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
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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 resistorcapacitor 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, R_{a} and R_{b}. 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, V_{out}, depends
upon the input voltage, V_{in}, and the values of the two
resistors, R_{a} and R_{b}. 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 (I_{series}) flows through the series combination
of R_{a} and R_{b}?
Once we find that current, then we can calculate the output voltage using
Ohm's Law. You have to notice that R_{a} and R_{b}
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.

I_{series}
= V_{in} /( R_{a} + R_{b })
because the two resistors are in series if no current is drawn from the
voltage divider circuit.

V_{out}
= V_{in} R_{b}/( R_{a} + R_{b})
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:

R_{a} =
1000 W

R_{b }=
500 W

V_{in }=
15 v
Determine the value of
the output voltage, V_{out}, 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 R_{a}, what value should be used for R_{b}?
We can plot the ratio of input and output voltages as a function of the
resistor across which we measure the output  R_{b}.
This plot is for R_{a} = 10W.
NOTE: The horizontal scale is R_{b}, and the vertical
scale is V_{out}/V_{in}.)
V_{out}
= V_{in} R_{b}/( R_{a} + R_{b})
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
R_{a} is zero
V_{out}
= V_{in} R_{b}/( R_{a} + R_{b})
= 0 if R_{b} = 0
That's what shown in
the plot above. Does it make sense?
If R_{a} = 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 R_{a} = 0 makes
sense.
If R_{a} 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 R_{b}
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
R_{t}.

A constant input voltage,
Vsource.
In the voltage divider circuit we can note some important facts.

As R_{t}
increases in value the output voltage increases in value. If R_{t}
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 25^{o}C. 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 (^{o}C)

R (W)

0

16,330

25

5000

50

1801

Clearly, in this circuit, assuming we make a wise choice for R_{a},
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 R_{a} and the source
voltage, V_{source}. 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 R_{a}. At this point, we're going to make
an arbitrary decision. We'll choose R_{a} so that
the output voltage,V_{out}, is half the source voltage,
i.e.2.5v, at room temperature (25^{o}C). That means R_{a}
= 5000W.

By choosing R_{a}
= 5000W
we will have an output voltage of 2.5v at room temperature (25^{o}C).
Now with a value for R_{a}, we can calculate the output
voltage at some different temperatures. First, we'll calculate the
output voltage at 0 ^{o}C where the resistance, R_{t},
is 16,330W.
Problems
& Questions
P4. Here
is the voltage divider circuit again. At 0^{o}C R_{t}
is 16,330W.
What is the value of V_{out} at 0 ^{o}C if
we have chosen R_{a} = 5000W.
Q3. The
temperature decreased from 25 ^{o}C to 0 ^{o}C.
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 ^{o}C
to 0 ^{o}C 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 R_{b}
= 5000W,
so that you can compare things with the previous situation.
Now, let's look at the situation at 50 ^{o}C where the resistance,
R_{t} is 1801W.
You should be able to use the voltage divider formula again to calculate
V_{out }when R_{a}, is 5000W,
and R_{t} 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 50^{o}C R_{t} is 1801W
 see the table above. What is the value of V_{out}
at 50
^{o}C if we have chosen R_{a} = 5000W.
Q5. The
temperature increased from 25 ^{o}C to 50 ^{o}C.
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 25^{o}C,
the voltage goes down by 1.176v. When the temperature goes down by
25^{o}C, 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, R_{L}. The light senstive resistor has
two values. In the dark it has a resistance, R_{dark}
= 500,000W
(500kW).
In the light it has a resistance, R_{light} = 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, R_{a}.
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 R_{a}.

The voltage across R_{a}is
given by:

V_{darkerror}
= V_{s} R_{a}/(R_{a} + R_{dark})

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 nonzero is an error.
In that situation, the voltage across the photoresistor is given by (and
remember we need to use the light resistance):

V_{lighterror}
= V_{s} R_{light}/(R_{a} +
R_{light})
Now, there is only one thing that can be adjusted
to optimize this situation, and the thing you can change is the resistor,
R_{a}.
Previously, we have noted that R_{a} 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 R_{a} so that we have the best
situation. Ideally, we would like to make both errors zero, but since
R_{a} has to be finite (nonzero 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 R_{a}.
The "best" value for R_{a} 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 R_{a}.

(Analytical Method)
Using
the analytical expression for TotalError as a function of R_{a},
differentiate and find the value of R_{a} 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:

V_{s} =
5v,

R_{dark}
= 500,000W
(500kW),

R_{light}
= 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 R_{a} + R_{ab}
= R_{total} some constant value that is the resistance you would
measure if you measured the resistance from endtoend 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
Send
your comments on these lessons.