1. Capacitors and Inductors

2.  A capacitor is a device that stores an electrical charge  It is made of two metallic plates separated by an insulator or dielectric (plastic, ceramic, air)

4.  A capacitor’s storage potential or capacitance is measured in Farads (f) – typically small values i.e. picofarads (pf 10 -12 ) or nanofarads (nf 10 -9 ) or microfarads (µf 10-6 )  Capacitors in a DC circuit will be charged and current stops when the capacitor is at the same charge as the dc source  In an AC circuit, capacitors charge and discharge as current fluctuates making it appear that the ac current is flowing

5.  Difference between capacitor and a battery  A capacitor can dump its entire charge in a fraction of a second  Uses of capacitors  Store charge for high speed use (flash, lasers)  Eliminate ripples or spikes in DC voltage – absorbs peaks and fills in valleys  As a filter to block DC signals but pass AC signals

6.  Parallel Plate Capacitor  C = k A / d  C - capacitance in Farads  k – dielectric constant  A –area in square meters  d – distance between the electrodes in meters

7.  Parallel Plate Capacitor  C = Q / V or Q = CV  C - capacitance in Farads (charge on plate)  Q – charge in Coloumbs  V – voltage in Volts  Q= CV  Charge on the plates of the capacitor  E = ½ QV = ½ CV 2  Energy stored on the plates of a capacitor  Measured in Joules (J)

8.  Capacitors return stored energy to the circuit  They do not dissipate energy and convert it to heat like a resistor  Energy stored in a capacitor is much smaller than energy stored in a battery so they cannot be used as a practical energy source

9.  τ = R x C  τ - Time constant (Tau)  R – Resistance in Ohms (Ω)  C – Capacitance in farads (F)  This is the time response of a capacitive circuit when charging or discharging

10.  τ = L / R  τ - Time constant (Tau)  R – Resistance in Ohms (Ω)  L – Inductance in Henrys (H)  This is the time constant for circuits that contain a resistor and an inductor

11.  The voltage across a capacitor at any given point in time is given by the equation above  Vc – Voltage across the capacitor  Vs - Supply voltage  t – Time elapsed since application of supply voltage  RC – Time constant of the RC charging circuit

12.  Time taken for charging or discharging current to fall to 1/e its initial value  e = 2.71828  So roughly the time constant is the time taken for the current to fall to 1/3 its value Charging Discharging

13.  After 5 time constants (5RC) the current has fallen to less than 1% of its initial value  For all intents and purposes the capacitor can be considered to be fully charged / fully discharged after 5RC  A large time constant means a capacitor charges /discharges slowly

14.  Charging current I = (Vs-Vc)/R  As Vc increases, I decreases  When charge builds up on the capacitor, the voltage across it increases  This reduces the voltage across the resistor and reduces charging current  Rate of charging becomes slower

16.  At first current is large because voltage is large and charge is lost quickly  As charge/voltage decreases, current becomes smaller so rate of discharging becomes progressively slower  After 5RC voltage across capacitor is almost 0

19.  Ceq = C1 + C2  Ceq = Q / V = (Q1 + Q2 ) / V = Q1 / V + Q2 /V = C1 + C2

20.  1/Ceq = 1/C1 + 1/C2  1/Ceq = V /Q = (V1 + V2) / Q = V1 / Q + V2 /Q = 1 /C1 + 1/C2

21.  Capacitors typically block DC signals but allow AC signals to pass  When the current in an AC circuit reverses direction, the capacitor discharges so it appears that current is flowing continuously  A capacitor will oppose the flow of an AC current  This opposition to current flow is called the reactance of the capacitor

22.  Xc = 1 / 2πfC  Xc – Reactance of the capacitor in ohms (Ω)  f – Frequency of the input signal in (Hz)  C – Capacitance of the capacitor in farads (F)

23.  The output sine wave has the same frequency as the input sine wave  Because of the reactance (opposition to the flow of ac current) of the capacitor the amplitude of the output is smaller than the input

24.  A capacitor and resistor in series as above functions as a voltage divider

25.  For a normal series circuit  Rt = R1 + R2  Total opposition to the flow of an electric current in a circuit containing a capacitor and a resistor in series is called the impedance  Z = √X 2 C + R 2 = (X 2 C + R 2 ) 1/2  Z – the impedance of the circuit in ohms  Xc – the reactance of the capacitor in ohms  R – the resistance of the resistor in ohms

26. 1. C = 530 µF, R = 12 ohms, Vin = 26 Vpp, f=60 Hz. Calculate the impedance and the current in the circuit. 2. C = 1.77 µF, R = 12 ohms, Vin = 150 Vpp, f=10 kHz. Calculate the impedance and the current in the circuit.

27. 10 Vpp f = 1kHz C = 0.32 µF R = 1 k ohm Find Xc Find Z Find Vout

28.  In the previous examples, the output signal is attenuated  The output amplitude is smaller than the input amplitude  The frequency of the input and output is the same

29.  RMS is the root mean square value  For a sign wave S of amplitude a  S RMS = a / √2 a 2a

30.  RMS – Root Mean Square  AVG – Practical Average  P-P – Peak to Peak

31.  As the frequency increases, the reactance decreases and the voltage drop becomes larger over the resistor

32.  As the frequency increases, the reactance decreases and the voltage drop across the capacitor becomes smaller  This circuit is used in many electronic devices

33.  fcutoff = 1/2 π RC  fcutoff – cutoff frequency in Hertz (Hz)  R – resistance of resistor in ohms (Ω)  C – capacitance of the capacitor in farads (F)  The cutoff or corner frequency is the frequency above which the output voltage falls to 70.7% (3 DB) (√ 1/2) of the source voltage  This is an important in the design of filters

34.  fcutoff = R/2 π L  fcutoff – cutoff frequency in Hertz (Hz)  R – resistance of resistor in ohms (Ω)  L – inductance of the inductor in Henrys (H)

35.  In an RC circuit, the current and the voltage across the resistor is in phase. They have no phase difference  The voltage across the capacitor lags the current through the capacitor by 90 degrees (they are 90 degrees out of phase)  The input voltage and the output voltage across a component of an RC circuit have the same frequency but the output is attenuated and is out of phase with the input

36.  Output voltage of the LP lags the input voltage  Output voltage of HP leads the input voltage Low Pass FilterHigh Pass Filter

37. HP FilterLP Filter Output leads Input VoltageOutput lags Input Voltage

38.  tan θ = V c / V R = 1 / 2 πfRC = Xc / R

39.  Essentially a coil of wire  When current flows, a magnetic field is created and the inductor will store the magnetic energy until released  A capacitor stores voltage as electrical energy while a conductor stores current as magnetic energy  The strength of an inductor is it’s inductance which is measured in Henrys (H)

40.  Capacitors block DC current and let AC pass  Inductors block AC and let DC pass  Capacitors oppose the change of voltage in a circuit while an inductor opposes the change in current

41.  Like capacitors, inductors cannot change the frequency of the sine wave but can reduce the amplitude of the output voltage  The opposition to the current flow is called the impedance  In many cases the DC resistance of the inductor is very low

42.  X L = 2 πfL  X L – reactance of the inductor in ohms (Ω)  f – frequency of the signal in Hertz (Hz)  L – inductance of the inductor in Henrys (H)

43. 1.Find DC Ouput Voltage if resistance of inductor is 0 2.Find the reactance of the inductor 3.Find the AC impedance 4.Find the AC output voltage 5.Draw actual output

44. 1. DC Vout = 10 V (full 10 V DC is dropped across Vout) 2. X L = 2 πfL= 2 X π x 1000 x 0.163 = 1 K Ω approx 3. Z 2 = X L 2 + R 2 = 1000 2 + 1000 2 1. Z = 1.414 k Ω 4. AC out = 1 k Ω / 1.414 k Ω X 2Vpp = 1.414 Vpp

45.  The current through an inductor lags the voltage across the inductor by 90 degrees LP Filter HP Filter Output lags Input VoltageOutput leads Input Voltage

46.  tan θ = V L / V R = X L / R =2 πfL/ R