Flyback Switch Mode Regulator Calculator

Flyback Switch Mode Regulator Calculator finds the most power a transformer passes in boundary conduction from its inductance and Isat, or the inductance that a load current needs.

Flyback Switch Mode Regulator Calculator Inputs and Modes

The Flyback Switch Mode Regulator Calculator models a flyback at boundary conduction. That is the point where the secondary current reaches zero just as the switch turns on again. Texas Instruments calls the same point critical conduction mode, or CRM, in its LM3481 flyback application report.

Voltage Primary is the DC rail across the primary winding and switch. Voltage Out 1 is the regulated output that the feedback loop watches. Turns Ratio is primary turns divided by secondary turns, so an entry of 1.5 means a 1.5 : 1 transformer.

Primary Inductance comes from the transformer data sheet in microhenries. The saturation limit goes in as Isat in amps or as a VT product in volt-microseconds. Diode and transistor drops sit in their own panel and start at 0.7 V and 0.5 V.

The Calculate menu sets the direction of the math. Transformer Max Power starts from a part already on the bench and reports the most it can deliver. Inductance for a Load starts from an output current and a switching frequency and reports the inductance and peak current to order.

From Isat to Maximum Power

The output voltage and diode drop appear on the primary as the reflected voltage while the diode conducts. The turns ratio scales them up, so a higher ratio means a higher reflected voltage.

$$V_{OR}=N\times (V_O+V_D)$$

Volt-second balance on the primary then fixes the duty cycle. The on-state voltage is the input minus the transistor drop, and the off-state voltage is the reflected voltage.

$$D=\frac{V_{OR}}{V_{OR}+V_{IN}-V_{TRAN}}$$

The switch can stay on only until the primary current reaches the saturation current. A VT product entry is the same limit, because VT equals Isat times the primary inductance.

$$T_{ON}=\frac{L_P\times I_{SAT}}{V_{IN}-V_{TRAN}}$$

Dividing the duty cycle by that longest on time gives the lowest frequency the transformer can run at. Below it the current would climb past Isat before the switch opens.

$$F_{MIN}=\frac{D}{T_{ON}}$$

Each cycle stores half the inductance times the peak current squared. Multiplying that energy by the frequency gives the power that crosses the transformer.

$$P=\frac{L_P\times I_{SAT}^2\times F_{MIN}}{2}$$

That power lands on the output plus the diode drop, so the output current divides by both. Output power is then the output current times Voltage Out 1, the same product the Electrical Power Calculator works out for any DC load.

$$I_O=\frac{P}{V_O+V_D}$$

The top result of the Flyback Switch Mode Regulator Calculator names the limit as core saturation and shows the maximum output power. The line under it gives the output voltage times current and the share of input power that reaches the load after the two drops.

The first card pairs the minimum frequency with the on time and duty cycle. The second card shows the microjoules stored per cycle, the power through the transformer and the VT product. If VT was the entry, that row shows the peak current instead.

12V Output From a 24V Input

The Flyback Switch Mode Regulator Calculator opens on a 12V output from a 24 V rail. The transformer is 1.5 : 1 with 50 µH on the primary and a 3 A saturation current. The reflected voltage is 1.5 × 12.7 V, or 19.05 V, and the primary sees 23.5 V while the switch is on.

Duty cycle is 19.05 ÷ 42.55, or 44.77%. The VT product is 3 A × 50 µH, or 150 V·µs, so the on time is 6.383 µs. That sets a minimum frequency of 70.14 kHz, with 225 µJ stored on every cycle.

Power through the transformer is 15.78 W. The 12.7 V at the secondary turns that into 1.243 A, and the load receives 14.91 W. The input supplies 16.12 W, so the diode and transistor drops take 1.21 W between them.

That 1.21 W ends up as heat in the diode and the switch, and the Heat Sink Calculator sizes the cooling for it. Real losses run higher. Analog Devices uses about 85% efficiency in the output power equation of its LT8302 data sheet, which would put this stage near 13.7 W.

The Operating Point by Input Voltage panel reruns the same transformer at half to double the entered rail. A 12 V supply and a 48 V supply behave very differently on the same part.

The same transformer passes 10.16 W from 12 V and 19.27 W from 48 V. Peak current is pinned at 3 A, so a higher rail reaches it sooner and the frequency rises. More cycles per second carry more power at the same energy per cycle.

On a 12 V battery the average input current is 935 mA at full output, half the 3 A peak times the 62.36% duty. That figure goes into the Battery Life Calculator to estimate runtime. If the minimum frequency falls below about 20 kHz, the Flyback Switch Mode Regulator Calculator warns that the transformer can whine.

Flyback Transformer Calculator for a Set Load

Inductance for a Load turns the Flyback Switch Mode Regulator Calculator into a flyback transformer calculator. The output current and switching frequency go in, and the primary inductance that puts that load exactly on the boundary comes out.

$$L_P=\frac{(V_{IN}-V_{TRAN}{)}^2\times D^2}{2\times F\times I_O\times (V_O+V_D)}$$

With 24 V in, 12 V at 1 A out and 100 kHz, the answer is 43.58 µH. The on time is 4.477 µs and the primary peak is 2.414 A. The VT product needed is 105.2 V·µs, and the secondary peak is 3.621 A through the diode.

That peak is the Isat the transformer must clear. Coilcraft's guide to flyback transformers says to pick a saturation current well above the expected primary peak. An Analog Devices flyback design note sets the controller current limit at 1.2 times the peak, which is 2.9 A here.

The input-voltage panel in this mode lists inductance and peak for each rail. At 12 V in, the same 1 A load needs 20.25 µH and peaks at 3.542 A. At 48 V in it needs 72.79 µH and peaks at 1.868 A. A design that must stay out of continuous conduction takes the low-line inductance and the low-line peak.

A flyback transformer design tool that outputs winding turns needs one more number, the AL value of the gapped core. Turns equal the square root of inductance divided by AL. A core with 100 nH per turn squared needs 21 primary turns for 43.58 µH, and 14 secondary turns at 1.5 : 1.

Wire gauge follows from RMS current, which for a triangle wave is the peak times the square root of duty over three. The 2.414 A primary peak works out to 0.93 A RMS. The peak also sets the current-sense resistor through Ohm's law, as sense threshold divided by peak current.

Switch and Diode Voltage Ratings

When the switch opens, the reflected voltage stacks on top of the input. The diode blocks the output plus the primary voltage divided by the turns ratio while the switch is on.

$$V_{DS}=V_{IN}+V_{OR}$$

$$V_R=V_O+\frac{V_{IN}-V_{TRAN}}{N}$$

The Flyback Switch Mode Regulator Calculator puts 43.05 V on the switch and 27.67 V across the diode in the 24 V example. Neither figure includes the leakage spike. The TI LM3481 report advises a MOSFET rated 1.5 to 2 times the calculated voltage and a diode rated 1.2 to 1.5 times.

That puts the switch at 65 V to 86 V, so an 80 V or 100 V part fits. The diode lands at 33 V to 42 V, which points to a 45 V or 60 V Schottky. At 48 V in, the switch voltage is 67.05 V before any margin.

The LT8302 data sheet shows the same habit on an integrated part. Its switch is rated 65 V, and the input plus reflected voltage is held under 50 V to leave 15 V for the spike. A higher turns ratio raises switch voltage and lowers diode voltage, so the two ratings trade against each other.

Output Capacitor Calculation for a Flyback Converter

The output capacitor carries the load alone while the switch is on. The LT8302 data sheet sizes it from the primary inductance, the peak switch current and the ripple allowed on the output.

$$C_{OUT}=\frac{L_P\times I_{PK}^2}{2\times V_O\times \Delta V_O}$$

The top of that fraction is twice the energy per cycle that the Flyback Switch Mode Regulator Calculator shows on its second card. For the 24 V example, 225 µJ sits on a 12 V output with 120 mV of ripple allowed, which is 1%. The result is 450 µJ ÷ (2 × 12 V × 0.12 V), or 156 µF.

Analog Devices adds that a ceramic capacitor can fall to 40% of its marked value at full rated voltage. A 156 µF target therefore needs more than 156 µF of marked capacitance. Ripple on a scope is read peak to peak, the same measure the Peak to Peak Voltage Calculator works with.

The capacitor also has to survive the ripple current. The secondary current is a triangle that peaks at 4.5 A in the example and flows for 55.23% of the cycle. Its RMS value is 1.93 A, against 1.243 A delivered to the load.

Flyback Switch Mode Regulator Calculator Excel Formulas

A Flyback Switch Mode Regulator Calculator Excel sheet needs eight formulas under seven entries. The cells below repeat the Step-by-Step Working panel line for line, with inductance in µH and time in µs.

A VT product entry replaces B5 with the VT figure divided by B4. For a flyback transformer design calculation in Excel, two more cells finish the job. Switch voltage is =B1+B8 and diode reverse voltage is =B2+(B1-B7)/B3.

Copying the sheet across a row of input voltages rebuilds the operating point table. Energy per cycle in B12 is in microjoules, and the Electrical Energy Calculator applies the same power-times-time relation to a whole circuit.

Flyback Converter Design With TI Tools

TI publishes flyback converter design tools that pick up where the Flyback Switch Mode Regulator Calculator stops. Power Stage Designer is a Java program that plots voltages and currents for 21 topologies, the flyback among them. It adds an RCD snubber calculator for the leakage spike and a FET loss comparison.

The Fly-Buck and Flyback Topology Selector is a downloadable calculation tool. It covers DCM and CCM flyback with up to 6 isolated outputs, recommends a controller and includes loop compensation. For flyback converter design in Excel, the LM5180 data sheet points to a quick-start spreadsheet for that part.

Those tools start from a load and a controller. The Flyback Switch Mode Regulator Calculator answers an earlier question, which is whether a given transformer can carry the load at all. Coilcraft places the flyback's usual range below about 100 W and 10 A of output, with forward and bridge topologies taking over above that.

Duty Cycle, Saturation and Flyback Voltage

Can a boundary mode flyback run above 50% duty?

Yes. The Flyback Switch Mode Regulator Calculator shows 62.36% duty in the 12 V row of the table above. In a TI E2E forum thread on this exact question, TI engineer Eric Faraci explains that a flyback can work anywhere from 0 to 100% duty with the right control method.

The familiar 50% ceiling comes from subharmonic oscillation in fixed-frequency peak current mode control without slope compensation. The LT8302 data sheet notes that boundary mode does not show that oscillation. Its own 5 V design table lists duty from 33% to 67% at a 3 : 1 ratio.

What happens when the peak current passes Isat?

The core saturates and the primary inductance drops, so current climbs faster for the same on time. Coilcraft warns that the load then stops receiving the energy it needs and regulation is lost. Analog Devices adds that energy injected after saturation is dissipated in the core instead of reaching the secondary.

In the 24 V example the limit is 3 A, reached after 6.383 µs at 23.5 V. Raising the rail to 48 V shortens that to 3.158 µs, because 150 V·µs is fixed by the part. A controller that holds a fixed on time across a wide input range will overshoot Isat at high line.

How is the flyback voltage calculated?

The flyback voltage on the primary is the reflected voltage, the turns ratio times the output plus the diode drop. For 12 V out, a 0.7 V diode and 1.5 : 1, that is 19.05 V. The Flyback Switch Mode Regulator Calculator reports it on the switch card as Reflected Voltage (Vor).

It can also be backed out of a measured duty cycle. Multiply the duty by the primary voltage and divide by one minus the duty. With 44.77% and 23.5 V that is 10.52 ÷ 0.5523, which returns the same 19.05 V and confirms the converter sits on the boundary.