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
Transformer William Sandqvist william@kth.se
Voltage ratio N 1 : N 2 Φ Φ d d = = U N U N 1 1 2 2 d t d t U N = 1 1 U N 2 2 William Sandqvist william@kth.se
Ideal transformer I 0 = 0 N 1 ⋅ I 0 = N 1 ⋅ I 1 – N 2 ⋅ I 2 Magnetisig current I 0 ≈ 0 is small compared to the work currents I 1 and I 2 . The transformer itself has a high inductance. William Sandqvist william@kth.se
Current ratio N 1 : N 2 = = P P ( P , I 0 ) 1 2 0 0 ⋅ = ⋅ � U I U I 1 1 2 2 I U N ≈ = 2 1 1 I U N 1 2 2 William Sandqvist william@kth.se
Eddy current losses Eddy currents – currents inside the iron core is prevented with lacquered ( = isolation ) sheet metal. William Sandqvist william@kth.se
E I -core • EI-core is very economical to manufacture ! William Sandqvist william@kth.se
E I -core William Sandqvist william@kth.se
Toroid Toroid core has a low leakage field – so it will not disturb nearby electronics! How do one wind such a transformer? William Sandqvist william@kth.se
Automatic Winding of toroidal core William Sandqvist william@kth.se
William Sandqvist william@kth.se
Transformer (15.4) William Sandqvist william@kth.se
Transformer (15.4) U 2 = � 1 U 1 2 1 I = 1 I 2 2 William Sandqvist william@kth.se
Transformer (15.4) U 2 = � 1 U 1 2 1 I = 1 I 2 2 − ⋅ − = � = − ⋅ = 10 R I U 0 U 10 0 , 2 10 8 1 1 1 1 William Sandqvist william@kth.se
Transformer (15.4) U 2 = � 1 U 1 2 1 I = 1 I 2 2 − ⋅ − = � = − ⋅ = 10 R I U 0 U 10 0 , 2 10 8 1 1 1 1 1 8 = U ⋅ = = U 4 2 1 2 2 William Sandqvist william@kth.se
Transformatorn (15.4) U 2 = � 1 U 1 2 1 I = 1 I 2 2 − ⋅ − = � = − ⋅ = 10 R I U 0 U 10 0 , 2 10 8 1 1 1 1 1 8 2 = U ⋅ = = = I ⋅ = U 4 I 0 , 4 2 1 2 1 2 2 1 William Sandqvist william@kth.se
Transformer (15.4) U 2 = � 1 U 1 2 1 I = 1 I 2 2 − ⋅ − = � = − ⋅ = 10 R I U 0 U 10 0 , 2 10 8 1 1 1 1 1 8 2 = U ⋅ = = = I ⋅ = U 4 I 0 , 4 2 1 2 1 2 2 1 U 4 = = = Ω 2 R 10 2 I 0 , 4 2 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Transforming impedances 2 � � N � � ⋅ 1 R � � � � N 2 N 1 U 2 � � 2 U U U N N U � � = = = = ⋅ 1 2 1 2 1 2 R R � � 1 2 N � � I I I N I 2 I 1 2 1 2 2 2 N 1 2 � � N � � = ⋅ 1 R R � � ← 1 2 2 � � N 2 William Sandqvist william@kth.se
Transforming impedances 2 � � N � � ⋅ 1 R � � � � N 2 N 1 U 2 � � 2 U U U N N U � � = = = = ⋅ 1 2 1 2 1 2 R R � � 1 2 N � � I I I N I 2 I 1 2 1 2 2 2 N 1 2 � � N � � = ⋅ 1 R R � � ← 1 2 2 � � N 2 William Sandqvist william@kth.se
Ex. Transforming impedances A transformer has the voltage ratio 240V/120V. We have two capacitors 1 µ F and 16 µ F. How should one connect to get 5 µ F ? William Sandqvist william@kth.se
Ex. Transforming impedances A transformer has the voltage ratio 240V/120V. We have two capacitors 1 µ F and 16 µ F. How should one connect to get 5 µ F ? 1 = � Z ω 2 C 1 1 = ⋅ = 2 Z 2 ← ω ω 1 2 C ( C / 4 ) William Sandqvist william@kth.se
Ex. Transforming impedances A transformer has the voltage ratio 240V/120V. We have two capacitors 1 µ F and 16 µ F. How should one connect to get 5 µ F ? 1 = � Z ω 2 C 1 1 = ⋅ = 2 Z 2 ← ω ω 1 2 C ( C / 4 ) µ µ 4 F 16 F William Sandqvist william@kth.se
William Sandqvist william@kth.se
Series and parallel connection of inductors (Ex. 15.6) Assuming that none of the coils parts magnetic lines of force with each other but are completely independent components, they can be treated series and parallel inductors just as if they were resistors. ⋅ � � 4 4 + ⋅ � � 4 6 + � � 4 4 = = L 3 H ⋅ ERS 4 4 + + 4 6 + 4 4 William Sandqvist william@kth.se
Series and parallel connection of inductors? We have previously studied serial and parallel coils as if they were completely independent components that do not share magnetic lines with each other. We are now treating coils with interconnected flow ? ? William Sandqvist william@kth.se
Inductive coupling Induction ϕ d = ⋅ + u r i d t A portion of the flow in the coil 1 is interconnected with flow from the coil 2. ϕ d = ⋅ + ϕ = ⋅ + ⋅ 1 u r i i L i M 1 1 1 1 1 1 2 d t In same ϕ d = ⋅ + ϕ = ⋅ + ⋅ way: 2 u r i i L i M 2 2 2 2 2 2 1 d t William Sandqvist william@kth.se
Inductive coupling ± M is called mutual inductance d i d i = ⋅ + + 1 2 u r i L M 1 1 1 1 d t d t d i d i = ⋅ + + 2 1 u r i L M 2 2 2 2 d t d t Coupling factor: j ω -method: M k = = ⋅ + ω + ω U r I j L I j MI 1 1 1 1 1 2 L 1 L = ⋅ + ω + ω 2 U r I j L I j MI 2 2 2 2 2 1 The coupling factor indicates An ideal transformer has how much of the flow a coil has coupling factor k = 1 (100%) in common with another coil William Sandqvist william@kth.se
Series with mutual inductance Derive: L M U I L M U I 1 12 L 1 L 1 2 21 L 2 L 2 = ω ± ω = ω ± ω U j L I j M I U j L I j M I L 1 1 L 1 12 L 2 L 2 2 L 2 21 L 1 Series connection has the same current = = = + = = � I I I U U U M M M L 1 L 2 L 1 L 2 12 21 = ⋅ ω ± + ± ( ) U I j L M L M 1 2 U = ω + ± j ( L L 2 M ) 1 2 I William Sandqvist william@kth.se
Series with mutual inductance M -dot M -dot M- dot M -dot Series connection has the same current I 1 = I 2 =I = + + = + − L TOT L L 2 M L TOT L L 2 M 1 2 1 2 M can can contribute or counter act to the flow, this gives ± sign. Therefore, coil winding polarity is usually indicated by a dot convention in schematics. William Sandqvist william@kth.se
”Dot” convention An increasing current in to a dot results in induced voltages with directions that would give increasing currents out of other dots. William Sandqvist william@kth.se
”Dot” convention An increasing current in to a dot results in induced voltages with directions that would give increasing currents out of other dots. William Sandqvist william@kth.se
In parallel with mutual inductance L L TOT TOT Anti paral conected coils Parallel connected coils ⋅ − ⋅ − 2 2 L L M L L M = = 1 2 1 2 L TOT L TOT + − + + L L 2 M L L 2 M 1 2 1 2 William Sandqvist william@kth.se
Ex. 15.7 Series connection 13 = M 1 [H] 23 = 12 = M 3 M 2 3 = 1 = 2 = L 5 L 10 L 15 William Sandqvist william@kth.se
Ex. 15.7 Series connection 13 = M 1 [H] 23 = 12 = M 3 M 2 3 = 1 = 2 = L 5 L 10 L 15 L TOT = L 1 + M 12 – M 13 + L 2 + M 12 – M 23 + L 3 – M 23 – M 13 = = 5 +2 –1 + 10 + 2 – 3 + 15 –3 –1 = 26 [H] William Sandqvist william@kth.se
William Sandqvist william@kth.se
Measuring the mutual inductance? L L + TOT − TOT = + + = + − L TOT L L 2 M L TOT L L 2 M + − 1 2 1 2 William Sandqvist william@kth.se
Measuring the mutual inductance? L L + TOT − TOT = + + = + − L TOT L L 2 M L TOT L L 2 M + − 1 2 1 2 + − L L = − TOT TOT M 4 William Sandqvist william@kth.se
Variometer (to an antique radio) = + ± 2 L TOT L L M 1 2 = α ( ) M f William Sandqvist william@kth.se
William Sandqvist william@kth.se
A bad actuator can become a good sensor ���� William Sandqvist william@kth.se
The industry's "rugged" position sensor William Sandqvist william@kth.se
Differential transformer LVDT Linear Variable Differential Transformer primary coil core secondary coil secondary coil The secondary coils are connected in series but with opposite polarity – when the core is in the middle U = 0. William Sandqvist william@kth.se
LVDT design William Sandqvist william@kth.se
LVDT principle The output voltage is relatively high – it makes this a popular sensor … William Sandqvist william@kth.se
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