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how does current flow in a flyback

here is a flyback dc/dc converter, it converts 5V to 100V/1A in cw mode. Now the load is a 50ohm, yes, 50ohm in series with a super fast on/off sw. The sw is normally off but it will be open/close very fast to get pulse charging to the 50ohm load. Then the peak current is 2A. This pulse is as long as 5 switching cycles of the flyback.

if there is no ovi/ovp, my understanding for current flow is: During flyback sw-on, the sec. inductor will supply current to charge output cap and 50ohm load. The peak switching current will exceed the limit by the chip, suppose the chip has an integrated switcher, the chip will enter "protection"mode by either blanking some cycles or change frequency or duty cycle. During flyback sw-off, the output cap will supply current to 50ohm load.

Am I right?

Reply to
John Lee
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here 50ohm could be much less, the test is aiming at exceeding IC's peak switching current.

The output cap could be large to lower the sagging and then prevent the feedback voltage from trigging some frequency foldback or similar protection mechanism.

Reply to
John Lee

No. This is on the Wikipedia page on flyback converters, for heaven's sake!

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The basic action of a flyback circuit is that during the switch on period, the current is building in the primary and no current is flowing in the secondary. When the switch turns off and the field starts to collapse, all* of the energy in the field goes to the output cap.

So during your five switching intervals, the voltage on the output cap will be dropping, but if you arrange your circuit right that drop will be the only bad thing happening.

  • With the exception of that in the leakage inductance, which will take out your switch if you don't have a snubber.
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

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ttdesign.com

I mistyped a thing: during switch on the primary has peak current limit and store the energy. The secondary inductor has no current. The load current 2A is from the output cap. during switch off the load current 2A is from the secondary inductor.

they need to follow volt-seconds. The switcher's peak current during switcher on is affected by the load because the energy released from the secondary actually should be the same energy stored during switching on.

So the bad thing is the switcher primary side will suffer from the peak current surge???

Reply to
John Lee

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ttdesign.com

and if you think your load is grounded for one pulse width like 100 usec, what happened to the primary side if switcher is working on a

100kHz?
Reply to
John Lee

To both of your questions: The primary side will not see the current on the secondary side. Only the voltage.

Reason from there.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

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ttdesign.com

I drew a picture, am I right? I checked equations to calculate the output cap, some equations use peak current some use ripple current.

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Reply to
Lee

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it's not the fact.

ttdesign.com

I got answer regarding the current flow: see simulation

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Reply to
John Lee

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