Introduction
to Strain Gages
Introduction
Voltage
Divider Circuit
Bridge
Circuit
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 Strain Gages
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A strain gage
is a resistor in which the resistance changes with strain. A strain
gage is a thin piece of conducting material that may look like the drawing
below. (Although, there are also semiconductor strain gages  not
covered here.) It is literally glued on to the device where you want
to measure strain. If you want more information on strain gages,
it is a good idea to check with manufacturers. Here are some good
links.

Omega
Engineering  A nice writeup about strain gage basics.

Vishay
 Links to catalogs

NMB
Products  a page on strain gage selection

Entran
 A page that gives good definitions of strain gage terms

efunda
 Go to this page if you want a more theoretical treatment of strain gages.

PennState
 Another learning resource.
Now, assume that you have a strain gage glued on a device and the device
is under stress. When the deviceundertest is put under stress it
may elongate or shrink, and the strain gage is sensitive to that small
change in geometry.
The small elongation
in the strain gage produces a small change in the resistance of the strain
gage. Small as it is, it is what we need to use to get a voltage
indicative of the strain in the bar. To convert that small change
in resistance into a usable signal is not impossible, but it takes a little
doing. Often, the strain gage is used is a bridge
circuit like this one.
What are you trying to do in this lesson?

Given a sensor  like
a strain gage  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.
Sensors
In Voltage Divider Circuits
The kind of sensor that we will examine is a resistive sensor, and to make
things specific we will look at using a strain gage to make mechanical
measurements of strain. Here's that sensor (R_{s})
in a voltage divider with another resistor, R_{a}.
Let's examine what
happens in this circuit. Some of the things that happen in this circuit
include the following.

When the sensor resistance
changes, the output voltage changes.

Although the voltage changes,
if the resistance change is small, then the voltage change will also be
small.

When the supply voltage
changes, the output voltage will also change.

Let's assume that we have
a typical strain gage. Normally a strain gage has a nominal resistance
of either 120W
or 350W.
Here is how a strain gage looks.

We need to remember that
strain is the fractional change in length in a material when the material
is stressed. It is normally measured in inches/inch (or you could
make that furlongs/furlong if you like.)

Normally, in most metals,
for instance, the strain will not exceed .005 inch/inch.

The material will elongate
no more than .5 inches in a 100 inch long piece of material.

If the maximum strain
is .005 (.5%) then the maximum fractional change in resistance will be
1%  and that is far larger than you would expect to see since it is an
extreme case.

We went into the laboratory
where we have some strain gages attached to a ten inch bar of .05" thickness.

That's shown at the top
of the picture, where the bar extends from the wooden block.

We had strain gages on
both the top of the bar and underneath the bar so that one would always
elongate and the other would compress.

The bar is clamped at
one end and the other end is free. Putting 75 cents (three US quarters)
on the free end of the bar produced no measurable change in resistance
with a 51/2 digit ohmmeter.

However an overloaded
wallet (about 1/2 pound) changed resistance from 350.550W
to 350.520W.
The change in resistance can be very small.
Now, if we have some sense of the resistance change, then we can think
about how we will sense that change. Because the change is very
very small we will have to worry about how we are going to use that
very small change in resistance We have at least a couple of options
for how we can make that measurement.

We can measure resistance
directly. But as we have seen, the resistance change is very small
and ohmmeters will have trouble showing us much.

We can put the strain
gage in a circuit like a voltage divider, letting the change in resistance
cause a voltage change and measure that change. We will need to check
to see if that makes our situation any better.

Our problem is complicated
by the fact that most devices that are used for electrical measurements
will measure DC voltage. The ohmmeter is an exception. But
if we want to get our measurements into a computer and we don't want to
type them in, we'll probably need a voltage.

So, we conclude that the
voltage divider (below) is a good idea in some ways.
Let's compute the output voltage for the voltage divider.
V_{1}
= V_{in} R_{s}/( R_{a} + R_{s})

The expression for the
output voltage is one we have seen many times before.

The output voltage from
the voltage divider increases as the sensor resistance increases.
Now, let's compute some typical values.

Let's use a source voltage
of 5 volts.

Assume we have a standard
strain gage sensor  a nominal resistance of 350W.

We'll choose the same
value, 350W,
for the other resistor in the voltage divider.
Now, we can check what happens when the resistance changes by a typical
small amount.

We will assume that the
resistance changes from 350 to 350.03W.
The typical small change we discussed earlier.

The question we need to
answer is "How much does the output voltage change when the resistance
changes from 350 to 350.03W?"
We can compute the output
for both cases.

When the sensor is unstrained
and has a resistance of 350W,
the output is:

V_{1} =
V_{in}/2. Remember we are assuming that R_{a}
= 350W.

When the strain gage sensor
changes to 350.03W,
the output changes to:

V_{1} =
V_{in}*350.03/(350+350.03) = 0.50002142*V_{in}.
That's a pretty small change in the output voltage.

When the sensor is unstrained
and has a resistance of 350W,
the output is:

V_{1} =
V_{in}/2. Remember we are assuming that Ra = 350W.

When the strain gage sensor
changes to 350.03W,
the output changes to:

V_{1} =
V_{in}*350.03/(350+350.03) = 0.50002142*V_{in}.
If the supply voltage
is 5v, that change becomes:

.00002142*V_{in}
= 0.0001071v.

That is about one tenth
of a millivolt. If we read that change with a voltmeter that goes
to 3 v  which we would need since V_{1} is around 2.5 volts
 we would need a 51/2 digit meter just to see the first significant figure
in the voltage change.
Now we can define what the problem really is.

If we use a voltage divider,
the voltage change is very small and occurs out in the fifth decimal place
in a typical example.

We need something that
will improve this situation.
Using
A Bridge Circuit
A bridge circuit can help with our problem. Here's a bridge circuit.
We will choose R_{a}
and R_{b} to have the same value. That will produce
2.5 volts at the middle of the left branch.

Since both R_{c}
and R_{s} are 350W,
the voltage at the midpoint of R_{c} and R_{s}
is also 2.5 volts.

That means Vout = 0 volts
when the strain gage is unstrained!
There are some implications of this result with the bridge circuit.

If the voltage is zero
when the gage is unstrained  the bridge is balanced  and the voltage
becomes 0.0001071v when the gage is strained, then the change is large
percentagewise.

That voltage may be small,
but we can amplify it  and we won't be amplifying it embedded in a DC
voltage. We'll need a differential amplifier  something like this.
Here the bridge output can be amplified to a usable level  depending upon
the gain of the amplifier  and the output can be made to be zero at zero
strain.
There are some other alternatives
also.

The output voltage from
the bridge can be amplified by a differential amplifier in a data acquisition
board. You don't necessarily have to build your own amplifier.
Most currently available data acquisition boards have differential amplifiers
that will amplify the difference between two input voltages.

You could measure the
output voltage with a good voltmeter with an ungrounded input.
An important consideration when using bridge circuits is choice of values
for those resistors that have values you can choose. In the bridge
we just considered, only the strain gage resistance was fixed. That
leads to a question.

How do you choose resistors
in a bridge circuit so that performance is optimized?
Let's imagine that you
have a strain gage. Let's also assume that you have measured the
thermistor, and you know the following.

Resistance at zero strain
= 350W.

You want to build a bridge
that has a zero output voltage at zero strain. In other words, you
want to build a balanced bridge.
The question is "How to
build the bridge?". We'll work on an answer to that question starting
next. We will assume that the supply voltage is five (5) volts.

The strain gage is Rs.
That means that Rs is 350W.
There are many ways that we can build a balanced bridge. Here are
a few.

Circuit 1:

R_{a} =
R_{b} = 10,000W.

R_{c} =
R_{s} = 350W.

Both voltages out of the
bridge (V_{out,+} and V_{out,}) are
half of the supply voltage, so, since they are equal, the output of the
bridge is zero volts.

Circuit 2:

R_{a} =
1,000W,
R_{b} = 10,000W.

R_{c} =
3500W,
R_{s} = 350W.

Both voltages out of the
bridge (V_{out,+} and V_{out,}) are
(1/11) of the supply voltage, so, since they are equal, the output of the
bridge is zero volts.

Circuit 3:

R_{a} =
10,000W,
R_{b} = 1,000W.

R_{c} =
35W,
R_{s} = 350W.

Both voltages out of the
bridge (V_{out,+} and V_{out,}) are
(10/11) of the supply voltage, so, since they are equal, the output of
the bridge is zero volts.
Let's look at the implications of one choice. We'll look at Circuit
1.

Unstrained, both sides
form voltage dividers with 350W
and 10,000W
 equal values, on both sides of the divider, so that the output from both
is 2.5v with a 5v supply.

Now, compute the changed
voltage from the sensor side of the bridge when the strain gage is strained.
We will assume that we have the load we discussed above, and that the strain
gage sensor's resistance changes to 350.03W.

When the strain gage sensor
changes to 350.03W,
the output on the sensor side of the bridge changes to:

V_{out,}
= 5*350.03/(350+350.03) = 0.50002142*5 = 2.500107.

With that voltage, the
output of the bridge becomes:

2.5  2.500107 = .000107
= 107mv.
What we have demonstrated
is that we get a very small voltage change with this choice of resistors
for the bridge. There is always the possibility that a different
choice of resistors would produce better results. Let's check that
out. Let's look at Circuit 2. Here is what we noted above for
Circuit 2.

Circuit 2:

R_{a} =
1,000W,
R_{b} = 10,000W.

R_{c} =
3500W,
R_{s} = 350W.

Both voltages out of the
bridge (V_{out,+} and V_{out,}) are
(1/11) of the supply voltage, so, since they are equal, the output of the
bridge is zero volts.
Now,
with no strain the bridge is balance, and with a 5 volt supply, we would
have 10/11 of five volts or 0.4545454 volts. When the load
is applied and the sensor resistance changes, the voltage from the sensor
side of the bridge is going to be:

V_{out,}
= 5*350.03/(350.03+3500) = 0.4545808734

The voltage has changed
by 35.4mv,
so that is the output voltage from the bridge.

The output voltage here
is much smaller than the output voltage from Circuit 1.
Problem
Compute the output voltage from the loaded bridge for Circuit 3.
What can we conclude from this?

When we chose resistors
that placed Vs and Vd near the "rails"  i.e. near zero/ground or near
the power supply voltage  the voltage didn't change very much when the
strain gage was strained. Maybe we should have expected that!

The best sensitivity 
in terms of voltage change for the same resistance change  seemed to come
when all the resistors were equal when unstrained.
What might we think about
now?

Investigating the sensitivity
mathematically is one thing we should do. That's another topic, and
there is a section in the
lesson on bridge circuits that covers senstivity.

For now, you have enough
information to do some interesting things in the lab, and you have some
idea of how to choose resistors when you use the bridge circuit with a
resistive sensor.
Links
to Other Lessons on Sensors
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to Related Lessons
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