Voltage Dividers
The voltage divider is almost certainly the most frequently used two-resistor circuit in all of electronics. Every transistor amplifier bias network, every attenuator pad, every volume control, every reference voltage generator, and every signal level-shifter is built from the same fundamental principle: two resistors in series, with the output taken from the midpoint. Master the voltage divider and you will recognise it instantly inside any schematic — and you will be able to calculate every quantity that matters in under a minute.
The voltage divider is simple to understand but it has one critical pitfall that even experienced engineers fall into: the loading effect. When you connect anything to the divider output, it changes the output voltage — sometimes dramatically. Understanding why this happens and how to prevent it is just as important as knowing the formula itself. This lesson covers both.
- The Voltage Divider Formula
- Where the Formula Comes From
- Worked Example 1: Basic Divider
- Designing a Voltage Divider for a Specific Output
- Worked Example 2: Bias Divider Design
- The Loading Effect — The Critical Pitfall
- Worked Example 3: Loading Effect with Numbers
- Potentiometers: The Adjustable Voltage Divider
- Voltage Divider Calculator
- Experiment: Build and Load a Voltage Divider
- Voltage Dividers in Ham Radio
The voltage divider: R1 and R2 in series, with the output taken across R2. The fraction of input voltage at the output is determined by the ratio R2/(R1+R2).
View LargerThe Voltage Divider Formula
When two resistors are connected in series across a voltage source Vin, the voltage across the lower resistor (R2), measured from the midpoint junction to ground, is:
This formula says that Vout is a fraction of Vin. The fraction is the ratio of the bottom resistor (R2) to the total resistance (R1 + R2). This fraction is always between 0 and 1, so Vout is always less than Vin — you can only divide voltage down, never up, with a passive resistor divider.
Two limiting cases make the formula intuitive: if R2 = 0 (short circuit to ground), Vout = 0. If R1 = 0 (direct connection from Vin to output node), Vout = Vin. Between these extremes, adjusting the ratio of R2 to the total gives any fraction between 0 and 1.
Similarly, you can calculate the voltage across R1 (the top resistor) as:
And as a check, Vout + VR1 = Vin (by KVL). The two formulas are symmetric — each gives the voltage across the resistor whose name appears in the numerator.
Where the Formula Comes From
The formula follows directly from Ohm's Law and the series circuit rules. The derivation is short but worth understanding fully — it shows you that the formula is not a special trick but just Ohm's Law applied twice.
Since R1 and R2 are in series, the same current I flows through both:
I = Vin / (R1 + R2)
The voltage across R2 is this current times R2:
Vout = I × R2 = [Vin / (R1 + R2)] × R2
Vout = Vin × R2 / (R1 + R2)
The word "divider" refers to the fact that the output is a divided fraction of the input. The ratio R2/(R1+R2) is the division factor.
This derivation also shows that the formula assumes the same current flows through both resistors — which is only true if nothing else connects to the midpoint node to draw additional current. The moment you connect a load at the output, a second current path exists and the simple formula no longer applies directly. This is the loading effect, explained in detail below.
Worked Example 1: Basic Divider
Vout:
Vout = 12 × 3900 / (8200 + 3900) = 12 × 3900 / 12100 = 3.868 V ≈ 3.87 V
Current through the divider (quiescent current):
IQ = Vin / (R1 + R2) = 12 / 12100 = 0.9917 mA ≈ 0.99 mA
Power dissipated:
PR1 = IQ² × R1 = (0.0009917)² × 8200 = 8.07 mW
PR2 = IQ² × R2 = (0.0009917)² × 3900 = 3.83 mW
Ptotal = Vin × IQ = 12 × 0.0009917 = 11.9 mW ✓
Check: VR1 = 12 − 3.87 = 8.13 V; VR1 + Vout = 8.13 + 3.87 = 12.00 V ✓
Designing a Voltage Divider for a Specific Output
Given Vin and Vout, you have two unknowns (R1 and R2) and one equation (the divider formula). You need one additional constraint. Common choices are:
Option 1: Fix the quiescent current IQ. IQ determines total power consumption and how stiff the divider is against loading. A typical rule: IQ should be at least 10× the expected load current to keep loading error below ~10%. Set Rtotal = Vin / IQ, then split into R2 = Rtotal × (Vout / Vin) and R1 = Rtotal − R2.
Option 2: Choose R2 first, then solve for R1. Rearranging the divider formula: R1 = R2 × (Vin / Vout − 1). Pick R2 from a standard value, calculate R1, and round to the nearest standard value. Verify with the forward calculation.
Option 3: Fix the resistance ratio. The ratio R2 / R1 = Vout / (Vin − Vout). Compute this ratio and choose resistor values that achieve it while staying within power and tolerance constraints.
Worked Example 2: Bias Divider Design
Design a voltage divider to produce a 3.3 V base bias voltage from a 12 V supply for an NPN transistor amplifier stage. The transistor's base current is expected to be approximately 20 µA. Design the divider so that the bleeder current is at least 10× the base current.
IQ(min) = 10 × IB = 10 × 20 µA = 200 µA
Step 2 — Find the total divider resistance:
Rtotal = Vin / IQ = 12 / 0.000200 = 60 kΩ
Step 3 — Find R2:
R2 = Rtotal × (Vout / Vin) = 60 kΩ × (3.3 / 12) = 60 kΩ × 0.275 = 16.5 kΩ → use 15 kΩ (nearest standard)
Step 4 — Find R1:
R1 = Rtotal − R2 = 60 kΩ − 15 kΩ = 45 kΩ → use 47 kΩ (nearest standard)
Step 5 — Verify:
Vout = 12 × 15000 / (47000 + 15000) = 12 × 15000 / 62000 = 2.903 V
This is close to 3.3 V (within 12%) — acceptable for a bias network. For tighter accuracy, use 43 kΩ for R1: Vout = 12 × 15 / 58 = 3.103 V. Or use 39 kΩ: Vout = 12 × 15 / 54 = 3.333 V ≈ 3.3 V ✓
The Loading Effect — The Critical Pitfall
The voltage divider formula is only exact when no current is drawn from the output. The moment any load connects to Vout, it draws current — current that also flows through R1. This extra current increases the drop across R1, reducing the voltage that remains at the output node. The output voltage sags below the formula prediction. This sag is called the loading effect.
Physically, the load resistor RL connects from the output node to ground — directly in parallel with R2. The parallel combination R2 ∥ RL is always less than R2 alone. Substituting the parallel combination into the divider formula:
Reff = R2 ∥ RL = (R2 × RL) / (R2 + RL)
Vout(loaded) = Vin × Reff / (R1 + Reff)
The loading effect: connecting a load (RL) puts it in parallel with R2, reducing the effective lower resistance and causing Vout to sag below the unloaded prediction.
View LargerThe practical design rule to minimise loading error is:
For loading error below 1%, choose RL ≥ 100 × R2.
This rule follows from the parallel combination formula: if RL = 10 × R2, then Reff = (10R2 × R2) / (10R2 + R2) = 10R2²/11R2 = 10R2/11 = 0.909 × R2 — about 9% less than R2. Substituting back into the divider formula shows approximately 9–10% sag in Vout.
When you cannot make the load impedance ten times larger than R2, you have two options: make R1 and R2 smaller (increase IQ) to reduce the source impedance, or use a buffer amplifier (voltage follower) between the divider output and the load. A voltage follower presents a very high impedance to the divider output (minimal loading) while providing a low-impedance drive to the load.
Worked Example 3: Loading Effect with Numbers
Unloaded Vout:
Vout = 9 × 4700 / (10000 + 4700) = 9 × 4700 / 14700 = 2.878 V
With RL = 47 kΩ (10× R2):
Reff = (4700 × 47000) / (4700 + 47000) = 220,900,000 / 51700 = 4273 Ω
Vout = 9 × 4273 / (10000 + 4273) = 9 × 4273 / 14273 = 2.694 V
Loading error = (2.878 − 2.694) / 2.878 = 6.4%
With RL = 4.7 kΩ (equal to R2 — severe loading):
Reff = (4700 × 4700) / (4700 + 4700) = 22,090,000 / 9400 = 2350 Ω
Vout = 9 × 2350 / (10000 + 2350) = 9 × 2350 / 12350 = 1.713 V
Loading error = (2.878 − 1.713) / 2.878 = 40.5%
With RL = 470 kΩ (100× R2 — negligible loading):
Reff = (4700 × 470000) / (4700 + 470000) = 4700 × 470000 / 474700 = 4653 Ω
Vout = 9 × 4653 / (10000 + 4653) = 9 × 4653 / 14653 = 2.858 V
Loading error = (2.878 − 2.858) / 2.878 = 0.7% — virtually negligible.
This example shows starkly how critically load impedance matters. The same divider, with loads ranging from 4.7 kΩ to 470 kΩ, produces outputs ranging from 1.71 V to 2.86 V — a range of 1.15 V even though Vin has not changed. This is why multimeter input impedance matters (the meter loads the circuit you are measuring), why buffer amplifiers are used between bias networks and following stages, and why power supply output impedance is specified.
Potentiometers: The Adjustable Voltage Divider
A potentiometer is a resistor with a sliding contact (the wiper) that moves along the resistive track. Connecting the wiper between the two end terminals creates an adjustable voltage divider: rotating the knob changes the ratio of the resistance below the wiper to the total resistance, continuously varying Vout from 0 V to Vin.
If the potentiometer has a total resistance of Rpot and the wiper is at fraction α (0 to 1) of the total travel from the ground end, then:
Rbelow wiper = α × Rpot
Rabove wiper = (1 − α) × Rpot
A potentiometer has two common taper characteristics. A linear taper gives a Vout that is directly proportional to shaft rotation — useful for fine-tuning adjustments where you want predictable incremental control. An audio (logarithmic) taper gives a Vout that increases logarithmically with rotation — matching the ear's perception of loudness (which is also logarithmic). Volume controls in receivers and audio amplifiers almost always use audio taper potentiometers, so that equal turns of the knob give equal changes in perceived volume.
Loading affects potentiometers as well. At the midpoint of a linear pot (α = 0.5), the equivalent Thevenin resistance is Rpot/4 — which can be significant for high-value pots. For a 100 kΩ pot, the output impedance at midpoint is 25 kΩ. Any load below 250 kΩ causes more than 10% loading. This is why audio volume controls are often followed by a buffer amplifier (in modern equipment) or rely on the high input impedance of the following stage (in older valve equipment).
Voltage Divider Calculator
Voltage Divider Calculator
Mode A: enter Vin, R1, and R2 to find Vout, quiescent current, and loading effect. Mode B: enter Vin, Vout, and R2 to find the required R1. Optionally enter a load resistance to see the loading effect.
⚖ Experiment: Build and Load a Voltage Divider
Build a voltage divider, measure the unloaded output, then add increasing loads to directly observe and quantify the loading effect. This experiment teaches the most important practical lesson about voltage dividers.
- 9 V battery with clip connector
- R1: 10 kΩ; R2: 4.7 kΩ
- Load resistors: 470 kΩ, 47 kΩ, 4.7 kΩ
- Breadboard and jumper wires
- Digital multimeter
- Predict first. Use the calculator (Mode A) with Vin = 9 V, R1 = 10 kΩ, R2 = 4.7 kΩ. Calculate Vout for: no load, RL = 470 kΩ, RL = 47 kΩ, RL = 4.7 kΩ. Record all four predicted values.
- Build the divider: battery + → 10 kΩ → node A → 4.7 kΩ → battery −. Measure battery terminal voltage Vbattery with the circuit connected.
- Measure unloaded Vout (red probe at node A, black to battery −). Record. Compare to prediction using your measured battery voltage.
- Add 470 kΩ load from node A to battery −. Measure Vout. Calculate loading error as (unloaded − loaded) / unloaded × 100%.
- Add 47 kΩ load (replace 470 kΩ). Measure and record Vout and loading error.
- Add 4.7 kΩ load (severe loading — RL = R2). Measure Vout. This is the dramatic case.
- Fill in the table below with measured values and compare to predictions.
The unloaded output is approximately 2.88 V (9 × 4700/14700). With 470 kΩ load (100× R2): virtually no change — <1% error. With 47 kΩ load (10× R2): about 6% sag, Vout drops to about 2.69 V. With 4.7 kΩ load (= R2): severe 40% loading error, Vout drops to about 1.71 V. This vividly demonstrates why the 10:1 rule exists and why load impedance always matters.
Voltage Dividers in Ham Radio
Transistor voltage-divider bias
The voltage-divider bias configuration is the standard method for biasing BJT common-emitter stages in nearly all modern transistor amplifiers. R1 (from VCC to base) and R2 (from base to ground) form the divider. The base voltage is VB = VCC × R2 / (R1 + R2), from which VE = VB − 0.65 V and the collector current IC ≈ VE / RE. The bleeder current (IQ) must be at least 5–10 times the maximum base current so that the transistor's loading of the divider output causes negligible shift in VB. If this condition is not met, the operating point drifts with temperature and transistor gain — a common fault in homebrew circuits.
RF attenuator pads
A resistive L-pad attenuator is a voltage divider with both components selected to achieve a specific attenuation ratio while presenting a correct impedance at both input and output ports. For a 50 Ω system requiring 6 dB attenuation (a voltage ratio of 2:1), the series arm and shunt arm are chosen so that the divider output is exactly half the input voltage while the input impedance is 50 Ω. The voltage divider formula directly gives the attenuation: Vout/Vin = Rshunt / (Rseries + Rshunt) = 0.5 for 6 dB.
Volume and AF gain controls
The AF gain control in any receiver is a potentiometer — an adjustable voltage divider. The incoming audio signal is applied across the pot's full resistance, and the wiper taps off the desired fraction. Audio taper (logarithmic law) pots are used so that equal angular rotation produces equal changes in perceived loudness. The pot's value is chosen to match the output impedance of the preceding stage and the input impedance of the following stage, with loading effects in mind. A typical audio pot is 10–50 kΩ, balanced between minimising loading of the source and avoiding self-loading at the wiper.
S-meter scaling and AGC voltage division
The S-meter in a receiver is driven from the AGC (Automatic Gain Control) voltage, which may range from −3 V to −12 V as signal strength changes. A voltage divider scales this range to the meter's full-scale deflection voltage (typically 100–500 µA through a defined resistance). The divider ratio is chosen to map the full AGC range to the meter's full-scale deflection, and the total divider resistance must be much higher than the meter resistance to avoid meter loading of the AGC line.
Frequently Asked Questions
Can a voltage divider be used to produce any voltage from any supply?
A passive voltage divider can only produce voltages between 0 V and Vin — it can only divide, not boost. To produce a voltage higher than Vin, you need an active circuit (DC-DC boost converter or charge pump). To produce a negative voltage from a positive supply, you need an inverting converter. For voltages between 0 and Vin, a divider works — but only if the output current requirement is small compared to the divider's quiescent current. For regulated outputs under variable load, use a voltage regulator IC instead.
What is the Thevenin equivalent of a voltage divider?
Thevenin's theorem (covered in a later lesson) says any linear network can be replaced by a single voltage source in series with a single resistance, as seen from the output terminals. For a voltage divider: Vth = Vin × R2/(R1+R2) — the unloaded output voltage. Rth = R1 ∥ R2 — R1 and R2 in parallel. This Thevenin resistance is the source impedance that any load "sees." If RL ≫ Rth, loading is negligible. If RL is comparable to Rth, the loaded output is Vth × RL / (Rth + RL) — another voltage divider, this time between Rth and RL.
How does temperature affect a voltage divider's output?
If both resistors are from the same material (e.g. both metal film with the same temperature coefficient), their individual resistances both change by the same fraction with temperature — so the ratio R2/(R1+R2) stays nearly constant. This is why precision reference voltage dividers use matched resistors, often in the same package where both resistors track together. A mismatch in temperature coefficient between R1 and R2 causes the division ratio to drift with temperature, and hence Vout drifts — a common cause of thermal instability in bias networks.
Why is there a "top" and "bottom" resistor by convention?
By convention, R1 is the resistor connected between the higher-potential terminal (Vin) and the output node, and R2 is connected between the output node and the lower-potential terminal (ground). The output voltage is taken across R2 — the lower, grounded resistor. The formula Vout = Vin × Rbottom / (Rtop + Rbottom) always applies regardless of naming convention. In some texts and schematics you may see the positions swapped or the labels reversed — always check which resistor connects to ground and which connects to the supply to apply the formula correctly.
What happens if I need a very precise output voltage from a divider?
Several factors limit divider accuracy: resistor tolerance (typically ±1% for metal film, ±5% for carbon film), temperature drift (if resistors are mismatched), and loading. For precision below 0.1%, use precision resistors (0.1% tolerance or better, matched pairs), keep the divider resistance low enough that any expected load does not disturb the output by more than your accuracy budget allows, and consider a trimmer (variable resistor in series with one arm) for final calibration. For better accuracy still, an active voltage reference (such as a bandgap reference IC) provides a thermally stable, load-independent output voltage.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.