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enter image description here

I was, "OK, this is feasible"... But then I traced how it worked and it simply blocked current through Drain and Source when a P and N pair are reverse biased; then when the other P and N pair are forward biased current flows through Forward Diodes; then alternatingly... Then it's the same, one is just using diodes to Bridge Rectify. Worse still, MOSFETs generally don't have a low diode voltage drop... Or maybe I'm missing something here....

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  • \$\begingroup\$ could be interesting on a 25VAC supply with a capacitor as load. it would pass current back out on the down slope of the sine, \$\endgroup\$ – Jasen Jan 1 '16 at 8:56
  • \$\begingroup\$ Down slope of positive or negative? \$\endgroup\$ – kozner Jan 1 '16 at 8:59
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    \$\begingroup\$ "down" from the peak \$\endgroup\$ – Jasen Jan 1 '16 at 9:05
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    \$\begingroup\$ the trick that circuit is relying on is using MOSFETs as perfect rectifiers because when based on they conduct in both directions \$\endgroup\$ – Jasen Jan 1 '16 at 9:07
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    \$\begingroup\$ You could consider this a type of synchronous rectification with the switch signal supplied by the input voltage. It only has low resistance if the voltage is high enough to turn the MOSFETs on. And the MOSFETs must tolerate the maximum peak voltage as Vgs (usually 8V-20V absolute maximum). Those are somewhat conflicting constraints- MOSFETs that turn on at a lower voltage tend to have lower Vgs(max). Of course you can add zeners and resistors to handle that, and if the input is square wave then the low voltage constraint doesn't matter much. \$\endgroup\$ – Spehro Pefhany Jan 1 '16 at 17:13
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Just look at how the biasing works: -

enter image description here

With positive on the top input rail the lower left N channel FET is switched on and, with negative on the bottom input rail the top right P channel FET is switched on.

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    \$\begingroup\$ @kosner: And when the FET is switched on it behaves like a very low value resistance bypassing the diode. If, for example, the 'on' resistance, Rds, was 0.1 Ω then at 1 A the voltage drop would be 0.1 V. For the diode it would be 0.7 V. Double these numbers for a bridge rectifier and you see the advantage, particularly in low voltage circuits. The diode, by the way, is a side-effect of the FET construction - it's not added in. \$\endgroup\$ – Transistor Jan 1 '16 at 13:00
  • \$\begingroup\$ Really? You, Andy, don't see that it doesn't work? Let's only discuss the positive first half period like you noted above... So what is the upper right PMOS and lower right NMOS are on? The AC positive rail is still higher than the cathode output (unless there is a capacitor filter afterwards). So it's going from Drain to Source terminals of the PMOS, but that never happens even if the PMOS is on (or unless voltage breakdown occurs). There is always a diode, however, so it will through there. The same goes symmetrically for the NMOS. Stop misleading the discussion. \$\endgroup\$ – kozner Jan 2 '16 at 6:45
  • \$\begingroup\$ @kozner I'm really trying to understand what you are talking about and why you think I'm misleading a discussion. \$\endgroup\$ – Andy aka Jan 2 '16 at 11:48
  • \$\begingroup\$ Alright, in an NMOS, where Source is higher than Drain (or for whichever MOSFET biased in reverse from what they are normally biased on), will the current flow from Source to Drain through the channel (or Drain to Source in a PMOS), if the MOSFET is turned on and not just the Body Diode? That is to say, is the current through the channel bidirectional? \$\endgroup\$ – kozner Jan 2 '16 at 12:42
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    \$\begingroup\$ @kozner, yes. When turned on it behaves as a RESISTOR. See my first comment. I think you need to take a little more care in your writing. Both Andy and I are struggling to interpret what you are saying due to poor writing. If you now understand Andy's answer you should retract your allegation that he is 'misleading the discussion'. \$\endgroup\$ – Transistor Jan 2 '16 at 12:58
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The rectifier has no voltage drop at no current.The availability of low RDs on mosfets means that the volt drop could be very low .It can be lower than a shottky diode .The effective resistance is the sum of the N chan and the P chan .I did this in a previous life but for production I used a dual schottky instead of the 2 P chan fets.P channel was a big penalty 25 years ago so I figured that 2 n chans and 1 dual schottky was better value for money.Everything was fine for 12V 10 Amp battery charger .Nowdays the p chan could be economic depending on your application .Remember that if you do the p chan into a big electrolytic cap you will have to do something about high reverse currents .Maybe a Fiode connection or some reverse current sense that shuts the gates down .

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  • \$\begingroup\$ Users: be aware that this circuit -- while it rectifies -- doesn't prevent backwards current flowing; you'll need something like a single schottky at the output to use this effectively in an AC-DC converter. \$\endgroup\$ – jp314 Jan 1 '16 at 17:23
  • \$\begingroup\$ @jp314: Why is this? My understanding is that if there is no AC and during the low-voltage portions of the AC cycle that the FETs aren't biased on (so they're high resistance) and the diodes are reverse biased. How does reverse current occur? \$\endgroup\$ – Transistor Jan 2 '16 at 12:48
  • \$\begingroup\$ If there is an output voltage (e.g. a reservoir capacitor), then this capacitor will discharge into the AC voltage source (e.g. a transformer secondary). \$\endgroup\$ – jp314 Jan 2 '16 at 15:54
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There have been a couple of comments and answers here about the failure of the MOSFET bridge rectifier: That it conducts in both directions, so if you have a capacitor-filtered power supply the capacitors will simply drain on the AC downslope, back to the source.

There are a couple of commercial solutions to this problem: at least two that I know of, the LT4320 and LM74670-Q1.

See https://www.analog.com/en/products/lt4320.html#product-overview and https://e2e.ti.com/blogs_/b/motordrivecontrol/archive/2016/01/11/a-novel-approach-to-full-wave-bridge-rectifier-design

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I tested this rectifier in LTSpice. Using only a resistive load it worked perfectly, generating a full-wave rectified current over the load resistor, with a very small voltage drop in the transistors (depending on on-resistance, not on the body diode forward voltage).

Then I added a capacitor to make it a continous DC current. In that case the rectifier faled totally. When there was a voltage over the capacitor, the MOSFET's was conducting in the wrong direction, making the current flow back to the AC source again.

If you replace the two P-MOS transistors with two diodes, it works, because the diodes will block any reverse current. That's why Autistic's solution worked (described in the last post).

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