1 / 89

ANALOGUE ELECTRONICS.

ANALOGUE ELECTRONICS. 2.1. Resistors. Resistors are components which resist the flow of electricity through a circuit for a given voltage. A resistor implements electrical resistance. Image of a resistor. Symbol of a resistor.

dreama
Télécharger la présentation

ANALOGUE ELECTRONICS.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ANALOGUE ELECTRONICS.

  2. 2.1. Resistors. Resistors are components which resist the flow of electricity through a circuit for a given voltage. A resistor implements electrical resistance. Image of a resistor Symbol of a resistor 1a.Remember the main electrical magnitudes and find the unit for each one: Watt (W) Volts (V) Ohms (Ω) Ampere (A)

  3. 2.1. Resistors. Resistors are components which resist the flow of electricity through a circuit for a given voltage. A resistor implements electrical resistance. Image of a resistor Symbol of a resistor 1a.Remember the main electrical magnitudes and find the unit for each one:

  4. OHM’S LAW connects resistance, voltage and current in an electrical circuit. a) Formula for finding the voltage across a resistor for a given current. b) Formula for finding the current through a resistor for a given voltage. 1b. Which formula represents these formulations of Ohm’s law better, a) or b)? [ _ ] The voltage (V) across a resistor is proportional to the current (I) passing through it, where the constant of proportionality is the resistance (R). [ _ ] When a voltage V is applied across the terminals of a resistor, a current I will flow through the resistor in direct proportion to that voltage. [ _ ] Voltage across a resistor equals the current through it multiplied by the resistance. [ _ ] Current through a resistor equals the voltage across it divided by the resistance.

  5. OHM’S LAW connects resistance, voltage and current in an electrical circuit. a) Formula for finding the voltage across a resistor for a given current. b) Formula for finding the current through a resistor for a given voltage. 1b. Which formula represents these formulations of Ohm’s law better, a) or b)? [ a ] The voltage (V) across a resistor is proportional to the current (I) passing through it, where the constant of proportionality is the resistance (R). [ b ] When a voltage V is applied across the terminals of a resistor, a current I will flow through the resistor in direct proportion to that voltage. [ a ] Voltage across a resistor equals the current through it multiplied by the resistance. [ b ] Current through a resistor equals the voltage across it divided by the resistance.

  6. slide 1c. Choose the right answer or answers. The higher the resistance, the lower the current. The higher the resistance, the higher the current. The lower the resistance, the higher the current. The lower the resistance, the lower the current.

  7. slide 1c. Choose the right answer or answers. The higher the resistance, the lower the current. The higher the resistance, the higher the current. The lower the resistance, the higher the current. The lower the resistance, the lower the current.

  8. slide 1d. In this circuit, R can be 0.5 Ω, 1 Ω or 2 Ω. Identify which resistance corresponds to each graph. . I I R= V R= + R= 1 Ω R= R V _ R= R= V I a) b) Construct a sentence that makes sense for graph a) and one for graph b). a) The ................................................................................................................. b) The ...............................................................................................................

  9. slide 1d. In this circuit, R can be 0.5 Ω, 1 Ω or 2 Ω. Identify which resistance corresponds to each graph. . I I R= 0.5 Ω V R= 2 Ω + R= 1 Ω R= 1 Ω R V _ R= 2 Ω R= 0.5 Ω V I a) b) Construct a sentence that makes sense for graph a) and one for graph b). a) The higher the resistance the lower the current for a given voltage. b) The higher the resistance the higher the voltage for a given current.

  10. slide The Ω is too small for many resistors. Then we use the MULTIPLES kilo (k) and mega (M). Sometimes, to avoid reading errors, the letters R,k and M substitute the decimal point. 4k7 = 4.7 kΩ = 4,700 Ω 5M6 = 5.6 MΩ = 5,600,000 Ω 3R3 = 3.3 Ω 2a.Give the value in Ω for the following resistors. a) 6k8 = b) 1M2 = c) 47R = d) 5R6 = Write the answers like the example:   5M6: five point six mega-ohms are five million six hundred thousand Ω. a) 6k8: b) 1M2: c) 47R: d) 5R6:

  11. slide The Ω is too small for many resistors. Then we use the MULTIPLES kilo (k) and mega (M). Sometimes, to avoid reading errors, the letters R,k and M substitute the decimal point. 4k7 = 4.7 kΩ = 4,700 Ω 5M6 = 5.6 MΩ = 5,600,000 Ω 3R3 = 3.3 Ω 2a.Give the value in Ω for the following resistors. a) 6k8 = 6,800 Ω b) 1M2 = 1,200,000 Ω c) 47R = 47 Ω d) 5R6 = 5.6 Ω Write the answers like the example:   5M6: five point six mega-ohms are five million six hundred thousand Ω. a) 6k8: six point eight kilo-ohms are six thousand eight hundred Ω. b) 1M2: one point two mega-ohms are one million two hundred thousand Ω. c) 47R: forty-seven Ω. d) 5R6: five point six Ω.

  12. slide 2b. Now apply Ohm’s law to calculate the current through the resistors as in the example. When you finish, check the answers with your partner without reading their workbook. I? I? + + 6M6 6M6 5V 5V Remember: 0.001 A = 1 mA and 0.000001 A = 1µA a) What result did you get for part a)?

  13. slide 2b. Now apply Ohm’s law to calculate the current through the resistors as in the example. When you finish, check the answers with your partner without reading their workbook. I? I? + + 6M6 6M6 5V 5V Remember: 0.001 A = 1 mA and 0.000001 A = 1µA a) What result did you get for part a)?

  14. slide 2b. b) I? I? I? + + + 6M6 6M6 6M6 5V 5V 5V c) d)

  15. slide 2b. b) I? I? I? + + + 6M6 6M6 6M6 5V 5V 5V c) d)

  16. slide 3a. Fill in the blanks looking at the table below. A lot of resistors have coloured rings on them instead of numbers. Each colour stands for a different unit: black is zero, brown is ___ , red is two; orange is three; yellow is ___; green is five; __ _ is six; violet is seven; grey is ____; white is nine, as you can see in the table below. The first band is for tens and the second band for units. The third band is the multiplier.  Ex.: red / violet / green stands for 2 / 7 / 00000, that is 2,700,000 Ω.

  17. slide 3a. Fill in the blanks looking at the table below. A lot of resistors have coloured rings on them instead of numbers. Each colour stands for a different unit: black is zero, brown is one, red is two; orange is three; yellow is four; green is five; blue is six; violet is seven; grey is eight; white is nine, as you can see in the table below. The first band is for tens and the second band for units. The third band is the multiplier.  Ex.: red / violet / green stands for 2 / 7 / 00000, that is 2,700,000 Ω.

  18. slide 3b. Obtain the value of these resistors: Brown / green / red: Orange / orange / brown: Green / grey / yellow: Yellow /violet / orange: Express the previous values with M or k if possible. For example, 27000 Ω= 27 kΩ

  19. slide 3b. Obtain the value of these resistors: Brown / green / red: 1/5/00= 1,500 Ω = 1.5 kΩ Orange / orange / brown: 3/3/0= 330 Ω Green / grey / yellow: 5/8/0000= 580,000 Ω = 580 kΩ Yellow /violet / orange: 4/7/000= 47,000 Ω = 47 kΩ Express the previous values with M or k if possible. For example, 27000 Ω= 27 kΩ

  20. slide Manufacturers of the resistors cannot guarantee an exact value. The fourth band expresses the TOLERANCE in %.   Red /violet / orange //silver R =27000 Ω ±10% 10% of 27000 = 27000·10/100=2700 R = 27000 Ω ± 2700 Ω Minimum value =27000-270=26730 Ω Maximum value = 27000+270=27270

  21. slide 3c. Calculate the minimum and maximum real values for these four resistors:

  22. slide 3c. Calculate the minimum and maximum real values for these four resistors:

  23. slide 3d. Work with your partner in turns. Choose 1 resistor from the pool and write down its colours. Then you have to tell your partner the colours and he has to find out the value. 330 kΩ 1kΩ 680 kΩ 1.8 MΩ 2200 Ω 1.5 kΩ 270 Ω 8.2 kΩ 1.2 kΩ 120Ω 270 kΩ 18 Ω 3.3 MΩ 390 Ω 820 Ω 5.6 kΩ 47 kΩ 4700 kΩ - My resistor is brown, black, red. - Yes, it is. You are right… - Is it 1000 Ω? - My resistor is …

  24. slide 3e. Your teacher will give you one real resistor. Note down the colours, calculate its value and write the text to describe your resistor to the class. The first band colour of my resistor is...... The quoted value is .......................... The tolerance is... The minimum…

  25. slide A A A 2 k 5 k 8 k C 10k C 10k 10k C _ k _ k _ k B B B Fixed resistors are the most common type of resistor. Variable resistors are also known as potentiometers. They are used to act on a circuit, for example to adjust sensitivity or to change gain. They have 3 legs. The resistance between the two outside legs (RAB) is fixed. By moving the middle leg or cursor, we adjust the resistance between the middle leg and the outside legs. The three values are linked like this: RAB= RAC + RCB. 4a.Can you get the values for RCB in these 10 kΩ potentiometers?

  26. slide A A A 2 k 5 k 8 k C 10k C 10k 10k C 5 k 8 k 2 k B B B Fixed resistors are the most common type of resistor. Variable resistors are also known as potentiometers. They are used to act on a circuit, for example to adjust sensitivity or to change gain. They have 3 legs. The resistance between the two outside legs (RAB) is fixed. By moving the middle leg or cursor, we adjust the resistance between the middle leg and the outside legs. The three values are linked like this: RAB= RAC + RCB. 4a.Can you get the values for RCB in these 10 kΩ potentiometers?

  27. slide Special resistors change resistance as a result of a change in other magnitudes. They are used in sensing circuits. _ +

  28. slide 4b.Explain how the special resistor works as in the model: NTC thermistors’ resistance changes according to the temperature. As temperature goes up, the resistance goes down. They are used in temperature-sensing circuits. PTC .... LDR ....

  29. slide 4b.Explain how the special resistor works as in the model: NTC thermistors’ resistance changes according to the temperature. As temperature goes up, the resistance goes down. They are used in temperature-sensing circuits. PTCthermistors’ resistance changes according to the temperature. As temperature goes up, the resistance goes up. They are used in temperature-sensing circuits. LDR’s resistance changes according to light. As light is brighter, the resistance goes down. They are used in light-sensing circuits.

  30. slide + - 4c. Complete the visual organizer. - ________ resistors. - Variable or _______________ Resistors - _____________ - _____________ - ______________ - _____________

  31. slide + - 4c. Complete the visual organizer. - Fixed resistors. - Variable or potentiometers. Resistors - NTC thermistors - PTC thermistors - Special resistors - LDR’s

  32. slide I Vin R1 + R2 Vout POTENTIAL or VOLTAGE DIVIDERS are used for dividing up the voltage, so that parts of a circuit receive only the voltage they require. They usually consist of two resistors connected in series across a power supply. Potential dividers are used, for example, with LDRs in circuits which detect changes in light.

  33. slide I Vin= 9V R1=20Ω + R2=10Ω Vout 5a.Calculate Vout by applying the formula of a voltage divider.

  34. slide I Vin= 9V R1=20Ω + R2=10Ω Vout 5a.Calculate Vout by applying the formula of a voltage divider.

  35. slide Vin R1 + R2 Vout Light goes up Light goes down R2 goes ___ R2 goes ___ Vout goes ___ Vout goes ___ 5b.When one of the resistors is a special resistor the circuit is a sensor. Predict how light changes will affect Vout. cause effect cause effect • What is the effect of light going down? .... • What is the cause of Vout going up? .....

  36. slide Vin R1 + R2 Vout Light goes up Light goes down R2 goes down R2 goes up Vout goes up Vout goes down 5b.When one of the resistors is a special resistor, the circuit is a sensor. Predict how light changes will affect Vout. cause effect cause effect • What is the effect of light going down? If light goes down, Vout goes up. • What is the cause of Vout going up? Vout goes up if light goes down.

  37. slide Vin= 9V 10kΩ + Vout 10kΩ 10kΩ 5c. Calculate the minimum and maximum values of Vout that we can get by adjusting the potentiometer.

  38. slide Vin= 9V 10kΩ + Vout 10kΩ 10kΩ 5c. Calculate the minimum and maximum values of Vout that we can get by adjusting the potentiometer. Cursor at the top end: R1=10k and R2=20k Cursor at the bottom end: R1=20k and R2=10k

  39. slide Symbol 2.2. Capacitors. 6a. Listen and fill in the gaps in this text about capacitors. u2a6.mp3 A capacitor is a discrete component which can store an electrical charge. The larger the __________ the more ________ it can store.Capacitors are used in _______ circuits, filter _______ and as _______ devices.

  40. slide Symbol 2.2. Capacitors. 6a. Listen and fill the gaps in this text about capacitors. A capacitor is a discrete component which can store an electrical charge. The larger the capacitance the more charge it can store.Capacitors are used in timing circuits, filter signals and as sensing devices.

  41. slide The unit of capacitance is the Farad. As this is a large amount, these submultiples are used. 6b.Convert these values to Farads as in the example. Check answers with your partner. Example: 33 nF = 0.000000033 F = 33·10-9 F 100 pF = 10 µF = 0.1 µF = 68 nF =

  42. slide The unit of capacitance is the Farad. As this is a large amount, these submultiples are used. 6b.Convert these values to Farads as in the example. Check answers with your partner. Example: 33 nF = 0.000000033 F = 33·10-9 F 100 pF = 0.0000001 F = 100·10-9 F 10 µF = 0.00001 F = 10·10-6 F 0.1 µF = 0.0000001 F =100·10-9 F 68 nF = 0.000000068 F = 68·10-9 F

  43. slide + 6c.Read the text and then answer the questions. The small capacitance capacitors are made of polyester (nF) and ceramic (pF). For large capacity values (µF) electrolytic capacitors are used. These are polarised and marked with the maximum voltage. Be careful not to connect electrolytic capacitors the wrong way or across a higher voltage. Polarised capacitor symbol Ceramic and plastic capacitors

  44. slide 6c.After reading the text answer the questions. • What kind of capacitor is this? • It’s an e___________ c__________. • Describe its characteristics? • Its value __________________ • _____________________Volts. • It can work between _______________ Discuss with your partner what will happen if we use them in a 50V circuit? I think it _____________ because ________________________

  45. slide 6c.After reading the text answer the questions. • What kind of capacitor is this? • It’s an electrolytic capacitor. • Describe its characteristics? • Its valueis 4700 µF. • Its maximum voltage is 25Volts. • It can work between -40º and 85 ºC. Discuss with your partner what will happen if we use them in a 50V circuit? I think it will explode because it can only stand 25 V.

  46. slide Vo S1 Charge Discharge Vc R Vc C S2 Vo 1 2 3 4 5 6 7 8 9 10 Time Usually we connect a CAPACITOR IN SERIES WITH A RESISTOR FOR TIMING purposes. The flow of current through a resistor into the capacitor charges it until it reaches the same voltage than the power supply. 7a .Analyse the diagrams and try to sequence the text with your partner putting order numbers in the empty cells.

  47. slide Vo S1 Charge Discharge Vc R Vc C S2 Vo 1 2 3 4 5 6 7 8 9 10 Time 7a .Put order numbers in the empty cells to sequence the text.

  48. slide Vo S1 Charge Discharge Vc R Vc C S2 Vo 1 2 3 4 5 6 7 8 9 10 Time 7a .Put order numbers in the empty cells to sequence the text.

  49. slide The time it takes to charge a capacitor depends on a time constant called tau. Tau depends on the resistor and the capacitor. τ = R · C The total charging time (tc) is approximately 4 times this time constant. tc = 4 τ

  50. slide 7b. • What % of the final voltage does the capacitor reach after τ? And after 4τ? • Calculate the time constant for R=100 kΩ and C=100µF. • What happens to the charging time if we halve the value of the resistor? • What happens to the charging time if we double the value of the capacitor?

More Related