Introduction to Bridge Circuits
What is a Bridge Circuit?
Balancing a Bridge
Sensitivity of Bridge Circuit Output Voltage
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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.

Other kinds of sensors might include:         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.

        There are some problems with using a voltage divider.  Consider the following. 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 (Ra and Rb constitute one arm here, and Rc and Rs 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?

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 Vtop.  Then, the voltage at the left terminal (labelled "+") is given by:

Vtop. [Rb/(Ra + Rb)]

Similarly, the voltage at the right terminal (labelled "-") is given by:

Vtop. [Rs/(Rc + Rs)]

The difference between these two voltages - the output voltage - is given by:

Vtop. [Rb/(Ra + Rb)] - Vtop. [Rs/(Rc + Rs)]

Setting the output voltage to zero (the condition for a balanced bridge), we get:

Vtop. [Rb/(Ra + Rb)] - Vtop. [Rs/(Rc + Rs)] = 0

Since Vtop is a common factor it can be removed.  Then, we get:

[Rb/(Ra + Rb)] - [Rs/(Rc + Rs)] = 0

[Rb/(Ra + Rb)] = [Rs/(Rc + Rs)]

Now, cross-multiply the denominators.

RbRc + RbRs= RsRa + RbRs

Note that the term RsRb appears on both sides of the equation and can be taken out on both sides.  That gives us:

RbRc = RsRa

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.


Q1  For these values of resistance would the bridge be balanced?


P1   For these values of resistance determine the value of Rs for which the bridge is balanced.

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P2 For these values of resistance determine the value of Ra for which the bridge is balanced.
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        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.
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        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.

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.)

Assume the following values for the source
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        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.

P5   For these values of resistance bridge would be unbalanced.

Assume the following values for the source Compute the bridge output voltage for the unbalanced bridge, and be sure you get the sign correct.
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        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. 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.
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        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.

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 mid-point 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:

Vout = Vtop. [Rb/(Ra + Rb)] - Vtop. [Rs/(Rc + Rs)]

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 Rs and Rc.  The derivative of output voltage (dVout/Rs.)with respect to the sensor resistance is:

dVout/Rs = VsRc/[(Rc +Rs)2]

Now, if you want the maximum slope, you would choose Rs = 0. However, you don't have control over the sensor resistance, but you can choose Rc 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(dVout/Rs)/dRc =  Vs(Rc - Rs)/[(Rc +Rs)3]

And, that tells us that the maximum slope occurs when you pick Rc = Rs.

        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.


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.


        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.

Vout = Vtop. [Rb/(Ra + Rb)] - Vtop. [Rs/(Rc + Rs)]

The only way that the output voltage will change linearly with the sensor resistance is if you have:

Rc >> Rs

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.

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