Memorial University of Newfoundland
Faculty of Engineering and Applied Science
ENGR 3844 - Basic Electrical Components and Systems
Part-II (Electronics)
EXP. 3844-1E
Voltage Divider and Wheatstone Bridge
PURPOSE
To introduce students to various voltage divider and Wheatstone bridge
circuits.
INTRODUCTION
1.0 Voltage Divider Circuits
The voltage divider circuit is used to reduce the supply voltage to the desired
level at the load. A simple voltage divider circuit is shown in Figure 1 where a
resistor is connected in series with the load.
+ E
+
VRS _ IL
+
VRS
RS
+
+
RS
E
_
RL
VL
_
≡
_
+
RL
Load
VL
IL
Figure 1
KVL around the circuit gives, E - VRS - VL = 0
⇒ VL = E - VRS
= E - ILRS
But, I L =
Ex RL
E
. Therefore, VL =
RS + RL
RS + RL
E 1.1
(1)
(2)
However, this type of voltage divider has a serious disadvantage as the value
of VL depends on the load current IL. If IL changes, the voltage drop across RS
also changes and hence, the voltage drop across the load RL will also change.
This change in voltage at the load is called voltage regulation. The load
voltage will remain relatively constant if the load is maintained constant.
In order to overcome the above limitation another type of voltage divider
circuit is used which is shown in Figure 2. In this case a bleeder resistor is
used in parallel with the load.
+ E
+
IT
IL
IT
_
+
VRS
_
R
S
+
+
IL
RS
≡
VL
+ VL
E
RL
RB
_
RB
IB
RL
_
Load
IB
Figure 2
Using voltage divider rule,
VL =
E (R B R L )
RS + RB RL
RB
RL =
(3)
RBRL
RB + RL
If RB << RL, then VL =
(4)
E RB
RS + RB
(5)
If the bleeder current IB is high in comparison with IL then IT will remain
more or less constant with changes in IL. Therefore, the load voltage will
change slightly and hence it will have a better regulation than the previous
circuit.
2.0 Wheatstone Bridge Circuit
The Wheatstone bridge circuit is mainly used to measure an unknown
resistance. However, the principle of Wheatstone bridge is used in many
transducers and instrumentation systems to measure the physical variable
E 1.2
such as strain, temperature, and pressure. A Wheatstone bridge circuit is
shown in Figure 3. In this figure R4 is a variable resistance, which models the
transducer. If a change in physical variable results in a change in resistance
of the transducer, the Wheatstone bridge can be used to convert the change in
resistance to a change in voltage, which can be measured.
Figure 3
Under balanced condition of the bridge the output voltage V0 will be zero.
The balance condition can be achieved by varying any one or more
resistances of the bridge until,
Va = Vb
Under balanced conditions, voltage divide rule for R1 and R2, and R3 and R4
gives
E R2
R 2 + R1
E R4
Va =
R3 + R4
Vb =
(6)
(7)
For Va = Vb,
E R4
E R2
=
R 3 + R4 R 2 + R1
(8)
∴ R1R 4 = R 2R 3
(9)
Equation (9) shows the balance condition for the Wheatstone bridge. When
the bridge is not balanced, the output voltage is given by,
E 1.3
 R4
R2 
−
V0 = Va − Vb = E 

 R 3 + R 4 R 2 + R1 
(10)
Equation (10) shows that if R1, R2 and R3 are fixed (or given) a change in R4
translates into a change in V0. Thus, a variation in V0 is a measure of the
variation in R4.
PRELAB
1. Read the section entitled “Introduction” carefully.
2. Read the handout entitled “Safety in the Laboratory” carefully.
3. Referring to Fig. 4, determine RL so that V0 = 7.5 V. If the load resistance
is increased by 25% calculate the new load voltage, V0.
+
E
+
VRS
RS
RS = 1 kΩ
E = 10 V
_
+
RL
V0
IL
Fig. 4
4. Referring to Fig. 5, design a voltage divider for the load resistance RL
obtained in Fig.4 so that it can operate at 7.5 V from a 10 V source.
Assume that the bleeder current, IB is 75mA. If the load resistance is
increased by 25% calculate the new load voltage. Compare this result with
that obtained in question (3) above.
+
E
IT
+
RS
_
VRS
IL
+
RB
IB
Fig. 5
E 1.4
RL
V0
EXPERIMENT
1. Voltage Divider
1.1 Construct the voltage divider circuit as shown in Fig. 6 on the op-amp
designer board using a nominal 1000 Ω resistor in series with a 1 kΩ
variable resistor. Make sure that the source voltage is 10 V after
connecting to the circuit. Adjust the variable resistor to get the output
voltage V0 = 7.5 V. Measure the value of R2 for this condition. Comment
on any discrepancies between the measured and calculated value in
question (3) of the Prelab.
+
E
+
RS
VRS
_
+
R2
V0
RS = 1 kΩ
E = 10 V
R2 = 2.2 kΩ pot
I0
Fig. 6
1.2 Obtain a plot of the output voltage V0 as a function of R2 for the circuit
shown in Fig.6. Use a range of resistor values between 680 Ω and 2.2 kΩ.
1.3 Construct the voltage divider circuit shown in Fig.5 and measure the load
voltages to verify your design in question (4) of the Prelab. (Use the
nearest available resistance values).
E 1.5
2. Wheatstone Bridge
2.1 Construct the Wheatstone bridge circuit shown in Fig.7 on the op-amp
designer board using three nominal 120 Ω resistors in three arms of the
bridge and a 200 Ω variable resistor (pot) in the fourth arm as shown.
Ensure that the source voltage is 10 V after it is connected to the circuit.
Fig.7
2.2 Adjust the pot until the bridge is balanced (i.e., V0 = 0). Measure the
value of R4 for this condition. Confirm your observation by calculation
and comment on any discrepancies.
2.3 Obtain a plot of the output voltage, V0 as a function of R4. Determine the
sensitivity of the bridge circuit.
E 1.6