Introduction
to Bridge Circuits
Introduction
What
is a Bridge Circuit?
Balancing
a Bridge
Sensitivity
of Bridge Circuit Output Voltage
Problems
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 Bridge Circuits
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An
Introduction To Bridge Circuits  Why Do You Need Them?
Making measurements with sensors is a common way in which many engineers
and scientists encounter electrical devices. There are many different
ways in which physical variables like temperature, light intensity, pressure
and numerous other physical variables can be measured electrically.
Devices used to measure a physical variable are called sensors. Some
different kinds of sensors include the following.

Sensors which change resistance
as the physical variable changes.

Thermistors for temperature
measurement

Photoresistors for light
measurement

Strain gages for measurement
of mechanical strain

Sensors which produce
a voltage change for a change in a physical variable.

Thermocouples for temperature

Solar cells for light.
Other kinds of sensors
might include:

Sensors which produce
some other sort of electrical change. Some examples might include
the following.

A tachometer that produces
a frequency proportional to rpm.

Sensors which produce
a set of signals in binary code proportional to pressure.

Sensors which produce
a voltage signal with a frequency proportional to flow rate.
First, let us consider what happens if we put a temperature sensor (a thermistor)
into a voltage divider circuit. Here's the circuit. Actually,
we examined this circuit in the lesson on voltage dividers, and you should
review that material if you don't remember it. Click
here for that material.
What we found in the
voltage divider lesson is the following.

At some nominal value
of the temperature (if we have a temperature sensor) the voltage divider
will have a nominal output voltage.

As the temperature changes
the voltage output of the voltage divider changes.

The voltage change will
probably be a nonlinear function of temperature as the temperature deviates
from the nominal value of the temperature.
There
are some problems with using a voltage divider. Consider the following.

You might want an output
signal that is zero at the nominal conditions.

You could then have an
output signal that was positive when the temperature deviated in one direction
and a negative output signal when the temperature deviated in the other
direction.

If the outut is something
other than temperature that becomes even more clear.

If the sensor is a strain
gage, the strain could be positive or negative, and you would want the
output voltage to be similarly positive or negative, and ideally it would
be proportional to strain.

If the sensor is a pressure
sensor, you could want a positive signal for a presssure above atmospheric
(or any other reference) and negative for a pressure below.
The point here is that it is common to want
a zero output signal under certain, known conditions. There are several
ways you could subtract out that pesky DC voltage. For example, you
could use an operational amplifier circuit to subtract out the DC voltage.
That's more complex than another solution  using a bridge circuit. Usually
a bridge circuit
is what is used in this situation, and in this lesson you're going to learn
about bridge circuits. Bridge circuits are simple circuits that permit
us to solve the problems noted above, and you need to learn about them.
Bridge
Circuits
What is a bridge circuit? It's easier to look at one than to try
to describe it. Here is a bridge circuit.
The bridge circuit
has two arms
(R_{a} and R_{b} constitute one arm here,
and R_{c} and R_{s} constitute the other
arm). Each arm is composed of two resistors in series, and you may
want to think of each arm as a voltage divider. The output is the
difference
between the outputs of the two voltage dividers. In the bridge circuit
above we have also included some source resistance for the source which
drives the bridge circuit. This is the circuit we want to understand.
What are you trying to do in this lesson?

Given a sensor that changes
resistance as some physical variable changes,

Be able to use the sensor
in a bridge circuit.

Be able to choose components
for the bridge circuit that will produce good performance.
Analysis
Of Bridge Circuits  Balancing The Bridge
We have noted that it might be possible to get a bridge output of zero
volts. That's true, but it only happens under certain conditions.
When the output of a bridge is zero, the bridge is said to be balanced.
The first thing we will do is to determine the conditions for a bridge
circuit to be balanced.
If the output voltage of a bridge circuit is zero, that will happen when
the outputs of both dividers is the same. Here's the bridge circuit
again. We'll probably have to be looking at it as we make this argument.
The first thing that we notice is that both voltage dividers have the same
voltage at the "top" of the bridge. Call that voltage V_{top}.
Then, the voltage at the left terminal (labelled "+") is given by:
V_{top}.
[R_{b}/(R_{a
}+
R_{b})]
Similarly, the voltage at the right terminal
(labelled "") is given by:
V_{top}.
[R_{s}/(R_{c
}+
R_{s})]
The difference between
these two voltages  the output voltage  is given by:
V_{top}.
[R_{b}/(R_{a
}+
R_{b})]
 V_{top}.
[R_{s}/(R_{c
}+
R_{s})]
Setting the output
voltage to zero (the condition for a balanced bridge), we get:
V_{top}.
[R_{b}/(R_{a
}+
R_{b})]
 V_{top}.
[R_{s}/(R_{c
}+
R_{s})]
= 0
Since V_{top}
is a common factor it can be removed. Then, we get:
[R_{b}/(R_{a
}+
R_{b})]
 [R_{s}/(R_{c
}+
R_{s})]
= 0
or
[R_{b}/(R_{a
}+
R_{b})]
= [R_{s}/(R_{c
}+
R_{s})]
Now, crossmultiply
the denominators.
R_{b}R_{c
}+
R_{b}R_{s}=
R_{s}R_{a}
+ R_{b}R_{s}
Note that the term R_{s}R_{b}
appears on both sides of the equation and can be taken out on both sides.
That gives us:
R_{b}R_{c
}=
R_{s}R_{a}
This is the condition for balance that we
were looking for. It is a very simple relationship that must be obeyed
by the resistors in the bridge portion of the circuit.
Question
Q1
For these values of resistance would the bridge be balanced?

R_{a
}=
1000 W

R_{b
}=
2000 W

R_{c
}=
3000 W

R_{s
}=
6000 W
Problems
P1 For
these values of resistance determine the value of R_{s
}for
which the bridge is balanced.

R_{a
}=
1000 W

R_{b
}=
2000 W

R_{c
}=
5000 W
P2 For
these values of resistance determine the value of R_{a
}for
which the bridge is balanced.

R_{b
}=
1000 W

R_{c
}=
2000 W

R_{s
}=
5000 W
If you answered the question and did the problems you should have a good
handle on how to choose resistors. However, there are some other
problems that could come up in a bridge circuit. To help you understand
the problems that can arise do this next problem and then we will help
you to think about what the results mean.
P3 For
these values of resistance determine the value of R_{s
}for
which the bridge is balanced.

R_{a
}=
10 W

R_{b
}=
10000 W

R_{c
}=
10 W
Now, if you did the problem above, and the earlier problems, we have a
few more questions for you. We'll put those in the form of a few
problems.
Problems
P4
For these values of resistance you should have found that the bridge would
be balanced. (These are the values from Question 1 above.)
For these values, what are the values of V_{+} and V_{}?
(And remember, both values of voltage are the same when the bridge is balanced.)

R_{a
}=
1000 W

R_{b
}=
2000 W

R_{c
}=
3000 W

R_{s
}=
6000 W
Assume the following values
for the source

V_{c
}=
10v

R_{o
}=
0 W
(Ideal voltage source)
Now,
we want to unbalance the bridge just a little bit. Let us imagine
that the sensor resistor changes by 10% to 6600 W.
In the next problem you need to compute the amount of unbalance by computing
the voltage of the left side (well, you just did that above) and then computing
the voltage on the right side (you'll have to do that because the resistance
changed on that side) and computing the difference.
Problems
P5 For
these values of resistance bridge would be unbalanced.

R_{a
}=
1000 W

R_{b
}=
2000 W

R_{c
}=
3000 W

R_{s
}=
6600 W
Assume the following values
for the source

V_{c
}=
10v

R_{o
}=
0 W
(Ideal voltage source)
Compute the bridge output
voltage for the unbalanced bridge, and be sure you get the sign correct.
Now,
you should have found a substantial change in the output voltage when the
resistance changed. Actually, we calculate that the percentage change
in the voltage out of the rightmost bridge is 3.1% when the sensor resistance
changed by 10%.
Now, we want you to check another situation. Here's the problem.
P6 For
these values of resistance the bridge is balanced.

R_{a
}=
10 W

R_{b
}=
10000 W

R_{c
}=
10 W

R_{s
}=
10000 W
Change the value of the
sensor resistance by 10% to 11000 W.
Compute the change in the voltage output of the right side of the bridge.
You should have found that there was only a very small voltage change and
that the percentage change is miniscule. (We calculate a .01% change
in the output voltage for a 10% change in sensor resistance. The
conclusion is that not all 10% sensor resistance changes are created equal.
There is an issue buried in here. By now you should realize that
it is possible to have the bridge balanced, but it might be unusable because
the voltages out of the voltage dividers in the two different arms fall
into one of these two categories.

The voltage out of the
voltage dividers can be very small (close to zero) if R_{s
}and
R_{b
}are
both small compared to the other resistors.

The voltage out of the
voltage dividers can be very close to the supply voltage if R_{s
}and
R_{b
}are
both large compared to the other resistors  the situation we contrived
above.
In both of those cases the voltage divider outputs
are close to one of the extremes, and in both of those cases the output
voltage will not change much even for substantial changes in the resistances.
Intuitively you expect that the best situation will be when the outputs
of the voltage dividers are close to the midpoint between zero volts (ground)
and the supply voltage. It would be nice if that were more than just
an intuitiona and if it could be shown mathematically. It's possible
to do that and we're going to take a shot at that.
Sensitivity
of Bridge Voltage Output
Earlier, we found that the output voltage from the bridge was given by:
V_{out}
= V_{top}. [R_{b}/(R_{a
}+
R_{b})]  V_{top}. [R_{s}/(R_{c
}+
R_{s})]
We want to determine
the conditions for which this output voltage changes the most for a given
change in a resistor  and we will choose the sensor resistance.
We start by computing the rate of change of the output voltage as a function
of R_{s}
and R_{c}.
The derivative of output voltage (dV_{out}/R_{s}.)with
respect to the sensor resistance is:
dV_{out}/R_{s}
= V_{s}R_{c}/[(R_{c} +R_{s})^{2}]
Now, if you want the maximum slope, you would
choose R_{s} = 0.
However, you don't have control over the sensor resistance, but you can
choose R_{c}
to get the maximum slope. And, to determine the maximum slop, we
differentiate the expression for the slope and set the result to zero.
Doing that, we have the expression below.
d(dV_{out}/R_{s})/dR_{c}
= V_{s}(R_{c}  R_{s})/[(R_{c}
+R_{s})^{3}]
And, that tells us that the maximum slope
occurs when you pick R_{c} = R_{s}.
Summing that up, if you have a resistive sensor and you want to use it
in a bridge circuit, the resistor in series with the sensor should be chosen
so that the resistance in series has the same value as the nominal resistance
of the sensor.
Example
E1 You
have a strain gage with a nominal resistance of 350W.
The resistor in series with the strain gage should have the same value,
i.e. 350W,
if you want the bridge to have maximum sensitivity.
Linearity
There may be times when you want the output voltage to change linearly
with resistance change in the sensor. In that case, recall the result
above.
V_{out}
= V_{top}. [R_{b}/(R_{a
}+
R_{b})]  V_{top}. [R_{s}/(R_{c
}+
R_{s})]
The only way that the
output voltage will change linearly with the sensor resistance is if you
have:
R_{c}
>> R_{s}
Then, if you have that
situation, the output from the right leg of the bridge will be almost zero,
and you will have sacrificed sensitivity for linearity. It will be
a tradeoff, and you will have to make the call.
Problems
Links
to Other Lessons on Resistors
Send
your comments on these lessons.