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schematic

simulate this circuit – Schematic created using CircuitLab

I'm new to the world of mosfets and am trying to learn them better. Would this rectifier IC be safe for input voltages of 24V if Vgs on these fets is 20V?

Vdss is 30V and reverse on D1 and D2 is 30V.

With L1_2 at 24V and L1_1 at 0V it seems that Q1 Vgs would be a smoke-inducing 24V.

Mosfet Fullwave Bridge

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  • \$\begingroup\$ I am confused. Why not just use a bridge rectifier or 4 diodes? Are you reinventing the wheel? This is not what mosfets are used for, and not how you use them. What are you trying to do? \$\endgroup\$ – Sparky256 Jun 10 '16 at 23:15
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    \$\begingroup\$ @Sparky256 It's semi-active rectification to minimize conduction losses. The low Rds(on) of the mosfets mean this 4mm square package can handle 3A of current without overheating. \$\endgroup\$ – joshperry Jun 10 '16 at 23:23
  • \$\begingroup\$ You need to remember that a low Rds(on) will also mean that you have a higher low end. You are assuming that your threshold of the MOSFET is lower than the diode drop of the resistor (which moves with V btw, whereas it does not with a MOSFET) The datasheet suggests as such, and your Vmax is 30V, so you should not induce smoke. Why are you expecting the magic smoke to escape? \$\endgroup\$ – b degnan Jun 10 '16 at 23:48
  • \$\begingroup\$ Why stick with D1 and D2. Make them active as well :) \$\endgroup\$ – KarlKarlsom Jun 11 '16 at 2:41
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    \$\begingroup\$ @bdegnan I'm not a trained engineer so I'm still trying to figure out how all this jives together. I'm going to have to read your explanation a few more times to get it. Like I said in my question, I was under the impression that when L1_2 is at 24V and L1_1 is at 0V that the Vgs of Q1 would be 24V. Will it actually be < 20V? \$\endgroup\$ – joshperry Jun 11 '16 at 3:29
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This is actually clever. There are issues in the real world, like Vgs EOS (Electrical over stress) but as a part of a thought process it's worth playing with it to see if it takes you somewhere else.

Frankly the MOSFET's are much more expensive than the diode's they replace so that is an expensive solution to gain back the one diode drop.

And to really run this properly, you should have a differential signal wrt to the ground point. So to me this looks more like a AC differential measurement circuit than a power conversion circuit. And with that, the additional MOSFET (Scaled appropriately) could be justified.

On EDIT, now that I realize there isn't a latent genius amongst us ... That this is a standard part. to answer the question of "Just trying to figure out if this bad boy will handle 24V input or if the Vgs of 20V will render it dead with that much voltage".

Just look to the table on the second page, included here for reference.

enter image description here

\$V_{GSS}\$ shows +/- 20 V Max.

That is how you read the table.

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    \$\begingroup\$ Thank you for the answer. This is actually a bridge rectifier replacement IC by DIODES INC (though Fairchild and others have very similar parts). Just trying to figure out if this bad boy will handle 24V input or if the Vgs of 20V will render it dead with that much voltage. diodes.com/_files/datasheets/DFBR030U3LP.pdf \$\endgroup\$ – joshperry Jun 11 '16 at 3:36
  • \$\begingroup\$ I've never seen that! So now I'm less impressed with someone "just learning" ;) \$\endgroup\$ – placeholder Jun 11 '16 at 3:38
  • \$\begingroup\$ I would just like to add that for a bench test or similar, feel free to run that Vgs up close to 20 V - no problem. But, if you would make this into a product which needs several years of life time, you will actually wear out the device if your Vgs hits the absolute maximum rating on every cycle in a way that Vds does not. I haven't cracked them open and studied them under a microscope, but the gate structure does weaken over time if you apply maximum voltage for a long time. \$\endgroup\$ – winny Jun 11 '16 at 9:16

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