understand the principle of a potential divider circuit

Potential Dividers

Learning Objective

By the end of this lesson you should be able to:

• Explain the principle of a potential divider circuit.
• Derive the expression for the output voltage of a two‑resistor divider.
• Apply the formula to calculate resistor values for a required output voltage.
• Identify the effect of loading the divider and how to minimise it.

1. Introduction

A potential divider (also called a voltage divider) is a simple linear circuit that produces a fraction of an input voltage. It is widely used in measurement, biasing of transistors, and as a reference voltage source.

2. Theory

Consider two resistors $R_1$ and $R_2$ connected in series across a source voltage $V_{\text{s}}$. The same current $I$ flows through both resistors because they are in series.

$$I = \frac{V_{\text{s}}}{R_1 + R_2}$$

The voltage drop across each resistor is given by Ohm’s law:

$$V_{R_1} = I R_1,\qquad V_{R_2} = I R_2$$

The output voltage $V_{\text{out}}$ is taken across $R_2$ (or $R_1$ depending on the application). Substituting for $I$ gives the classic divider formula:

$$V_{\text{out}} = V_{\text{s}} \frac{R_2}{R_1 + R_2}$$

Similarly, if the output is taken across $R_1$:

$$V_{\text{out}} = V_{\text{s}} \frac{R_1}{R_1 + R_2}$$

3. Derivation Step‑by‑Step

1. Write the total resistance of the series combination: $R_{\text{total}} = R_1 + R_2$.
2. Apply Kirchhoff’s voltage law (K \cdot L) around the loop: $V_{\text{s}} = V_{R_1} + V_{R_2}$.
3. Express each voltage drop using Ohm’s law: $V_{R_1}=IR_1$, $V_{R_2}=IR_2$.
4. Solve for the current: $I = V_{\text{s}}/(R_1+R_2)$.
5. Substitute $I$ back into $V_{R_2}=IR_2$ to obtain $V_{\text{out}} = V_{\text{s}}R_2/(R_1+R_2)$.

4. Example Calculations

Suppose $V_{\text{s}} = 12\ \text{V}$ and we require $V_{\text{out}} = 5\ \text{V}$ across $R_2$. Choose $R_1 = 1.0\ \text{k}\Omega$. Find $R_2$.

$$5 = 12 \frac{R_2}{1.0\text{k} + R_2}$$

Rearranging:

$$5(1.0\text{k}+R_2)=12R_2$$ $$5\,000 + 5R_2 = 12R_2$$ $$5\,000 = 7R_2$$ $$R_2 \approx 714\ \Omega$$

Resulting divider:

Component	Value	Purpose
$R_1$	1.0 kΩ	Series resistor limiting current
$R_2$	≈ 714 Ω	Provides the required 5 V output

5. Loading Effect

If a load resistance $R_{\text{L}}$ is connected across the output, the effective resistance across $R_2$ becomes the parallel combination:

$$R_{\text{eq}} = \frac{R_2 R_{\text{L}}}{R_2 + R_{\text{L}}}$$

The output voltage then becomes:

$$V_{\text{out}} = V_{\text{s}} \frac{R_{\text{eq}}}{R_1 + R_{\text{eq}}}$$

To minimise loading, design the divider so that $R_{\text{L}} \gg R_2$ (typically at least ten times larger).

6. Practical Considerations

• Use resistors with a tolerance of 1 % or better for precise voltage references.
• Consider power dissipation: $P = I^2 R$ for each resistor.
• For high‑impedance loads, buffer the divider with an op‑amp voltage follower.
• Temperature coefficients can cause drift; select low‑TC resistors for stable applications.

7. Common Mistakes

• Assuming the output voltage is independent of the load – it is not unless the load is much larger than $R_2$.
• Using resistors that are too small, leading to excessive power loss and heating.
• Confusing the positions of $R_1$ and $R_2$ in the formula; remember $V_{\text{out}}$ is taken across the resistor that is *below* the output node.

8. Summary

The potential divider is a fundamental circuit that produces a predictable fraction of an input voltage. Its output is given by $V_{\text{out}} = V_{\text{s}}\,R_2/(R_1+R_2)$. The accuracy of the output depends on resistor values, tolerance, and the effect of any load connected to the output. Proper design ensures the divider provides a stable reference voltage for a wide range of A‑Level physics experiments and electronic applications.

9. Practice Questions

1. Design a divider that gives $3.3\ \text{V}$ from a $9\ \text{V}$ supply using standard resistor values. Show your calculations.
2. A divider uses $R_1 = 2.2\ \text{k}\Omega$ and $R_2 = 1.0\ \text{k}\Omega$. What is the output voltage? If a $5\ \text{k}\Omega$ load is connected, what is the new output voltage?
3. Explain why a voltage follower (buffer) is often placed after a potential divider in precision circuits.