Question

How to calculate winding of transformer?

Answer 1

Part 1: The Core Theoretical Principles

The design of a transformer is governed by fundamental electromagnetic principles. Understanding these is key to the calculations.

1. The Universal Transformer EMF Equation

This is the cornerstone of transformer design. It links the voltage in a winding to the physical characteristics of the core and the AC supply.

$$
E = 4.44 \times f \times N \times B_{max} \times A_c
$$

Where:
$E$ = RMS Voltage in the winding (Volts)
$f$ = Frequency of the AC supply (Hertz, Hz)
$N$ = Number of turns in the winding
$B_{max}$ = Maximum Magnetic Flux Density the core can support (Tesla, T)
$A_c$ = Cross-Sectional Area of the core's center limb (square meters, m²)
The constant $4.44$ is the form factor, derived for a pure sine wave as: $4.44 = \frac{2\pi}{\sqrt{2}}$.

2. Core Material and Flux Density ($B_{max}$)

The core material determines the maximum magnetic flux density ($B{max}$) before it saturates. Saturation leads to high magnetizing currents and inefficiency.
* Silicon Steel Laminations (for 50/60 Hz): A typical design value for $B{max}$ is between 1.0 T and 1.4 T. A lower value (e.g., 1.1 T) results in a larger transformer with lower core losses, while a higher value (e.g., 1.3 T) allows for a smaller transformer at the cost of higher losses.
* Ferrite Cores (for high frequency): $B_{max}$ is much lower, typically 0.2 T to 0.4 T.

3. Core Geometry and Area ($A_c$)

You must measure the physical dimensions of your core. The most important is the cross-sectional area of the central limb where the windings are placed.

For a square or rectangular core limb with dimensions $l$ and $w$:
$$
A_c = l \times w
$$

Important: All calculations must use SI units. Ensure your core area $A_c$ is converted to square meters (m²).
Example: A core limb of 3 cm by 4 cm has an area of $12 \, \text{cm}^2$.
$$
12 \, \text{cm}^2 = 12 \times (10^{-2} \, \text{m})^2 = 12 \times 10^{-4} \, \text{m}^2 = 0.0012 \, \text{m}^2
$$

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Part 2: Step-by-Step Calculation Process

Let's walk through the design of a transformer from start to finish.

Step 1: Define Requirements

• Primary (Input) Voltage: $V_p$
• Secondary (Output) Voltage: $V_s$
• Secondary (Output) Current: $I_s$
• Supply Frequency: $f$

Step 2: Calculate Power Handling

1. Secondary Apparent Power ($P_s$): This is the power delivered to the load.
   $$
   P_s = V_s \times I_s \quad (\text{in Volt-Amps, VA})
   $$

2. Required Primary Power ($P_p$): This must account for transformer inefficiency ($\eta$). A good estimate for $\eta$ is 85-95% (0.85-0.95).
   $$
   P_p = \frac{P_s}{\eta}
   $$

Step 3: Calculate Turns Per Volt (TPV)

This is the most critical parameter for your specific core and frequency. We rearrange the EMF equation to solve for $N/E$.

$$
\text{TPV} = \frac{N}{E} = \frac{1}{4.44 \times f \times B_{max} \times A_c}
$$

This value tells you how many turns are needed to induce or support a single volt.

Step 4: Calculate Winding Turns

Now, use the TPV to find the number of turns for each winding.

1. Primary Turns ($N_p$):
   $$
   N_p = V_p \times \text{TPV}
   $$

2. Secondary Turns ($N_s$): First, calculate the ideal number of turns.
   $$
   N_{s, \text{ideal}} = V_s \times \text{TPV}
   $$
   Then, add 5% to 10% to compensate for voltage drop under load (regulation). For a 5% compensation:
   $$
   Ns = N{s, \text{ideal}} \times 1.05
   $$

Step 5: Calculate Wire Gauge (Diameter)

The wire size is determined by the current it must carry.

1. Calculate Primary Current ($I_p$):
   $$
   I_p = \frac{P_p}{V_p}
   $$

2. Determine Required Wire Area ($A_w$): This is based on Current Density ($J$). A standard, conservative value for $J$ in transformers is 2.5 to 4.0 A/mm².
   $$
   Aw = \frac{\text{Current (I)}}{J} \quad (\text{in mm}^2)
   $$
   Calculate this for both primary and secondary windings:
   $$
   A{w,p} = \frac{Ip}{J} \quad \text{and} \quad A{w,s} = \frac{I_s}{J}
   $$

3. Select Wire Gauge: Use a standard wire table (AWG or SWG) to find the wire gauge whose cross-sectional area is equal to or just larger than your calculated $A_w$.

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Part 3: Worked Example

Design a transformer to convert 230V AC to 18V AC with a 3A output.

1. Requirements:
$V_p = 230 \, \text{V}$
$V_s = 18 \, \text{V}$
$I_s = 3 \, \text{A}$
$f = 50 \, \text{Hz}$

2. Power Calculation:
$P_s = 18 \, \text{V} \times 3 \, \text{A} = 54 \, \text{VA}$
Assume efficiency $\eta = 90\% = 0.90$.
* $P_p = \frac{54 \, \text{VA}}{0.90} = 60 \, \text{VA}$

3. Core and Properties:
Let's use a core with a center limb cross-section of $3.2 \, \text{cm} \times 3.5 \, \text{cm}$.
$Ac = 3.2 \times 3.5 = 11.2 \, \text{cm}^2 = 0.00112 \, \text{m}^2$
* We'll choose a conservative flux density: $B{max} = 1.2 \, \text{T}$

4. Turns Per Volt (TPV):
$$
\text{TPV} = \frac{1}{4.44 \times 50 \times 1.2 \times 0.00112} = \frac{1}{0.2983} \approx 3.35 \, \text{Turns/Volt}
$$

5. Winding Turns:
Primary: $N_p = 230 \, \text{V} \times 3.35 = 770.5 \rightarrow \mathbf{771 \, \text{turns}}$
Secondary: $N_{s, \text{ideal}} = 18 \, \text{V} \times 3.35 = 60.3 \rightarrow 61 \, \text{turns}$
* Add 5% for load compensation: $N_s = 61 \times 1.05 = 64.05 \rightarrow \mathbf{65 \, \text{turns}}$

6. Wire Gauge:
Let's use a current density of $J = 3 \, \text{A/mm}^2$.
Primary Current: $Ip = \frac{60 \, \text{VA}}{230 \, \text{V}} \approx 0.26 \, \text{A}$
* Primary Wire Area: $A{w,p} = \frac{0.26 \, \text{A}}{3 \, \text{A/mm}^2} = 0.087 \, \text{mm}^2$

*   From a wire table, **28 AWG** ($0.081 \, \text{mm}^2$) is slightly small. **27 AWG** ($0.102 \, \text{mm}^2$) is a safer choice.

• Secondary Current: $I_s = 3 \, \text{A}$
• Secondary Wire Area: $A_{w,s} = \frac{3 \, \text{A}}{3 \, \text{A/mm}^2} = 1.0 \, \text{mm}^2$
  • From a wire table, 17 AWG ($1.04 \, \text{mm}^2$) is a perfect fit.

Final Design Specification:

• Primary Winding: 771 turns of 27 AWG enameled copper wire.
• Secondary Winding: 65 turns of 17 AWG enameled copper wire.
• Insulation: Use appropriate insulating tape (Mylar, Kapton, or paper) between the core and primary, between primary and secondary, and between layers as needed.

SAFETY NOTICE: Working with mains voltage is extremely dangerous and can be fatal. Ensure you have a solid understanding of electrical safety practices. Always use proper insulation, and never work on a live circuit. Test your completed transformer carefully, starting with a current-limited supply if possible.

Answer 2

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Part 1: The Core Concepts (The "Why")

Before we jump into calculations, let's understand the key principles.

1. The Universal EMF Equation

This is the heart of all transformer design. It relates voltage to the transformer's physical properties.

E = 4.44 × f × N × B_max × A_c

Where:
E = RMS Voltage of the winding (Volts)
4.44 = A constant derived from the sine wave form factor (2π / √2).
f = Frequency of the AC supply (in Hertz, e.g., 50 Hz or 60 Hz).
N = Number of turns in the winding.
B_max = Maximum Magnetic Flux Density the core material can handle (in Tesla, T). This is a critical value you get from the core's datasheet.
A_c = Cross-Sectional Area of the transformer core (in square meters, m²).

2. Core Material (Determining B_max)

The core channels the magnetic field. Different materials have different limits for magnetic flux density before they "saturate" (can't hold any more magnetic flux), which causes massive inefficiency and distortion.
Silicon Steel (Laminations): The most common for 50/60 Hz transformers. A typical design value for B_max is between 1.0 T and 1.4 T. Using a lower value results in a larger, heavier transformer but with lower core losses.
Ferrite Cores: Used for high-frequency transformers (like in SMPS). They have a much lower B_max (e.g., 0.2 T to 0.4 T).

3. Core Geometry (Determining A_c)

You need to measure the central "limb" or "leg" of the core that the windings will go around.
* Core Area (A_c): For a square core, it's length × width. For a round core, it's π × r². Remember to convert this to square meters for the formula! (e.g., 5 cm x 4 cm = 20 cm² = 0.0020 m²).

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Part 2: Step-by-Step Calculation Guide

Let's design a transformer.

Step 1: Define Your Requirements

First, you must know what you want the transformer to do.
Input Voltage (V_p): e.g., 120V or 230V.
Output Voltage(s) (V_s): e.g., 12V.
Output Current(s) (I_s): e.g., 5A.
Frequency (f): e.g., 60 Hz.

Step 2: Calculate Power

• Secondary Apparent Power (P_a): This is the power the load requires.
  • P_a = V_s × I_s
  • Example: 12V × 5A = 60 VA (Note: We use VA, or Volt-Amps, for transformers).
• Primary Apparent Power (P_in): The primary winding needs to handle the secondary power plus the transformer's own losses. Assume an efficiency (η) of about 80-90% for a small transformer.
  • P_in = P_a / η
  • Example (90% efficiency): 60 VA / 0.90 = 66.7 VA

Step 3: Select a Core

For a new design, you select a core based on its power handling capability. A key metric is the Area Product (A_p), which is A_p = A_c × W_a (Core Area × Window Area). For simplicity here, let's assume you already have a core and have measured its Core Area (A_c).

Step 4: Calculate Turns Per Volt (TPV)

This is the most useful intermediate calculation. We rearrange the EMF equation to find how many turns are needed to support one volt.

TPV = 1 / (4.44 × f × B_max × A_c)

This value is a "magic number" for your specific core and frequency.

Step 5: Calculate Winding Turns

Now, simply use the TPV to find the number of turns for your primary and secondary windings.

• Primary Turns (N_p):
  • N_p = V_p × TPV
• Secondary Turns (N_s):
  • N_s = V_s × TPV
• Practical Tip: Add 5-10% extra turns to the secondary winding. This compensates for voltage drop that will occur when the transformer is under full load (this is called "regulation").
  • N_s_final = N_s × 1.05 (for a 5% increase)

Step 6: Calculate Wire Gauge (Diameter)

The wire size depends on the current it must carry.

1. Calculate Primary Current (I_p):
   • I_p = P_in / V_p
   • Example: 66.7 VA / 120V = 0.56 A
2. Determine Required Wire Area (A_w): This is based on Current Density (J), which is how much current can safely flow through a given cross-section of copper. A typical value is 2.5 to 4 Amps per square millimeter (A/mm²). A lower value means less heat and a more efficient transformer, but requires thicker wire.
   • A_w = Current / J
3. Calculate for both windings:
   • A_w_primary = I_p / J
   • A_w_secondary = I_s / J
4. Look up the Wire Gauge: Use the calculated wire area (A_w) to find the corresponding wire gauge (AWG or SWG) from a standard wire table. Choose the gauge with an area that is equal to or slightly larger than your calculated value.

Step 7: Check if it Fits (The Sanity Check)

Will all that copper wire actually fit in the core's window?
1. Find the total area occupied by each winding (Number of turns × Area per turn of insulated wire).
2. Sum the areas of all windings.
3. This total area must be less than the core's available Window Area (W_a). Practically, you can only fill about 30-40% of the window area due to insulation, air gaps, and the round shape of the wire. If it doesn't fit, you need a bigger core.

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Part 3: Worked Example

Let's design a 120V to 12V, 2A transformer.

1. Requirements:
V_p = 120 V
V_s = 12 V
I_s = 2 A
f = 60 Hz

2. Power:
P_a = 12V × 2A = 24 VA
Assume η = 90%. P_in = 24 / 0.90 = 26.7 VA

3. Core Selection & Properties:
Let's say we have a salvaged E-I lamination core. We measure the central leg and find its cross-section is 2.5 cm × 3.0 cm.
A_c = 7.5 cm² = 0.00075 m²
* We choose a conservative B_max of 1.2 Tesla for our silicon steel core.

4. Turns Per Volt (TPV):
TPV = 1 / (4.44 × 60 Hz × 1.2 T × 0.00075 m²)
TPV = 1 / (239.76)
* TPV = 4.17

5. Winding Turns:
Primary (N_p): 120V × 4.17 TPV = 500.4 -> Round up to 501 turns.
Secondary (N_s): 12V × 4.17 TPV = 50.04 -> Round to 50 turns.
* Add 5% for regulation: 50 turns × 1.05 = 52.5 -> Round up to 53 turns.

6. Wire Gauge:
Primary Current (I_p): 26.7 VA / 120V = 0.22A
Secondary Current (I_s): 2A
Let's use a Current Density (J) of 3 A/mm².
Primary Wire Area (A_w_p): 0.22A / 3 = 0.073 mm²

*   Looking at an AWG chart, **29 AWG** (0.064 mm²) is a bit small, **28 AWG** (0.081 mm²) is perfect. Let's choose **28 AWG**.

• Secondary Wire Area (A_w_s): 2A / 3 = 0.667 mm²
  • Looking at an AWG chart, 19 AWG (0.653 mm²) is very close, but 18 AWG (0.823 mm²) is a safer choice to reduce heat. Let's choose 18 AWG.

Final Design Summary:

• Primary Winding: 501 turns of 28 AWG copper wire.
• Secondary Winding: 53 turns of 18 AWG copper wire.
• Insulation: Use paper or Mylar tape between the primary and secondary windings, and between layers if needed.

Safety Warning: Building transformers involves mains voltage, which can be lethal. Always ensure proper insulation between the primary and secondary windings and between the windings and the core. If you are a beginner, start with low-voltage projects first.