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.
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.
Pop QuizI 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.