After a long hiatus I think it’s about time to get back into some
circuits concepts with these EE Fundamentals posts. Today’s topic is voltage
dividers/voltage division. I know last time I promised to go over RC circuits (“next
time, we solve RC circuits”), but when I started writing that entry I realized
this one would make more sense to put first. We will start easy since we are
just getting back into circuits stuff.

Voltage dividers are not so much a special type of circuit
but more a property of voltage itself. In any closed loop circuit, the sum of
the voltage drops around the loop must equal the voltage being supplied by the
voltage source. For example, if your circuit contains a 10V source and two
resistors in series, the voltage drops across the two resistors will be equal
to 10 volts. Knowing this is a helpful tool for quick circuit analysis because
it lets you determine the voltages, and subsequently the currents, at every
point along the closed loop. As it turns out, voltage tends to split up
according to predictable ratios based on the impedance of the elements in the
circuit under analysis. Engineers use the simple equation below to determine
the voltage across multiple series circuit elements:

where Vx is the voltage, Vs is the source voltage, Rx is the
impedance of the element in question, and the denominator is the sum of the all
the impedances in the circuit. For reference, let’s look at the circuit in
Figure 1 to delve into this concept a little deeper.

**Figure 1. Sample series circuit**

The circuit above shows a 10V source in series with a 2 kohm
and 1 kohm resistor, making the total series resistance 3 kohms. If we want to
find the voltage across the 2 kohm resistor, we use the equation and find that
it is equal to:

You could apply the
same process to find the voltage on the 1 kohm resistor or subtract the voltage on the 2 kohm
resistor from the source voltage and arrive at the same answer. You could also scale
these resistors in any direction and end up with the same **voltage **on each provided that the ratio between the two remains the
same. In other words, there would still be 6.67 volts across a 200k resistor
provided the other resistor is 100k, but the current through the circuit would
be 100 times lower (Ohm’s Law).

In case you still are not convinced, I
built this circuit to show the voltage split between the resistors. As you can
see in picture below, a 10V source places 6.65 volts on the 2 kohm resistor and 3.351
volts on the 1 kohm resistor. Yes, these numbers are slightly different from the
predicted values, but the reason for that is a combination of the voltage
source not being exactly at 10V, the resistors not being precision values, and
the measurement error of the multimeter I am using. For all practical purposes, however, voltage division can be easily proven.

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So how do circuit designers making use of
voltage dividers? Often times you will see voltage dividers used to set
reference voltages or create DC bias conditions in various places throughout a schematic, like in
Figure 2.

**Figure 2. Voltage divider setting comparator input voltage**

Conceptually, the voltage divider resistor network is
very simple and easy to implement. When put into practice though these sorts of
voltage references often suffer from excess power loss and can drift out of
spec with temperature variations. Still, they are widely used and offer
hobbyists an easy way to create a specific voltage for their circuit.

But voltage divider circuits can only be
used as references provided that the load using the reference has a
significantly higher impedance than the divider itself. I believe this is
one of the least talked about subjects in electronics education, but it has a
huge impact on the functionality of analog circuitry and introduces the concept
of loading/attenuation.

Look back at the circuit in Figure 2. Knowing
nothing else about electronics other than what has been discussed in these EE
Fundamentals posts, you would probably expect the voltage applied to the input terminals
of the comparator to be 3.33 volts.
While I am not disputing this assumption, I want to point out that there are
other issues to consider. First, think back to what happens to a circuit when
two resistors are put in parallel: the equivalent resistance drops to a value
lower than either of the original two resistors. If that is true, then consider
Figure 3.

**Figure 3. Voltage divider with low impedance load**

This new circuit is virtually identical
the divider used as a reference input to the comparator. However, instead of
the high impedance of the comparator in parallel with the 1kohm resistor, we
have now put a 10 ohm load instead. Without this load attached to
the voltage divider, we would see roughly 3.33 volts on the
output, but with the 10 ohm load attached the output drops to about 0.05 volts. The
higher the load resistance on the divider, the less change we should see in the
expected value. This, my friends, is an example of loading down a circuit and I
am going to show you how.

************************************

Above is a picture of my build
of the circuit from Figure 3. I have placed a 10 ohm resistor in parallel with
1 kohm resistor. As you can tell from the measurements, the voltage across the 1 kohm
resistor is significantly lower than it was in the previous tests. This is
because we have loaded down the circuit with the 10 ohm load resistor and made
the voltage divider equation:

There is clearly a strong correlation between the expected value and measured result on the 1kohm resistor. The voltage across the 2 kohm resistor
is equal to the source voltage (10V) minus the measured value (0.05V), or roughly 9.95 volts with measurement error.

Conversely, placing a 1megaohm resistor
in parallel with the 1kohm resistor does not significantly affect the
equivalent resistance because it is three orders of magnitude greater than the 1 kohm resistor. For
reference, I measured the effects of this new load resistance. You can see them
in the picture below.

************************************

This demonstration has attempted to make
clear two distinct and often undervalued concepts: loading and attenuation. I “loaded”
the circuit when I placed an additional load resistor onto the output of the
circuit such that it had a significant impact on the circuit behavior. When the circuit was loaded down with the 10 ohm resistor, the output voltage decreased severely compared to the intended output of the voltage divider, which we call attenuation. Hence, when you hear someone say the signal is being "attenuated", it means that there is a loss of fidelity at some point along the circuit resulting in a loss of information or signal strength. This same concept applies to AC waveforms like music streams. If, for instance, you loaded an amplifier circuit with a resistance too small for its output impedance, it would result in a weaker signal and thus a quieter song in your ear.

__Pop Quiz__

I am going to post the quiz for this subject in a separate entry because this one has run a little long and I want to mock up my quiz circuit to make sure it does what I think it should. This quiz is all my own and it requires knowledge from my previous entry level tutorials to figure out. Look for the question before the end of the year.