Voltage Dividers

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^oC). That means R_a = 5000W

.

• By choosing R_a = 5000W

  we will have an output voltage of 2.5v at room temperature (25^oC).

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 ^oC where the resistance, R_t, is 16,330W

.

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 R_b = 5000W

, so that you can compare things with the previous situation.

Now, let's look at the situation at 50 ^oC 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.)

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^oC, the voltage goes down by 1.176v. When the temperature goes down by 25^oC, 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_ais 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 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):

• 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 (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 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 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.