IE1206 Embedded Electronics PIC-block Documentation, Seriecom Pulse sensors Le1 Le2 I , U , R , P , serial and parallel KC1 LAB1 Pulsesensors, Menu program Ex1 Le3 • Start of programing task • • • Kirchhoffs laws Node analysis Two-terminal R2R AD Ex2 Le4 Two ports, AD, Comparator/Schmitt KC2 LAB2 Ex3 Le5 Transients PWM KC3 LAB3 Le6 Ex4 Le7 Step-up, RC-oscillator Phasor j ω PWM CCP CAP/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 • • Trafo, Ethernet contact Le13 Written exam William Sandqvist william@kth.se
Two port circuits – Black box ? = ! William Sandqvist william@kth.se
The power supply Buttons to select VOLTAGE the display of knob to set the voltage or constant voltage. current. Coarse and fine Voltage / Amps adjustments. C.V. Continuous Voltage. Led indicating that the unit operates as a voltage generator. + and – poles - + ( GND is to connect the metal casing to +/- to suppress interference ). William Sandqvist william@kth.se
The power supply To set the current CURRENT limit you show knob to set the “Amps” and then current limit. short voltage poles. Coarse and fine adjustments. The set current then becomes the C.C. Continuous Current. maximum current Led indicating that the that can occur. unit operates as a current generator. William Sandqvist william@kth.se
Voltage and Current generator (Ex. 8.1) What value will the U get in these idealized and usually unrealistic circuits? William Sandqvist william@kth.se
Voltage and Current generator (Ex. 8.1) What value will the U get in these idealized and usually unrealistic circuits? William Sandqvist william@kth.se
Voltage and Current generator (Ex. 8.1) What value will the U get in these idealized and usually unrealistic circuits? William Sandqvist william@kth.se
William Sandqvist william@kth.se
Simplify … (8.2) William Sandqvist william@kth.se
Simplify … (8.2) − = − 7 10 3 ⋅ 3 6 = 2 + 3 6 William Sandqvist william@kth.se
Simplify … (8.2) − = − 7 10 3 ⋅ 3 6 = 2 + 3 6 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Equvalents step by step … Electronics prefix [V] [k Ω ] [mA] (8.4) William Sandqvist william@kth.se
Equvalents step by step … Electronics prefix [V] [k Ω ] [mA] (8.4) 10 ⋅ 5=50 William Sandqvist william@kth.se
Equivalents step by step … Electronics prefix [V] [k Ω ] [mA] (8.4) 10 ⋅ 5=50 William Sandqvist william@kth.se
Equvalents step by step … Electronics prefix [V] [k Ω ] [mA] (8.4) 10 ⋅ 5=50 ⋅ + 2 ( 5 8 ) = 1 , 73 + + 2 5 8 2 ⋅ = 50 6 , 67 + + 2 5 8 William Sandqvist william@kth.se
At last … Voltage divider: 0 , 5 = ⋅ = U 6 , 67 1 , 49 V + 0 , 5 1 , 73 William Sandqvist william@kth.se
William Sandqvist william@kth.se
( Wheatstone bridge equivalent ) Determine the Wheatstone bridge Thevenin equivalent. William Sandqvist william@kth.se
( Determine R I ) Voltage turned down to zero ⋅ ⋅ 6 3 12 4 = + = Ω R 5 + + I 6 3 12 4 William Sandqvist william@kth.se
( Determine E 0 ) 6 = ⋅ = U 72 48 + 1 6 3 12 = ⋅ = U 72 54 + 2 12 4 = − = E 54 48 6 V 0 William Sandqvist william@kth.se
( Determine R I E 0 ) 6 = ⋅ = U 72 48 + 1 6 3 12 = ⋅ = U 72 54 + 2 12 4 = − = E 54 48 6 V 0 Done! William Sandqvist william@kth.se
William Sandqvist william@kth.se
Equivalent circuits (instead of mesh analysis)! William Sandqvist william@kth.se
Equivalent circuits (instead of mesh analysis)! William Sandqvist william@kth.se
Equivalent circuits (instead of mesh analysis)! ⋅ ⋅ 1 2 3 5 = = 0 , 67 1 , 88 + + 2 1 3 5 1 5 = = 10 3 , 33 6 3 , 75 + + 1 2 5 3 William Sandqvist william@kth.se
Equivalent circuits (instead of mesh analysis)! ⋅ ⋅ 1 2 3 5 = = 0 , 67 1 , 88 + + 2 1 3 5 − 1 5 3 , 33 3 , 75 = = 10 3 , 33 = = − I 6 3 , 75 0 , 064 A + + 1 2 5 3 + + 0 , 67 4 1 , 88 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Ex. current generator at node analysis (7.2) − − + = + = I I I I 1 0 1 1 2 1 2 U U = = I 2 R 12 2 − − U E U 24 = = I 1 R 6 1 − ⋅ − + U U U U 24 2 48 = + = ⇔ = ⋅ − U 1 12 3 48 12 6 12 = U 20 V William Sandqvist william@kth.se
Node analysis – the currents 20 = = I 1 , 67 2 12 − 20 24 = = − I 0 , 67 1 6 + = � − + = I I 1 0 , 67 1 , 67 1 1 2 William Sandqvist william@kth.se
William Sandqvist william@kth.se
Example (8.9) a) Derive a Thevenin’s equivalent, E 0 R I , to the circuit with the two current sources. b) Calculate how big the current I would be if you connected a resistor R 4 = 2 k Ω to the circuit (or it’s equivalent). William Sandqvist william@kth.se
Example (8.9) William Sandqvist william@kth.se
William Sandqvist william@kth.se
Example (8.10) a) Derive a Thevenin’s equivalent, E 0 R I , to the circuit with the two voltage sources and the three resistors. b) How big is the voltage drop U AB over 1 k Ω resistor in the original circuit? William Sandqvist william@kth.se
Example (8.10) Let’s calculate the voltage drop U AB over the 1 k Ω resistor in the circuit, from the Thevenin’s equivalent, as then U AB will be the same as the E 0 ! R I is the equivalent resistance when the both voltage sources are turned down to zero: 1 1 = = Ω R k I 1 1 1 3 + + Ω Ω Ω 1 k 1 k 1 k Suppose A and B short circuited. The third 1 k Ω resistor will then be without current and can be ignored. The short cicuit current will come from the two votage sources through their 1k Ω resistors: 12 V 6 V = + = I 18 mA K Ω Ω 1 k 1 k William Sandqvist william@kth.se
Example (8.10) The Thevenin equivalent will have the same short circuit current I K = 18 mA. This makes it easy to calculate E 0 : E 1 = � = ⋅ = ⋅ = I E I R 0 18 6 V K K I 0 R 3 I And the voltage drop U AB is the same E 0. U AB = 6 V. William Sandqvist william@kth.se
William Sandqvist william@kth.se
Example (8.11) a) Derive a Thevenin’s equivalent, E 0 R I , to the circuit with the voltage source and the current source and the three resistors. (The 6 k Ω resistor is not includes in the circuit). b) Calculate how big current I would flow in a resistor R = 6 k Ω connected to A-B? What direction will the current have? William Sandqvist william@kth.se
Example (8.11) The current source with the 1 k Ω resistor can be transformed to a voltage source. The circuit then becomes a 1 V voltage source with a voltage divider. ⋅ 2 3 2 = = = = Ω E R 1 0 , 4 V 1 , 2 k I + + 0 3 2 3 2 The open circuit voltage is 0,4 V, and the internal resistance 3k Ω ||2k Ω = 1,2 k Ω . Note. The voltage source 0,4V is opposite to the definition of the figure. William Sandqvist william@kth.se
William Sandqvist william@kth.se
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