Flyback Switch Mode Regulator Calculator estimates ideal BCM output power, switching frequency, duty cycle, transformer energy, current ratings and semiconductor stresses.
Why T_on and T_off Are Both Set by the Same Inductor Equation
A flyback transformer’s primary winding behaves like any other inductor: current through it ramps in proportion to the voltage across it and the time that voltage is applied. This calculator runs on that one relationship, applied twice — once while the switch is on and charging the core, once while it’s off and discharging into the output.
It’s boundary conduction mode, sometimes called critical conduction mode, where the next switching cycle starts the instant primary current reaches zero, with no dead time and no fixed clock.
During the on-time, the primary sees the input voltage minus whatever the switch drops across itself, and that drives current up to a peak:
$$T_{on} = \frac{L_P \times I_{pk}}{V_{IN} – V_{TRAN}}$$
Once the switch opens, that same stored current has to unwind through the secondary, and the voltage driving it down is the output voltage and diode drop reflected back through the turns ratio:
$$V_{OR} = N \times (V_{OUT} + V_D)$$
$$T_{off} = \frac{L_P \times I_{pk}}{V_{OR}}$$
Here $L_P$ is primary inductance, $I_{pk}$ is peak primary current, $V_{IN}$ is the input rail, $V_{TRAN}$ is the switch’s on-state voltage drop, $N$ is the primary-to-secondary turns ratio, $V_{OUT}$ is the regulated output, and $V_D$ is the output diode’s forward drop.
Add the two times together and invert the sum for switching frequency; divide on-time by the total for duty cycle. Neither is fixed — both move every time the input or the load changes, which is the defining trait of boundary conduction mode.
Sizing a 24V-to-12V Flyback Stage With a 3A Peak Current Limit
Take a 24V input stepping down to 12V, a 50 µH primary inductance, a peak current limit set directly at 3A, a 2:1 turns ratio, a 0.7V diode drop, and a 0.5V switch drop.
The voltage actually charging the inductor is 24 – 0.5 = 23.5V, so on-time works out to $$(50\text{µH} \times 3\text{A}) / 23.5\text{V} = 6.38$$ µs. Reflected voltage is $$2 \times (12 + 0.7) = 25.4$$V, giving an off-time of $$(50\text{µH} \times 3\text{A}) / 25.4\text{V} = 5.91$$ µs. Total period is 12.29 µs, which inverts to a switching frequency of 81.38 kHz at a duty cycle of 51.94%.
| Quantity | Value |
|---|---|
| Energy stored per cycle | 225.00 µJ |
| Output power | 17.30 W |
| Average output current | 1.44 A |
| Average input current | 0.78 A |
| Secondary peak current | 6.00 A |
| Switch V_DS stress | 49.40 V |
| Diode PIV | 24.00 V |
The energy figure comes from $\frac{1}{2} \times 50\text{µH} \times 3\text{A}^2 = 225$ µJ, stored and fully released once per cycle. Multiply that by the 81.38 kHz switching frequency and you get roughly 18.3W flowing into the diode-and-output combination; divide by the 12.7V that current sees on its way out and you land on 1.44A average output current, which at 12V is 17.30W delivered. None of this accounts for conduction or switching losses, so real efficiency will trail the number here.
When “Physics Error” Means the Numbers You Entered Can’t Exist
The calculator refuses to run if the switch’s voltage drop is equal to or greater than the input voltage, and it refuses again if the reflected voltage works out to zero or less. Neither is a soft warning — both describe a circuit that can’t physically operate.
If $V_{TRAN} \geq V_{IN}$, there’s no voltage left to push current into the primary at all. If $V_{OR} \leq 0$, the transformer has nothing pulling current back down during the off-time, so it never resets, and boundary conduction mode simply doesn’t exist for that combination of turns ratio and output voltage.
The switch and diode stress numbers, by contrast, don’t trigger any error at all — they just report what the semiconductors will see. A 49.4V V_DS and a 24V PIV in the example above aren’t pass or fail figures on their own; they’re the minimum a component has to survive.
Common practice in switch-mode design is to pick a MOSFET and diode rated with at least 20 to 25% margin above these calculated peaks, since leakage inductance spikes and line transients push the real voltage higher than this idealized model predicts.
Peak Current Squared: The Single Biggest Lever on Output Power
Output power tracks energy per cycle times switching frequency, and energy per cycle is $\frac{1}{2}L_PI_{pk}^2$. That squared term means a modest bump in peak current moves output power more than almost anything else in the circuit — double the peak current and stored energy quadruples, even before accounting for the frequency shift that comes with it.
Turns ratio pulls in the opposite direction on two different stresses at once. Raise it and reflected voltage climbs, which shortens off-time and adds directly onto the switch’s V_DS rating — but it also multiplies straight through to secondary peak current, since $I_{sec,pk} = N \times I_{pk}$.
There’s no turns ratio that improves both sides simultaneously; every increase trades switch voltage stress for higher secondary current demand, and every decrease does the reverse.
Inductance mainly sets the time scale rather than the power level. For a fixed peak current, a larger inductor stretches both on-time and off-time proportionally, which lowers switching frequency without changing the energy stored per cycle — useful for cutting switching losses, at the cost of a physically larger core.
One input changes which variable is fixed and which is derived rather than changing the physics: the saturation limit can be entered directly as a peak current in amps, or as a volt-microsecond rating, which is how many transformer datasheets actually specify core saturation, independent of whatever inductance ends up wound onto it.
Switch to that mode and the calculator flips the math around, deriving peak current from the volt-microsecond figure divided by primary inductance instead of taking it as a direct input.
Common Questions About Flyback BCM Design
What is boundary conduction mode in a flyback converter?
It’s an operating mode where the next switching cycle begins exactly when primary current decays to zero, with no gap and no fixed switching frequency. It sits right at the edge between continuous conduction, where current never reaches zero, and discontinuous conduction, where there’s a dead period before the next cycle starts.
How do you calculate the switching frequency of a flyback converter?
Add the on-time and off-time together to get the full switching period, then take the reciprocal. In boundary conduction mode that frequency isn’t a design input — it’s an output that shifts with line voltage, output current, and every other variable in the on-time and off-time equations.
What determines the duty cycle in a flyback converter?
Duty cycle is on-time divided by total period, and both of those depend on the same peak current but different driving voltages — input voltage minus switch drop on the way up, reflected output voltage on the way down. A higher reflected voltage relative to the input pulls duty cycle down; a lower one pushes it up.
How do you calculate peak primary current in a flyback transformer?
Either it’s set directly as a current limit from the controller or current-sense resistor, or it’s derived from a volt-microsecond saturation rating divided by primary inductance. Both routes describe the same physical ceiling: the current at which the core either saturates or trips the current-sense threshold.
What is reflected voltage (VOR) in a flyback converter?
It’s the output voltage plus the diode drop, multiplied by the turns ratio, representing how the secondary-side discharge voltage appears back on the primary side during the off-time. It adds directly onto the input voltage to set the switch’s peak voltage stress.
How do you size the MOSFET voltage rating for a flyback converter?
Start from input voltage plus reflected voltage, which gives the ideal peak drain-source stress, then add margin — commonly 20 to 25% — because leakage inductance produces a voltage spike this idealized calculation doesn’t include. A snubber or clamp circuit is usually what keeps that spike from exceeding the margin in a real build.
What is diode PIV in a flyback converter?
Peak inverse voltage is the reverse voltage the output diode blocks while the switch is on, equal to output voltage plus input voltage divided by turns ratio. It’s the number that sets the minimum voltage rating for the rectifier on the secondary side.
Why does flyback switching frequency vary with load in BCM?
Because on-time and off-time are both derived from the peak current needed to deliver the load, and that peak current changes as load changes. There’s no external clock forcing a fixed frequency the way there is in a fixed-frequency PWM controller — the period is whatever the volt-second math produces for that operating point.
These equations assume an ideal transformer and switch, so they don’t include leakage inductance voltage spikes, snubber losses, conduction and switching losses, or core loss — a real build will run hotter and less efficient than this idealized energy-balance model predicts.