Strain Gages |
If the supply voltage is 5v,
that change becomes:
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.00002142*Vin
= 0.0001071v.
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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 V1 is around 2.5 volts - we would
need a 5-1/2 digit meter just to see the first significant figure in the voltage
change.
Now we can define what the problem really is.
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If we use a voltage divider,
the voltage change is very small and occurs out in the fifth decimal place in a
typical example.
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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 Ra
and Rb to have the same value. That will produce 2.5
volts at the middle of the left branch.
There are some implications of this result with the bridge circuit.
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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 percentage-wise.
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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.
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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.
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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.
Let's imagine that you have a
strain gage. Let's also assume that you have measured the thermistor, and
you know the following.
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.
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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.
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Circuit 1:
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Ra = Rb
= 10,000W.
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Rc = Rs
= 350W.
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Both voltages out of the bridge
(Vout,+ and Vout,-) are half of the
supply voltage, so, since they are equal, the output of the bridge is zero
volts.
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Circuit 2:
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Ra = 1,000W,
Rb = 10,000W.
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Rc = 3500W,
Rs = 350W.
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Both voltages out of the bridge
(Vout,+ and Vout,-) are (1/11) of the
supply voltage, so, since they are equal, the output of the bridge is zero
volts.
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Circuit 3:
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Ra = 10,000W,
Rb = 1,000W.
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Rc = 35W,
Rs = 350W.
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Both voltages out of the bridge
(Vout,+ and Vout,-) 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.
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Unstrained, both sides form
voltage dividers with 350Wand 10,000W- equal values, on both sides of the divider, so that the output from both is
2.5v with a 5v supply.
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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.
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.
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Circuit 2:
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Ra = 1,000W,
Rb = 10,000W.
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Rc = 3500W,
Rs = 350W.
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Both voltages out of the bridge
(Vout,+ and Vout,-) 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:
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Vout,-
= 5*350.03/(350.03+3500) = 0.4545808734
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The voltage has changed by 35.4mv,
so that is the output voltage from the bridge.
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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?
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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!
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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?
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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.
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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.
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