Voltage Divider Calculator
Calculate output voltage, current, and power dissipation for resistor divider circuits.
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Voltage Divider
VOUT VS R2
Vout vs R2 (R1 = 10 kΩ fixed)
How does a Voltage Divider work?
When to use this: Use this to design a resistive voltage divider for level-shifting signals, setting bias points in transistor circuits, or scaling voltages for ADC inputs. It is not suitable for powering loads — use a voltage regulator for that.
A voltage divider is two resistors in series, with the output taken at the midpoint. The formula is Vout = Vin × R2 / (R1 + R2). What makes it useful is that the ratio depends only on the proportion between R1 and R2, not their absolute values. A 10kΩ/10kΩ pair and a 100kΩ/100kΩ pair both give exactly 50% of the input — the absolute values only affect how much current the divider draws.
A common use case: your microcontroller's ADC accepts 0–3.3V, but your sensor outputs 0–5V. With R1 = 20kΩ and R2 = 33kΩ you get Vout = 5V × 33/53 ≈ 3.1V, safely within range. This kind of level-shifting is everywhere in mixed-voltage systems. Another example: biasing a BJT base. With R1 = 82kΩ and R2 = 33kΩ on a 12V supply, Vout = 12 × 33/115 ≈ 3.44V, which sets a Q-point for a common-emitter stage.
The load effect is the most common trap. If you connect a load in parallel with R2 — say, a 10kΩ input impedance on a sensor module — it lowers the effective resistance of that branch. The divider now sees a different R2 than you designed for, and Vout shifts down. Rule of thumb: the load impedance should be at least 10× higher than R2. If it isn't, your output will be lower than calculated.
This is exactly why voltage dividers cannot power circuits. Every milliamp drawn by a load changes the effective R2 and therefore changes Vout. A voltage regulator solves this by actively adjusting to maintain constant output regardless of load current. Use dividers for signal scaling, biasing, and sensing — not for powering things.
Designing a divider: start with R2 (typical choice: 10kΩ for a general-purpose divider). Then calculate R1 = R2 × (Vin/Vout − 1). For a 5V to 3.3V converter: R1 = 10k × (5/3.3 − 1) ≈ 5.15kΩ. Round to the nearest E24 value, 5.1kΩ, which gives 3.32V. Close enough for a logic-level interface.
Output Voltage (Vout)
Vout = Vin × R2 / (R1 + R2)Current
I = Vin / (R1 + R2)Key Points
- Vout is always less than Vin (for positive R1, R2)
- The ratio depends only on the resistor values, not absolute values
- Load resistance in parallel with R2 changes the output voltage
- Higher resistance = less current = less power dissipation
Applications
- Level shifting (e.g. 5V to 3.3V logic)
- Biasing transistor base voltage
- ADC reference voltage generation
- Potentiometers and sensor interfaces
Practical Examples
Convert 5V GPIO signal to 3.3V for microcontrollers or ADC inputs. R values in kΩ.
Vout = 5 × 33 / (20+33) = 3.11 V — suitable for 3.3V logic
Scale a 12V supply down to ~5V for ADC reference voltage measurement.
Vout = 12 × 10 / (14+10) = 5.0 V — precise 5V reference
Set base bias voltage for BJT common-emitter amplifier from 12V supply. R values in kΩ.
Vout = 12 × 33 / (82+33) = 3.45 V — sets Q-point at Vbe + Ve
Common Values Reference
| Use Case | Vin | Vout | R1 | R2 | Ratio |
|---|---|---|---|---|---|
| 5V → 3.3V (logic) | 5 V | 3.30 V | 20 kΩ | 33 kΩ | 0.66 |
| 5V → 2.5V (ADC midpoint) | 5 V | 2.50 V | 10 kΩ | 10 kΩ | 0.50 |
| 12V → 5V (MCU input) | 12 V | 5.00 V | 14 kΩ | 10 kΩ | 0.42 |
| 12V → 3.3V | 12 V | 3.30 V | 27 kΩ | 8.2 kΩ | 0.28 |
| Precision 10:1 divide | 10 V | 1.00 V | 90 kΩ | 10 kΩ | 0.10 |
The load effect: why your measured voltage is lower than calculated
When you connect anything to the output of a voltage divider, it becomes part of the circuit. The load resistance RL appears in parallel with R2, lowering the effective bottom resistance. The real output voltage is:
Vout_loaded = Vin × (R2 ∥ RL) / (R1 + (R2 ∥ RL))Where R2 ∥ RL = (R2 × RL) / (R2 + RL). As a rule of thumb: if RL is 10× greater than R2, the error is about 9%. If RL is 100× greater, the error is about 1%. For critical measurements, buffer the divider output with an op-amp voltage follower — its input impedance is typically >1MΩ and has essentially no loading effect.
Why you can't use a voltage divider to power a circuit
It seems logical: you have 12V and you need 5V — just use a voltage divider. The problem is that a voltage divider is a fixed-ratio circuit, not a regulated source. Any current drawn by the load changes the effective R2, which changes Vout. The more current the load draws, the more Vout drops.
Example: you set up a 12V → 5V divider with R1=14kΩ and R2=10kΩ. No-load Vout = 5.0V. Now you connect a 1kΩ load (5mA): R2 effective = 10k∥1k = 909Ω → Vout drops to 3.5V. Connect a 500Ω load: Vout drops to 2.6V. The output collapses. Use a 78L05 or LM317 linear regulator instead — these actively maintain constant output.
Practical application examples
Scale a 5V signal down to 3.3V for a 3.3V microcontroller ADC input. Keep total resistance high enough not to load the source.
R1=1kΩ, R2=2kΩ → Vout = 5V × 2/(1+2) = 3.33V ✓
Interface a 5V sensor output with a 1.8V GPIO on a low-power MCU without a dedicated level-shifter IC.
R1=1.8kΩ, R2=1kΩ → Vout = 5V × 1/2.8 = 1.79V ✓
Monitor a 12V battery pack with a 3.3V ADC. Full scale (12V) maps to just under 3.3V, preserving the full ADC range.
R1=27kΩ, R2=8.2kΩ → Vout = 12V × 8.2/35.2 = 2.79V (within 3.3V range) ✓
Design tip: impedance and loading
Keep R1+R2 below 10% of the load resistance to avoid significant loading errors. For high-impedance ADC inputs (>1MΩ), resistor values of 10kΩ–100kΩ work well — high enough to limit quiescent current, low enough to drive the ADC input capacitance cleanly.
Step-by-step design guide
- Define Vin, Vout, and the minimum expected load impedance (RL_min).
- Choose R2 such that the divider current is at least 10× the maximum load current. If your load draws up to 100µA, the divider should source at least 1mA, so R1+R2 ≤ Vin/1mA. With 5V input: R1+R2 ≤ 5kΩ.
- Calculate R1 = R2 × (Vin/Vout − 1). Example: Vin=5V, Vout=3.3V, R2=10kΩ → R1 = 10k × (5/3.3 − 1) = 5.15kΩ → use 5.1kΩ (E24).
- Verify power dissipation: P_total = Vin² / (R1+R2). For 5V with 15.1kΩ total: P = 25/15100 = 1.66mW — well within 1/4W limits.
- Use 1% resistors if accuracy matters. Two 5% resistors in a divider can give up to ±10% error in Vout.
Did you know? Voltage dividers are the backbone of ADC input conditioning in every microcontroller project. They are also used inside every digital multimeter — a 10 MΩ input impedance is built from a precision resistor divider network.