Flyback Converter Calculator

Input Voltage (\( V_{\text{in}} \)):

Output Voltage (\( V_{\text{out}} \)):

Rectified Voltage (\( V_{\text{rect}} \)):

Windings Ratio (\( N \)):

Output Current (\( I_{\text{out}} \)):

Switching Frequency (\( f_s \)):

Duty Cycle (\( D \)):

Secondary Peak Current (\( I_{S,\text{peak}} \)):

Secondary Inductance (\( L_S \)):

Primary Peak Current (\( I_{P,\text{peak}} \)):

Primary Inductance (\( L_P \)):

1. What is Flyback Converter Calculator?

Definition: This calculator computes the duty cycle (\( D \)), peak currents (\( I_{S,\text{peak}}, I_{P,\text{peak}} \)), and inductances (\( L_S, L_P \)) of the primary and secondary windings of a flyback transformer in a flyback converter.

Purpose: It is used in power electronics to design flyback converters, which are commonly found in low-power DC-DC conversion applications such as power supplies for laptops, USB chargers, and LED drivers.

2. How Does the Calculator Work?

The calculator uses the following formulas:

Duty Cycle: \( D = \frac{V_{\text{out}} + V_{\text{rect}}}{(V_{\text{out}} + V_{\text{rect}}) \cdot N + V_{\text{in}}} \cdot N \)
Secondary Peak Current: \( I_{S,\text{peak}} = \frac{2}{1 - D} \cdot I_{\text{out}} \)
Secondary Inductance: \( L_S = \frac{V_{\text{out}} + V_{\text{rect}}}{2 \cdot I_{\text{out}} \cdot f_s \cdot (1 - D)^2} \)
Primary Peak Current: \( I_{P,\text{peak}} = \frac{I_{S,\text{peak}}}{N} \)
Primary Inductance: \( L_P = L_S \cdot N^2 \)

Where:

\( D \): Duty cycle (unitless, between 0 and 1)
\( V_{\text{in}} \): Input voltage (V)
\( V_{\text{out}} \): Output voltage (V)
\( V_{\text{rect}} \): Voltage rectified by diode \( D_1 \) (V)
\( N \): Windings ratio (\( N_p / N_s \), unitless)
\( I_{\text{out}} \): Output current (A)
\( f_s \): Switching frequency (Hz)
\( I_{S,\text{peak}} \): Peak current through the secondary winding (A)
\( L_S \): Inductance of the secondary winding (H)
\( I_{P,\text{peak}} \): Peak current through the primary winding (A)
\( L_P \): Inductance of the primary winding (H)

Steps:

Enter the input voltage (\( V_{\text{in}} \)), output voltage (\( V_{\text{out}} \)), rectified voltage (\( V_{\text{rect}} \)), windings ratio (\( N \)), output current (\( I_{\text{out}} \)), and switching frequency (\( f_s \)) with their units.
Convert all inputs to base units (V, A, Hz).
Calculate the duty cycle, peak currents, and inductances using the formulas.
Convert the results to the selected output units.
Display the results: if a value is less than 0.001 in the selected unit, use scientific notation; otherwise, display with 4 decimal places.

3. Importance of Flyback Converter Calculation

Calculating the parameters of a flyback converter is crucial for:

Power Supply Design: Ensuring the flyback converter can deliver the required output voltage and current for applications like chargers or LED drivers.
Efficiency: Optimizing the duty cycle to improve the efficiency of the converter and reduce losses.
Component Selection: Determining the appropriate inductance values and current ratings for the transformer windings to prevent saturation or overheating.

4. Using the Calculator

Example 1: Calculate the parameters for a flyback converter with \( V_{\text{in}} = 12 \, \text{V} \), \( V_{\text{out}} = 5 \, \text{V} \), \( V_{\text{rect}} = 0.7 \, \text{V} \), \( N = 2 \), \( I_{\text{out}} = 1 \, \text{A} \), and \( f_s = 50 \, \text{kHz} \):

Input Voltage (\( V_{\text{in}} \)): 12 V
Output Voltage (\( V_{\text{out}} \)): 5 V
Rectified Voltage (\( V_{\text{rect}} \)): 0.7 V
Windings Ratio (\( N \)): 2
Output Current (\( I_{\text{out}} \)): 1 A
Switching Frequency (\( f_s \)): 50 kHz = \( 50 \times 10^3 \) Hz
Duty Cycle (\( D \)): \( \frac{5 + 0.7}{(5 + 0.7) \cdot 2 + 12} \cdot 2 = \frac{5.7}{5.7 \cdot 2 + 12} \cdot 2 = \frac{5.7}{11.4 + 12} \cdot 2 = \frac{5.7}{23.4} \cdot 2 \approx 0.4872 \)
Secondary Peak Current (\( I_{S,\text{peak}} \)): \( \frac{2}{1 - 0.4872} \cdot 1 \approx \frac{2}{0.5128} \approx 3.8997 \, \text{A} \)
Secondary Inductance (\( L_S \)): \( \frac{5 + 0.7}{2 \cdot 1 \cdot 50 \times 10^3 \cdot (1 - 0.4872)^2} = \frac{5.7}{2 \cdot 50 \times 10^3 \cdot (0.5128)^2} \approx \frac{5.7}{100 \times 10^3 \cdot 0.2630} \approx 2.1673 \times 10^{-4} \, \text{H} = 0.2167 \, \text{mH} \)
Primary Peak Current (\( I_{P,\text{peak}} \)): \( \frac{3.8997}{2} \approx 1.9498 \, \text{A} \)
Primary Inductance (\( L_P \)): \( 2.1673 \times 10^{-4} \cdot 2^2 \approx 2.1673 \times 10^{-4} \cdot 4 \approx 8.6692 \times 10^{-4} \, \text{H} = 0.8669 \, \text{mH} \)
Result: \( D = 0.4872 \), \( I_{S,\text{peak}} = 3.8997 \, \text{A} \), \( L_S = 0.2167 \, \text{mH} \), \( I_{P,\text{peak}} = 1.9498 \, \text{A} \), \( L_P = 0.8669 \, \text{mH} \)

Example 2 (Demonstrating Scientific Notation): Calculate the parameters for a flyback converter with \( V_{\text{in}} = 100 \, \text{V} \), \( V_{\text{out}} = 5 \, \text{V} \), \( V_{\text{rect}} = 0.7 \, \text{V} \), \( N = 10 \), \( I_{\text{out}} = 0.01 \, \text{mA} \), and \( f_s = 1 \, \text{MHz} \):

Input Voltage (\( V_{\text{in}} \)): 100 V
Output Voltage (\( V_{\text{out}} \)): 5 V
Rectified Voltage (\( V_{\text{rect}} \)): 0.7 V
Windings Ratio (\( N \)): 10
Output Current (\( I_{\text{out}} \)): 0.01 mA = \( 0.01 \times 10^{-3} = 10^{-5} \, \text{A} \)
Switching Frequency (\( f_s \)): 1 MHz = \( 1 \times 10^6 \) Hz
Duty Cycle (\( D \)): \( \frac{5 + 0.7}{(5 + 0.7) \cdot 10 + 100} \cdot 10 = \frac{5.7}{5.7 \cdot 10 + 100} \cdot 10 = \frac{5.7}{57 + 100} \cdot 10 = \frac{5.7}{157} \cdot 10 \approx 0.3631 \)
Secondary Peak Current (\( I_{S,\text{peak}} \)): \( \frac{2}{1 - 0.3631} \cdot 10^{-5} \approx \frac{2}{0.6369} \cdot 10^{-5} \approx 3.1392 \times 10^{-5} \, \text{A} \), in mA: \( 3.1392 \times 10^{-5} \times 10^3 = 3.1392 \times 10^{-2} \, \text{mA} = 0.0314 \, \text{mA} \), in µA: \( 3.1392 \times 10^{-5} \times 10^6 = 31.392 \, \text{µA} \)
Secondary Inductance (\( L_S \)): \( \frac{5 + 0.7}{2 \cdot 10^{-5} \cdot 1 \times 10^6 \cdot (1 - 0.3631)^2} = \frac{5.7}{2 \cdot 10^{-5} \cdot 10^6 \cdot (0.6369)^2} \approx \frac{5.7}{20 \cdot 0.4056} \approx 0.7027 \, \text{H} \)
Primary Peak Current (\( I_{P,\text{peak}} \)): \( \frac{3.1392 \times 10^{-5}}{10} \approx 3.1392 \times 10^{-6} \, \text{A} \), in mA: \( 3.1392 \times 10^{-6} \times 10^3 = 3.1392 \times 10^{-3} \, \text{mA} \), in µA: \( 3.1392 \times 10^{-6} \times 10^6 = 3.1392 \, \text{µA} \)
Primary Inductance (\( L_P \)): \( 0.7027 \cdot 10^2 \approx 70.27 \, \text{H} \)
Result: \( D = 0.3631 \), \( I_{S,\text{peak}} = 0.0314 \, \text{mA} \) (or \( 31.3920 \, \text{µA} \)), \( L_S = 0.7027 \, \text{H} \), \( I_{P,\text{peak}} = 3.1392 \times 10^{-3} \, \text{mA} \) (or \( 3.1392 \, \text{µA} \)), \( L_P = 70.2700 \, \text{H} \)

5. Frequently Asked Questions (FAQ)

Q: What is a flyback converter?
A: A flyback converter is a type of DC-DC converter that uses a transformer to step up or step down voltage. It is commonly used in low-power applications due to its simplicity and ability to provide galvanic isolation.

Q: Why must the duty cycle be between 0 and 1?
A: The duty cycle represents the fraction of time the switch is on during one switching cycle. A value greater than 1 would imply the switch is on for longer than the cycle period, which is physically impossible.

Q: How does the windings ratio affect the converter design?
A: The windings ratio (\( N \)) determines the voltage transformation between the primary and secondary sides of the transformer. A higher \( N \) increases the step-up or step-down ratio, affecting the duty cycle and the currents in the windings.