Wheatstone Bridge

The Wheatstone Bridge
The Wheatstone Bridge consists of a dc voltage source, four resistors and a detector. The
detector is a type of ammeter called a galvanometer.
The galvanometer is used to detect the condition i g = 0 . When the circuit satisfies the condition
i = 0
g
we say that “the bridge is balanced”.
Because the galvanometer is a type of ammeter, v g = 0 . (It’s always true that v g = 0 , whether
the bridge is balanced or not. When the bridge is balanced it is also true that i = 0 .) Apply KVL
to the top mesh of the bridge to get
R i −v− Ri = 0 ⇒ Ri = Ri
(1)
2 2 g 1 1 1 1 2 2
Apply KVL to the bottom mesh of the bridge to get
vg + Rmim − R3i3 = 0 ⇒ R3i3 = Rmim When the bridge is balanced i = 0 . Apply KCL to node b of the balanced bridge to get
g
i = i + i = ⇒ i = i
(3)
1 g 3 0
Apply KCL to node c of the balanced bridge to get
1 3
i + i = i ⇒ i = i
(4)
2 g m 2 m
Using equations 3 and 4 to substitute for the currents in equation 2 gives
g
(2)
R3i1= Rmi2 (5)
1

Dividing equation 5 by equation 1 gives
Now and solving for R m we get
R3 m R
= (6)
R R
1 2
R2
Rm = R3
(6)
R
Typically, 1 R and 2 R are fixed resistors and 3 R is a variable resistor. R m is the resistance that
is being measured. R 3 is adjusted until the detector indicates that the bridge is balanced. Then
the value of R m is determined using equation 6.
Example
Consider using a Wheatstone bridge having R 1 = 200 Ω and R 2 = 2000 Ω to measure a
resistance R m . The bridge is balanced by adjusting 3 R until R 3 = 250 Ω . What is the value of
R m ?
Solution
From equation 6
R2
2000
Rm= R3=
250 = 2500 Ω
R 200
1
Example
Consider using a Wheatstone bridge having R 1 = 200 Ω and R 2 = 2000 Ω to measure a
resistance, R m , of a temperature sensor. Suppose the resistance of the temperature sensor, R m ,
in Ω, is related to the temperature T, in °C, by the equation
R = 1500 + 25T
m
The bridge is balanced by adjusting 3 R until R 3 = 250 Ω . What is the value of the temperature?
Solution
From equation 6
Next, the temperature in °C is given by
R2
2000
Rm= R3=
250 = 2500 Ω
R 200
1
m 1500 R − 2500 −1500
1000
T = = = = 40 ° C
25 25 25
1
2

Example
Consider using a Wheatstone bridge having R 1 = 200 Ω and R 2 = 2000 Ω to measure a
resistance, R m , of a temperature sensor. Suppose the resistance of the temperature sensor, R m ,
in Ω, is related to the temperature T, in °C, by the equation
R = 1500 + 25T
m
The temperature is expected to vary over the range 0 to 100 °C. Over what range must R 3 vary
in order for the bridge to measure temperature over the range 0 to 100 °C?
Solution:
Solve equation 6 for R 3 :
When T = 0 °C, Ω and 1500 R =
m
When T = 100 °C, ( )
m
2
R1
R3 = Rm
(7)
R
R1
200
R3= Rm=
1500 = 150 Ω
R 2000
R = 1500 + 25 100 = 4000 Ω and
R1
200
R3= Rm=
4000 = 400 Ω
R 2000
2
R 3 could be implemented as a 150 Ω resistor in series with a 250 Ω potentiometer:
2
3