Voltage Dividers
Why Do People Use Voltage Dividers?
What Does a Voltage Divider Do?
Using a Voltage Divider to Measure Temperature
Using a Voltage Divider to Sense a Light Signal
Practical Voltage Dividers
Problems
You are at:  Elements - Resistors - Voltage Dividers
Imagine:
• You're designing a piece of audio equipment, and you need a volume control. You insert a "pot" - short for potentiomenter - into your circuit.  The pot has a knob that will eventually let you adjust the volume.
• You're designing a measurement system for measuring small deflections in a cantilevered stairway in a performing arts center.  You buy a strain gage and use it to measure the deflection as loads are applied to the stairway to test the stairway.
• You're measuring rotational velocity in a motor, and you use a resistor-capacitor combination to "filter" out the noise you experience in the tachometer measuring velocity.
These three situations have something in common.  The circuitry you use in each of these situations will be some form of a common circuit - a voltage divider.

Goals

The goals for this lesson unit are:

Given a voltage divider,
Be able to  predict the output voltage from the divider given the input voltage and values for the two resistors in the divider.
Be able to determine how the output voltage from the divider
changes when any of the resistance values change.

Voltage Divider Analysis

Just above, we presented three different places where voltage dividers were used.

• The first - the potentiometer - is a variable voltage divider and it is often used in volume controls, gain controls and other situations where an operator wants to be able to control some variable in a continuous fashion.
• The second example involved a strain gage and that sensor is very often used in a bridge circuit which is really two voltage dividers.
• The third example - a "filter" is an example of an AC voltage divider.  You're going to need much more background in AC analysis before we can discuss filters meainingfully, but it gives us something to work toward.
In this lesson we will examine the simplest possible voltage divider, but let's remember that it is just the starting point for a whole family of circuits derived from it including bridges and filters.  Here's a circuit diagram of a very basic voltage divider.  It consists of two resistors, Ra and Rb.  Those two resistors are in series.  An input voltage is applied to the series combination and an output voltage is measured across one of the two resistors.

The output voltage will be some fraction of the input voltage, and the fraction is controlled by the values of the two resistors.  Our first goal is to understand how the output voltage, Vout, depends upon the input voltage, Vin, and the values of the two resistors, Ra and Rb.  We'll examine this problem and solve for the output voltage by using what we know about resistors and Kirchhoff's Laws.  We'll guide you through this with a few questions.

Question

Q1.   How much current (Iseries) flows through the series combination of Ra and Rb?

Once we find that current, then we can calculate the output voltage using Ohm's Law.  You have to notice that Ra and Rb are two resistors in series and divide the input voltage by the series equivalent.

Then, once you have the current you can find the voltage across either of the resistors because you know Ohm's Law.  Doing that you should find the following expression for the output voltage of the voltage divider.

• Iseries = Vin /( Ra + Rb ) because the two resistors are in series if no current is drawn from the voltage divider circuit.
• Vout = Vin Rb/( Ra + Rb) because the voltage across the resistor can be obtained using Ohm's law.

Problems & Questions

P1.   In this voltage divider circuit, the values for the parameters are given by:

• Ra = 1000 W
• Rb = 500 W
• Vin = 15 v
Determine the value of the output voltage, Vout, in volts.

P2.   In the same voltage divider circuit, the two resistors are of equal value.  What is the ratio of output voltage to input voltage?

Q2.   Will it ever be possible to choose resistance values so that the output voltage is larger than the input voltage?

P3.   You have a nine (9) volt source, and you need 5 volts.  Using 400 W for Ra, what value should be used for Rb?

We can plot the ratio of input and output voltages as a function of the resistor across which we measure the output - Rb.  This plot is for Ra = 10W.  NOTE:  The horizontal scale is Rb, and the vertical scale is Vout/Vin.)

Vout = Vin Rb/( Ra + Rb)

How the voltage divider output varies is important.  Another important point is that we should be sure that the expression and the variation are what we expect.  The expression for the output voltage is zero when Ra is zero

Vout = Vin Rb/( Ra + Rb)  =  0 if  Rb  =  0

That's what shown in the plot above.  Does it make sense?

If Ra  =  0, that's the same as having a short circuit (zero resistance) between the output terminals.  It makes sense that the output voltage is zero if it's taken across a short circuit, so our value of zero volts for Ra  =  0 makes sense.

If Ra becomes larger - tending toward infinity - that's the same as having an open circuit between the output terminals.  For an open circuit, no current will flow through Rb so the output voltage will be the same as the input voltage.

• There are a few points to note about the expression we have for the voltage divider.
• The expression we derived holds when no current is drawn from the divider.  At the very beginning we didn't take any output current flowing from the output node, and our analysis won't be valid if current flows out of the output node.  Our analysis can be modified in that case, but the calculations will be somewhat more complex - but not impossible.
• We've applied some limiting checks to our result.  You should realize that's always a wise thing to do.  If you have an expression that purports to tell you how something changes as a parameter varies, it's wise to let the parameter take extrememe values and check your results in that situation, just to increase your confidence in your work.
• A voltage divider is a very basic small circuit that is often found embedded in larger circuits.  In another lesson,  we point out that being able to recognize patterns like voltage dividers and resistor combinations is the mark of the true expert.  Click here to go to that lesson.
Now, you're ready to see how to use these results in some more practical circuits.

Using a Voltage Divider to Measure Temperature

Let's say you want to measure a temperature.  There are lots of devices there that vary somehow as temperature changes.  One device is called a thermistor.  (Click here for some material about the thermistor.)  It's a resistor that changes value as its' temperature changes.

There are other devices somewhat like a thermistor - except that they measure other physical variables - not temperature.

• A photoresistor changes value when incident light intensity changes.  It's used to measure light intensity in cameras, for example.
• A strain gage changes values when it is stretched.  It's used to measure strain (small elongations due to stress) in bridges, buildings, parts of aircraft and many other areas.
Anyhow, you can use a thermistor to measure temperature.  However, often you don't want to measure resistance.  A voltage change is often easier to measure and display - compared to a change in resistance value.  A voltage divider is a circuit that can be used to convert a resistance change to a voltage change.

Here's a voltage divider circuit.

In this circuit we have the following:

• A thermistor, shown as Rt.
• A constant input voltage, Vsource.
In the voltage divider circuit we can note some important facts.
• As Rt increases in value the output voltage increases in value.  If Rt decreases in value, the output voltage decreases in value.
• There are some temperature sensitive resistors that increase in value as temperature increases, but a thermistor has smaller resistance as it heats up.
Let's look at a typical situation to get started.  We'll assume that we have a thermistor that has a resistance of 5000W at 25oC.  That turns out to be a typical thermistor.  For this thermistor, we have the following values of resistance at the temperatures shown in the table below.

 T (oC) R (W) 0 16,330 25 5000 50 1801

Clearly, in this circuit, assuming we make a wise choice for Ra, we can produce a circuit with an output voltage that depends upon temperature.  We can conclude that we could use this circuit to measure temperature, or at least to obtain an output voltage that is temperature dependent and which can be translated into a temperature value.

Let's look at the issue of choosing Ra and the source voltage, Vsource.  Let's start by working with a 5v source.  That's a common value for a source voltage, and there's lots of power supplies that can supply 5v.  Next, we need to choose a value for Ra.  At this point, we're going to make an arbitrary decision.  We'll choose Ra so that the output voltage,Vout, is half the source voltage, i.e.2.5v, at room temperature (25oC).  That means Ra = 5000W.

• By choosing Ra = 5000W we will have an output voltage of 2.5v at room temperature (25oC).
Now with a value for Ra, we can calculate the output voltage at some different temperatures.  First, we'll calculate the output voltage at 0 oC where the resistance, Rt, is 16,330W.

Problems & Questions

P4.   Here is the voltage divider circuit again.  At 0oC Rt is 16,330W.  What is the value of Vout at 0 oC if we have chosen Ra = 5000W.

Q3.   The temperature decreased from 25 oC to 0 oC.  At the temperature decreased did the output voltage increase or decrease?

What actually happens is that the thermistor resistance goes up as the temperature goes down, and when the thermistor resistance goes up, the output voltage goes up in this voltage divider configuration.
Q4.   In this circuit, as the temperature decreases from 25 oC to 0 oC will the output increase or decrease?  The only difference is that the thermistor - the temperature dependent resistor - has been put in the other position in the voltage divider circuit.  For purposes of argument, you might want to assume that Rb = 5000
W, so that you can compare things with the previous situation.

Now, let's look at the situation at 50 oC where the resistance, Rt is 1801W.  You should be able to use the voltage divider formula again to calculate Vout when Ra, is 5000W, and Rt is 1801W.  And remember that we are back to discussing the original placement of the thermistor, i.e. the circuit below.  (Shown just to refresh your memory.)

Problems & Questions

P5.   Here we go again.  At 50oC Rt is 1801W - see the table above.  What is the value of Vout at 50 oC if we have chosen Ra = 5000W.

Q5.   The temperature increased from 25 oC to 50 oC.  At the temperature increased did the output voltage increase or decrease?

Let's draw a few conclusions.
• This circuit can be used to measure temperature.  The output voltage is clearly dependent upon the temperature of the thermistor.
• There are a few problems with this circuit.
• The output voltage probably varies in the wrong direction.
• Secondly, the change in voltage isn't consistent.  When the temperature goes up by 25oC, the voltage goes down by 1.176v.  When the temperature goes down by 25oC, the voltage goes up by 1.328v.  The amount of voltage change isn't the same for the same magnitude of temperature change in the two different directions.
• In other words, the response - change in voltage for change in temperature - is not linear.
There are many variations on the measurement circuit we've discussed.  You can stop here, or you may wish to examine circuits which have two voltage dividers embedded in them.  Those are bridge circuits - a subject of another lesson.

Using a Voltage Divider to Sense a Light Signal

Here's another voltge divider circuit.  In this case there is a light sensitive resistor, RL. The light senstive resistor has two values.  In the dark it has a resistance, Rdark =  500,000W (500kW).  In the light it has a resistance, Rlight =  1000W (1kW).  It is exposed to both light and dark at different times, and it is desired to have a circuit that will give a large signal - close to 5 v - when the sensor is exposed to a dark situation, and a small signal - close to 0 v - when the sensor is in the light.

The problem here is to produce the best possible logic signal at the output voltage terminals.

A typical set of values for a logic signal is zero volts (0v) for a logical "0" and five volts (5v) for a logical "1".  Let's assume that we use a souce voltage of 5v.  Then, in the dark, the voltage will be relatively large - because the sensor dark resistance is large.  In the light, the voltage will be relatively small - because the sensor light resistance is small.  Here is the situation shown in a graph.

In this situation, there are two errors.  One error is the amount by which the dark signal misses being a logical "1" when it is dark.  The other error is the amount by which the light signal misses being a logical "0" when it is light.

There's the possibility of a dilemma here.  We might minimize one error and end up making the other error very large.  Clearly, we want to make both errors as small as possible.  The one thing we can choose is the value of the constant resistance, Ra.

In order to start somewhere, let's look at the possibility of adding up the size of both errors and minimizing the total.  We'll express our terms symbolically as much as possible as we do this.  Note the following.

• The error in each case can be interpreted as the voltage across a resistor.
• When we have the dark situation we want five (5) volts across the output - the photoresistor - but if the entire five voltage does not appear there, the difference - the error - appears across Ra.
• The voltage across Rais given by:
• Vdarkerror = Vs Ra/(Ra + Rdark)
• When the sensor is in the light, the error in the voltage output is actually the voltage across the sensor, since any voltage there that is non-zero is an error.  In that situation, the voltage across the photoresistor is given by (and remember we need to use the light resistance):
• Vlighterror = Vs Rlight/(Ra + Rlight)
Now, there is only one thing that can be adjusted to optimize this situation, and the thing you can change is the resistor, Ra.  Previously, we have noted that Ra is constant.  Once you choose it, it does not change value, but you are free to choose a value for it.  The photoresistor will change resistance depending upon the light it "sees".

So, our problem is to adjust Ra so that we have the best situation.  Ideally, we would like to make both errors zero, but since Ra has to be finite (non-zero and not infinite) we are never going to have a situation without some sort of error.  We have to decide on a measure of how well we are doing, and we can't just take one error or the other.  One possible definition of what to minimize is to add the two errors together and minimize the sum.  There's nothing that specifically points to that, but it seems reasonable to choose that as a measure of the total error.  Here's what we want to minimize.

• TotalError = DarkError  +  LightError
At this point we have defined a problem.  While there are other ways we might have defined the problem, we have this definition to a point where it is possible to determine a "best" value for Ra.  The "best" value for Ra is the value that minimizes the "TotalError", as we have definted it.

There are a number of ways that you can minimize TotalError.  Here are two ways.

• (Numerical Method)  Assume values for the light resistance, dark resistance and source voltage.  Then plot TotalError as a function of Ra.
• (Analytical Method) Using the analytical expression for TotalError as a function of Ra, differentiate and find the value of Ra that produces minimum TotalError, then evaluate the minimum.
Solving the problem using the first method isn't too hard.  Shown below is a MathCad plot of the TotalError function for:
• Vs = 5v,
• Rdark =  500,000W (500kW),
• Rlight =  1000W (1kW).

There's a pretty broad minimum in this function, and it looks like the minimum is somewhere a little above 20kW.  To get the minimum more accurately will require more details of the calculation (more points, and expanding the plot around the minimum so you can see it better) or an analytical approach.

You're pretty much done with this problem.  You need to consider a few other things that might be important.

• Maybe it's more important to get close to zero volts than to five volts.  How would you modify the problem to take that into account?
• What would you do if you chose the best possible resistance, and you still couldn't get close enough to the logic values to make it work?
• What if you have a thermistor instead of a light sensor?  Would you still be able to produce a circuit that gives the right values of logical zeros and ones for situations when the thermistor is hot or cold?
• What happens if the measuring device attached to the circuit draws a current?

Practical Voltage Dividers

One of the most common voltage divider is the one used in volume and tone controls in stereos, radios and televisions.  That form of the voltage divider is based on a rotary potetiometer.  Most of the time when a rotary control is used for something it is a variable voltage divider that is being used.  For example, the following are often voltage dividers in action:

• A volume control on a radio, stereo or television
• A dimmer switch on a light
• The intensity control on a CRT

Here is a sketch of a rotary potentiometer.  It has several important parts.  In this sketch, each

• A circular piece of resistive material.  It might be a conducting polymer, but it could be wound wire.  In the sketch above you might imagine the conducting material (black) to be carbon embedded in something so that you get a smooth conducting film
• There are connections to both ends of the conducting material - which does not make a complete circle.
• A slider arm that makes contact with the resistive material.  The slider arm can be rotated around a pivot at the center, and has a separate connection wire.
This is what is inside many volume controls, intensity controls, etc.  Here is a sketch of a battery connected to a rotary potentiometer

The schematic symbol for a rotary potentiometer is the same as for a linear potentiometer, and is shown below.

• The resistive material is represented by a resistor.
• The slider arm is represented with an arrow that taps into the resistive material.
• The arrow is intended to represent something that can be set to various positions.
• The slider arm could run from the "bottom" of the potentiometer to the top.
Let's consider a practical example of a potentiometer.  Here's a sketch of a battery attached to a rotary potentiometer and a voltmeter attached between the slider arm and one end of the potentiometer.

You should be able to see that when the slider is all the way clockwise (in the 8 o'clock position), there will be no voltage across the voltmeter.  When you crank the slider completely the other way (in the 10 o'clock position) the full battery voltage will appear across the voltmeter (about 1.5 volts).  In the position shown, a fraction of the battery voltage is measured by the voltmeter.

You have seen rotary potentiometers used many times as volume controls in audio equipment, radios, etc.  Instead of a battery, the source might be a microphone or some other audio signal, and the size of the signal that is transmitted - ultimately to be heard - depends upon the setting of the potentiometer.

Finally, you should notice that the total resistance always stays the same in a potentiometer, so that if you want to model the potentiometer as an adjustable voltage divider you will always have Ra + Rab = Rtotal some constant value that is the resistance you would measure if you measured the resistance from end-to-end with nothing else connected to the potentiometer.  Our conclusion:

• A potentiometer is an adjustable voltage divider that has numerous uses.

What Does A Voltage Divider Look Like?

Here is a drawing of a voltage divider built on a circuit board.

This circuit corresponds to the circuit in this circuit diagram - assuming that the bottom (negative) end of the input voltage source is grounded.  The output voltage is take from the midpoint between the two resistors above.  And, remember, that each group of five terminals is really a node.

Problems
Some Things To Do

ToDo 1  Name as many examples of rotary potentiometers as you can, and identify which ones you think might be voltage dividers.

ToDo 2  Find a potentiometer on something broken beyond repair, buy one from an electronic supply store, or get a used one somewhere.  Take it apart and describe it, including the electrical connections.

Links to Other Lessons on Resistors