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Flyback Equation with a Series Multiplier

What's the power equation for a DCM flyback with a series multiplier?

I know a DCM flyback without a multiplier is P = 0.5 Li^2 * freq

Also, if there in no multiplier and the main switch is closed there is no energy transferred from the source to the load, energy is transferred from the source to the xfrm core.

However, when a series multiplier is used on the output and the main switch is closed there is energy transferred from the source to the half of the caps in the multiplier in addition to the xfrm core.

I was thinking something like:

P = Energy * freq = (0.5 Li^2 + 0.5CV^2 * n/2)

Where C is the value of the caps in the multiplier, V is the peak voltage on the caps in the multiplier and n is the number of the caps in the multiplier.

I thought about this a while back when I was working with a voltage doubler. If you put a doubler on a flyback the output voltage would double (in theory) and the output power would go up 4 times. However, if the input power was 0.5Li^2 * freq then the input power doesn't change so the output power doesn't change so something is missing.

Maybe it's not a flyback any more when a multiplier is added to the output?

Thanks,

Jason

Reply to
Kooner
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You mean like the circuit below?

First, it won't really double -- it'll add VCC * (turns ratio) to the output. Depending on your flyback ratio this may be good, bad, or immaterial.

Second, because current will flow in the secondary side when the switch is closed, it will increase the power taken from the primary side.

Third, because of the nature of the waveform, the current into the 'doubler' cap will be controlled by the leakage inductance of the transformer, and if this leakage inductance is not large will tend to have a very short, high-current (and therefor inefficient) peak. To prevent that from happening you'd have to design the transformer for more leakage, or you'd have to put an auxiliary inductor in there someplace.

Since you'd have to get that transformer custom-made anyway, why not just order it with a few more secondary turns, and save money on diodes, caps and coils?

VCC + | Vout '--. ,---------o--->|----o--------o * )|( | | )|( * | | .--' '--. - --- | | ^ --- | | | | o --- | | '\ --- === === \ | GND GND o \ | | === | GND | === GND (created by AACircuit v1.28.6 beta 04/19/05

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Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
Reply to
Tim

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Tim, thanks for the response.

Yes, thats the circuit.

I simulated a flyback with out a doubler and one with a doubler and I can see what you mentioned.

However, when I go to 4 caps and 4 diodes in the multiplier I can't figure out the math. Lets assume the transformer ratio is 1:1 then the very first cap on the secondary still charges to Vcc. How do I calculated the voltages on the other 3 caps? I can see the 0.5 l*i^2 energy still transfers to the output.

I can see the output power increases for every stage of the multiplier that is added and it looks like that the reflected voltage on the primary decreases for every stage that I add.

-Kooner

Reply to
Kooner

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