Chapter 5 – Part 2

AC – Bridges

Comparison Bridges

Capacitance

Measurements

1

Dr. Wael Salah

5.5 AC - BRIDGES

AC - Bridges enable us to perform precise

measurements for the following :

Reactance (capacitance and inductance)

measurements.

Determining the (Q-factor OR D-factor).

Frequency measurements.

Testing and analyzing of antenna and

transmission line performance

2

Dr. Wael Salah

AC BRIDGE

3

The basic circuit of an ac bridge is exactly the same as the Wheatstone

bridge circuit except that impedances are used instead of resistances, and

the supply is an ac-source. Also, the null detector must be an ac instrument.

Structure

VAC

Z1

Z3

Z2

Z4

Principle

V V V VZZ

Z

ZZ

ZV

42

4

31

3

The bridge offset voltage can be found as:

For balancing condition: (V=0)

Therefore, Z1.Z4 = Z2.Z3

For AC-bridges, the null-condition is only valid if Va and

Vb are equal in amplitude and also same in-phase.

* First balance condition

* 2nd balance condition

=amplitude

in-phase Dr. Wael Salah

EXAMPLE : VA AND VB

4

Va and Vb are equal

in amplitude but

not in-phase.

Va Va Va

Vb Vb Vb Vb

Va and Vb are

equal in-phase

but not equal in

amplitude.

Va and Vb are

equal in

amplitude and

in-phase.

Va – Vb 0 Va - Vb 0 Va - Vb = 0

Dr. Wael Salah

EXAMPLE 7

5

The impedances of the bridge arms are

;

;

Determine the values of the unknown arm.

302001Z 01502Z

402503Z unknown4 ZZx

VAC

Z1

Z3

Z2

Z4

Dr. Wael Salah

6

The first condition for balance requires that

321 ZZZZ x ;

5187

200

250150

1

32 .Z

ZZZ x

The second condition for balance requires that the sum of the phase angles of opposite

arms should be equal

321 x ; 7030400132 )(x

Hence the unknown arm impedance Zx can be written as

)... 191761364( 705187 jZx

Where, Zx= x + j y = R + j X

x = r cos φ = 187.5 cos (-70) = 64.13

y = r sin φ = 187.5 sin (-70) = -176.19

Solution 7

Dr. Wael Salah

CAPACITORS EQUIVALENT CIRCUITS

7

R S

C S

R P C

P

sss

wC

jRZ The series impedance is

Capacitor Equivalent Circuits:

The equivalent circuit of a capacitor consists of a pure capacitance C and

a resistance R .

Where, Cp represents the actual capacitance value,

and Rp represents the resistance of the dielectric or leakage resistance.

Capacitors with a high-resistance dielectric are best represented by a series RC

circuit,

Capacitors with a low-resistance dielectric represented by the parallel RC

circuit .

pp

p wCjR

Y 1

The parallel admittance is

Dr. Wael Salah

D-FACTOR OF A CAPACITOR

8

The quality of a capacitor can be expressed in terms of its

power dissipation.

A very pure capacitor has a high dielectric resistance (low

leakage current) and virtually zero power dissipation.

The dissipation factor or D-factor is simply the ratio of the

component reactance (at a given frequency) to the resistance

measurable at its terminals.

ppp

p

RwCR

XD

1For a parallel equivalent circuit R

S

C S

R P C

P

ss

s

s RwCX

RD For a series equivalent circuit

Dr. Wael Salah

INDUCTORS EQUIVALENT CIRCUITS

9

Inductor Equivalent Circuits:

• The series equivalent circuit represents an inductor as a pure

inductance LS , in series with the resistance of its coil RS .

• This series equivalent circuit is normally the best way to

represent an inductor, because the actual winding resistance

is involved and this is an important quantity.

• The parallel RL equivalent circuit for an inductor can also be

used.

sss jwLRZ The series impedance is

R S

L S

R P

L P

ppp

wL

j

RY

1The parallel admittance is

Dr. Wael Salah

Q - FACTOR OF AN INDUCTOR

10

The quality of an inductor can be defined in terms of its power

dissipation.

An ideal inductor should have zero winding resistance, and

therefore zero power dissipated in the winding.

The quality factor or Q-factor, of the inductor is the ratio of its

inductive reactance to its resistance at the operating frequency.

p

p

p

p

wL

R

X

RQ For a parallel equivalent circuit :

s

s

s

s

R

wL

R

XQ For a series equivalent circuit:

R S

L S

R P L P

Dr. Wael Salah

5.6.1. CAPACITANCE COMPARISON BRIDGES

11

i) Series-Resistance Capacitance Bridge (also known as Similar-angle bridge)

The unknown capacitance CS is represented in series with resistance RS.

A standard adjustable resistance R1 is connected in series with C1.

To obtain a bridge balance, resistors R1 and either C1 are adjusted

alternately.

341

1

3241

RwC

jRR

wC

jR

ZZZZ

ss

When the bridge is balanced:

442

331

11

;

;

RZwC

jRZ

RZwC

jRZ

ss

The impedances are:

Dr. Wael Salah

1. CAPACITANCE BRIDGES

12

s

swC

RjRR

wC

RjRR 3

3

1

441

Then the equation is simplified as:

341 RRRR s 3

41

R

RRRS

After equating the real terms for both sides, we get:

4

31

R

RCCS

swC

Rj

wC

Rj 3

1

4

And by equating the Imaginary terms both sides, we get: :

Dr. Wael Salah

EXAMPLE

13

A series resistance-capacitance bridge as shown below has a

standard capacitor for C1, and R3= 10k. Balance is achieved with

100Hz supply frequency when R1=125 and R4=14.7k. Calculate

the resistive and capacitive components of measured capacitor and

its dissipation factor.

F10.

Z1 Z2

Z4

C1 CS

Z3

R1 RS

R3 R4

G

SOLUTION

1 3

4

0.1 100.068

14.7S

C R kC F

R k

1 4

3

125 14.7183.75

10S

R R kR

R k

62 100 0.068 10 183.75 0.00785S SD wC R

Dr. Wael Salah

II) PARALLEL-RESISTANCE CAPACITANCE BRIDGE (ALSO

KNOWN AS SIMILAR-ANGLE BRIDGE)

14

In this bridge the unknown capacitance is represented by its parallel equivalent

circuit; Cp in parallel with Rp, and Z3 and Z4 are resistors.

The bridge balance is achieved by adjustment of R1 and C1.

3411

1

3241

11

RRjwC

RR

RjwC

R

ZZZZ

pp

p

When the bridge is balanced:

4433

21

111

;

1

1 ;

11

RZRZ

jwCRZ

YjwCRZ

Y pp

p

The load impedances are:

V

Z1

Z3

Z2

Z4

Cp

C1

Rp R1

R4 R3

After equating the real & Imaginary terms:

4

31

3

41 and R

RCC

R

RRR PP

Dr. Wael Salah

EXERCISE 6

15

Find the parallel equivalent circuit for CS and RS values

determined in Example 8. Also determine the component

values of R1 and R4 required to balance the calculated CP

and RP values in a parallel resistance-capacitance bridge.

Assume that R3 remains at 10k and C1 at . F10.

Dr. Wael Salah

16

SOLUTION EXERCISE 6

Dr. Wael Salah

EXAMPLE

17

A parallel-resistance capacitance bridge has a standard capacitance

value of 0.1F. Balance is achieved at a supply frequency of 100Hz

when R3= 10k, R1=375k , and R4=14.7k.

Determine the resistive and capacitive components of the measured

capacitor and its dissipation factor (D-factor).

Dr. Wael Salah

18

For the given parallel-resistance capacitance bridge, find Rp , Cp and the D-factor :-

Solution

Fk

kF

R

RCCP

068.0

7.14

101.0

4

31

k

k

kk

R

RRRP 3.551

10

7.14375

3

41

3

36105.42

103.55110068.01002

11

PPRwCD

The dissipation factor is the D-factor

Dr. Wael Salah