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I was recently looking at a 500W PSU schematic, and found out they used step-down converters on 12v and 5v output. Schematic is attached(pls see hi-res at https://ibb.co/nbDRJPq)

500W PSU output stage

The D2 diode charges the L2 inductor and then as the power removed from the L2 its voltage reverses and passes thru the D2A diode to recharge the C5 capacitor. Which appears to be a typical step-down. Now, my questions are as follows:

  1. Why did they use it? Wouldn't it just make sense to use a standard 2-diode rectifier with center-tapped secondary coil?

  2. Does that mean that negative half-wave is not used at all? Why did they use a Q2-Q3 half-bridge to drive the transformer then?

  3. If it really is a step-down, does that mean that output voltage on secondary coil of the T1 transformer is higher than 12v?

  4. How exactly does the 3.3v output work? I understand the Q6 is some sort of regulator (U6 is TL431 chip), so does the transistor here work in linear mode or PWM? Is the transistor used to discharge the L7 inductor and control output voltage?

  5. What kind of inductor is the L6 ? What does the mark next to it mean?

Thank you!

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  • \$\begingroup\$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. \$\endgroup\$
    – Community Bot
    Commented Oct 2, 2021 at 8:53

1 Answer 1

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The D2 diode charges the L2 inductor and then as the power removed from the L2 its voltage reverses and passes thru the D2A diode to recharge the C5 capacitor. Which appears to be a typical step-down.

Actually, the whole power supply is a 2-switch forward converter (a.k.a. double ended forward converter):

enter image description here

ImgSrc: Researchgate (Jindrich Windels)

As can be seen from the principal schematic of a double ended forward converter above, the secondary side of the converter is no different than a buck converter: The inductor stores energy when the primary switches are on, then releases its energy when they are off.

You can find a lot of articles and application notes explaining detailed operation.

  1. Without the inductor, what you'll observe at the output will be pulsating DC. So the required output capacitor will be extremely large to make the output DC. That's why flyback converters are not preferred at the power (or output current) levels that forward or bridge-type converters are suitable for. To understand the operation of the secondary better, I personally recommend you to study switching inductor behavior and buck converters.

  2. Don't mistake this converter for traditional half-bridge or push-pull converter. Primary side is not center-tapped here, and there's no divider capacitors. The negative voltage across the primary is only used to reset the core.

  3. The peak voltage at the secondary is higher than the output voltage. The average voltage is the peak voltage times the duty-cycle, just like in a buck converter.

  4. That regulator scheme is called MagAmp. This is completely a different topic, so I'll not dive into the details here. But basically, the core (L6) is used as a saturating device allowing some portion of the volt-seconds to "pass" to the output.

  5. It's a saturated core.

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  • \$\begingroup\$ Thank you very much for your answer. You are right, I overlooked the fact that Q2 and Q3 are not connected as half-bridge, they actually do turn on and off together. I am pretty much familiar with how buck converters work, and there's still something I don't understand. In a typical buck converter supply voltage is higher than output voltage because while the inductor charges output voltage is 0. It starts to rise only after inductor is 100% charged. To achieve this, we need more volts at source side. Otherwise, we have no time for inductor's reverse voltage to kick in. Am I wrong? \$\endgroup\$ Commented Oct 2, 2021 at 8:38
  • \$\begingroup\$ @JimmyFalcon ah yes, you are correct. I forgot to talk about the duty cycle. Fixed it. By the way, for the inductor's behaviour, you need to think steady state. This will make you understand easier. \$\endgroup\$ Commented Oct 2, 2021 at 9:25

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