magnetizing energy in Flyback converter

The energy stored in the inductor charges all primary and secondary capacitances. These include the inter winding capacitances of the coils, as well as the mosfet and diode capacitances. The MOSFET output capacitance may start out as the larges component of the total, but it falls as voltage rises, so at peak voltage, it may no longer be dominant.

Be careful with your units. Watts is unit of power, joules is unit of energy. I think you need to find out what the other capacitances in the circuit are. And the nonlinearity of Coss makes it a bit rough to come up with a precise energy stored at any given voltage rise.

The secondary rectifier becomes forward biased because the MOS is off which reverses all the voltages on the transformer windings. It is the reversed voltage that forward biases the diode. Of course the secondary voltage has to be higher then the value on the output capacitor for the diode to be forward biased.

Tell us how did you measure those values .

Andy

As stated elsewhere, there are other capacitances. Note also that the voltage rises as the mosfet current falls. If the transition period is long, the fet will be absorbing energy during that time.

The more common complaint is that leakage energy, that is not transferable has significant effects in clamp overshoot and dissipation. They're both LI^2 /2, in joules.

RL

I'm not certain about your explanation of what you are missing in terms of where the extra (loss) is from. However Mr Legg mentions energy that is stored in the leakage inductance, as LleakIpk^2/2, which ends up being lost to the clamp (snubber).

It is really a clamp because it limits the overvoltage from leakage inductance during flyback.

There is more to it than that though and it's a bit subtle.

As the leakage inductance is being reset it is sitting on top of the reflected secondary voltage. Not only do you get the energy from LleakIpk^2/2 dumped into the snubber energy is also removed from that reflected voltage.

Let's say you have a 100uH primary with 10uH leakage inductance and your peak primary current is 5 amps. The energy stored in the leakage inductance is 125uJ. If your supply is operating at 100KHz then the power in the clamp is 12.5W.

BUT.... If the reflected secondary voltage is 200V and your clamp voltage is

300V then the leakage inductance is being reset through 100V. That takes 500nS, T=dIL/V. The current waveform is triangular and you can work out the associated charge from its area as being 1.25uC.

With your supply operating at 100KHz the average current recovered from the reflected secondary voltage is that charge multiplied by the frequency or

125mA. Multiply that by the reflected 200V and you get the additional power in the clamp as 25W......... !!!!!!!!!!

So the actual loss that the clamp has to deal with is not 12.5W it is 37.5W.

If you don't know about it then you can spend a lot of time scratching your head wondering why the resistor in your clamp/snubber is smoking. If you do then you can design for it.

Having minimised leakage inductance the next thing you should do is maximise the clamp voltage. That reduces the time taken to reset the leakage inductance and the energy lost.

DNA

Thanks for all your valued input.

I measure the primary current by a current probe between the Source of the MOSFET and the series current sensing resistor.

When the Vds starts to increase, the primary current is 1.06A. And it reduces to 0.83A when Vds reaches 302V to let the secondary rectifier start conducting current. The magnetizing inductance is 600uH and the leakage inductance is 20uH (both measured by LCR meter). And the operating freq. is 60KHz. Therefore, the power difference during this period is:

1/2 * (600+20)*10^(-6)*(1.06^2-0.83^2)*60*10^3=3D8.09W---(1) The Coss of MOSFET is 135pF, which is measured at VDS=3D25V and freq=3D1MHz. Can anyone tell me how to transfer this data to meet the condition that my circuit actually works? If I still use this 135pF to calculate, the power would be: 1/2*135*10^(-12)*302^2*60*10^3=3D0.37W---(2) And the conduction loss during this period (around 400nsec): ((1.06 + 0.83)/2) * (302/2)* 40*10^(-9)*60*10*3=3D0.34W-(3)

So, where is the difference between (1) - ( (2) + (3) ) =3D 7.38W

Thanks!! Webber.

I assume you have no load on otherwise you'd be losing it there. You should keep in account your measured values don't take into account the spikes in currents and voltages which add considerably to Power losses. Just taking a rapid look at your calculations. I would think subtracting 2*1.06*0.83W in the power expression for primary inductance loss would be more accurate as energy remains stored in the primary inductances, due to the fact that the current does not drop to zero during the transition period. What is the gate drive impedance on the MOS? If too high you may have a bounce on-off effect during transitions. Most losses, generally more then a half are in the diode, so check on that too. Core losses in the transformer windings , capacitor , error control and switching circuitry could account for the difference.

Andy

What MOSFET are you using?

I haven't followed the convoluted arguments being put forth in this thread, but I can say I'm uncomfortable with the various assumptions being made to allow cranking through some perhaps oversimplified formulas. Completely missing circuit elements (e.g., core losses, winding capacitances, and continuing MOSFET drain-source conduction after supposed "turnoff"), is one serious issue. Misuse of "known" parameters is another.

For example, power MOSFET capacitances are certainly not fixed values, as John has pointed out, but instead vary by 10 to 40x over the full operating range. The datasheet specs and plots are nice to have, but I've found substantial variation in bench measurements of actual parts against the datasheet values, which can exceed 3x under some circumstances. I'm convinced that datasheet plots are sometimes either oversimplifications intended to convey a concept rather reality, or figments of a draftsman's mind. For example, actual MOSFETs often show a dramatic change in capacitance, 2x to even 5x over a few volts, somewhere in the 7 to 12V region. This effect is completely absent from the manufacturer's plots for the same part. My favorite RCA engineering MOSFET model (a low-voltage MOSFET in cascode with a high-voltage depletion-mode JFET) accurately handles this dramatic condition, but that model does not lend itself well to the classic parameters we see on the datasheet.

My suggestion is that you take a suite of bench measurements on your components before attempting to accurately model and calculate the power losses. You'll need specialized test jigs to accurately measure some of the more difficult things, like delayed incomplete MOSFET turnoff as a function of say gate drive, drain current and time.

Good luck. Let us know what you learn.

'reset'ing the leakage inductance is an interesting use of terms.

The dI/dt in the leakage inductance is seen by the clamp rectifier. The higher the dI/dt in this part, the higher the peak reverse current spike in this rectifier before it turns off. As the total reverse charge increases with both peak forward current and reversing dI/dT, you are working with some interesting relationships which can result in more reverse recovery charge than initial forward charge transfer - effectively 'regulating' the clamp voltage without appreciable dissipation in the clamp bleeder.

This is possibly why Power Integrations specifies an ancient and only moderately fast rectifier in this position, in most of their application circuits. Crunching the numbers using their recommended bleeder resistor values reveals quite low power levels being anticipated for dissipation, despite calculably higher potential losses when the 'ideal' parasite leakage term gets to work.

If you ignore the noise consequences, and the intentional stress of the appreciably higher clamp voltages (ameliorated somewhat by PI's intentional use of IET continuous current topologies), it's an interesting and cheap technique - but only, in my opinion, if you know you're doing it. Otherwise it's just plain fool's liuck.

RL

Sir,

Your circuit will either work, or not, in spite of the accuracy of your calculations - not because of it.

RL

LOL!

well put.

then suitable measurements can tell you whats really happening, and can be used to verify (or falsify) your calculations.

Cheers Terry