When working through an example of autotransformer sizing, the most critical realization is that the transformer's kVA rating is determined only by the transformed voltage and current, not the total load. Unlike a standard two-winding isolation transformer where the core and windings must handle 100% of the load's apparent power, an autotransformer shares a common magnetic and electrical path. This allows a physically small, inexpensive unit to control a massive load. Below is a complete, decision-forward walkthrough of a classic exam and field-sizing problem.
The Problem Statement: Sizing a Buck Autotransformer
You have a single-phase 240V AC supply. You need to power a single-phase 208V AC resistive heating bank rated at 12 kVA. You decide to use a buck autotransformer to step the voltage down.
Tasks:
1. Calculate the minimum required kVA rating of the autotransformer itself.
2. Identify the standard commercial size you must procure.
3. Select a specific physical part number for the job.
Method Selection and The Classic Exam Trap
Which method applies and why? We must use the Autotransformer Apparent Power formula combined with Kirchhoff's Current Law (KCL) at the tap node. Because the primary and secondary share a physical winding, the total power transferred is split into two components: conductive power (transferred directly via the electrical connection) and transformed power (transferred via magnetic induction). We only need to size the transformer for the transformed power.
The Trap: The most common mistake in this example of autotransformer calculation is sizing the unit for the full 12 kVA load. If this were a standard isolation transformer, you would indeed need a 15 kVA unit (the next standard size up from 12 kVA). However, sizing an autotransformer for the full load kVA results in massive overspending, unnecessary weight, and wasted panel space. The autotransformer only needs to handle the voltage difference multiplied by the input current.
Step-by-Step Algebraic Solution
Let's solve this with every algebraic step shown. No skipped math.
- Identify the Known Variables:
- Source Voltage ($V_{in}$) = 240V
- Load Voltage ($V_{out}$) = 208V
- Load Apparent Power ($S_{load}$) = 12,000 VA (12 kVA)
- Calculate the Load Current ($I_{load}$):
The current drawn by the 208V heating bank.
$I_{load} = S_{load} / V_{out}$
$I_{load} = 12,000 / 208 = 57.692 \text{ A}$ - Calculate the Source Current ($I_{in}$):
The current drawn from the 240V mains (assuming ideal efficiency).
$I_{in} = S_{load} / V_{in}$
$I_{in} = 12,000 / 240 = 50.000 \text{ A}$ - Determine the Series Winding Voltage and Current:
In a buck configuration, the source connects across the full winding, and the load connects to a tap. The 'series' portion of the winding drops the excess voltage.
$V_{series} = V_{in} - V_{out} = 240 - 208 = 32 \text{ V}$
The current flowing through this series winding is exactly the source current entering the unit.
$I_{series} = I_{in} = 50.000 \text{ A}$ - Calculate the Transformed kVA (Winding Rating):
$S_{winding} = V_{series} \times I_{series}$
$S_{winding} = 32 \text{ V} \times 50.000 \text{ A} = 1,600 \text{ VA}$
Result: The required autotransformer rating is 1.6 kVA.
$S_{auto} = S_{load} \times \frac{V_{in} - V_{out}}{V_{in}}$
$S_{auto} = 12 \text{ kVA} \times \frac{240 - 208}{240} = 12 \times 0.1333 = 1.6 \text{ kVA}$.
Sanity Check and Independent Verification
Order of Magnitude and Units: The voltage reduction is 32V out of 240V, which is exactly 13.33%. Therefore, the autotransformer should only need to handle 13.33% of the total load power. 13.33% of 12 kVA is 1.6 kVA. The units are in kVA (apparent power), which is correct for transformer sizing regardless of the load's power factor.
How to verify the answer independently: We can verify this by checking the common winding (the 208V portion of the coil). By KCL at the tap node, the current in the common winding is the difference between the load current and the source current: $I_{common} = 57.692\text{A} - 50.000\text{A} = 7.692\text{A}$.
Calculating the VA of the common winding: $V_{common} \times I_{common} = 208\text{V} \times 7.692\text{A} = 1,599.9\text{ VA}$.
Both the series winding (1600 VA) and the common winding (1600 VA) yield the exact same transformed power requirement. The math is perfectly balanced.
Decision Path: Selecting the Physical Hardware
We have a calculated requirement of 1.6 kVA. Transformers are manufactured in standard NEMA sizes (e.g., 0.5, 1.0, 1.5, 2.0, 3.0 kVA). You must always round up to the next standard size to account for ambient temperature derating and inrush currents.
| Calculated kVA Range | Action | Standard Size to Select |
|---|---|---|
| ≤ 1.0 kVA | Select base size | 1.0 kVA |
| 1.01 to 1.5 kVA | Round up to next tier | 1.5 kVA |
| 1.51 to 2.0 kVA (Our Case: 1.6) | Round up to next tier | 2.0 kVA |
| > 2.0 kVA | Evaluate 3-phase or parallel units | 3.0 kVA+ |
The Concrete Pick: Based on the decision tree, we require a 2.0 kVA single-phase buck-boost transformer. A highly reliable, industry-standard choice for this exact application is the Hubbell Acme T-1-83093 (or the equivalent Hammond Manufacturing 184C2). These units feature dual 120/240V primary and 12/24V secondary windings that are internally jumpered or externally wired as an autotransformer to achieve the exact 32V buck required to drop 240V to 208V safely. For deeper theory on how these windings interact magnetically, refer to standard texts on autotransformer operating principles.
Frequently Asked Questions
A: Yes, electrically it will work perfectly. However, a 15 kVA isolation transformer costs roughly 4 to 6 times more than a 2 kVA autotransformer, weighs over 100 lbs more, and requires significantly more physical space in your enclosure. Unless your local AHJ specifically mandates galvanic isolation for this specific heating circuit (which is rare for fixed resistive loads), the autotransformer is the correct engineering choice. See this comparative breakdown of isolation vs. autotransformers for more context.
A: The kVA calculation remains exactly the same because kVA represents apparent power, independent of power factor. However, if the motor has a high starting inrush current (e.g., 6x locked rotor current), you must verify that the 2.0 kVA autotransformer's thermal mass and winding wire gauge can survive the brief inrush without tripping upstream breakers or degrading the insulation. In high-inrush scenarios, moving up one size to a 3.0 kVA unit provides a necessary thermal buffer.
A: No. Because the primary and secondary share a physical electrical connection, a fault on the 208V load side is directly reflected to the 240V source. You must install properly sized overcurrent protection (fuses or breakers) on both the 240V supply side and the 208V load side, calculated based on the 1.6 kVA winding limits and the 12 kVA load limits, respectively.
