AC to DC Power Supply Design Questions

Hi,
I ran across this in (original thread).

I'm familiar with simpler power supplies so I mostly understand the diode bridge connecting to the transformer and the rectification/filtering/zener occurring in the bottom right output stage.

There are a number of elements that I don't get and was hoping for some insight on:

1) From what I can tell the IR2153 just provides an alternating pulse to the 2 MOSFETs. What is the benefit of doing this pulsing action? Why does the output incorporate the caps C5 and C11? Just smoothing?

2) The C6/C12 structure splitting the transformer surprised me. I expected just a single cap value leading to ground for transients. Purpose?

3) L4 leading into the diode bridge. I'd never seen this before. Inductors impede AC. Don't we want this AC entering the transformer?

4) I see the use of the first MOSFET (T1). I don't understand T2. It seems to me current should always flow out the left side of the diode bridge. If T1 isn't on, there's no path for current to enter T2. From the IR2153 datasheet LO and HO (pins 5 and 7) are mutually exclusive. I must be missing something.

5) The lower left quadrant. Not sure what its purpose is or how he's getting 18v AC down there.

Appreciate any help you can provide. Thanks.

C6-C12 are switched alternately by T1-T2. It's an easy simple way to send AC through the transformer.

Lower left quadrant. The winding marked '18V' is a third winding on the transformer (indicated by the dotted line connecting the transformer symbols).

C5-C11 have led's installed across them. This may cause the led's to light briefly, as a visual cue telling you the IC is sending bias pulses to T1-T2.

L4 probably limits inrush current on powerup.

Any thoughts on the C6/C12 split?

Also could you explain a little more about the T1/T2 interaction? When I look at the diagram I think the AC is rectified to a varying, but always positive current leaving the bridge. I'm not sure how that current interacts with T2 when T1 is off.

Thanks

The left side is a half-bridge, one transistor is On while the other is Off. The capacitors charge to 1/2 supply voltage each. The node between them is 1/2 supply voltage. Operation might be described as single-ended class B.

True AC (at 1/2 supply voltage) goes through the transformer primary. This allows a reduced step-down ratio.



A capacitor charges during one half of the cycle, then discharges through a transistor during the other half of the cycle.

Notice the Ampere value through the transistors. Some current comes from the supply, some from a capacitor discharging.

The 10m (I'm assuming milli?) ohm resistor at the bottom, what's its purpose?

rfb2 said:

The 10m (I'm assuming milli?) ohm resistor at the bottom, what's its purpose?

Click to expand...

Capacitor's charge. There can not be any loop without resistance.

CataM said:

Capacitor's charge. There can not be any loop without resistance.

Click to expand...

Correct. If there is not some resistance (low ohm), the simulator generates an error. Something about two capacitors in series by themselves. Also connected to supply leads by themselves.

This is only a low power converter, (10 -20W) so don't be tempted to scale it up... [Bang]

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without a soft start there will be some fairly large currents in the mosfets and o/p diodes at hard power up...