Topic 4 Transients in Electrical Circuits part I QOL 2spp PDF

Preview

Summary

Download Topic 4 Transients in Electrical Circuits part I QOL 2spp PDF

Description

Module FOE1006 Electrical Engineering Stage 1

Topic 4. TRANSIENTS IN ELECTRICAL CIRCUITS Part I CR Networks Part II LR Networks

4. TRANSIENTS IN ELECTRICAL CIRCUITS Topic Aim Introduce transient behaviour of circuits, to enable the analysis of circuits containing inductor, capacitor and resistor networks. Learning Objectives: • Analyse circuits that contain capacitors or inductors • Understand the transient behaviour CR & LR circuits • Calculate the transient behaviour of capacitors and Inductors

1

4.1 Capacitors A capacitor typically consists of two conducting surfaces or plates separated by an insulator or dielectric. It may be defined as a circuit element that stores energy in an electric field but does not dissipate it. • When a voltage is applied between the plates, an electric field is established in the dielectric and energy is stored in the dielectric. • The capacitance (symbol C, unit farad F) is a measure of the ability of the capacitor to store energy.

4.1 Capacitors Capacitor - A device which holds electric charge in the electric field surrounding the capacitive element for short periods of time. International Symbol

2

4.1 Capacitance

t  0 VC (t) 

RL

C

•

Capacitor charged to a potential Vc, prior to t=0 // Reservoir filled with water

•

At t=0, the switch is closed and the capacitor discharges // Fluid flow analogy

•

Capacitor discharging through a resistance R // Water in pipe time vs pipe cross-section

4.1 Capacitance We know from Topic 1.1 that electric field strength E between the plates is the potential drop per unit length:

E

Vc d

(volts per metre)

Electric flux density D between the plates is: D  So the ratio of flux density to electric field is:

C

Q A

(coulombs per square metre)

Q DA A   V c Ed d

Where:



=permittivity and is the ratio of electric flux density to electric field strength A = Area of the dielectric d = Distance between plates of the capacitor

3

4.1 Capacitance

• Absolute permittivity

 0  8 . 8 5  1 0 1 2



where

is usually defined as

   0 r

 r = relative permittivity of the medium and

is generally in the range 1 to 10.

• Therefore for a parallel plate capacitor:

C 

A A   0 r d d

• That is, capacitance can be varied by altering: the permittivity of the dielectric the area of the dielectric the distance between the plates

4.2 Capacitor Networks When capacitors are connected in parallel, the total capacitance is equal to the sum of individual capacitances.

C1

C2

CN

CE  C1  C2  ...  CN

C1

C2

CN

1 1 1 1    .... C E C1 C2 CN When capacitors are connected in series, the reciprocal of the equivalent capacitance of the circuit is equal to the sum of the reciprocals of the individual capacitances in the circuit.

4

4.1 Capacitors – Series & Parallel

Example C1 1F

C2

C3 10 F

1 0 0 F

1F

C1

1 0 F

C2

100 F

CN

4.3 Capacitor – Resistor Networks

CR Network Capacitor Resistor Circuit

5

4.3 Capacitor – Resistor Networks (Charging / Discharging) Transient and Steady State • Immediately after a switching action takes place in such a circuit, an unsettled state exists during which currents and voltages vary with time, - this is termed the transient state. • After sufficient time has elapsed, the transient state disappears and all currents and voltages (dc) are constant, - this is termed the steady state condition.

4.4 Charging - Discharging Understanding the circuit conditions for charge and discharge

 Ic

V

a

b

 Id

R C

 VC

• If the switch in the circuit is in position b the capacitor discharges any energy it stores, and the voltage across the capacitor is zero. (Zero initial conditions) • If the position of the switch blade is changed from contact b to contact a at t=0, a charging current i will flow in the circuit.

6

4.4 Charging – Discharging: Time Constant Deriving the time constant for the Capacitor

Q  C VC

We know that

If we differentiate the above equation with respect to time:

dQ dV C C dt dt

Voltage across the capacitor varies with time

Capacitance of a device remains constant

Electric flux varies with time

Remembering that current is the rate of change of electric flux

iC

dV C dt

OR

VC 

1  idt C

i

dQ dt

4.4 Charging – Discharging : Proof Derivation of Time Constant

a

 Ic

R b  VR

 

V

KVL around the loop (with switch in ‘a’ position):

 Id C

 VC

 emf   pd  0 V  VR  VC  0

V  VR  VC VR  I c R

Substituting IC in VR

V  CR

dVC  VC dt

IC  C

dV C dt

Rearranging

V  VC  CR

dVC dt

The product CR is known as the time constant of the circuit and it is given the symbol .



  CR

7

4.4 Charging – Discharging : Current Charging Current

i

V  CRt .e R t 

i  I 0e 

Calculate charging current within a CR circuit at any instant of time

1K 

Switch

a

 Ic

V e R

t 

1. Calculate the Time Constant.

 VR

b

5V

1000F

VC

1k

2. Calculate the Charging Current (Switch position?, t=0) 3. Calculate the Current (t=5τ)

4.4 Charging – Discharging : Current Discharging Current - Switch Position B Switch

a

 Ic

5V

1k

b  

1000 F

Note: the direction of current flow & voltage across the resistor

 VC 1k

Q. Calculate the Time Constant?

• If the switch is in position b the capacitor discharges any energy it stores, and the voltage across the capacitor is zero. (Zero initial conditions) • If the position of the switch blade is changed from contact ‘b’ to contact ‘a’ at t=0, a charging current ‘Ic’ will flow in the circuit.

8

4.4 Charging – Discharging : Current Time constant,  =CR= 100010 6 110 3  1 s It is the time it takes for the capacitor to charge or discharge (through R) by 2/3 (63%) of the difference of the initial and final values.

e



t 

 e 1  0 . 3 7

t  

1  e  0 .63

Fundamental Transient Circuit

 Ic

V

a

b

 Id

R C

 VC

9

4.2 Charging and Discharging VC VC

Steady State 5 periods+ The voltage across the capacitor rises from time T=0 to approx. 5 periods

The voltage across the capacitor tends to Zero from switch time T=0 to approx. 5 periods

4.5 Transients – Capacitor Equations

Charging A= Voltage B= Current

Discharging A = -Current B= Voltage

>

11

Summary of Equations for Capacitor Charging

i  I 0e

Discharging t 

V  e R

VR  IC R  Ve

VC  V

t 

t 

t (1  e 

t

V i  e  R

t

VR  Ve  t

)

VC  Ve 

NOTE 1: All the above is based on a simple loop circuit. If we are faced with a more complex arrangement, then a reduction to its Thévenin equivalent will enable the same form of transient analysis to apply. NOTE 2: Care must be exercised when selecting instruments for the measurement of voltage and current under transient conditions. Every instrument has some internal resistance and when it is connected into a circuit it may give results which differ from those expected from the theory of the circuit.

ENERGY Storage Suppose that an ideal capacitor C is charged to a voltage V by connecting it to a source of energy and this source is then suddenly removed. The voltage V will be maintained across C with the energy drawn from the source being stored in the electric field of C. Further, if a resistor is now connected across C, current will flow and the stored energy in the capacitor will be dissipated as heat in the resistor.

12

4.6 Energy Storage in Capacitors

Instantaneous value of power to capacitor  IV watts

dV

 VC

Since I  C dt

dV dt

 VC

Energy supplied during dt

watts

dV dt = CV·dV dt

joules

Hence total energy supplied to capacitor when voltage is increased from 0 to V volts

W   CVdV

1 W  CV 2 2

4.7 Worked Example 9.5 k

1 S 2 a

500  RF

155 F

375 V d.c.

b Figure 1

13

4.5 WORKED EXAMPLE (a) Figure 4 represents a simple model of a photographic flash unit. If the capacitor initially holds zero charge and switch S is closed to position 1, determine the value of the operating voltage Vab when the switch has been closed for 5 seconds. [6] Solution: Position 1. Charging Equations

t

Vab  VC  V (1 e )

(b) Switch S is then moved to position 2 by activating the camera shutter release and the flash tube produces light output on condition that the voltage across its equivalent resistance RF is at least 150 V. If RF = 7.4 , calculate the duration of the flash.

14

9.5 k  1

S

2

t a

VR  Ve 

500 

RF

155 F

Natural Log ln(x)

375 V

t V  ln  R     V 

d.c.

b

Figure 4

Rearrange equation for time t

V   150   t   ln  R   1.147 10 3 ln    1ms V   360  Energy Transferred =

1 1 C (V 1  V 2 ) 2  155  10  6 ( 360  150 ) 2  8 . 3 J 2 2

4. TRANSIENTS IN ELECTRICAL CIRCUITS Learning Outcomes: After completion of Topic 4, you should be able to: • Analyse circuits that contain capacitors or inductors • Understand transient behaviour CR & LR circuits • Analyse circuits that contain charging and discharging capacitors and Inductors, calculating their transient behaviour as a function of time.

15...