Magnetising current in a Full-Bridge SMPS? - diyAudio
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Old 14th December 2011, 12:52 PM   #1
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Default Magnetising current in a Full-Bridge SMPS?

Hello,

When a pair of FETs in a full-bridge SMPS turns off......the primary current ceases immediately (apart from a short discharge of leakage inductance through the anti-parallel diodes of the other FET pair).

This must mean that the magnetising current transfers to the secondary.

Would you agree?


(...the thing is, magnetising current definetely does not power the load in any way, so it cannot flow out into the load.....it must somehow act to diminish the secondary current....one things for sure, the magnetising flux, and the current associated with it, cannot just disappear)

(This LTspice simulatiojn shows it:

2shared - download Full-Bridge _CCM +-40V 4A.asc
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Old 14th December 2011, 02:08 PM   #2
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The instant both switches turn off (but before the voltage reversal), the current through the leakage inductance will be equal to the magnetizing current plus the reflected load current. Assuming there is not path for current in the secondary after the voltage reversal, the current in the leakage inductance becomes equal to the magnetizing current. In other words, the magnetizing current is not transfered to the secondary.

I'd like to see your simulation, but I'm not going to install the software necessary to download it. You can attach it to a message you post here, but you need to change the extension from .asc to .txt in order to do so. People who dowload it will have to change it back to .asc.
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Old 14th December 2011, 02:24 PM   #3
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i have attached the simulaiton here.
Attached Files
File Type: txt Full-Bridge _CCM +-40V 4A.txt (7.1 KB, 21 views)
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Old 14th December 2011, 04:06 PM   #4
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The circuit you posted does not have any leakage inductance. You can either add it in series with the primary of the transformer, or make K less than 1. If you add the leakage and run the circuit, I think you will see what I was trying to say. Look at the current through the primary and the voltage across both the magnetizing inductance and the leakage inductance. When the switches turn off, the voltage reversal takes place and the primary current drops quickly due to the large reverse voltage. Note that there will now be lots of ringing due to the leakage inductance and MOSFET capacitances. (I added 22 uH leakage.)

I have a few comments on your circuit.
1. You don't need to specify the number of cycles for the gate drive signals. Of course you can, if you want to.
2. You don't need to include all the .0001 ohm resistors. If you want to plot a current in a wire, place your cusor on the wire, depress the alt key, and then right click on the wire.
3. The catch diodes don't do much, because the reverse current flows through the MOSFET diode.
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Old 14th December 2011, 04:31 PM   #5
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Sawyerw: Thanks for the Spice tips.

The magnetising current builds up due to the V.us the primary received when the fet was on.....it should take the same V.us to bring the magnetising current back to zero.......and since V is the same in each case.....the magnetising current should still be flowing for tON useconds after the FET turns off..

but i can't see it.

I appreciate i did not include leakage....though i am just trying to see what happens to the magnetising current when the fet switches off.
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Old 14th December 2011, 04:56 PM   #6
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If I understand you correctly, I agree with you. If you change your load resistors to 10K, the primary current will be essentially equal to the magnetizing current, after the start up transient. You will see that the voltage across the primary is essentially a square wave and the current through the primary is a triangle.

Another way to see the magnetizing current is to add the magnetizing inductance in parallel with the existing transformer and increase the transformer inductances a couple of orders of magnitude. Keep the same inductance ratios. This make the transformer look (almost) like an ideal transformer.

You are quite right, you don't need to include leakage inductance to see what you are looking for.
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Old 14th December 2011, 05:10 PM   #7
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Going to a 10K load was a good idea.............that shows it like i think it should be.

....the magnetising current keeps flowing after the fet switches off and just goes through the anti-parallel diode of the opposite fet pair....dying down as it does so.

None of the magnetising current is seen in the secondary.

With the heavier 10R load, however.....the magnetising current just vanishes when the fet switches off, which puzzles me.
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Old 14th December 2011, 05:18 PM   #8
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I just realized you are correct about you initial comment. In the case of no load on the secondary, there is no place for the stored energy to go in the seconday so it flows back into the primary supply. In the case where there is a secondary load, the magnetizing energy is released into the secondary.
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Old 14th December 2011, 05:24 PM   #9
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I didn't see your post #7 before I sent my previous message. Essentially, what happens is if the filter inductor currents are continuous (if the load current is high enough), then the the stored energy will be transfered to the secondary. Otherwise, the primary voltage will reverse so the energy can go back into the supply.
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Old 14th December 2011, 06:30 PM   #10
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Thankyou Sawreyrw,

You have helped to smash a misconception which i had carried with me for >10 years..........before today, i had always thought that magnetising energy is entirely confined to the primary.

-of course, i can see the error in my ways now, flux in a core is just flux in a core, and doesn't "know" if its "attahed" to primary or secondary windings.
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