Free · Physics models
Boundary-Mode Flyback Transformer Estimate
Calculate ideal duty, peak winding currents and referred inductances at the discontinuous-conduction boundary.
- Formula & worked example
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Calculator inputs
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How to use this calculator
- Enter the known values in the units shown. Results update as you type.
- Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
- Use the worked example to check the method. Reset restores the starting fields.
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Formula and method
At the boundary with 20 V input, 9 V output, 1 V rectifier drop, 1 A output and 100 kHz switching, a 2:1 transformer gives 50% duty and 50 µH primary inductance.
D = n (Vout + Vrect)/(Vin + n (Vout + Vrect)); Is,pk = 2 Iout/(1 − D).
Worked example
Enter these known values and leave the other values blank.
- Input voltage (**V==in==**)
- 20 V
- Output voltage (**V==out==**)
- 9 V
- Voltage on the rectifier (**V==rect==**)
- 1 V
- Output current (**I==out==**)
- 1 A
- Switching frequency (**f==s==**)
- 100 kHz
- Windings ratio (**N**)
- 2
- Duty cycle (**D**)
- 50 %
- Peak current on the secondary side (**I==s,Peak==**)
- 4 A
- Inductance on the secondary side (**L==s==**)
- 12.5 μH
- Peak current on the primary side (**I==p,Peak==**)
- 2 A
- Inductance on the primary side (**L==p==**)
- 50 μH
Assumptions and limitations
- Peak currents assume triangular pulses with no idle interval: the boundary between continuous and discontinuous conduction.
- Inductance scales as the square of the turns ratio. Leakage, saturation, switching loss and current ripple in continuous operation are omitted.
- Enter known quantities and clear the unknown to use available inverse branches. Locks retain supplied values. Consistency messages cover fully visible equations; identities with hidden dependencies may require a manual consistency check. Preset values are illustrative approximations, and unit selectors preserve the underlying quantity.
- The secondary-inductance inverse selects the physical minus-square-root duty branch, correcting the captured root above 100% duty.
Common questions
Why is secondary peak current more than output current?
Only the off-time pulse supplies the output; its triangular average is half its peak over that interval.
How do I solve for a different quantity?
Clear that quantity and supply enough independent known values. Some outputs are one-way or tied to a selected preset, so they cannot determine every input.
References
The secondary-inductance inverse must use the root D = 1 − sqrt(2 Iout f Ls/(Vout + Vrect)); the captured plus root always exceeds one and violates the physical duty domain. Explicit physical duty domain accompanies the corrected secondary-inductance inverse root.