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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. When the other P and N pair are forward biased current flows through forward fiodes, 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. 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\$ Commented Jan 1, 2016 at 8:56
  • \$\begingroup\$ Down slope of positive or negative? \$\endgroup\$
    – kozner
    Commented Jan 1, 2016 at 8:59
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    \$\begingroup\$ "down" from the peak \$\endgroup\$ Commented Jan 1, 2016 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\$ Commented Jan 1, 2016 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\$ Commented Jan 1, 2016 at 17:13

5 Answers 5

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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
    Commented Jan 1, 2016 at 13:00
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    \$\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
    Commented Jan 2, 2016 at 11:48
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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
    Commented Jan 2, 2016 at 12:58
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    \$\begingroup\$ "I think it's conceptually wrong to think that MOSFETs behave like "resistors" when turned "ON"." -> But, for practical purposes they do when saturated. | " ... is the current through the channel bidirectional?" -> A MOSFET requires Vgs to be positive for an N Channel device to be turned on BUT is a bipolar switch wrt Vds. Because of the parasitic reverse polarity DS diode this means that the MOSFET appears as essentially a resistor of Rdson when on and as a poor diode when off. \$\endgroup\$
    – Russell McMahon
    Commented Jun 25, 2016 at 16:29
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    \$\begingroup\$ @RussellMcMahon not true - when a FET has a reverse biased drain/source AND it is turned on by a significant gate-source voltage then the channel shunts the parasitic diode in exactly the same way as it does when drain and source are fed in the conventional forward biased way. Look at SSRs - they use back-to-back MOSFETs; one is conventionally forward conducting and the other is reverse conducting. One of the MOSFETs doesn't show itself as a forward biased diode - it still conducts through the channel but the pinch-off region shifts to the source end. \$\endgroup\$
    – Andy aka
    Commented Jun 25, 2016 at 20:18
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The rectifier has no voltage drop at no current. The availability of low RDs on MOSFETs means that the voltage drop could be very low. It can be lower than a Schottky 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 ampere 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 capacitor then you will have to do something about high reverse currents. Maybe a diode 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
    Commented Jan 1, 2016 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
    Commented Jan 2, 2016 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
    Commented Jan 2, 2016 at 15:54
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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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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 stumbled across this in search of low-loss rectification for harvesting. The basic problem with the circuit shown is that any phase shift introduced by the load (e.g., with a filter cap) will cause the FETs to fire when you don't want them to (simulate it here):

enter image description here

With no cap, you get something like fullwave rectification. With it, the FET switching is messed up, allowing the cap to discharge back into the line-in.

Even when it's working, it still has forward drop. First, the FETs won't be conducting near the zero-cross points, as neither the body diode nor the Vgs voltages are met. Note also that the no-cap the peak rectified voltage is less than the input by two body diode drops. This is because during that low voltage time, the Vgs threshold isn't met so the enhancement path is off, so only the forward-biased body diodes (two of them) are conducting. Result? It has forward drop!

In short, it's a pointless circuit by itself. Doesn't work with a real load, and doesn't solve the forward drop.

How to make it work then? Use active control of the FETs to make ideal diodes. The FET based ideal diode lets the body diode behave like diode, and turns the FET on when Vs > Vd (for n-FET), using that FET party trick of conducting in either direction. A comparator monitors Vgs, and turns on the gate for the 'reverse' (drain-to-source) direction, shorting out the body diode and eliminating that pesky forward drop. Voila! Ideal diode.

For a bridge, as it so happens we need just two comparators to monitor Vds of the high-side FETs (simulate it here):

enter image description here

I've left out some details, like powering the comparators and getting them to swing high enough (via bootstrap?) to turn the high-side FETs fully like the commercial bridge ICs do (e.g., LT4320.) But this illustrates the principle. (The low-side FETs actually only need logic-level drive, but will have the same polarity as the high-side drive. This is the case for the LT4320.)

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