The fundamental formula for average voltage over a continuous time period is Vavg = (1/T) ∫0T v(t) dt. For discrete digital samples, it is Vavg = (V1 + V2 + ... + Vn) / n. In power electronics, the most frequently applied variant is the formula for average voltage of a full-wave rectified sine wave: Vavg = (2 × Vp) / π. This guide breaks down the exact mathematical derivations, defines every variable, and walks through bench-tested worked examples with strict unit tracking.
The Core Formulas for Average Voltage
Average voltage represents the DC equivalent value of a time-varying signal. It is the area under the voltage-time curve divided by the time period. Depending on your signal source—analog oscilloscope traces, microcontroller ADC readings, or rectified mains power—you will use one of three primary mathematical models.
1. Continuous Time (Calculus Definition):
Vavg = (1 / T) × ∫0T v(t) dt
2. Discrete Samples (Digital/ADC Definition):
Vavg = (1 / n) × Σi=1n Vi
3. Full-Wave Rectified Sine Wave:
Vavg = (2 × Vp) / π ≈ 0.6366 × Vp
4. Half-Wave Rectified Sine Wave:
Vavg = Vp / π ≈ 0.3183 × Vp
Symbol Definition Table
| Symbol | Definition | Standard Unit |
|---|---|---|
| Vavg | Average DC voltage over the defined period | Volts (V) |
| T | Time period of one complete waveform cycle | Seconds (s) |
| v(t) | Instantaneous voltage as a function of time | Volts (V) |
| n | Total number of discrete sample points | Dimensionless (count) |
| Vi | Voltage value of the i-th discrete sample | Volts (V) or ADC counts |
| Vp | Peak amplitude of the sine wave (zero to max) | Volts (V) |
| π | Mathematical constant Pi | ≈ 3.14159 |
Application Rules and Core Assumptions
Applying the wrong formula to the wrong waveform is a primary cause of design failures in power supplies and signal conditioning circuits. Here is when each formula applies and the assumptions baked into the math.
Continuous Integral Formula: Applies to any periodic analog waveform. The core assumption is strict periodicity; the signal must repeat exactly over interval T. If the signal contains sub-harmonics or drift, T must be extended to cover the lowest common frequency.
Discrete Sum Formula: Applies to microcontroller ADC readings (e.g., Arduino analogRead() or ESP32 SAR ADC). The critical assumption is a uniform sampling interval (Δt). If your interrupt service routine suffers from jitter, the arithmetic mean will skew. Furthermore, this formula assumes the samples span an integer multiple of the signal period; otherwise, you will calculate the average of a partial wave, yielding an incorrect DC offset.
Rectified Sine Formulas: Applies strictly to pure sinusoidal inputs post-rectification. The textbook derivation assumes ideal diodes with zero forward voltage drop and zero reverse recovery time. In practical 2026 bench work, silicon diodes (like the 1N4007) drop ~0.7V, and Schottky diodes drop ~0.3V. A full-wave bridge uses two diodes in series per half-cycle, meaning the real-world formula must be adjusted to: Vavg = (2 × (Vp - 2Vf)) / π. Always state your diode assumption when documenting calculations.
Worked Examples with Strict Unit Tracking
Abstract math fails on the workbench. Below are two solved problems demonstrating how to track units from raw data to final voltage.
Problem 1: Discrete ADC Samples from an ESP32
Scenario: You are reading a filtered PWM signal using an ESP32's 12-bit ADC (0-4095 counts) with a 3.3V reference. You take 5 rapid sequential readings: 2048, 2060, 2055, 2050, 2042. Find Vavg.
Step 1: Sum the discrete samples (Unit: counts)
Σ Vi = 2048 + 2060 + 2055 + 2050 + 2042 = 10,255 counts
Step 2: Divide by sample count n (Unit: counts)
Vavg_counts = 10,255 counts / 5 = 2,051 counts
Step 3: Convert to Volts using the ADC resolution multiplier
Resolution = 3.3 V / 4095 counts = 0.0008058 V/count
Vavg = 2,051 counts × 0.0008058 V/count = 1.652 V
Problem 2: Full-Wave Rectified 120V Mains
Scenario: You are designing a linear power supply fed by a 120VRMS 60Hz wall outlet, passing through a KBPC5010 bridge rectifier. Calculate the ideal average voltage, then calculate the real-world average voltage accounting for diode drops at a 5A load.
Step 1: Convert RMS to Peak Voltage (Unit: Volts)
Vp = VRMS × √2
Vp = 120 V × 1.4142 = 169.7 V
Step 2: Calculate Ideal Vavg (Unit: Volts)
Vavg_ideal = (2 × 169.7 V) / π
Vavg_ideal = 339.4 V / 3.14159 = 108.03 V
Step 3: Calculate Real-World Vavg with Diode Drops
The KBPC5010 datasheet specifies a forward voltage drop (Vf) of ~0.8V per diode at 5A. A bridge rectifier conducts through two diodes simultaneously.
Total drop = 2 × 0.8 V = 1.6 V
Adjusted Vp = 169.7 V - 1.6 V = 168.1 V
Vavg_real = (2 × 168.1 V) / π = 336.2 V / 3.14159 = 107.01 V
Rearranged Forms and Algebraic Manipulation
When debugging power supplies or reverse-engineering a transformer, you often know the DC output (Vavg) and need to find the AC peak or RMS. Here are the rearranged forms solving for each critical variable based on the full-wave rectified sine model.
- Solve for Peak Voltage (Vp):
Vp = (Vavg × π) / 2 - Solve for RMS Voltage (VRMS):
Since VRMS = Vp / √2, substituting Vp yields:
VRMS = (Vavg × π) / (2 × √2) ≈ Vavg × 1.1107 - Solve for Peak-to-Peak Voltage (Vpp):
Vpp = 2 × Vp = Vavg × π - Solve for Sample Sum (Discrete Model):
Σ Vi = Vavg × n
Common Unit Mistakes and Magnitude Sanity Checks
Which unit mistakes break the formula? The most fatal error is confusing Peak Voltage (Vp) with Peak-to-Peak Voltage (Vpp). Oscilloscopes default to displaying Vpp. If your scope reads 34Vpp, your Vp is 17V. If you blindly plug 34 into the Vp slot of the rectified formula, your calculated Vavg will be exactly double the true value. Always divide Vpp by 2 before calculating.
Another common failure is mixing ADC counts and Volts in the discrete formula. If you sum raw counts but divide by a time-based denominator instead of the sample count n, the resulting unit is counts/second, which is physically meaningless for DC voltage.
What does a realistic answer magnitude look like?
For any pure sine wave or rectified sine wave, the average voltage must be strictly lower than the RMS voltage. The ratio of RMS to Average is known as the Form Factor. For a sine wave, the form factor is π / (2√2) ≈ 1.11. Therefore, a realistic Vavg magnitude should always be roughly 90% of the VRMS. If you measure 120VRMS and calculate a Vavg of 135V, your math is inverted. According to foundational waveform theory documented by All About Circuits, verifying the form factor is the fastest sanity check on the bench.
Furthermore, for a symmetrical AC waveform centered on zero (like raw mains before a rectifier), the true mathematical average over a full cycle is exactly 0V. This is why we use RMS for AC power calculations, and why the formula for average voltage is almost exclusively applied to rectified or DC-biased signals in practical electronics, as detailed in Electronics Tutorials.
Frequently Asked Questions
What is the formula for average voltage of a square wave?
For a square wave that swings between 0V and Vp with a 50% duty cycle, the formula for average voltage is simply Vavg = Vp × Duty Cycle. At exactly 50%, Vavg = Vp / 2. If the square wave swings symmetrically between -Vp and +Vp, the average voltage over a full period is 0V, regardless of the duty cycle, provided the positive and negative pulse widths are equal.
Why is the average voltage of a pure AC sine wave zero?
The continuous integral of a pure sine wave (v(t) = Vpsin(ωt)) over one full 2π period evaluates to zero because the positive half-cycle area perfectly cancels the negative half-cycle area. The formula for average voltage is only useful for pure AC if you restrict the integration limits to a half-cycle (0 to π), which yields Vavg = (2 × Vp) / π, identical to the full-wave rectified result.
How do I calculate average voltage from an oscilloscope screenshot?
Modern digital storage oscilloscopes (DSOs) calculate this automatically using the discrete sum formula on the sampled screen buffer. If you must do it manually from a grid screenshot, count the number of vertical divisions the waveform occupies above the zero-line, subtract the divisions below the zero-line, multiply by the Volts/Division setting, and integrate geometrically. However, relying on the DSO's built-in "Vavg" measurement button is vastly more accurate as it samples thousands of points per cycle.
Is average voltage the same as DC offset?
Yes, in signal processing, the average voltage of a waveform is exactly equal to its DC offset. If you pass a signal through a perfect series capacitor (a high-pass filter), the capacitor blocks the Vavg (DC offset), re-centering the remaining AC waveform around exactly 0V. This is why coupling capacitors are rated based on the peak voltage, but the downstream biasing resistors must be calculated using the Vavg to establish the correct quiescent operating point.
