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IE1206 Embedded Electronics
Le1
Le2
Le3
Ex1
PIC-block Documentation, Serial com Pulse sensors
I, U, R, P, serial and parallell
KC1 LAB1
Pulse sensors, Menu program
 Start of programing task
Le4
Ex2
Le5
Ex3
Kirchhoffs laws Node analysis Two ports R2R AD
Le6
Ex4
Le7
Le8
Ex5
Le9
Ex6
Le10
Le11
Le12
Ex7
Le13
Two ports, AD, Comparator/Schmitt
KC2 LAB2
KC3 LAB3
Transients PWM
Step-up, RC-oscillator
Phasor j PWM CCP KAP/IND-sensor
KC4 LAB4
Display
Written exam
LC-osc, DC-motor, CCP PWM
LP-filter Trafo
 Display of programing task
Trafo, Ethernet contact
William Sandqvist [email protected]
Closed circuit?
 discussion.
Current can only flow
through a circuit on
condition there is a
closed circuit.
Describe in words the
action of circuits a) …
d) when when you
operate the two
switches.
220 V
220 V
a)
b)
220 V
220 V
c)
( All the circuits are perhaps not as useful … )
William Sandqvist [email protected]
d)
103
William Sandqvist [email protected]
Series resistors
William Sandqvist [email protected]
Series resistors
no current =
not included
in circuit!
William Sandqvist [email protected]
Series resistors
no current =
not included
in circuit!
RERS  4  2  3  8  17
William Sandqvist [email protected]
William Sandqvist [email protected]
Two resistors in Parallell
William Sandqvist [email protected]
Equivalent resistance (1.2)
R1 = 1 
R2 = 21 
R3 = 42 
R4 = 30 
RERS = 30//(1+21//42)
21  42
 14  (1  21 // 42)  15
21  42
30 15
30 // 15 
 10  RERS  10 
30  15
21 // 42 
William Sandqvist [email protected]
William Sandqvist [email protected]
N same value in parallell
R1  R2    RN  R
1
1 1
N
   
RERS R R
R
RERS ( N ) 
R
N
William Sandqvist [email protected]
OK to move …
Redrawn:
William Sandqvist [email protected]
William Sandqvist [email protected]
Equivalent resistance (1.6)
RTOT = 2+(12//12)//(24//24)
// means parallell connection
12 // 12  6 24 // 24  12
6 12
(12 // 12) //( 24 // 24) 
4
6  12
RTOT  2  4  6 
William Sandqvist [email protected]
Equivalent resistance (1.1)
RTOT = 1//(0,5+0,5) +1//(0,5+0,5) =1//1+1//1= 0,5+0,5 = 1
William Sandqvist [email protected]
William Sandqvist [email protected]
Equivalent resistance (1.8)
RTOT = (2+20//5)//(20//5+2)
20  5
(2  20 // 5)  2 
 24  6
20  5
RTOT  3 
William Sandqvist [email protected]
6 // 6  3
William Sandqvist [email protected]
Potentiometer
Appearance at our labs.
William Sandqvist [email protected]
Equivalent resistance (1.10)
William Sandqvist [email protected]
Equivalent resistance (1.10)
a) RERS = 10/2 = 5 k
William Sandqvist [email protected]
Equivalent resistance (1.10)
a) RERS = 10/2 = 5 k
b) RERS = 5/2 + 5/2 = 5 k
William Sandqvist [email protected]
Equivalent resistance (1.10)
a) RERS = 10/2 = 5 k
b) RERS = 5/2 + 5/2 = 5 k
c) RERS = 0  !
William Sandqvist [email protected]
William Sandqvist [email protected]
Voltage divider
Voltage
division
Divided Total
Voltage Voltage factor
According to the voltage divider formula tyou get a divided voltage, for example U1
across the resistor R1, by multiplying the total voltage U with a voltage division
factor. This voltage division factor is the resistance R1 divided by the sum of all the
resistorss that are in the series connection.
William Sandqvist [email protected]
Resistive sensors, rotate and slide
resistances
RTOT
RTOT
R = RTOTx
x
x
R
R
x relative movement/rotation 0 < x <1
William Sandqvist [email protected]
Potentiometer with load (1.11)
Without RB
U  E  x {0 ... x ... 1}
William Sandqvist [email protected]
Potentiometer with load (1.11)
10
9
8
7
6
5
4
3
2
1
U [V]
x
0,2
0,4
At x = 0 and x = 1 then U = 0 and U = 5V.
At x = 0,5 the load RB draws current from the
voltage divider and this ” reduce” U.
William Sandqvist [email protected]
0,6
0,8
1,0
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 R2!
William Sandqvist [email protected]
William Sandqvist [email protected]
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 [email protected]
William Sandqvist [email protected]
Serial – parallel circuits (3.4)
4
Calculate current I = ? And voltage
U = ? for the serial-parallel circuit in
the figure.
I =?
4
E
Calculate the equivalent resistance:10 V
RERS = 2//(4//4) = 2//2 = 1
149
1
2
4 1
Current I = URERS/4 = 2/4 = 0,5 A
Voltage
U 2
4
1,5 
Calculate voltage over the equivalent resistor URERS
U RERS  10
0,5 
0,5
 0,5 V
1,5  0,5
William Sandqvist [email protected]
+
U =?
-
William Sandqvist [email protected]
Serial – parallel circuits (3.3)
R 1 24 
Calculate current I = ? And voltage U = ?
for the serial-parallel circuit in the figure.
We start by calculating two equivalent
resistances:
R 2 12 
E
+U - R
3
9
I
R4
18 
R5
6
12V
203
R1// 2 
24 12
8
24  12
1
R3 // 4 // 5

1 1 1 2 1 3 6
18
  

 R3 // 4 // 5   3
9 18 6
18
18
6
Voltage divider:
U  12
8
 8,73  U 3 // 4 // 5  E  U
83
 I
William Sandqvist [email protected]
U 3 // 4 // 5 12  8,73

 0,55 A
R5
6
William Sandqvist [email protected]
Serial – parallel circuits (3.5)
Calculate current I = ? And voltage U
= ? for the serial-parallel circuit in the
figure.
E
We calculates a equivalent
36V
resistance:
R3 // 4,5 
R 2 4
R 4 1
I
R1
R3
R5
1
6
2
217
6  (1  2)
2
6 1 2
UR1 = 36 V. UR3 = UR3//4,5 can be calculated by voltage division:
U R3  E
R3 // 4,5
R3 // 4,5  R2
 36
U
2
12
 12  I  R 3   2 A
24
R3
6
U can be calculated by voltage division:
U  U R 3 // 4,5
R5
2
 12
8V
R4  R5
1 2
William Sandqvist [email protected]
+
- U
William Sandqvist [email protected]
(3.2) OHM’s law are often enough!
I
a) Calculate the resultant
resistance RERS for the three
parallel connected branches.
b) Culculate current I and
voltage U.
1 
+
U
-
I1
I2
I3
1 
8 
E
8 
3 
+
U1
-
12 V
102
c) Calculate the three currents I1 I2 and I3 together with the
voltage U1 over 3 -resistor.
William Sandqvist [email protected]
OHM’s law …
I
1 
+
U
-
I1
I2
I3
1 
8 
E
8 
3 
+
U1
-
12 V
102
William Sandqvist [email protected]
OHM’s law …
RERS
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
1 
+
U
-
I1
I2
I3
1 
8 
E
8 
3 
+
U1
-
12 V
102
William Sandqvist [email protected]
OHM’s law …
RERS
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
1 
E
12

4
1  RERS 1  2
+
U
-
I1
I2
I3
1 
8 
E
8 
3 
+
U1
-
12 V
102
William Sandqvist [email protected]
OHM’s law …
RERS
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
1 
E
12

4
1  RERS 1  2
I1
I2
I3
1 
8 
E
U  I  RERS  4  2  8
+
U
-
8 
3 
+
U1
-
12 V
102
William Sandqvist [email protected]
OHM’s law …
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
1 
E
12

4
1  RERS 1  2
I1
I2
I3
1 
8 
E
U  I  RERS  4  2  8
+
U
-
8 
3 
+
U1
-
12 V
102
I1 
U 8
 1
8 8
William Sandqvist [email protected]
OHM’s law …
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
1 
E
12

4
1  RERS 1  2
+
U
-
I1
I2
I3
1 
8 
E
U  I  RERS  4  2  8
8 
3 
+
U1
-
12 V
102
I1 
U 8
 1
8 8
I2 
U
8

2
1 3 1 3
William Sandqvist [email protected]
OHM’s law …
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
+
U
-
1 
E
12

4
1  RERS 1  2
I1
I2
I3
1 
8 
3 
E
U  I  RERS  4  2  8
8 
+
U1
-
12 V
102
I1 
U 8
 1
8 8
I2 
U
8

2
1 3 1 3
I3 
U 8
 1
8 8
William Sandqvist [email protected]
OHM’s law …
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
+
U
-
1 
E
12

4
1  RERS 1  2
I1
I2
I3
1 
8 
3 
E
U  I  RERS  4  2  8
8 
+
U1
-
12 V
102
I1 
U 8
 1
8 8
I2 
U
8

2
1 3 1 3
I3 
U 8
 1
8 8
U1  I 2  3  2  3  6
William Sandqvist [email protected]
OHM’s law …
1
1
1
1 4
8
 
 
 RERS   2
RERS 8 1  3 8 8
4
I
I
+
U
-
1 
E
12

4
1  RERS 1  2
I1
I2
I3
1 
8 
3 
E
U  I  RERS  4  2  8
8 
+
U1
-
12 V
102
I1 
U 8
 1
8 8
I2 
U1  I 2  3  2  3  6
U
8

2
1 3 1 3
I3 
U 8
 1
8 8
OHM’s law was enough!
William Sandqvist [email protected]
William Sandqvist [email protected]