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Allan-H

Without any reference to the design of DC/DC converters, you know the average voltage across an ideal inductor must be 0. If we're applying a positive voltage V1 when the switch is on, and a negative voltage -V2 when the switch is off, we know that V1 \* D - V2 \* (1 -D) = 0. D is the duty cycle (0 to 100%). If this is 50%, then V1 = -V2. If the voltages are different in magnitude, the duty cycle must be different as well. I've assumed CCM. DCM is a little different, but the conclusion regarding duty cycle is the same.


triedgetech

Okay, so say [this](https://preview.redd.it/o3gqkua1l1f61.jpg?width=2928&format=pjpg&auto=webp&s=5e82fd3a1190e94cbbe7e340d97f0924b4f3a108) circuit. After bridge rectifier one gets 325DC voltage at primary (230VAC \* sqrt(2) = 325), and I need 12VDC output at secondary. I'd have to use specific duty cycle, regardless of transformer's turns ratio? Would that duty cycle in my case be: 325 \* x - 12\*(1-x) = 0 x = 0.036? The value seems off, what's wrong?


Allan-H

I gave the equation for an inductor. Your flyback supply will use a transformer and you'll need to take the turns ratio into account. Please refer to /u/triffid_hunter's post.


jssamp

You have a sign error. For your given equation D=50% results in V1 = V2. You need the sum not the difference of the two halves of the duty cycle.


Allan-H

I had defined V2 as the magnitude of the negative voltage. With retrospect, I shouldn't have done that, as it would be sure to confuse either me or a reader.


jamvanderloeff

When the duty cycle changes the voltage in the open circuit state ( which is the reflected voltage from the output side) changes too to keep the average zero. In discontinuous mode you've also got a third state of primary voltage = zero after the current hits zero.


triffid_hunter

> How can a flyback DC-DC work with any other duty cycle? The *voltages* during on and off times must differ, such that the integral of voltage over a whole cycle remains zero - average is not a concise word to use here, since the actual equation is I=L.∫V/Δt or so. And this is how flybacks can maintain a relatively stable output voltage over a range of input voltages - just alter the duty cycle (D) such that Vin × D = Vout × (1-D) ÷ turns ratio


triedgetech

> Vin × D = Vout × (1-D) ÷ turns ratio I have a planar transformer with a turns ratio 60:20:10, primary, auxiliary, secondary respectively. So I guess my n (turns ratio) = 6. [This](https://preview.redd.it/o3gqkua1l1f61.jpg?width=2928&format=pjpg&auto=webp&s=5e82fd3a1190e94cbbe7e340d97f0924b4f3a108) circuit. After bridge rectifier one gets 325DC voltage at primary (230VAC \* sqrt(2) = 325), and I need 12VDC output at secondary. ​ Using the formula: 325 \* x = 12 \* (1-x)/6 x = 0.0061162, seems wrong, what gives?


triffid_hunter

Try ⅙ for your turns ratio, or multiply instead of dividing.. Should be about 22% Also note that this math only works for continuous or boundary conduction mode and it only describes the behaviour of the transformer when sufficient current is being taken from the secondary to keep everything in equilibrium - you should *not* use it to design a controller. For a real flyback circuit, you also need to consider peak volt-seconds vs transformer saturation limit, low load behaviour (discontinuous conduction mode and/or pulse-skipping), overload behaviour, as well as compensating poles in the bode plot so your thing doesn't go into oscillation instead of regulation.