Voltage Divider Circuit Diagram
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A voltage divider circuit diagram shows two resistors connected in series across a supply voltage, with the output taken from the junction between them. The output voltage is a fixed fraction of the input, set by the ratio of the two resistors. Voltage dividers are fundamental building blocks used for biasing transistors, setting reference voltages, interfacing sensors, and attenuating signals.
The resistor voltage divider is one of the simplest and most widely used circuits in electronics. Two resistors, R1 (top, connected to Vin) and R2 (bottom, connected to GND), are placed in series. The output voltage Vout is measured across R2.
Core formula: Vout = Vin × R2 / (R1 + R2). This equation shows that Vout is always less than Vin (for a two-resistor divider with no negative supply) and is independent of supply current — it depends only on the resistance ratio.
Example: Vin = 12 V, R1 = 8 kΩ, R2 = 4 kΩ. Vout = 12 × 4/(8+4) = 12 × 4/12 = 4 V. The circuit divides the supply in the ratio 2:1.
Loading effect: The formula above assumes no current is drawn from Vout. When a load resistance RL is connected, the effective lower resistance becomes R2 in parallel with RL: R2_eff = R2×RL/(R2+RL). The loaded output voltage is Vout(loaded) = Vin × R2_eff / (R1 + R2_eff). To minimise loading error, choose R1 and R2 values at least 10× smaller than RL. In practice, a rule of thumb is: the current through the divider (Vin/(R1+R2)) should be at least 10× the maximum load current.
Potentiometer as a variable voltage divider: A potentiometer is a three-terminal resistor with a sliding wiper. Connecting both ends to Vin and GND and taking the wiper output gives a continuously variable Vout from 0 V to Vin. Potentiometers are used for volume controls, brightness adjustment, and manual calibration.
Design considerations: For a given Vout/Vin ratio, the absolute values of R1 and R2 affect power dissipation and loading. High-value resistors waste less power but are more susceptible to loading and noise. Low-value resistors drive loads well but consume more quiescent current from the supply.
Power dissipation: Total power dissipated = Vin² / (R1+R2). For a 12 V supply with R1+R2 = 1 kΩ, P = 144/1000 = 144 mW. For battery-powered designs, increase R1+R2 to 100 kΩ or more to save power.
Frequency response: At high frequencies, stray capacitance across R2 causes Vout to rise (capacitive bypassing). Adding a small capacitor across R1 creates a compensated attenuator (as used in oscilloscope probes) where C1/C2 = R2/R1, making the response flat across all frequencies.
Common applications: ADC input scaling (e.g. reducing a 12 V sensor output to the 3.3 V range of a microcontroller ADC), transistor base bias (R1 from VCC and R2 to GND set the quiescent base voltage), resistive sensor interfaces (replacing R2 with an NTC thermistor or LDR produces a temperature- or light-sensitive voltage output), and reference voltage generation.
Sensor interface example: Replace R2 with an NTC thermistor. At 25°C, NTC resistance equals a nominal value (e.g. 10 kΩ). As temperature rises, the thermistor resistance falls, Vout falls, and a microcontroller ADC reads a lower voltage corresponding to higher temperature. Calibration maps ADC counts to temperature.
Three-resistor dividers and R-2R ladders: For more precise fractions or digital-to-analog conversion (DAC), multiple resistors in series or the R-2R ladder network extend the divider concept to produce multiple reference voltages or weighted output currents.
Build and experiment with a voltage divider on circuitdiagrammaker.com — the free browser-based circuit editor lets you place resistors, adjust values, and see Vout calculations instantly.
How to wire voltage divider circuit diagram
- Define the required output voltage Determine Vout and Vin. Calculate the required ratio k = Vout/Vin.
- Choose resistor values Select R2 = k × (R1+R2) and R1 = (1−k) × (R1+R2). For no loading error, ensure R1+R2 is at least 10× smaller than the load resistance RL.
- Check power dissipation Calculate P = Vin²/(R1+R2) and verify it is within the resistors' power rating (typically 250 mW for 1/4 W resistors).
- Assemble the circuit Connect R1 between Vin and the output node, and R2 between the output node and GND on a breadboard or PCB.
- Measure unloaded Vout Use a voltmeter to verify Vout matches the formula. Adjust resistor values if measurement deviates by more than the resistor tolerance.
- Connect the load and re-measure Attach the intended load. If Vout drops significantly (> 5%), reduce R1 and R2 proportionally to stiffen the divider.
- Add compensation capacitor if needed For high-frequency or oscilloscope probe applications, add C1 across R1 such that C1 = C2 × R2/R1 to achieve a frequency-flat attenuator.
Specifications
| Output voltage formula | Vout = Vin × R2 / (R1 + R2) |
|---|---|
| Loaded output voltage | Vout = Vin × (R2||RL) / (R1 + R2||RL) |
| Total current (no load) | I = Vin / (R1 + R2) |
| Power dissipation | P = Vin² / (R1 + R2) |
| Ratio rule for loading | R1, R2 ≤ RL / 10 for < 10% error |
| Potentiometer wiper output range | 0 V to Vin (continuously variable) |
| Compensated attenuator condition | C1/C2 = R2/R1 (flat frequency response) |
| Typical divider resistor range | 1 kΩ – 1 MΩ depending on load and power budget |
| Accuracy depends on | Resistor tolerance (use 1% metal-film for precision) |
| Common use: ADC scaling | e.g. 12 V to 3.3 V: R1 = 8.7 kΩ, R2 = 3.3 kΩ |
Safety warnings
- Ensure the total power dissipated by R1 and R2 does not exceed their wattage rating — use P = Vin²/(R1+R2) and select resistors with at least 2× the calculated dissipation.
- Never use a resistor voltage divider as a power supply rail for a circuit; use a proper voltage regulator to avoid unpredictable behaviour under varying load conditions.
Tools needed
- Two resistors (values selected by calculation)
- Digital multimeter (voltage measurement)
- DC regulated power supply
- Breadboard and connecting wires
- Resistor colour-code chart or LCR meter (for value verification)
Common mistakes
- Using the voltage divider to supply power to a load with significant current draw — the output voltage will collapse under load.
- Ignoring the loading effect when connecting an ADC or transistor base — always verify that the input impedance of the load is at least 10× the parallel combination R1||R2.
- Choosing excessively high resistor values (e.g. 1 MΩ) in environments with electrical noise, causing the output to pick up interference.
- Swapping R1 and R2 positions so the higher resistor is on the bottom, resulting in Vout = Vin × R1/(R1+R2) instead of the expected ratio.
Troubleshooting
- Vout is lower than the formula predicts
- Cause: Load resistance is pulling down the output voltage (loading effect). Fix: Measure RL with a multimeter and recalculate using Vout = Vin × (R2||RL)/(R1 + R2||RL). Decrease R1 and R2 to stiffen the divider, or add a unity-gain op-amp buffer.
- Vout drifts with temperature
- Cause: Resistors with high temperature coefficients (carbon composition) change value with temperature. Fix: Replace with 1% metal-film resistors which have a temperature coefficient of ±50–100 ppm/°C or better.
- Output voltage is noisy at high frequencies
- Cause: Stray capacitance across R2 causes a rise in Vout at high frequencies. Fix: Add a small capacitor (e.g. 10 pF) across R1 to compensate, satisfying C1×R1 = C2×R2 where C2 is the estimated stray capacitance.
Frequently asked questions
What is the voltage divider formula in a circuit diagram?
The output voltage is Vout = Vin × R2 / (R1 + R2), where R1 is the top resistor (connected to the supply) and R2 is the bottom resistor (connected to ground).
Why does a voltage divider not work well under heavy loads?
A load connected to Vout appears in parallel with R2, reducing the effective lower resistance. This shifts the voltage ratio and pulls Vout lower than the formula predicts. Use a buffer op-amp (voltage follower) after the divider if the load is low-impedance.
Can a voltage divider be used to step down 5 V to 3.3 V for a microcontroller?
Yes. Use R1 = 1.7 kΩ and R2 = 3.3 kΩ (ratio 3.3/5 = 0.66). Ensure the ADC or GPIO input draws negligible current compared to the divider current. For signal lines, this works well; for power supply rails, use a linear regulator instead.
What is the difference between a voltage divider and a potentiometer?
A fixed voltage divider uses two discrete resistors for a set ratio. A potentiometer is a single component with a movable wiper that varies the ratio continuously between 0 and 1, producing an adjustable Vout.
How do I choose resistor values for a voltage divider?
First determine the ratio from Vout/Vin. Then pick absolute values based on: load (make divider current ≥ 10× load current), power dissipation budget, and noise immunity (higher resistance is noisier). A typical starting point is R1+R2 = 10 kΩ to 100 kΩ.
What is a compensated voltage divider?
A compensated divider adds capacitors C1 across R1 and C2 across R2 such that C1×R1 = C2×R2. This cancels the frequency-dependent error caused by stray capacitance, producing a flat response at all frequencies. Oscilloscope 10× probes use this principle.
Is a voltage divider suitable for powering circuits?
No. A voltage divider is only suitable for low-current signal levels. Any significant load current will shift Vout unpredictably. For regulated power, use a linear regulator (e.g. LM7805, LM1117) or a DC-DC converter.