Introduction to Strain Gages
Introduction
Voltage Divider Circuit
Bridge Circuit
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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 device-under-test 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 (Rs) in a voltage divider with another resistor, Ra. 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 5-1/2 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.

V1 = Vin Rs/( Ra + Rs)

• 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:
• V1 = Vin/2.  Remember we are assuming that Ra = 350W.
• When the strain gage sensor changes to 350.03W, the output changes to:
• V1 = Vin*350.03/(350+350.03) = 0.50002142*Vin.
That's a pretty small change in the output voltage.
• When the sensor is unstrained and has a resistance of 350W, the output is:
• V1 = Vin/2.  Remember we are assuming that Ra = 350W.
• When the strain gage sensor changes to 350.03W, the output changes to:
• V1 = Vin*350.03/(350+350.03) = 0.50002142*Vin.
If the supply voltage is 5v, that change becomes:
• .00002142*Vin = 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 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.
• 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 Ra and Rb to have the same value.  That will produce 2.5 volts at the middle of the left branch.

• Since both Rc and Rs are 350W, the voltage at the mid-point of Rc and Rs 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 percentage-wise.
• 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:
• Ra = Rb = 10,000W.
• Rc = Rs = 350W.
• 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.
• Circuit 2:
• Ra = 1,000W,  Rb = 10,000W.
• Rc = 3500W,  Rs = 350W.
• 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.
• Circuit 3:
• Ra = 10,000W,  Rb = 1,000W.
• Rc = 35W,  Rs = 350W.
• 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.
• 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:
• Vout,- = 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:
• Ra = 1,000W,  Rb = 10,000W.
• Rc = 3500W,  Rs = 350W.
• 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:
• Vout,- =  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.

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