Download UNIT 1 - THE WHEATSTONE BRIDGE

Digital and Interfacing
Systems
Ceng 306.
Wheatstone Bridge
Supplement
Prepared by Joe Enekes Sept 1999 (Rev by Mike Crompton 29 May 2002)
UNIT 1 - THE WHEATSTONE BRIDGE
(Written by: J. Enekes edited by Mike Crompton)
Text reference: pages 330-334 in Introductory DC/AC Electronics 4TH Edition by Cook
pages 268-272 in Introductory DC/AC Electronics 5TH Edition by Cook
Introduction.
The Wheatstone bridge (invented by S. H. Christie in 1833 but ignored until it was further
developed by Charles Wheatstone in 1850) served for a long time as the most accurate way of
measuring resistance. In today's information age the Wheatstone bridge has gained additional
significance as part of data acquisition subsystems for supplying information such as temperature,
pressure, force, and flow values to computer interface circuitry in a variety of applications. To
appreciate the need for measuring these quantities just think of the car you are driving. The type of
information any driver takes for granted, such as speed, engine rpm, amount of fuel remaining in
the tank, oil pressure, engine temperature, battery voltage level, tire pressure, etc. all have to be
sensed and supplied to the car's computer for display or control purposes. The Wheatstone bridge
forms an indispensable part of the circuitry performing these functions.
Without further ado, let us look at the simple Wheatstone bridge circuit shown in Fig. 1.1.
To make the analysis simple, we assume that the voltmeter
resistance in Fig. 1.1 has an ideal internal resistance of
infinity, thus it does not draw any current and can be
regarded as an open circuit. Now you can think of the bridge
as two series circuits: one consisting of resistors R1 and R3
and the other consisting of R2 and R4. The two series
circuits are connected in parallel with each other. As we
shall see later, depending on the ratio of resistor R1 to R3
with respect to the ratio of resistor R2 to R4 the voltmeter
reading may be positive, negative, or zero. For now, let us
consider the special case when the voltmeter reading is zero.
This condition is called the balanced bridge condition.
Fig. 1.1 Basic Wheatstone bridge.
1.1 Analysis of the balanced bridge.
By using the voltage divider formula, the voltage across resistor R3 can be expressed as:
VR3 = (VT/RT) * R3
or
VR3 = VT * (R3 / R1+R3)
Similarly, the voltage across resistor R4 can be determined as:
VR4 = (VT/RT) * R4
or
VR4 = VT * (R4 / R2+R4)
At balance, by definition, the voltmeter reading must be 0. This is only possible if voltages VR3 and
VR4 are equal. Thus by equating the expression for VR3 and VR4 above we get:
VT * (R3 / R1+R3) = VT * (R4 / R2+R4)
After eliminating E and cross-multiplying terms over the equal sign we get:
R3(R2+R4) = R4(R1+R3)
R3 * R2 + R3 * R4 = R4 * R1 + R4 * R3
Subtracting R3 * R4 from both sides gives:
R3 * R2 = R4 * R1
Equation (1-1)
Equation (1-1) tells us that at balance the products of the resistance values of the diagonally
opposite arms of the Wheatstone bridge are equal.
Equation (1-1) also tells us that if one of the four resistors is unknown, its value can be calculated
in terms of the other three. However, before embarking on how the Wheatstone bridge is used for
measuring unknown resistance values, we look at the unbalanced bridge.
1.2 Analysis of the unbalanced bridge.
The Wheatstone is said to be unbalanced when the voltmeter reading in Fig. 1.1 is not zero.
Whether the voltmeter reading is positive or negative depends on the relative magnitudes of the
voltages at the two voltage divider points in the circuit where the meter terminals are connected.
The following example illustrates the calculations involved in an unbalanced bridge.
Example 1.2.1
Given the resistance values of the unbalanced Wheatstone bridge shown in Fig. 1.2, calculate
(a) the voltmeter reading,
(b) currents flowing in both series branches,
(c) the power dissipated in each resistor.
Fig. 1.2 Unbalanced Wheatstone bridge for Example 1.2.1
Solution:
(a)
VR3 = 30V * [10k / (10k + 10k)] = 15V
VR4 = 30V * [20k / (10k + 20k)] = 20V
Thus the voltmeter reading is 20V - 15V = 5V.
(b)
The current flowing through resistor R3 is
IR3 = VR3/R3 = 15V/10k = 1.5 mA
Because R3 is in series with R1 and no current flows through the voltmeter, the current through R1
is also 1.5 mA.
The current flowing through resistor R4 is
IR4 = VR4/R4 = 20V/20K = 1.0 mA
Because R4 is in series with R2 and no current flows through the voltmeter, the current
through R2 is also 1.0 mA.
(c)
The power dissipated in each resistor can be calculated by using the power formula:
P = I2R Thus:
PR1 = (1.5x10-3)2 * (10x103) = 22.5x10-3 W = 22.5 mW
PR3 = (1.5x10-3)2 * (10x103) = 22.5x10-3 W = 22.5 mW
PR2 = (1.0x10-3)2 * (10x103) = 10.0x10-3 W = 10.0 mW
PR4 = (1.0x10-3)2 * (20x103) = 20.0x10-3 W = 20.0 mW
1.3 Determining unknown resistance.
In order to make it easier to understand how a Wheatstone bridge can be used for measuring
resistance, consider the following example:
Example 1.3.1
Let us assume that in the bridge circuit of Fig. 1.2 R4 is unknown and R2 is a resistor decade box.
A resistor decade box is a variable resistor equipped with calibrated dials such that for each
resistance setting the value of its resistance can be read off the dials. The connection diagram of
the revised Wheatstone bridge is shown in Fig. 1.3
At this point we use our knowledge that at balance the
products of the opposite arm resistances of the
Wheatstone bridge are equal. Thus by substituting into
equation (1-1) we obtain the following result:
(R3) * (R2) = (R4) * (R1)
(10 x 103) * RBOX = Rx * (10 x 103)
which, after dropping the 10k values, becomes
RBOX = Rx
Fig. 1.3 Bridge circuit for Example 1.3.1.
i.e. The unknown resistance value is the same as the value of the decade box at balance.
How do we know that the bridge is at balance? We balance the bridge by adjusting the decade box
variable resistance dials until the voltmeter reads 0 volt. At that point the bridge is balanced, and all
the dial readings of the decade box will be equal to the unknown resistance.
To allow for a wide range of resistance value measurements, one arm of the Wheatstone bridge is
usually arranged such that it serves as the ratio arm while the other consists of the adjustable
decade box and the unknown resistor. To see the function of the ratio arm, we rearrange equation
(1-1) as follows:
R3 * R2 = R4 * R1
to
R4 = R2 * (R3/R1)
where R3/R1 is a constant multiplier and R2 is the decade box.
If R4 is the unknown resistor, its value in terms of the multiplier and decade box resistor can be rewritten as:
Rx = multiplier x RBOX
Typical multiplier values are 1, 10, 100, etc.