100nF

Flyback transformer calculator

A flyback stores energy in its transformer's gap while the switch is on and delivers it through the secondary when the switch is off, so the turns ratio sets the duty cycle, the reflected voltage on the switch and the reverse voltage on the rectifier all at once. TI's SNVA866 lays out the continuous-conduction design in order — turns from the maximum duty cycle at minimum input, primary inductance from a 30–70 % ripple ratio at maximum input, then the ripple and peak currents — and the calculator follows it, adding the reflected voltage and the drain and diode stresses with the clamp margin you allow.

0.0 A3.8 A7.5 A0D·TTone period at V_in min = 18.0 V, D = 35.7 %, T = 4.00 µsprimary2.5 A → 3.8 Asecondary: ×N_P/N_S, 7.51 A peak
Fig 1 — One period at the minimum input in continuous conduction. During the on-time the primary current ramps by 1.22 A to 3.75 A (SNVA866 eq 7 and 8) while energy is stored in the gap; during the off-time the same ampere-turns appear in the secondary, 7.51 A peak for a 0.500 turns ratio, and the output voltage reflects back onto the primary as 10.0 V.
Output power · N_S/N_P for 40 % at 18.0 V · used
20.2 W · 0.417 · 0.500 (2.00:1), aux 1.00
Duty cycle at V_in min · at V_in max
35.7 % · 21.7 %
Primary inductance for 60 % ripple at 36.0 V · used · ripple ratio with it
20.2 µH · 21.0 µH · 58 %
At 18.0 V: primary ripple · peak · secondary peak
1.22 A · 3.75 A · 7.51 A
Reflected voltage V_OR · drain at 36.0 V with clamp · rectifier reverse
10.0 V · 61.0 V · 23.0 V

The turns ratio trades the switch against the rectifier: V_OR = 10.0 V adds to the input on the drain (61.0 V at 36.0 V including the 15.0 V leakage spike the clamp allows), while the rectifier sees the output plus the input divided by the same ratio, 23.0 V. SLUP254: the clamp voltage "is maximum at full load and minimum input voltage" and is a "tradeoff between efficiency, peak drain voltage, output current limit and cross regulation".

The peak of 3.75 A sizes the current-sense resistor and the transformer's saturation current — SNVA866's example chose a 6 A part for a 3.75 A peak. The leakage inductance is not in these numbers: SLUP254 puts it at "a function of winding geometry, number of turns and separation between primary and secondary", reduced by interleaving, and "not lowered with a high permeability core".

How this is calculated

Standard: TI SNVA866; TI SLUP254

D=(NP/NS) VLOADVSUPPLY+(NP/NS) VLOAD,NS=VLOAD(1−Dmax) NPVSUPPLY,min DmaxD = \frac{(N_P/N_S)\,V_{LOAD}}{V_{SUPPLY} + (N_P/N_S)\,V_{LOAD}}, \qquad N_S = \frac{V_{LOAD}(1 - D_{max})\,N_P}{V_{SUPPLY,min}\,D_{max}}
SNVA866 eq 2–4: the duty cycle the turns impose, and the turns for a chosen maximum duty cycle at minimum input.
LM=NP2 Vmax2 VLOAD2ILRR fSW POUT (NSVmax+NPVLOAD)2L_M = \frac{N_P^2\,V_{max}^2\,V_{LOAD}^2}{I_{LRR}\,f_{SW}\,P_{OUT}\,(N_S V_{max} + N_P V_{LOAD})^2}
SNVA866 eq 6: primary inductance for a ripple ratio I_LRR at maximum input; 30–70 % is the note's range.
ΔIL=VSUPPLY,min DmaxLMfSW,IPEAK=POUTVSUPPLY,min Dmax+ΔIL2\Delta I_L = \frac{V_{SUPPLY,min}\,D_{max}}{L_M f_{SW}}, \qquad I_{PEAK} = \frac{P_{OUT}}{V_{SUPPLY,min}\,D_{max}} + \frac{\Delta I_L}{2}
SNVA866 eq 7 and 8: the ripple and peak primary current at minimum input.
VOR=NPNSVLOAD,VDS=VSUPPLY,max+VOR+Vclamp,Vdiode=VLOAD+NSNPVSUPPLY,maxV_{OR} = \frac{N_P}{N_S} V_{LOAD}, \qquad V_{DS} = V_{SUPPLY,max} + V_{OR} + V_{clamp}, \qquad V_{diode} = V_{LOAD} + \frac{N_S}{N_P} V_{SUPPLY,max}
The reflected voltage and the switch and rectifier stresses; the clamp margin is the leakage spike the RCD clamp allows (SLUP254).

Assumptions

What sets a flyback transformer

A flyback transformer is an inductor with two windings. SLUP254: "energy is stored in flyback transformer" while the switch is on, and when it turns off the "stored energy [is] transferred to output" — through the secondary, at a voltage set by the turns. Because the two windings never conduct at once, the turns ratio does what a buck's duty cycle does and more. SNVA866's design order is the calculator's. The duty cycle follows from the ratio and the input: D = (NP/NS· VLOAD) / (VSUPPLY + NP/NS · VLOAD), largest at the minimum input. So choosing the maximum duty cycle chooses the turns: "the maximum duty cycle occurs when the supply voltage is at the minimum value. By selecting the maximum duty cycle, the number of turns on the secondary winding is calculated", and the note keeps it under 50 % because that "reduces the need for slope compensation" and "the right-half plane zero of the modulator is pushed to high frequencies".

The primary inductance sets the ripple. "Three main parameters are considered when selecting the inductance value of primary winding: primary winding current ripple ratio, falling slope of the transformer current and the right half plane zero frequency", and the note's balance is "a maximum ripple ratio between 30 % and 70 %": more ripple means "the core losses increase and the copper losses decrease", less means a larger inductance and a lower right-half-plane zero. The ripple is largest at maximum input, so the inductance is set there (eq 6), and the peak current, which "occurs at the minimum supply voltage", is the pedestal POUT/(VSUPPLY_min·D) plus half the ripple (eq 7 and 8). That peak "is used to properly size the current sense resistor" and the transformer's saturation rating.

Two voltages come with the ratio. When the switch is off the output reflects onto the primary as VOR = (NP/NS)·VLOAD, which sits on top of the input at the drain — and above that the spike from leakage inductance, which the RCD clamp limits: SLUP254's "during commutation primary-to-secondary, the leakage energy is absorbed by the clamp circuit", "Vclamp is maximum at full load and minimum input voltage", and the level is a "tradeoff between efficiency, peak drain voltage, output current limit and cross regulation". The rectifier, during the on-time, sees the output plus the input divided by the ratio. A ratio that eases the switch loads the diode.

Worked example: SNVA866's 18–36 V to 5 V, 4 A at 250 kHz

The defaults are the note's: 18–36 V in, 5 V at 4 A plus a 10 V, 20 mA bias winding (20.2 W), 250 kHz, a 40 % target duty cycle, 60 % ripple, and the rounded values it chose — 0.5 turns per primary turn and 21 µH — with a 15 V clamp margin.

secondary turns   V_LOAD (1 − D_max) / (V_min D_max) = 5 × 0.6 / (18 × 0.4) = 0.417 per turn   (eq 3)
chosen 0.5        D_max = (2 × 5) / (18 + 2 × 5)      = 35.7 %                                  (eq 4)
auxiliary         0.5 × 10 V / 5 V                     = 1 turn                                  (eq 5)
inductance        36² × 5² / (0.6 × 250 kHz × 20.2 W × (0.5 × 36 + 5)²) = 20.2 µH               (eq 6; note: 20.6 µH)
chosen 21 µH      ripple 18 V × 0.357 / (21 µH × 250 kHz) = 1.22 A                              (eq 7)
peak              20.2 W / (18 V × 0.357) + 1.22 / 2   = 3.75 A                                  (eq 8)
reflected         2 × 5 V                              = 10 V;  drain 36 + 10 + 15 = 61 V;  diode 5 + 36 / 2 = 23 V

The note's transformer: "turns ratio 1:0.5:1 (2:1:2), primary winding inductance 21 µH, saturation current 6 A". The 6 A against a 3.75 A peak is the margin for the current limit and the leakage-shifted duty cycle SLUP254 warns of — "higher duty cycle and magnetizing current than expected" — and the 61 V drain is why a 100 V MOSFET is the usual choice on a 36 V input.

Where the flyback model stops being valid

Common flyback mistakes

Further reading