Analog and Digital Electronics Design

Voltage Divider Design

In this posting, we’re going to learn to design voltage dividers. Typically taught as a shortcut for analysis in introductory circuits, the method is often used to quickly find the voltage drops across series-connected resistors. However, voltage dividers are often used in circuit designs to provide fractions of power supply voltages or to attenuate signal voltages for applications such as biasing or creating ranges for a voltmeter. Therefore, it makes sense for us to develop a synthesis method to quickly design a voltage divider circuit given an input voltage, a desired output voltage, and resistors within a limited range of resistance.

Using Figure 1 as a reference, let’s first analyze the voltage divider, particularly, to find \(V_2\) as a function of \(V_S\), \(R_1\), and \(R_2\). Note, we can easily find \(V_1\) by subtracting \(V_2\) from \(V_S\). First off, we can use Kirchhoff’s Voltage Law (KVL) to sum the voltages around the loop:

\(-V_S+V_1+V_2=0\) or \(V_S=V_1+V_2\).

Knowing that in a series circuit the element current, \(I\), is the same for all elements, and using Ohm’s Law, we can write \(V_S\) as a function of \(I\), \(R_1\), and \(R_2\) and then solve for \(I\):

\(V_S=IR_1+IR_2=I\left(R_1+R_2\right)\) or \(I=\Large{\frac{V_S}{R_1+R_2}}\).

Now, by Ohm’s Law we also know:

\(V_1=IR_1\) and \(V_2=IR_2\).

Let’s now solve for \(V_2\) as a function of \(V_S\):

\( V_2=IR_2 = \left( \Large{\frac{V_S}{R_1+R_2}} \right) R_2\) \(= V_S \Large{\frac{R_2}{R_1+R_2}} \).

We can similarly find \(V_1\) as a function of \(V_S\).

Now, let’s turn things around and rework the equation to synthesize a voltage divider given \(V_S\) and \(V_2\). First let’s go back to the equation for \(V_2\):

\(V_2=V_S\Large{\frac{R_2}{R_1+R_2}}\).

Dividing both sides by \(V_S\):

\(\Large{\frac{V_2}{V_S}}=\Large{\frac{R_2}{R_1+R_2}}\),

and flipping:

\(\Large{\frac{V_S}{V_2}} = \Large{\frac{R_1+R_2}{R_2}} = \Large{\frac{R_1}{R_2}}\)\( +1\).

Solving for the ratio of resistors, we get our synthesis equation:

\(\Large{\frac{R_1}{R_2}}=\Large{\frac{V_S}{V_2}}\)\(-1\).

All we need do is find the values of \(R_1\) and \(R_2\) such that their ratio is \(V_S/V_2\ –1\).

Let’s try an example. Suppose we have \(V_S =\) 12 V and we want \(V_2 =\) 3 V. Then

\(\Large{\frac{R_1}{R_2}}\) = (12 V / 3 V) – 1 = 3.

If we let \(R_1\) = 30 k\(\Omega\), then \(R_2\) = 10 k\(\Omega\). Let’s check our design:

\(V_2\) = 12 V (10 k\(\Omega\)) / (30 k\(\Omega\) + 10 k\(\Omega\)) = 12 V \(\times\) 0.25 = 3 V!

A few details to consider when designing voltage dividers:

1. \(V_2\) must obviously be less than \(V_S\).
2. Standard resistance values of appropriate tolerance can be used for \(R_1\) and \(R_2\) depending on how much error in \(V_2\) is acceptable.
3. The magnitudes of the resistors matter. Smaller resistors conduct more current which negatively affects the power supply budget but the divider is less sensitive to the effects of loading. For instance, in the previous example, a load impedance roughly equal to 10 k\(\Omega\) would change the effective value of \(R_2\) to 5 k\(\Omega\) resulting in:

   \(V_2 = \) (12 V)(5/35) \(\approx \) 1.71 V.

   This is a substantial change. On the other hand, if the load is 1 M\(\Omega\), then \(V_2 \approx\) 2.98 V, which is closer to the desired 3 V.

4. The process can be automated to find the “best” pair of resistors by doing a brute-force search using a Python script for instance. In a later post, I’ll supply a documented script that will find sum and parallel combinations of resistor values to find a desired equivalent resistance and will also design voltage dividers.

Until the next post,

Dr. Dave

14 Oct 2023