Kalkulator Pembahagi Voltan – Voltan Keluaran & Nilai Perintang
Kira voltan keluaran, nilai perintang, dan arus untuk litar pembahagi voltan. Kalkulator elektronik dalam talian percuma untuk keputusan segera.
What Is a Voltage Divider?
A voltage divider is one of the simplest and most useful circuits in electronics — two resistors in series that split a voltage into a smaller fraction. If you need 3.3V from a 5V supply, or you want to scale a sensor output for an ADC input, a voltage divider is your go-to solution.
The output voltage formula is: Vout = Vin × R2 / (R1 + R2)
Where R1 is the top resistor (between Vin and Vout), and R2 is the bottom resistor (between Vout and ground). The ratio R2/(R1+R2) determines what fraction of the input appears at the output.
Voltage dividers are fundamental building blocks taught in every introductory electronics course. They appear in countless applications — from simple reference voltage generation to complex analog signal conditioning chains. Understanding their behavior, limitations, and design tradeoffs is essential for anyone working with circuits.
Voltage Divider Formulas
| Find | Formula | Given |
|---|---|---|
| Output voltage | Vout = Vin × R2 / (R1 + R2) | Vin, R1, R2 |
| R1 (top resistor) | R1 = R2 × (Vin/Vout − 1) | Vin, Vout, R2 |
| R2 (bottom resistor) | R2 = R1 × Vout / (Vin − Vout) | Vin, Vout, R1 |
| Current through divider | I = Vin / (R1 + R2) | Vin, R1, R2 |
| Power dissipated | P = Vin² / (R1 + R2) | Vin, R1, R2 |
Example: Vin = 12V, R1 = 10kΩ, R2 = 5kΩ: Vout = 12 × 5000/(10000+5000) = 12 × 0.333 = 4V. Current = 12/15000 = 0.8mA. Power = 12²/15000 = 9.6mW.
Deriving the Voltage Divider Formula
The voltage divider formula comes directly from Ohm's Law and Kirchhoff's Voltage Law (KVL). Here's the derivation:
- R1 and R2 are in series, so the same current flows through both: I = Vin / (R1 + R2)
- The voltage across R2 (which is Vout) follows Ohm's Law: Vout = I × R2
- Substituting: Vout = (Vin / (R1 + R2)) × R2 = Vin × R2 / (R1 + R2)
This assumes no current is drawn from the output (unloaded divider). In practice, any load connected to Vout draws current, which changes the effective resistance and reduces Vout — this is the loading effect, covered below.
Choosing Resistor Values
Resistor value selection involves tradeoffs between power consumption, loading effects, and noise:
- Low resistance values (100Ω – 1kΩ): Minimal loading effect, but high current and power dissipation. Good for power-critical circuits.
- Mid-range values (1kΩ – 100kΩ): Best balance for most applications. Common for sensor interfaces and logic level shifting.
- High resistance values (100kΩ – 1MΩ): Very low current draw, but susceptible to noise and loading from downstream circuits.
For microcontroller ADC inputs, use 10kΩ–100kΩ total resistance. If your load draws significant current (< 10× the divider current), a voltage divider alone won't maintain stable output — use a voltage regulator or op-amp buffer instead.
Standard Resistor Values (E24 and E96 Series)
Real resistors come in standard values defined by the IEC 60063 standard. You cannot buy an arbitrary resistance — you must select from the available series or combine resistors.
| Series | Tolerance | Values per Decade | Examples (1kΩ decade) |
|---|---|---|---|
| E12 | ±10% | 12 | 1.0k, 1.2k, 1.5k, 1.8k, 2.2k, 2.7k, 3.3k, 3.9k, 4.7k, 5.6k, 6.8k, 8.2k |
| E24 | ±5% | 24 | Adds 1.1k, 1.3k, 1.6k, 2.0k, 2.4k, 3.0k, 3.6k, 4.3k, 5.1k, 6.2k, 7.5k, 9.1k |
| E96 | ±1% | 96 | Fine-grained: 1.00k, 1.02k, 1.05k, 1.07k, 1.10k, … |
When designing a voltage divider, pick the closest standard values and verify the output is within your tolerance. For example, to get 3.3V from 5V:
- Ideal ratio: R2/(R1+R2) = 3.3/5 = 0.66
- With E24 values: R1 = 5.1kΩ, R2 = 10kΩ → Vout = 5 × 10/(5.1+10) = 3.311V (0.3% error) ✓
- Alternative: R1 = 1kΩ, R2 = 2kΩ → Vout = 5 × 2/3 = 3.333V (1% error) ✓
Common Voltage Divider Applications
| Application | Example | Notes |
|---|---|---|
| Logic level conversion | 5V to 3.3V for Arduino/ESP32 | Use resistors; consider bidirectional level shifter for signals |
| Sensor scaling | 0–10V sensor → 0–3.3V ADC | Ratio should match sensor range to ADC range |
| Battery voltage monitoring | 12V battery → 3.3V ADC pin | Add decoupling capacitor at output |
| Bias voltage generation | Setting op-amp reference voltage | Use large capacitor for low noise |
| Potentiometer | Variable voltage divider | R2 is variable, allows Vout adjustment |
Design Example: Battery Voltage Monitor
A common project is monitoring a 12V car battery with a 3.3V microcontroller ADC. Here's the complete design process:
- Requirement: Scale 0–15V (max battery voltage during charging) to 0–3.3V
- Divider ratio: 3.3/15 = 0.22
- R2/(R1+R2) = 0.22, so R1/R2 = (1−0.22)/0.22 = 3.545
- Choose R2 = 10kΩ, then R1 = 35.45kΩ → nearest E96 = 35.7kΩ
- Verify: Vout = 15 × 10/(35.7+10) = 3.279V ✓ (under 3.3V, safe for ADC)
- Current: I = 15/45.7kΩ = 0.328mA (negligible drain on battery)
- Add 100nF ceramic capacitor across R2 to filter noise
- Add Zener diode (3.3V) across R2 for overvoltage protection
The capacitor and Zener are essential safety measures. Without the Zener, a voltage spike on the car's electrical system could destroy the microcontroller's ADC pin. The capacitor filters high-frequency noise from the alternator and ignition system.
Design Example: 5V to 3.3V Logic Level
Converting a 5V output to a 3.3V input is the most common voltage divider application in embedded systems:
| R1 | R2 | Vout at 5V | Error | Current | Suitability |
|---|---|---|---|---|---|
| 1kΩ | 2kΩ | 3.333V | +1.0% | 1.67mA | Good for most GPIO/I2C |
| 3.3kΩ | 5.6kΩ | 3.146V | −4.7% | 0.56mA | OK, but on the low side |
| 5.1kΩ | 10kΩ | 3.311V | +0.3% | 0.33mA | Excellent for ADC/slow signals |
| 10kΩ | 20kΩ | 3.333V | +1.0% | 0.17mA | Low power, check signal speed |
| 47kΩ | 100kΩ | 3.401V | +3.1% | 0.034mA | Ultra-low power, slow signals only |
Signal speed consideration: A voltage divider combined with the input capacitance of the receiving pin forms an RC low-pass filter. With R1 = 10kΩ and 15pF input capacitance, the time constant is 0.15μs, limiting clean signal transfer to roughly 1MHz. For SPI (10+ MHz) or fast UART (1+ Mbps), use a dedicated level shifter IC (TXB0108, 74LVC245) instead.
The Loading Effect Explained
When you connect a load resistance (RL) to the output of a voltage divider, RL appears in parallel with R2. The effective bottom resistance becomes:
R2eff = (R2 × RL) / (R2 + RL)
This always reduces Vout. The error depends on the ratio of RL to R2:
| RL / R2 Ratio | Vout Error | Acceptable? |
|---|---|---|
| 100× | <1% | Excellent — negligible loading |
| 10× | ~9% | Marginal — recalculate or buffer |
| 3× | ~25% | Unacceptable — use op-amp buffer |
| 1× | ~50% | Severe — divider is useless here |
Rule of thumb: The load resistance should be at least 10× R2 for the voltage divider to maintain reasonable accuracy. Microcontroller ADC inputs typically have >1MΩ input impedance, so a 10kΩ voltage divider works perfectly. But driving an LED (which draws milliamps) from a voltage divider will fail completely.
Resistor Tolerance and Output Accuracy
Resistor tolerance directly affects the accuracy of your voltage divider output. Worst-case analysis:
| Resistor Tolerance | Worst-Case Vout Error | Series | Cost |
|---|---|---|---|
| ±5% (E24) | Up to ±10% | Carbon film, common | $0.01–0.03 |
| ±1% (E96) | Up to ±2% | Metal film, standard | $0.02–0.05 |
| ±0.1% (E192) | Up to ±0.2% | Precision thin film | $0.10–0.50 |
| ±0.01% | Up to ±0.02% | Ultra-precision | $1.00–5.00 |
For most hobbyist and general-purpose applications, 1% (E96) resistors are the best cost-performance tradeoff. For precision analog design (instrumentation amplifiers, reference circuits), use 0.1% or better, or use a dedicated precision voltage divider IC like the Analog Devices LT5400.
Temperature Effects on Voltage Dividers
Resistor values drift with temperature, characterized by the Temperature Coefficient of Resistance (TCR), measured in ppm/°C:
| Resistor Type | Typical TCR (ppm/°C) | Effect over 50°C range |
|---|---|---|
| Carbon composition | ±1,500 | ±7.5% resistance change |
| Carbon film | ±200 to ±500 | ±1–2.5% |
| Metal film (standard) | ±50 to ±100 | ±0.25–0.5% |
| Precision thin film | ±5 to ±25 | ±0.025–0.125% |
| Wirewound (precision) | ±5 to ±10 | ±0.025–0.05% |
If R1 and R2 are the same type and in thermal contact (both on the same PCB area), their resistances drift together, and the ratio stays relatively constant even as absolute values change. This is called ratio tracking and is a key advantage of matched resistor pairs. For precision work, buy resistor arrays (multiple resistors in one package) that guarantee tight ratio tracking.
Capacitive Voltage Dividers
Just as resistors divide DC voltage, capacitors divide AC voltage. In a capacitive divider, the formula is inverted because capacitive reactance is inversely proportional to capacitance:
Vout = Vin × C1 / (C1 + C2)
Note that C1 is in the numerator (not C2 as in the resistive formula), because the capacitor with less capacitance drops more voltage. A capacitive divider draws no DC current (ideal), making it useful in high-impedance AC measurement circuits.
| Property | Resistive Divider | Capacitive Divider |
|---|---|---|
| Works with DC? | Yes | No (blocks DC) |
| Works with AC? | Yes (frequency-independent) | Yes (frequency-independent ratio) |
| Power dissipation | I²R losses (continuous) | Ideally zero (reactive power) |
| Loading effect | R_load in parallel with R2 | C_load in series with C2 |
| Typical application | DC reference, signal scaling | HV measurement, oscilloscope probes |
Oscilloscope probes combine both: a compensated voltage divider uses matched resistive and capacitive dividers so the attenuation ratio is constant across all frequencies. The adjustable trimmer capacitor on the probe compensates for the input capacitance of the oscilloscope.
Multi-Stage Voltage Dividers
You can cascade voltage dividers for larger attenuation ratios. However, each stage loads the previous one, so the overall ratio is not simply the product of individual ratios unless buffered:
Unbuffered two-stage divider: The second stage's R1+R2 acts as a load on the first stage. The actual output must be calculated using Thévenin equivalent circuits for each stage.
Buffered two-stage divider: An op-amp voltage follower (unity gain buffer) between stages eliminates loading. Stage 1 output feeds the buffer input (essentially infinite impedance), and the buffer output drives Stage 2. This guarantees the overall ratio = Stage 1 ratio × Stage 2 ratio.
For most practical designs, a single-stage divider is sufficient. Multi-stage dividers are mainly used in precision measurement instruments, high-voltage probing, and programmable gain circuits using switched resistor networks.
Common Design Mistakes and How to Avoid Them
| Mistake | Consequence | Solution |
|---|---|---|
| Using voltage divider to power a device | Vout drops under load; device may malfunction or undervolt | Use a voltage regulator (LDO or switching) |
| Ignoring load impedance | Output voltage lower than calculated | Ensure R_load ≥ 10 × R2; add buffer if needed |
| Using 5% resistors for precision | Output may be 10% off target | Use 1% (E96) or better; measure actual resistance |
| No input protection | Voltage spikes destroy downstream IC | Add Zener diode and/or TVS diode across R2 |
| High-value resistors near noisy sources | Noise pickup on Vout | Use lower resistance; add decoupling capacitor |
| Forgetting power rating | Resistor overheats or burns | Calculate P = Vin²/(R1+R2); use appropriately rated parts |
💡 Did you know?
- The potentiometer (variable resistor) is essentially a voltage divider with an adjustable tap — that's why it's also called a "pot."
- A voltage divider only works well as a power source when the load resistance is much larger than the divider resistance. When loaded heavily, Vout drops significantly.
- Voltage dividers are used in virtually every circuit that interfaces high-voltage signals with low-voltage microcontrollers and sensors.
- The Wheatstone bridge — the basis of most strain gauges, load cells, and precision measurement circuits — is essentially two voltage dividers compared against each other. A change in one resistor unbalances the bridge, producing a measurable voltage difference.
Frequently Asked Questions
Can I use a voltage divider to power a device?
Generally no — not for anything that draws significant current. A voltage divider's output voltage drops when current is drawn, because the load effectively becomes R2 in parallel. For powering devices, use a linear voltage regulator (like LM7805) or switching regulator. Voltage dividers are best for setting reference voltages for high-impedance inputs like ADC pins or op-amp inputs.
How do I convert 5V logic to 3.3V?
Use R1 = 1kΩ and R2 = 2kΩ: Vout = 5 × 2/(1+2) = 3.33V. This works for slow signals like I2C or GPIO. For fast signals (SPI, UART at high baud rates), use a dedicated level-shifter chip (like TXB0108 or 74LVC245). The voltage divider method introduces RC delay that can corrupt fast signals.
What resistor tolerance should I use?
For most voltage divider applications, 1% tolerance (E96 series) resistors are recommended. Standard 5% (E24) resistors have enough variation that the actual output voltage could differ by up to 10% from calculated. For precision circuits, use matched 0.1% resistors or a dedicated precision voltage divider IC.
What is the loading effect?
When you connect a load to a voltage divider's output, the load resistance appears in parallel with R2, reducing the effective R2 and thus lowering Vout. The error is negligible when the load resistance is at least 10× larger than R2. Always check that your circuit's input impedance is much higher than R2 for accurate voltage division.
Can I use a voltage divider for AC signals?
Yes, resistive voltage dividers work for AC signals at low frequencies. At higher frequencies, parasitic capacitance becomes significant and you may need a compensated divider (adding capacitors in parallel with the resistors so the AC ratio matches the DC ratio). Oscilloscope probes are a perfect example — a 10:1 probe contains a compensated voltage divider adjusted via a trimmer capacitor.
How do I calculate power dissipation in each resistor?
The current through both resistors is I = Vin/(R1+R2). Power in R1 = I²×R1, power in R2 = I²×R2. Total power = Vin²/(R1+R2). For example: 12V with 10kΩ+5kΩ → I=0.8mA, P_R1=6.4mW, P_R2=3.2mW, total=9.6mW. Standard ¼W (250mW) resistors handle this easily. But at lower resistance (100Ω total on 12V = 1.44W), you need appropriately rated resistors.