Transformer Secondary Voltage Calculator

Transformer turns relationship

For an ideal transformer, the secondary voltage is proportional to the turns ratio: Vs = Vp × (Ns / Np). This is commonly used with RMS voltages in AC systems.

If voltage regulation is provided, the loaded secondary voltage is estimated as: Vs(load) = Vs(ideal) × (1 − Reg/100). Regulation represents the approximate drop from no-load to rated load.

When secondary current is entered, power estimates use: Pout ≈ Vs(load) × Is × PF. If efficiency is entered, input power is approximated by Pin ≈ Pout / (Eff/100).

Steps

1. Select a method: turns count or turns ratio.
2. Enter the primary voltage using consistent RMS units.
3. Provide Np and Ns, or enter the ratio Ns/Np directly.
4. Optionally add regulation to estimate loaded voltage.
5. Optionally add current, power factor, and efficiency for power estimates.
6. Press Calculate. Use CSV or PDF export from the results panel.

Sample input and output

Examples are illustrative; real devices vary with load, frequency, and winding resistance.

This guide explains how the calculator estimates secondary voltage, and how regulation, efficiency, and load characteristics influence real measurements.

1) Turns ratio and expected output

The core estimate is Vs = Vp × (Ns/Np). For example, 230 V with a 0.12 ratio targets about 27.6 V at no-load. Step-up applications work the same way; a 4× ratio turns 24 V into about 96 V.

2) RMS voltage and frequency considerations

Use RMS values when working with sinusoidal power. Common supplies are 110–120 V or 220–240 V at 50/60 Hz. Frequency changes can affect magnetizing current and heating, especially near the transformer’s rated limits.

3) Regulation explains load drop

Real windings have resistance and leakage reactance, so voltage falls under load. Many small power units show about 3–10% regulation at rated current. Using a 5% setting converts a 12.0 V ideal output into about 11.64 V loaded.

4) Efficiency ranges and what they imply

Efficiency depends on size and design. Small wall adapters may be 80–90%, while larger power units can exceed 95%. With 92% efficiency, a 50 W output typically requires about 54.35 W input, meaning roughly 4.35 W is lost as heat.

5) Current, power factor, and real power

When you enter secondary current, the calculator estimates output power as Pout ≈ Vs × Is × PF. Resistive loads use PF≈1. Inductive loads (motors, coils) often sit around 0.7–0.95, reducing real power for the same amps.

6) Troubleshooting with measured data

If measured Vs is lower than expected, compare against regulation and load current first. A heavy load, undersized winding, or long leads can add extra drop. If no-load voltage is low, confirm the primary voltage and the intended turns ratio.

7) Safety and rating checks

Verify insulation class and temperature rise for your application. Keep secondary current below the rated value. For bench testing, start with a light load and increase gradually while monitoring output voltage, input current, and heating.

8) Practical workflow for this calculator

First compute the ideal secondary voltage from turns or ratio. Next apply regulation to approximate loaded output. Add current and power factor to estimate real output power, then apply efficiency to infer input power and primary current. Export CSV/PDF to document designs or test reports.

Frequently asked questions

1) Should I use peak or RMS voltage?

Use RMS for standard AC ratings. Peak values will overstate expected secondary voltage by about 41% for a sine wave.

2) Why is the loaded voltage lower than the ideal value?

Winding resistance and leakage reactance cause voltage drop under load. Regulation summarizes that drop as a percentage at rated current.

3) What turns ratio should I enter if I only know voltage ratings?

Use the rated secondary-to-primary voltage ratio as an approximation for Ns/Np. Then refine with regulation if you have load data.

4) Does the calculator handle step-up transformers?

Yes. If Ns/Np is greater than 1, the secondary voltage increases proportionally, assuming an ideal relationship and the chosen regulation.

5) How do I choose a typical regulation value?

Small power units often fall in the 3–10% range. Larger, better-regulated designs may be closer to 1–5% at rated load.

6) What power factor should I use?

Use 1.0 for resistive heaters or lamps. For motors and inductive loads, a practical range is 0.7–0.95, depending on the device and loading.

7) Why does efficiency change the estimated primary current?

Lower efficiency means more input power is needed for the same output. The calculator uses efficiency to scale input power, which then increases the estimated primary current.