Active clamp flyback clamp voltage calculation

Question

Does anyone know how to calculate the clamp voltage? I read some papers and documents. Some of them use \$n \times Vo\$ to calculate the voltage, but when I use simulation to verify it I find the voltage is not \$n \times Vo\$.

• Turn ratio Np:Ns is 1:200m.
• Output voltage is 24V
• Lleakage is 5uH
• Lmag is 205uH
• Vin is 330V
• Fsw is 100kHz
• Po: 575W

If \$C_{clamp}\$ is 22uF, the clamp voltage is around 135V (this is not n*Vo)

If \$C_{clamp}\$ is reduced to 5u, the clamp voltage is around 133V.

If \$C_{clamp}\$ is reduced to 0.22u, the clamp voltage is weird, I don't know why.

Answer 1

What matters for the active-clamp flyback or ACF is the capacitor value to ensure the following equality: \$C_c\geqslant\ \frac{(1-D_{min})^2}{F_{sw}^2\pi^2l_{leak}}\$. It ensures that for a proper operation, one half of the \$l_{leak}C_c\$ resonant period is not shorter than the maximum off-time duration which is observed in high-line conditions.

Now, what matters for the voltage stress of the clamp capacitor and its type selection is not to exactly know the voltage across its terminal within the mV. Determining the exact voltage mathematically requires complicated equations (see this paper) and that is the reason why designers often choose a value of \$\approx\frac{V_{out}+V_f}{N} \$ which is then affected by a safety margin naturally including the ripple.

I have proposed the formula shown below in my book on SPICE simulations and I included another formula from another paper. Please note that these are approximate formulas which point to a 200-V capacitor (you need of course to check worst-case voltage excursion for the final selection):

But the point is that mathematical analysis to obtain the exact same-looking waveforms from the solver and simulations represents a good exercise - that I like too :-), but, in the end, keep in mind that reliable design with adequate components is the ultimate goal. This the founding rule of design-oriented analysis or D-OA.

Answer 2

Does anyone know how to calculate the clamp voltage? I read some papers and documents. Some of them use \$n×Vo\$ to calculate the voltage

\$n×Vo\$ tells you the secondary-to-primary reflected voltage in normal operation and, for the ideal case of zero leakage inductance. The clamp voltage cannot be lower than this.

Using your numbers, if Vs is 330 volts and Vo is 24 volts, when the transformer flies-back, the voltage seen at the drain of the main MOSFET is 330 + (24 × n).

You have specified Np:Ns as being a 5:1 step-down and therefore, the output voltage of 24 volts is stepped up by 5 (to produce a peak of 120 volts) and, this is superimposed on the 330 volts at the drain. Hence, the peak voltage at the MOSFET drain is 450 volts. You can't clamp this; it is a fact of life; a fundamental aspect of fly-back converters.

So, don't clamp this voltage because, this is how a fly-back circuit works.

What you do need to clamp are voltages in excess of 450 volts because, these are formed by the primary leakage inductance (adding to the fly-back voltage) and, may be detrimental to your MOSFET. Ideal scenario (no leakage): -

Image from my basic website.

So, your clamp voltage needs to be higher than 450 volts (with respect to ground) but lower than the MOSFET maximum DS voltage.