Analogue Electronics (Aero) 1
4.47
§4 Wheatstone Bridge
(Hambley sect 2.8)
This is a circuit which can be used in principle
to find the value of an unknown resistor very
accurately.
Often used for Platinum resistance thermometer,
or strain gauge to measure temperature or
weight.
I1
+
Vs
-
C
R1
A
R2
G
I2
R3
B
R4
D
In operation the resistance of three of the
resistors is known accurately, and the 4th is the
unknown. A sensitive galvanometer G
(ammeter) is used to measure when the bridge is
balanced, that is when no current flows from A
to B.
Analogue Electronics (Aero) 1
4.48
At this point the the voltages VA and VB are the
same and we have
VA = VB or VAB = 0
Because, for this condition, the same current I1
flows through the resistors R1 and R2 we may
use the potential divider rule and write.
VA
I1R2
R2
=
=
VS I1 ( R1 + R2 ) R1 + R2
Similarly
VB
I2 R4
R4
=
=
VS I2 ( R3 + R4 ) R3 + R4
Note again: this is only true if no current flows
A to B!
In this case
R2
R4
=
R1 + R2 R3 + R4
So R2(R3+R4) = R4(R1+R2)
R2R3 + R2R4 = R1R4 + R2R4
R2R3 = R1R4
R3 R1
=
R4 R2
That is: the ratio of the resistances in both
branches is the same, if the bridge is balanced.
€
Analogue Electronics (Aero) 1
4.49
This makes sense: the fraction of the voltage
dropped in the first resistor in each branch must
be the same.
The power of this method is that, since no
current flows in the Ammeter, its physical
properties (such as internal resistance) do not
matter. It is effectively not part of the circuit.
But now suppose the bridge is out of balance:
how much current would flow in the meter?
This will depend on the meter properties as we
will see.
Look at the Thevenin equivalent of the circuit.
Divide the circuit up into the component under
study, The ammeter (Galvanometer) G, and the
rest of the circuit.
First calculate the Thevenin resistance RT. Short
circuit the voltage source - this means that nodes
C and D are connected together.
A
B
R1
R2 R
3
R4
R1R2
R3 R4
RT =
+
R1 + R2 R3 + R4
Analogue Electronics (Aero) 1
4.50
Now for the Thevenin voltage
If we open circuit the load (meter) between A
and B we would get
R2
R4
VS
− VS
VT = VAB = VA – VB =
R1 + R2
R3 + R4
Suppose the set up we had was
€
C
R1
+
Vs
-
60Ω
A
R2
30Ω
R3
60Ω
B
R4
40Ω
G
D
R1R2
R3 R4
60 × 30 60 × 40
RT =
+
=
+
R1 + R2 R3 + R4 60 + 30 60 + 40
= 20Ω + 24Ω = 44Ω
So
⎛ 30
⎛ 1 2 ⎞
40 ⎞
2
VT = VS ⎜
−
⎟ = 10V ⎜ − ⎟ = − V
⎝ 30 + 60 40 + 60 ⎠
⎝ 3 5 ⎠
3
Analogue Electronics (Aero) 1
4.51
Thévenin equivalent circuit plus load is
therefore:
44 Ω
-2/3V
A
10 Ω
B
Current flowing is V/R = 2/3 × 1/54 = 12.4 mA
Voltage across 10Ω resistor is
10Ω×12.35mA = 0.124V
In this example we could not have reduced the
circuit to resistors in series or in parallel until
the whole circuit was represented by a single
equivalent resistor. This worked here, but we
require a general technique which could be used
in all circumstances.
One such general technique is Nodal Analysis.
Briefly, we write a set of equations, describing
the current into key Nodes, and solve them
simultaneously. The hard part is making sure all
the signs are right!
Analogue Electronics (Aero) 1
4.52
Nodal analysis for the Wheatstone bridge.
C
R1
+
Vs
-
R3
60Ω
60Ω
A
B
R 2 10Ω
R4
30Ω
40Ω
D
Because Ohm's law only tells us about potential
differences, we have to assume one of the
potentials and work relative to that. Here we
will assume, as is conventional, that the lowest
potential is 0V. That is:VD = 0V
Node C is then at +10V, i.e. VC = +10V
Now apply Kirchhoff’s current law at the
remaining nodes. Watch the signs! We must
always be consistent. One method is to imagine
standing at the node and looking at potentials
from that node to each adjacent node. Use this
to work out currents.
Analogue Electronics (Aero) 1
4.53
Here we go: Node A first
Current flowing from A to C is (VA-VC) / 60
Current flowing from A to B is (VA-VB) / 10
Current flowing from A to D is (VA-VD) / 30
€
Since the total current into the node must equal
the total current out of the node, the sum of
these three currents must be zero.
∑Inode = ∑Iin + ∑Iout = 0
So
VA − VC VA − VB VA − VD
+
+
=0
60
10
30
We can do this since we are using a consistent
sign convention: all the voltages are with respect
to VA. All the currents are written as currents
flowing out of the node.
Setting VC = +10V and VD = 0V we have:
VA −10 VA − VB VA
+
+
=0
60
10
30
⎛ 1
⎛ 1 ⎞ 10
1
1 ⎞
VA ⎜ + + ⎟ + VB ⎜− ⎟ −
=0
⎝ 60 10 30 ⎠
⎝ 10 ⎠ 60
Analogue Electronics (Aero) 1
4.54
⎛ 1+ 6 + 2 ⎞
⎛ 1 ⎞ 10
VA ⎜
⎟ + VB ⎜− ⎟ =
⎝ 60 ⎠
⎝ 10 ⎠ 60
⎛ 9 ⎞
⎛ 1 ⎞ 10
VA ⎜ ⎟ + VB ⎜− ⎟ =
⎝ 60 ⎠
⎝ 10 ⎠ 60
Equation1
9VA – 6VB = 10
Node B
Similarly we take the algebraic sum of the
currents flowing from node B we have:
VB − VA VB − VC VB − VD
+
+
=0
10
60
40
Setting VC = +10V and VD = 0V we have:
VB − VA VB −10 VB
+
+
=0
10
60
40
⎛ 1 ⎞
⎛ 1
1
1 ⎞ 10
VA ⎜ − ⎟ + VB ⎜ + + ⎟ −
=0
⎝ 10 ⎠
⎝ 60 10 40 ⎠ 60
⎛ 1 ⎞
⎛12 + 2 + 3 ⎞ 10
VA ⎜ − ⎟ + VB ⎜
=0
⎟ −
⎝ 10 ⎠
⎝ 120 ⎠ 60
⎛ 1 ⎞
⎛ 17 ⎞ 10
VA ⎜ − ⎟ + VB ⎜
⎟ =
⎝ 10 ⎠
⎝120 ⎠ 60
-12VA+17VB = 20
Equation2
Previously we had
9VA – 6VB = 10
Equation1
Analogue Electronics (Aero) 1
4.55
4 times this is
36VA – 24VB = 40
3 times Equation 2 is
-36VA+51VB = 60
Adding gives
0VA+(51-24)VB = 40+60
100
so VB = 27 = 3.704 V
From Equation 1
€
9VA – 6VB = 10
6VB + 10 6 × 3.704 + 10
VA =
=
= 3.580 V
9
9
VBA = 3.704-3.580 = 0.124 V
Current flow from B to A is
0.124 = 0.0124A = 12.4 mA.
10
Just as before!
Analogue Electronics (Aero) 1
4.56
Another quick example :
80 Ω
40Ω 2
10 Ω
1
3
+
60V
-
0
60Ω
20 Ω
V2 − 60 V2 − V3 V2
Node 2: 40 + 10 + 60 = 0
Multiply by 120 gives:
3V2 -180 + 12V2 -12V3+ 2V2 = 0 or
Eq1
17V2 -12V3 = 180
€
V3-0 + V3-V2 + V3-60 = 0
Node 3: 20
10
80
Multiply by 80 gives:
4V3 +8V3 -8V2 +V3-60 = 0 or
Eq2
13V3 -8V2 = 60
Multiply Eq 1 by 13 and Eq 2 by 12 gives
Eq1a
221V2 -156V3 = 2340
Eq2
156V3 -96V2 = 720
Adding 125V2 = 3060 so V2 = 24.48V
Substituting in either Eq 1 or Eq 2 gives
V3 = 19.68V
Analogue Electronics (Aero) 1
4.57
Nodal analysis: a more general example.
I1
V1
1
+
-
R
10
0
R12
R13
R30
2
R
23
R 20
3
Writing an equation for node 1 in the form
VA − VB
RAB we have;
V1 − 0 V1 − V2 V1 − V3
+
+
= I1
R10
R12
R13
Look at this carefully. (V1-VX)/R1X is the current
leaving node 1 in the direction of node X over
resistance R1X. The current leaving the node on
€ these three paths must equal the current coming
into the node on the other path which is I1. One
could just as easily have written the same
equation entirely in terms of currents leaving the
node. Signs!
Analogue Electronics (Aero) 1
€
€
€
€
€
4.58
V1 − 0 V1 − V2 V1 − V3
+
+
− I1 = 0
R10
R12
R13
Writing equations for the other nodes we have:
V − 0 V2 − V1 V2 − V3
Node 2 : 2
+
+
=0
R20
R12
R23
V3 − 0 V3 − V1 V3 − V2
Node 3 :
+
+
=0
R30
R13
R23
Rewriting these equations gives
⎛ 1
1
1 ⎞
1
1
V1⎜
+
+
− V3
= I1
⎟ − V2
R12
R13
⎝ R10 R12 R13 ⎠
⎛ 1
1
1
1 ⎞
1
−V1
+ V2 ⎜
+
+
=0
⎟ − V3
R12
R23
⎝ R20 R12 R23 ⎠
⎛ 1
1
1
1
1 ⎞
−V1
− V2
+ V3 ⎜
+
+
⎟ = 0
R13
R23
⎝ R30 R13 R23 ⎠
Note that, while the order of current and voltage
labels matter, the order of resistance labels does
not
⎛ 1
1
1 ⎞
1
1
V1⎜
+
+
−
V
−
V
= I1
⎟ 2
3
R12
R13
⎝ R10 R12 R13 ⎠
⎛ 1
1
1
1 ⎞
1
−V1
+ V2 ⎜
+
+
−
V
=0
⎟ 3
R21
R23
⎝ R20 R21 R23 ⎠
€
Analogue Electronics (Aero) 1
4.59
⎛ 1
1
1
1
1 ⎞
−V1
− V2
+ V3 ⎜
+
+
⎟ = 0
R31
R32
⎝ R30 R31 R32 ⎠
This can be written in matrix form as:
⎡⎛ 1
⎤
1
1 ⎞
1
1
+
+
−
−
⎢⎜
⎥
⎟
R
R
R
R
R
⎝
⎠
12
13
12
13
⎥ ⎡V ⎤
⎡ I1 ⎤ ⎢ 10
⎛ 1
⎥ ⎢ 1 ⎥
1
1
1 ⎞
1
⎢ ⎥ ⎢
−
+
+
−
⎜
⎟
⎥ × ⎢V2 ⎥
⎢I2 ⎥ = ⎢
R
R
R
R
R
⎝ 20
21
21
23 ⎠
23
⎥ ⎢⎣V ⎥⎦
⎢⎣I3 ⎥⎦ ⎢
3
⎛ 1
1
1
1
1 ⎞⎥
⎢
−
−
+
+
⎜
⎟⎥
⎢
R
R
R
R
R
⎝
⎣
31
32
30
31
32 ⎠⎦
This means that we can work on these
equations, and solve them, using standard
matrix techniques. For big complicated circuits
this proves to be (by far!) the easiest technique.
For such purposes it is often more convenient to
work in terms of the inverse resistance, called
the conductance, G.
For a simple resistance, G = 1/R so we can
rewrite Ohm's law as either V = IR or I = GV
Here we have a matrix of inverse resistances,
that is an conductance matrix.
⎡G11 G12 G13 ⎤
⎢
⎥
G = ⎢G21 G22 G23 ⎥
⎢⎣G31 G32 G33 ⎥⎦
Analogue Electronics (Aero) 1
4.60
For the full equation we can write, simply.
I = G ×V
Where G is the admittance matrix, V is the
voltage vector and I is the current vector. This
means that,
€ for example Gjk is the conductance
j and k and Vj is the
€ connected between nodes €
voltage at node j.
€
⎡ I1 ⎤ ⎡I1⎤
⎢ ⎥ ⎢ ⎥
I = ⎢I2 ⎥ = ⎢0 ⎥
⎢⎣I3 ⎥⎦ ⎢⎣0 ⎥⎦
where I1 is the current injected by a source at
node 1. Since there are no sources connected to
nodes 2 and 3, I2 = I3 = 0. Notice that we have a
€ four nodes, but we end up with the
circuit with
equations. This is because we choose one node
to be at 0V.
Remember: we only know about potential
differences. It is always the voltage with respect
to something, and not the absolute voltage. We
choose the analysis reference voltage for our
convenience.
Analogue Electronics (Aero) 1
4.61
As an aside, it is unusual to see G used in this
way very often in real analysis.
This is because in more realistic circuit
analysis we often deal with a.c. signals and the
resistance R and conductance G are replaced
by terms which are more general, the
impedance Z and the admittance Y.
It is therefore more usual to see Y×V=I as the
matrix equation, where,
y11 y12 y13
Y = y21 y22 y23
y31 y32 y33