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ELECTRICAL MACHINERY

Transformer Design & Calculator

Compute transformer efficiency, leakage inductance, and peak flux density from turns ratio, core area, iron loss, and copper loss in real time. Build intuition for power supply and switching converter design.

Transformer Design Calculator

Voltage & Power

10 V1 kV100 kV
1 V100 V10 kV
10 VA1 kVA10 MVA

Frequency

Core Material

Formulas

Key Formulas
$a = N_1/N_2 = V_1/V_2$
$E = 4.44 \cdot f \cdot N \cdot B_{max} \cdot A_c$
$I_1 = S/V_1,\quad I_2 = S/V_2$
Results
Turns Ratio a = N₁/N₂
turns ratio
Primary Turns N₁
turns (A_c=10 cm² assumed)
Secondary Turns N₂
turns
1. B-H Hysteresis Loop (Selected Core Material)
2. Voltage & Current Waveforms (2 Cycles)
3. Efficiency η vs Load Factor (%)

What is Transformer Design?

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What exactly is a "turns ratio" and why is it the first thing I need to set in this simulator?
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Basically, the turns ratio is the heart of a transformer. It's the ratio of the number of wire turns in the primary coil to the secondary coil ($a = N_1 / N_2$). It directly determines how much the voltage is stepped up or down. In this tool, when you move the "Primary Voltage" and "Secondary Voltage" sliders, you're defining this ratio. For instance, setting 120V primary and 12V secondary gives a 10:1 step-down ratio.
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Wait, really? So if the voltage changes, the current must change too to conserve power. How does the simulator handle the "Apparent Power" slider with that?
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Exactly right! The apparent power (S in Volt-Amps) is what the transformer is designed to handle. The simulator uses it to calculate the currents. Try it: set your voltages, then slide the "Apparent Power" up. You'll see the calculated primary and secondary currents change instantly. The relationship $S = V_1 I_1 = V_2 I_2$ is always maintained. A common case is a 1 kVA transformer stepping 240V down to 24V, which would give you about 4.2 A on the primary and 41.7 A on the secondary.
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Okay, that makes sense for the "big picture" design. But what about the "Core Loss" and "Copper Loss" sliders? They seem to affect efficiency. What's happening there?
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Great question! Those sliders model the real-world energy losses that prevent a transformer from being 100% efficient. Core loss (or iron loss) is roughly constant—it's energy lost to magnetizing the core. Copper loss depends on the load current—it's heat from resistance in the windings. The key insight is that maximum efficiency happens when these two losses are equal. Play with the sliders: set a high core loss and low copper loss, then check the efficiency. Now adjust the load until they balance—you'll see the efficiency peak!

Physical Model & Key Equations

The fundamental relationship governing voltage transformation is the turns ratio, which assumes an ideal, lossless magnetic coupling.

$$a = \frac{N_1}{N_2}= \frac{V_1}{V_2}= \frac{I_2}{I_1}$$

Where $a$ is the turns ratio, $N$ is the number of coil turns, $V$ is voltage, and $I$ is current. Subscripts 1 and 2 denote primary and secondary sides, respectively.

For real transformers, we account for power and losses. The apparent power rating defines the transformer's capacity, and efficiency is determined by balancing two main loss mechanisms.

$$S = V_1 I_1 = V_2 I_2, \quad \eta = \frac{P_{out}}{P_{in}}= \frac{P_{out}}{P_{out}+ P_{core}+ P_{cu}}$$

Here, $S$ is the apparent power (VA), $\eta$ is efficiency, $P_{core}$ is the constant core/iron loss, and $P_{cu}$ is the copper loss (proportional to the square of the load current, $I^2R$). Maximum efficiency occurs when $P_{core}= P_{cu}$.

Real-World Applications

Power Distribution Grids: Massive step-up transformers at power plants increase voltage to hundreds of kilovolts for efficient long-distance transmission, minimizing $I^2R$ losses in the lines. Substations then use step-down transformers to reduce voltage for local distribution and finally to 120/240V for homes.

Consumer Electronics Power Supplies: The small "wall wart" charger for your phone contains a miniaturized switch-mode transformer. It steps down mains voltage (e.g., 120V AC) to a much lower voltage (e.g., 5V DC), with design focused on high efficiency and low heat generation in a tiny package.

Industrial Machinery: Heavy equipment like arc welders or large motors often require specific, high currents at lower voltages. Custom-designed transformers provide this, with robust windings to handle the high secondary current and cooling systems to manage the significant copper losses.

Audio Equipment: Audio transformers are used for impedance matching between components (like a microphone and an amplifier). The turns ratio is carefully designed to maximize power transfer and signal clarity, not just to change voltage, making loss management critical for sound quality.

Common Misconceptions and Points to Note

First, it's crucial not to confuse "Rated Capacity" with "Output Power". Rated capacity is apparent power [VA], not active power [W]. When you connect a load with a power factor other than 1 (like a motor), the available active power from the same capacity is reduced. Even if you set the capacity to 1000VA, with a load of power factor 0.8, you can only draw 800W. Next, a larger "Core Cross-Sectional Area" is not simply better. While increasing the cross-sectional area tends to reduce the number of turns and lower copper loss, a larger core increases iron loss, material cost, and weight. For instance, using a massive power transformer core for a small household transformer isn't practical, right? Finding the optimal balance is where design skill shines. Finally, consider the number of turns calculated by the simulator as a "theoretical minimum". In practice, you'll almost always wind slightly more turns to meet specifications like voltage regulation and impedance. Think of this tool's output as your "starting point".