 0 , 5 = = U Voltage 2 0 , 5 V + 1 , 5 0 , 5 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Serial – parallel circuits (3.3) R 1 24 Ω Calculate current I = ? And voltage U = ? for the serial-parallel circuit in the figure. R 12 Ω 2 I U + - R R R We start by calculating two equivalent 3 4 E 5 18 Ω Ω Ω 9 6 resistances: 12V 203 ⋅ + + 24 12 1 1 1 1 2 1 3 6 18 = = = + + = = � = = R R 8 3 + R 1 // 2 3 // 4 // 5 24 12 9 18 6 18 18 6 3 // 4 // 5 Voltage divider: − U 8 12 8 , 73 = = � = − � = = = U U E U I 3 // 4 // 5 12 8 , 73 0 , 55 A + R 3 // 4 // 5 8 3 6 5 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Serial – parallel circuits (3.5) Calculate current I = ? And voltage U R 4 R 2 Ω Ω 4 1 = ? for the serial-parallel circuit in the I figure. R R 3 R 5 E 1 + - U We calculates a equivalent 1 Ω Ω Ω 6 2 36V resistance: 217 ⋅ + 6 ( 1 2 ) = = R 2 + + 3 // 4 , 5 6 1 2 U R1 = 36 V. U R3 = U R3//4,5 can be calculated by voltage division: R U 2 12 = = = � = = = U E I R 3 // 4 , 5 3 36 12 2 A R + + 3 R R R 2 4 6 3 // 4 , 5 2 3 U can be calculated by voltage division: R 2 = = = U U 5 12 8 V R + + R R 3 // 4 , 5 1 2 4 5 William Sandqvist william@kth.se
William Sandqvist william@kth.se
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