IE1206 Embedded Electronics PIC-block Documentation, Seriecom Pulse - - PowerPoint PPT Presentation

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IE1206 Embedded Electronics PIC-block Documentation, Seriecom Pulse - - PowerPoint PPT Presentation

Nov 20, 2023 390 likes •811 views

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

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SLIDE 1

IE1206 Embedded Electronics

Transients PWM Phasor jω PWM CCP CAP/IND-sensor

Le1 Le3 Le6 Le8 Le2 Ex1 Le9 Ex4 Le7

Written exam

William Sandqvist william@kth.se

PIC-block Documentation, Seriecom Pulse sensors I, U, R, P, serial and parallel

Ex2 Ex5

Kirchhoffs laws Node analysis Two-terminal R2R AD Trafo, Ethernet contact

Le13

Pulsesensors, Menu program

Le4

KC1 LAB1 KC3 LAB3 KC4 LAB4

Ex3 Le5

KC2 LAB2

Two ports, AD, Comparator/Schmitt Step-up, RC-oscillator

Le10 Ex6

LC-osc, DC-motor, CCP PWM

LP-filter Trafo

Le12 Ex7

Display

Le11

  • Start of programing task
  • Display of programing task
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SLIDE 2

William Sandqvist william@kth.se

Two port circuits – Black box

? = !

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SLIDE 3

The power supply

William Sandqvist william@kth.se

VOLTAGE knob to set the constant voltage. Coarse and fine adjustments.

  • +

Buttons to select the display of voltage or current. Voltage / Amps

+ and – poles ( GND is to connect the metal casing to +/- to suppress interference ).

C.V. Continuous

  • Voltage. Led indicating

that the unit operates as a voltage generator.

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SLIDE 4

The power supply

William Sandqvist william@kth.se

CURRENT knob to set the current limit. Coarse and fine adjustments. To set the current limit you show “Amps” and then short voltage poles. The set current then becomes the maximum current that can occur.

C.C. Continuous Current. Led indicating that the unit operates as a current generator.

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SLIDE 5

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?

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SLIDE 6

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?

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SLIDE 7

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?

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SLIDE 8

William Sandqvist william@kth.se

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SLIDE 9

William Sandqvist william@kth.se

Simplify … (8.2)

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SLIDE 10

William Sandqvist william@kth.se

Simplify … (8.2)

2 6 3 6 3 3 10 7 = + ⋅ − = −

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SLIDE 11

William Sandqvist william@kth.se

Simplify … (8.2)

2 6 3 6 3 3 10 7 = + ⋅ − = −

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SLIDE 12

William Sandqvist william@kth.se

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SLIDE 13

William Sandqvist william@kth.se

Equvalents step by step …

Electronics prefix [V] [kΩ] [mA] (8.4)

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SLIDE 14

William Sandqvist william@kth.se

Equvalents step by step …

10⋅5=50

Electronics prefix [V] [kΩ] [mA] (8.4)

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SLIDE 15

William Sandqvist william@kth.se

Equivalents step by step …

10⋅5=50

Electronics prefix [V] [kΩ] [mA] (8.4)

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SLIDE 16

William Sandqvist william@kth.se

Equvalents step by step …

10⋅5=50

67 , 6 8 5 2 2 50 73 , 1 8 5 2 ) 8 5 ( 2 = + + ⋅ = + + + ⋅

Electronics prefix [V] [kΩ] [mA] (8.4)

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SLIDE 17

William Sandqvist william@kth.se

At last …

V 49 , 1 73 , 1 5 , 5 , 67 , 6 = + ⋅ = U

Voltage divider:

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SLIDE 18

William Sandqvist william@kth.se

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SLIDE 19

William Sandqvist william@kth.se

( Wheatstone bridge equivalent )

Determine the Wheatstone bridge Thevenin equivalent.

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SLIDE 20

William Sandqvist william@kth.se

( Determine RI )

Ω = + ⋅ + + ⋅ = 5 4 12 4 12 3 6 3 6

I

R

Voltage turned down to zero

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SLIDE 21

William Sandqvist william@kth.se

( Determine E0 )

V 6 48 54 54 4 12 12 72 48 3 6 6 72

2 1

= − = = + ⋅ = = + ⋅ = E U U

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SLIDE 22

William Sandqvist william@kth.se

Done!

( Determine RI E0 )

V 6 48 54 54 4 12 12 72 48 3 6 6 72

2 1

= − = = + ⋅ = = + ⋅ = E U U

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SLIDE 23

William Sandqvist william@kth.se

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SLIDE 24

William Sandqvist william@kth.se

Equivalent circuits (instead of mesh analysis)!

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SLIDE 25

William Sandqvist william@kth.se

Equivalent circuits (instead of mesh analysis)!

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SLIDE 26

William Sandqvist william@kth.se

Equivalent circuits (instead of mesh analysis)!

33 , 3 2 1 1 10 = + 67 , 1 2 2 1 = + ⋅ 88 , 1 5 3 5 3 = + ⋅ 75 , 3 3 5 5 6 = +

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SLIDE 27

William Sandqvist william@kth.se

Equivalent circuits (instead of mesh analysis)!

33 , 3 2 1 1 10 = + 67 , 1 2 2 1 = + ⋅ 88 , 1 5 3 5 3 = + ⋅ 75 , 3 3 5 5 6 = +

A 064 , 88 , 1 4 67 , 75 , 3 33 , 3 − = + + − = I

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SLIDE 28

William Sandqvist william@kth.se

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SLIDE 29

William Sandqvist william@kth.se

  • Ex. current generator at node analysis

V 20 48 3 12 12 48 2 6 24 12 1 6 24 12 1 1

1 1 2 2 2 1 2 1

= − ⋅ = ⇔ + − ⋅ = − + = − = − = = = = + = + − − U U U U U U U R E U I U R U I I I I I

(7.2)

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SLIDE 30

William Sandqvist william@kth.se

Node analysis – the currents

1 67 , 1 67 , 1 67 , 6 24 20 67 , 1 12 20

2 1 1 2

= + −

  • =

+ − = − = = = I I I I

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SLIDE 31

William Sandqvist william@kth.se

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SLIDE 32

William Sandqvist william@kth.se

Example (8.9)

a) Derive a Thevenin’s equivalent, E0 RI, to the circuit with the two current sources. b) Calculate how big the current I would be if you connected a resistor R4 = 2 kΩ to the circuit (or it’s equivalent).

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SLIDE 33

William Sandqvist william@kth.se

Example (8.9)

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SLIDE 34

William Sandqvist william@kth.se

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SLIDE 35

William Sandqvist william@kth.se

Example (8.10)

a) Derive a Thevenin’s equivalent, E0 RI, to the circuit with the two voltage sources and the three resistors. b) How big is the voltage drop UAB over 1 kΩ resistor in the original circuit?

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SLIDE 36

William Sandqvist william@kth.se

Example (8.10)

Ω = Ω + Ω + Ω = k 3 1 k 1 1 k 1 1 k 1 1 1

I

R

Suppose A and B short circuited. The third 1 kΩ 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:

mA 18 k 1 V 6 k 1 V 12 = Ω + Ω =

K

I

Let’s calculate the voltage drop UAB over the 1 kΩ resistor in the circuit, from the Thevenin’s equivalent, as then UAB will be the same as the E0! RI is the equivalent resistance when the both voltage sources are turned down to zero:

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SLIDE 37

William Sandqvist william@kth.se

Example (8.10)

V 6 3 1 18 = ⋅ = ⋅ =

  • =

I K I K

R I E R E I

The Thevenin equivalent will have the same short circuit current IK = 18 mA. This makes it easy to calculate E0: And the voltage drop UAB is the same E0. UAB = 6 V.

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SLIDE 38

William Sandqvist william@kth.se

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SLIDE 39

William Sandqvist william@kth.se

Example (8.11)

a) Derive a Thevenin’s equivalent, E0 RI, to the circuit with the voltage source and the current source and the three resistors. (The 6 kΩ resistor is not includes in the circuit). b) Calculate how big current I would flow in a resistor R = 6 kΩ connected to A-B? What direction will the current have?

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SLIDE 40

William Sandqvist william@kth.se

Example (8.11)

The current source with the 1 kΩ resistor can be transformed to a voltage source. The circuit then becomes a 1 V voltage source with a voltage divider.

Ω = + ⋅ = = + = k 2 , 1 2 3 2 3 V 4 , 2 3 2 1

I

R E

The open circuit voltage is 0,4 V, and the internal resistance 3kΩ||2k Ω = 1,2 k Ω. Note. The voltage source 0,4V is opposite to the definition of the figure.

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SLIDE 41

William Sandqvist william@kth.se