DIY Transformer Design for Small Electronics: Primary and Secondary Coil Calculations

The inductances of a transformer are typically derived from its design parameters, such as core material, winding turns, and operating frequency. Here a typical design calculation for a 50 Hz transformer based on voltage levels and power rating requirement for a small electronics circuit board that will power +12V motor and actuator and +5V for IC and microcontroller will be explained. Here's how I approached it:

The following shows simple AC to DC converter circuit, that is power supply circuit.

In the above power supply design, the transformer TR1 receives 220V AC @50Hz signal at its primary and the primary and secondary coils are wound such that the secondary output voltage is 20V RMS. Now we will calculate the primary and total secondary inductances, the number of turns required to make the DIY power supply transformer with the said input and output voltage requirement. For details on the operation of the transformer see What Is Step-Down Transformer and How It Works?.

1. Primary Inductance ($L_p$)

The primary inductance can be approximated using the formula:

Where:

• $N_p$: Number of turns in the primary winding
• $\mu$: Permeability of the core material
• $A_e$: Cross-sectional area of the core (in $m^2$)
• $l_c$: Mean magnetic path length of the core (in $m$)

For a 50 Hz transformer:

• $L_p$ typically falls in the range of $1\,\text{H} - 10\,\text{H}$, depending on the transformer's power rating.
• Small transformers ($<50VA$): $L_p \approx 1 - 3\,\text{H}$
• Medium transformers ($50-500VA$): $L_p\approx 3-7\text{H}$

Approximation for Your Transformer:

Assuming a power rating around 100 VA and good magnetic coupling, I selected $L_p \approx 5\,\text{H}$ as a reasonable estimate the design.

2. Secondary Inductance ($L_s$)

The secondary inductance is related to the primary inductance by the square of the turns ratio:

The turns ratio is determined by the voltage levels of the primary and secondary windings:

For your transformer:

• Primary voltage $V_p = 230\,\text{V$
• Secondary voltage $V_s = 20\,\text{V RMS}$

Using the selected $L_p = 5\,\text{H}$:

3. DC Resistances ($R_p$ and $R_s$)

The DC resistance of the windings depends on the wire gauge (AWG), length of the wire, and resistivity of copper ($1.68 \times 10^{-8}\,\Omega\,\text{m}$).

Formula:

Where:

• $\rho$: Resistivity of copper
• $l$: Length of the wire
• $A$: Cross-sectional area of the wire

For a typical transformer:

• Primary Resistance ($R_p$): Small transformers have primary resistances in the range $0.1 - 1\,\Om$.For your case, $R_p \approx 0.1\,\Omega$
• Secondary Resistance ($R_s$): Based on the lower voltage and higher current, $R_s \approx 0.5\,\Omega$

Final Values Recap:

• Primary Inductance ($L_p$): $5\,\text{H}$
• Secondary Inductance ($L_s$): $37.8\,\text{mH}$
• Primary DC Resistance ($R_p$): $0.1\,\Omega$
• Secondary DC Resistance ($R_s$): $0.5\,\Omega$