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Module 5 — Electrical Systems  ·  LEO Technical Academy
Lesson 5.11 — Transformers and Power Supplies
L2 — Guided Practice 🟡 Yellow Risk ⚡ Electrical ⏱ 60 min LEO-ACE-05-011 v1.0 · 2026-06-14

In This Lesson

§00 Safety §01 Overview §02 Objectives §03 Prerequisites §04 How a Transformer Works §05 Turns Ratio Calculations §06 Transformer Types §07 Nameplate §08 CPT in an MCC §09 DC Power Supplies §10 Interactive Calculator §11 Diagnostics §12 Assessment §13 Summary
§00

Safety Briefing

⚠ Yellow Risk — L2 Guided Practice

This lesson covers understanding, measuring, and diagnosing transformers. Working on the primary side (480V) is Red risk and requires QEP status. The L2 scope here is limited to secondary-side measurement and diagnostic activities with QEP oversight.

Hazards Present When Working Near Transformers

🔴 Stop — Primary Side is Red Risk

If your diagnosis points to a primary winding fault, blown primary fuse, or any 480V-side work, stop and escalate to your QEP or foreman. Do not open the primary side as an L2 technician.

✔ L2 Scope for This Lesson

Understand transformer theory, read nameplates, measure secondary voltage, identify CPT secondary fuses, diagnose no-control-power conditions from the secondary side, and understand power supply operation. All energized work performed with appropriate PPE and QEP oversight.

§01

Overview

Transformers are among the most common and most important devices in an industrial facility. Walk through any Motor Control Center (MCC) and you will find a Control Power Transformer (CPT) in nearly every bucket — stepping 480V down to 120VAC to power the START/STOP buttons, pilot lights, PLC inputs, and relay coils that control each motor.

Beyond the MCC, transformers appear in every DC power supply powering PLC racks and instrumentation, in variable frequency drives, in welding machines, and at the utility entrance where the service transformer steps distribution voltage down to usable levels.

Understanding how a transformer works enables you to:

§02

Learning Objectives

§03

Prerequisites

You should be comfortable with the following lessons before starting:

5.3 — AC Fundamentals 5.6 — Series/Parallel Circuits 5.7 — Three-Phase Power

Key concepts needed: AC voltage and frequency, Ohm's Law, apparent vs. real power (VA vs. watts), and a basic understanding of magnetic fields around conductors.

§04

How a Transformer Works

Mutual Inductance

A transformer operates on the principle of electromagnetic induction. When alternating current flows through the primary winding, it creates a continuously changing magnetic field in the iron core. That changing flux passes through the secondary winding and induces a voltage in it — with no direct electrical connection between the two windings.

This is called mutual inductance. The iron core provides a low-reluctance (easy) path for magnetic flux, coupling nearly all of the primary flux to the secondary.

Why Transformers Require AC — Not DC

Electromagnetic induction requires a changing magnetic field. A DC current creates a constant field — after the initial transient, flux is stable and no voltage is induced in the secondary winding. Faraday's Law: V = −N × dΦ/dt. If dΦ/dt = 0 (constant DC flux), then V = 0.

Applying DC to a transformer primary will cause it to overheat rapidly, because the only resistance limiting primary current is the low DC resistance of the winding itself — the inductive reactance (XL = 2πfL) that normally limits AC current drops to zero at 0 Hz.

The Turns Ratio

The ratio of primary turns (Np) to secondary turns (Ns) determines how voltage is transformed. This is the fundamental transformer relationship:

Vp / Vs = Np / Ns Voltage ratio equals turns ratio

Current behaves inversely — if voltage steps down, current steps up by the same ratio:

Ip / Is = Ns / Np Current ratio is the INVERSE of the turns ratio

Conservation of Power

A transformer cannot create energy. For an ideal (100% efficient) transformer, apparent power in equals apparent power out:

Vp × Ip = Vs × Is Power in = Power out (ideal transformer)

This is why stepping voltage down means stepping current up by the same ratio. A 4:1 step-down transformer that receives 1A at 480V will deliver 4A at 120V — the same 480VA on both sides.

✔ Real-World Efficiency

Real transformers are 95–99% efficient, losing some power as heat in the windings (copper losses: I²R) and core (iron losses: eddy currents and hysteresis). For field sizing and diagnostics, assume ideal transformer behavior — the efficiency difference is negligible at the L2 level.

§05

Turns Ratio Calculations

Vp / Vs = Np / Ns Turns ratio / voltage
Vs = Vp × (Ns / Np) Solve for secondary voltage
Is = Ip × (Np / Ns) Solve for secondary current
kVAprimary = kVAsecondary Power balance (ideal)

Worked Example 1 — Find Secondary Voltage

Problem

A transformer has a 480V primary and a 4:1 turns ratio. What is the secondary voltage?

Vs = Vp × (Ns / Np) = 480 × (1/4) = 480 ÷ 4 = 120V

This is the most common industrial CPT — 480V primary, 120V secondary, 4:1 turns ratio.

Worked Example 2 — Find Primary Current and Verify kVA

Problem

The secondary of the transformer above draws 10A. What current flows in the primary? Verify the kVA balance.

Ip = Is × (Ns / Np) = 10 × (1/4) = 2.5A

Primary kVA = 480V × 2.5A = 1,200VA = 1.2 kVA

Secondary kVA = 120V × 10A = 1,200VA = 1.2 kVA ✓

Power is conserved — both sides show 1.2 kVA of apparent power.

Worked Example 3 — Find Required Turns Ratio

Problem

A transformer must step 480V down to 24V for a solenoid valve circuit. The secondary draws 5A. Find the turns ratio, primary current, and kVA.

Turns ratio = Vp / Vs = 480 / 24 = 20:1

Primary current = Is × (Ns / Np) = 5 × (1/20) = 0.25A

kVA = 24V × 5A = 120VA = 0.12 kVA (a small transformer)

Verify: 480V × 0.25A = 120VA ✓

§06

Transformer Types

TypeDescriptionApplication
Step-down Np > Ns; Vp > Vs. More primary turns than secondary. Most industrial uses — 480→120V CPT, utility distribution, DC supply inputs
Step-up Np < Ns; Vs > Vp. Fewer primary turns than secondary. Generator output to transmission voltage; high-voltage testing equipment
Isolation Np = Ns (1:1 turns ratio). No electrical connection between windings — galvanic isolation. Voltage ratio is 1:1. Medical equipment (patient safety), sensitive instrumentation (breaks ground loops, reduces noise), GFCI-type protection
Autotransformer Single winding with taps. Primary and secondary share winding conductor — there IS an electrical connection. Smaller and cheaper, but no isolation. Reduced-voltage motor starters; buck-boost voltage adjustment (e.g., 480V→460V)
Control Power Transformer (CPT) Small step-down, typically 480→120VAC. Often includes primary and secondary fuse clips built into the unit. Every MCC bucket; panel control power for starters, PLCs, and instruments
Current Transformer (CT) Ring or window-type; primary is the conductor passing through the ring (1 turn). Secondary produces a scaled-down current proportional to primary. Never open CT secondary under load. Metering (kWh), protection relays, overload relay coils, power monitoring
Potential Transformer (PT) Precision step-down for measuring high voltages safely. Scales thousands of volts to 120V or 69V for meters and relays. Medium-voltage metering and protection (4.16kV systems and above)
🔴 Critical — Never Open a CT Secondary Under Load

A current transformer secondary must never be open-circuited while the primary conductor carries current. With no secondary load, the primary magnetomotive force drives the core into saturation, producing extremely high voltages (potentially thousands of volts) at the secondary terminals — a lethal hazard. Always short the CT secondary before removing a burden or secondary lead.

§07

Transformer Nameplate

Every transformer carries a nameplate with critical information. Being able to read and interpret it is an essential field skill.

ACME TRANSFORMER CORPORATION
DRY TYPE — GENERAL PURPOSE DISTRIBUTION
kVA Rating
75 kVA
Phase
HV (Primary)
480V
LV (Secondary)
120 / 240V
Frequency
60 Hz
Impedance Z%
2.5%
Cooling Class
AN (Air Natural)
Temperature Rise
80°C
Insulation Class
Class H (180°C max)
Serial Number
2024-ACE-77341
HV Tap Connections
FCAN: +5%  •  A: +2.5%  •  B: Nominal  •  C: −2.5%  •  FCBN: −5%

Reading Each Nameplate Field

FieldWhat It MeansWhy It Matters
kVA Apparent power capacity — 75,000 VA maximum continuous load Exceeding the kVA rating causes overheating and shortened life. Size the transformer to the load.
HV / LV Voltage High voltage (primary) and low voltage (secondary) ratings. "120/240V" secondary means a center-tap provides two 120V circuits or one 240V circuit. Must match facility voltage. Wrong voltage connection = wrong secondary output and potential damage.
Z% (Impedance) Percentage of rated voltage needed to drive rated current through the transformer's internal impedance. Determines available fault current at the secondary. Lower Z% = higher available fault current = larger AIC rating required on downstream overcurrent devices. Critical for system coordination studies.
Cooling Class AN = Air Natural (convection cooled, no fan). ONAN = Oil Natural Air Natural (oil-immersed distribution transformers). Never block ventilation on an AN transformer. Restricted airflow raises hot-spot temperature and shortens insulation life dramatically.
Temperature Rise Maximum winding temperature rise above ambient (40°C standard). 80°C rise + 40°C ambient + 15°C hot-spot allowance = 135°C maximum hot-spot temperature. Facilities with ambient above 40°C (e.g., un-cooled buildings in summer) require derating the transformer's kVA capacity.
Insulation Class Class H = 180°C maximum winding temperature. Class F = 155°C. Class B = 130°C. Insulation degrades exponentially above its rated class. Running hot drastically shortens life.
Taps Alternate primary connections that adjust the effective turns ratio by ±2.5% or ±5%. Used to compensate for off-nominal primary voltage. If facility runs 504V instead of 480V, use the +5% tap so the secondary sees the correct voltage.

Calculating Available Fault Current from Z%

The impedance percentage determines how much short-circuit current the transformer can deliver to a bolted fault on the secondary:

Isc = (kVA × 1,000 / Vsecondary) / (Z% / 100) Available fault current at secondary terminals
Example — 75 kVA Transformer

kVA = 75  |  Vs = 120V  |  Z% = 2.5%

Rated secondary current = 75,000 / 120 = 625A

Isc = 625A / 0.025 = 25,000A available fault current

Any overcurrent devices on this secondary must be rated for at least 25 kAIC interrupting capacity. This is why residential and light-commercial panels are commonly rated "25,000 AIC."

§08

Control Power Transformer (CPT) in an MCC

The CPT is the transformer an MCT encounters most often. In a standard MCC bucket, the CPT taps 480VAC from the motor power circuit (before the contactor) and produces 120VAC for the control circuit.

MCC BUCKET (480V / 1Ø MOTOR STARTER) 480VAC BUS CONTACTOR M MOTOR Load CPT 480→120V 2A FU 120VAC CTRL STOP NC START NO COIL (M) 480V tap Sec. fuse

Simplified MCC bucket: CPT taps 480V before the contactor, steps down to 120V for the control circuit. The secondary fuse (2A FU) protects the CPT and all control wiring.

CPT Sizing

The CPT must handle all control loads simultaneously. Sum up the VA requirement of every device on the secondary:

Control DeviceTypical VA Draw
Contactor coil (NEMA 00–1, small)10–20VA
Contactor coil (NEMA 2–3, medium)30–60VA
Pilot light (LED)2–5VA
Pilot light (incandescent)7–15VA
Timer relay coil10–25VA
Control relay coil5–15VA
PLC digital input card (per card)Per mfr. spec — typically 20–60VA

After summing, select the next standard CPT size up from the calculated total:

Standard CPT Sizes

50VA • 100VA • 150VA • 250VA • 500VA • 750VA • 1,000VA • 1,500VA • 2,000VA

Minimum recommendation for a typical one-motor starter with START/STOP/RUN lights: 150VA

CPT Secondary Fusing

Ifuse = CPT VA rating / Vsecondary Maximum secondary fuse amperage
⚠ Most Common "No Start" Troubleshooting Find

Blown CPT secondary fuse = loss of ALL control power = motor won't start. The contactor coil won't energize. Pilot lights go dark. Nothing responds to START.

First check: Verify CPT primary has voltage, then measure secondary voltage. If you have primary but no secondary, inspect the secondary fuse. Replace with same rating. If it blows again immediately, there is a short circuit in the control wiring — chase the fault before re-fusing.

CPT Sizing Example

Calculation

Motor starter bucket: 1× NEMA-2 contactor coil (40VA) + 3× pilot lights at 8VA each + 1× timer relay coil (20VA)

Total VA = 40 + (3 × 8) + 20 = 40 + 24 + 20 = 84VA

Next standard size up from 84VA = 100VA CPT

Secondary fuse = 100VA / 120V = 0.83A → use 1A or 2A fuse

Note: Many designers always use 150VA minimum to allow for future additions and easier troubleshooting.

§09

DC Power Supplies

PLCs, HMIs, sensors, solenoid valves, and most modern control components run on 24VDC. The device that converts AC line power to regulated 24VDC is a power supply — and understanding how it works helps you diagnose failures and specify replacements.

A linear power supply converts AC to DC through four sequential stages. It is older, heavier technology — but extremely reliable and produces very clean DC with minimal electrical noise.

🔄
1. Transformer
120VAC → 18VAC
Steps AC down
2. Rectifier
4-diode bridge
AC → pulsing DC
🌊
3. Filter Cap
Large electrolytic
smooths pulsing DC
4. Regulator
78xx IC or transistor
holds output steady
  1. Transformer: A conventional iron-core transformer steps 120VAC down to the AC voltage required for the desired DC output — typically 15–18VAC for a 12VDC supply, to allow headroom for diode drops and regulation.
  2. Full-wave bridge rectifier: Four diodes in a bridge arrangement convert AC to pulsating DC. On each half-cycle, two diodes conduct; the result is a waveform that is always positive but pulses at 120 Hz (twice per 60 Hz cycle).
  3. Filter capacitor: A large electrolytic capacitor (hundreds to thousands of microfarads) charges during each pulse and discharges slowly into the load between pulses. The result is near-constant DC with only a small ripple voltage remaining.
  4. Linear voltage regulator: An IC regulator (e.g., LM7812 for 12V, LM7805 for 5V) or discrete transistor circuit maintains a precise output voltage regardless of load or input variations. It dissipates excess voltage as heat — which is why linear supplies have large heatsinks and are heavy.
CharacteristicValue / Description
Efficiency40–65% (regulator wastes excess as heat)
Output noiseVery low — excellent for analog instrumentation
Weight / sizeHeavy and large — big iron core and heatsinks
Input rangeNarrow — designed for one nominal line voltage
Best industrial useSensitive analog instruments; specialty applications requiring clean DC

A switching mode power supply (SMPS) achieves voltage conversion using high-frequency switching — operating at 20,000–500,000 Hz instead of 60 Hz. This allows magnetic components to be dramatically smaller and lighter.

1. Input Rect.
AC line → ~340VDC
High-voltage bus
🔀
2. Switcher
MOSFET at 20–500kHz
Chops HV DC
🔄
3. HF Transformer
Tiny core steps
down HF signal
🌊
4. Output Filter
Diodes + inductor
+ cap → smooth DC
🔁
5. Feedback
Senses output, adjusts
duty cycle
  1. Input rectification (no transformer first): The AC line (85–264VAC worldwide) is rectified directly to high-voltage DC. From 240VAC this gives approximately 340VDC. A bulk capacitor stores this high-voltage DC.
  2. Switching stage: Power MOSFETs switch the high-voltage DC on and off at 50–200 kHz. This very high frequency is why the transformer and filter components can be tiny.
  3. High-frequency transformer: Because the switched signal is at HF (not 60 Hz), the core only needs to be a few cubic centimeters for hundreds of watts. This is the core reason SMPS units are so compact and light. The transformer also provides galvanic isolation between input and output.
  4. Output rectification and filtering: Fast rectifier diodes and a small LC filter convert the HF AC from the transformer secondary to smooth DC. The high switching frequency means the filter components are very small.
  5. Feedback loop: An optocoupler senses the output voltage (isolated from the HV input side) and adjusts the MOSFET duty cycle to maintain regulation. If load increases and output sags, the feedback increases duty cycle within microseconds.

Industrial 24VDC SMPS Features (DIN-Rail Mount)

Common RatingOutput CurrentOutput PowerTypical Width
24VDC / 5A5A120W40mm
24VDC / 10A10A240W60mm
24VDC / 20A20A480W80mm
24VDC / 40A40A960W120mm
FeatureLinear SupplySMPS
Efficiency40–65%90–95%
Size / WeightHeavy, large (iron core, heatsinks)Compact, light
Heat generatedHigh — wastes ~40% as heatLow — loses only 5–10%
EMI / NoiseVery low — inherently cleanHigher — switching creates EMI (filtered internally)
Input voltage rangeNarrow (one nominal voltage)Wide (85–264VAC, universal)
Transient responseModerateVery fast (feedback adjusts in microseconds)
ReliabilityVery high (few components)High (modern units: 100,000+ hours MTBF)
CostLow for small power, high for largeCost-effective at all power levels
Typical industrial useOlder analog instrumentation; specialty applicationsPLC racks, HMI, field devices, all modern 24VDC control
✔ Field Takeaway

In modern industrial facilities, virtually all 24VDC supplies are SMPS — DIN-rail-mounted units with a green LED on the front. When one fails: check the DC OK LED, measure output under load, check input voltage. If output sags under load, the supply is undersized or failing internally. Replace with same or next size up. Common brands: Phoenix Contact, PULS, Siemens SITOP, Murr, Wago.

§10

Interactive Transformer Calculator

Turns Ratio & Power Calculator

Enter primary voltage and either the turns ratio or secondary voltage. Provide primary current, secondary kVA, or secondary current to fully solve both sides.

CPT Sizing Tool

Add your control loads. The tool sums the VA requirements and recommends a CPT size and secondary fuse rating.

Device Description VA
Recommended CPT Size
§11

Common Transformer Failures and Diagnostics

FailureSymptomDiagnostic TestCommon Cause
Open primary winding No secondary voltage; primary fuse intact Measure primary voltage (present). Measure secondary (0V). De-energize and measure primary winding resistance — open winding reads infinite ohms. Sustained overload; voltage surge; age-related insulation failure
Open secondary winding No control power; all control devices dead; CPT primary fuse OK Verify primary has voltage. Measure secondary (0V). De-energize and measure secondary winding resistance — open reads infinite ohms. Sustained overload on secondary; short followed by fuse clearing that damaged the winding
Shorted turns Transformer runs abnormally hot; higher than normal primary current; primary fuse blows repeatedly; burning varnish smell Measure secondary voltage under no load — shorted turns often pull it below nameplate rating. Use IR gun to check enclosure temperature. Measure no-load primary current vs. nameplate. Replace CPT if suspected. Internal insulation breakdown — caused by heat cycling, moisture, over-voltage transients, rodent damage
CPT secondary fuse blown No control power; motor won't start; contactor won't pull in; all pilot lights dark 1) Verify CPT primary has correct voltage. 2) Measure secondary — reads 0V if fuse is blown. 3) Inspect or test fuse continuity. 4) Replace fuse — if it holds, test operation; if it blows again immediately, chase the short in control wiring. Control circuit short; pinched wiring; overloaded secondary; incorrect fuse size
Wrong tap selected Secondary voltage consistently high or low by a fixed percentage (e.g., 5%); equipment runs hot or malfunctions Check tap connections on CPT primary terminal block. Compare nameplate voltage to actual facility primary voltage. If facility runs 504V and tap is at "Nominal" (480V), secondary will read 126V instead of 120V. Tap set incorrectly; facility primary voltage changed after installation; wrong tap set during replacement
24VDC SMPS output low / sagging PLC behaves erratically; sensors unreliable; SMPS DC OK LED orange or flashing Measure output under full load with calibrated meter. Check DC OK relay status. If output sags below 22V, supply is undersized or failing. Verify input voltage is within specified range. Load growth exceeded supply rating; supply internal failure; input voltage low (brownout); electrolytic capacitors degraded (end of life)
✔ Standard Troubleshooting Order — No Control Power
  1. Check primary voltage at CPT input terminals (480V present?)
  2. Check primary fuse continuity (measure voltage drop across fuse)
  3. Check secondary voltage at CPT output terminals (120V present?)
  4. Check secondary fuse — this is the most common failure
  5. Replace secondary fuse — if it holds, test operation; if it blows again, chase the short
  6. If primary is present but secondary reads 0V with fuse intact: CPT has open secondary winding — replace CPT (QEP required for primary-side de-energization)
§12

Knowledge Assessment

Answer all five questions, then check each one. Review incorrect answers before proceeding to Lesson 5.12.

1. A transformer has a 480V primary and a 4:1 turns ratio. What is the secondary voltage?
2. The secondary of a 480/120V transformer draws 8A. What current flows in the primary?
3. A motor starter control circuit has: 1 contactor coil at 40VA, 3 pilot lights at 8VA each, and 1 timer relay at 20VA. What is the minimum recommended CPT rating?
4. An industrial 24VDC switching power supply enters current limit mode when a new PLC I/O card is installed. This indicates:
5. Why will a standard two-winding transformer not pass DC from primary to secondary?
§13

Lesson Summary

Key Takeaways — Lesson 5.11
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