 Other AC Bridges # Other AC Bridges

In previous chapter, we discussed about two AC bridges which can be used to measure inductance. In this chapter, let us discuss about the following two AC bridges.

• Schering Bridge
• Wien’s Bridge

These two bridges can be used to measure capacitance and frequency respectively.

## Schering Bridge

Schering bridge is an AC bridge having four arms, which are connected in the form of a rhombus or square shape, whose one arm consists of a single resistor, one arm consists of a series combination of resistor and capacitor, one arm consists of a single capacitor & the other arm consists of a parallel combination of resistor and capacitor.

The AC detector and AC voltage source are also used to find the value of unknown impedance, hence one of them is placed in one diagonal of Schering bridge and the other one is placed in other diagonal of Schering bridge.

Schering bridge is used to measure the value of capacitance. The circuit diagram of Schering bridge is shown in the below figure. In above circuit, the arms AB, BC, CD and DA together form a rhombus or square shape. The arm AB consists of a resistor, R2R2. The arm BC consists of a series combination of resistor, R4R4 and capacitor, C4C4. The arm CD consists of a capacitor, C3C3. The arm DA consists of a parallel combination of resistor, R1R1 and capacitor, C1C1.

Let, Z1Z1Z2Z2,Z3Z3 and Z4Z4 are the impedances of arms DA, AB, CD and BC respectively. The values of these impedances will be

Z1=R1(1jωC1)R1+1jωC1Z1=R1(1jωC1)R1+1jωC1

Z1=R11+jωR1C1⇒Z1=R11+jωR1C1

Z2=R2Z2=R2

Z3=1jωC3Z3=1jωC3

Z4=R4+1jωC4Z4=R4+1jωC4

Z4=1+jωR4C4jωC4⇒Z4=1+jωR4C4jωC4

Substitute these impedance values in the following balancing condition of AC bridge.

Z4=Z2Z3Z1Z4=Z2Z3Z1
1+jωR4C4jωC4=R2(1jωC3)R11+jωR1C11+jωR4C4jωC4=R2(1jωC3)R11+jωR1C1

1+jωR4C4jωC4=R2(1+jωR1C1)jωR1C3⇒1+jωR4C4jωC4=R2(1+jωR1C1)jωR1C3

1+jωR4C4C4=R2(1+jωR1C1)R1C3⇒1+jωR4C4C4=R2(1+jωR1C1)R1C3

1C4+jωR4=R2R1C3+jωC1R2C3⇒1C4+jωR4=R2R1C3+jωC1R2C3

By comparing the respective real and imaginary terms of above equation, we will get

C4=R1C3R2C4=R1C3R2Equation 1

R4=C1R2C3R4=C1R2C3Equation 2

By substituting the values of R1,R2R1,R2 and C3C3 in Equation 1, we will get the value of capacitor, C4C4. Similarly, by substituting the values of R2,C1R2,C1 and C3C3 in Equation 2, we will get the value of resistor, R4R4.

The advantage of Schering bridge is that both the values of resistor, R4R4 and capacitor, C4C4 are independent of the value of frequency.

## Wien’s Bridge

Wien’s bridge is an AC bridge having four arms, which are connected in the form of a rhombus or square shape. Amongtwo arms consist of a single resistor, one arm consists of a parallel combination of resistor and capacitor & the other arm consists of a series combination of resistor and capacitor.

The AC detector and AC voltage source are also required in order to find the value of frequency. Hence, one of these two are placed in one diagonal of Wien’s bridge and the other one is placed in other diagonal of Wien’s bridge.

The circuit diagram of Wien’s bridge is shown in the below figure. In above circuit, the arms AB, BC, CD and DA together form a rhombus or square shape. The arms, AB and BC consist of resistors, R2R2 and R4R4 respectively. The arm, CD consists of a parallel combination of resistor, R3R3 and capacitor, C3C3. The arm, DA consists of a series combination of resistor, R1R1 and capacitor, C1C1.

Let, Z1,Z2,Z3Z1,Z2,Z3 and Z4Z4 are the impedances of arms DA, AB, CD and BC respectively. The values of these impedances will be

Z1=R1+1jωC1Z1=R1+1jωC1
Z1=1+jωR1C1jωC1⇒Z1=1+jωR1C1jωC1

Z2=R2Z2=R2

Z3=R3(1jωC3)R3+1jωC3Z3=R3(1jωC3)R3+1jωC3
Z3=R31+jωR3C3⇒Z3=R31+jωR3C3

Z4=R4Z4=R4

Substitute these impedance values in the following balancing condition of AC bridge.

Z1Z4=Z2Z3Z1Z4=Z2Z3
(1+jωR1C1jωC1)R4=R2(R31+jωR3C3)(1+jωR1C1jωC1)R4=R2(R31+jωR3C3)

(1+jωR1C1)(1+jωR3C3)R4=jωC1R2R3⇒(1+jωR1C1)(1+jωR3C3)R4=jωC1R2R3

(1+jωR3C3+jωR1C1ω2R1R3C1C3)R4=jωC1R2R3⇒(1+jωR3C3+jωR1C1−ω2R1R3C1C3)R4=jωC1R2R3

R4(ω2R1R3C1C3)+jωR4(R3C3+R1C1)=jωC1R2R3⇒R4(ω2R1R3C1C3)+jωR4(R3C3+R1C1)=jωC1R2R3

Equate the respective real terms of above equation.

R4(1ω2R1R3C1C3)=0R4(1−ω2R1R3C1C3)=0

1ω2R1R3C1C3=0⇒1−ω2R1R3C1C3=0

1=ω2R1R3C1C3⇒1=ω2R1R3C1C3

ω=1R1R3C1C3ω=1R1R3C1C3

Substituteω=2πfω=2πf in above equation.

2πf=1R1R3C1C3−−−−−−−−−√⇒2πf=1R1R3C1C3

f=12πR1R3C1C3⇒f=12πR1R3C1C3

We can find the value of frequency, ff of AC voltage source by substituting the values of R1,R3,C1R1,R3,C1 and C3C3 in above equation.

If R1=R3=RR1=R3=R and C1=C3=CC1=C3=C, then we can find the value of frequency, ff of AC voltage source by using the following formula.

f=12πRCf=12πRC

The Wein’s bridge is mainly used for finding the frequency value of AF range. 