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Good day

I'm busy designing a small DC UPS to run some appliances when the power goes out. I am, however, having trouble with an element in the battery switch-over circuit.

I decided on using a circuit configuration similar to those that were discussed here, here and here.

I learned that replacing a diode with two opposite facing series-connected MOSFETS has the great advantage of a much smaller voltage drop, while still being able to block reverse currents when the devices are in the off state.

My question is: Does it matter which way around the MOSFETS are connected (e.g. source-to-source or drain-to-drain)? I have seen them being used in both configurations. Could somebody please explain the difference to me?

Please refer to the attached images to clarify my answer.enter image description here enter image description here

I am aware that the voltage difference between the two sources will allow me to just use or'ing diodes, but I need the voltage drop advantage on the battery side, unless there is a simple way to construct "ideal" diodes that can be used in stead. Unfortunately I am limited to through-hole components.

I am looking forward to any answers and responses. Reinforcing critics on my circuit design is also most welcome.

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I don't think there will be any difference between the circuits.

In principle a MOSFETs drain and source are equal, like this: enter image description here See, no physical difference !

But many discrete (power) MOSFETs are build such that the Drain can take a higher voltage. To keep Rds_on low, the same changes are not applied to the source.

Then there is the body, bulk or substrate (different names but same thing), both source and drain have a diode to substrate (N+ to p junction as in picture). In discrete MOSFETs the bulk is almost always connected to the source resulting in another difference ! The result is that only the drain-substrate diode remains but since substrate is connected to the source, the diode is now between drain and source ! When you would use only one NMOS in your circuit, how you connect the drain and the source would matter.

But you have two mosfets in (anti) series one drain-source diode will conduct but the other diode will be in reverse and make the open circuit.

So in my opinion, both circuits will work equally well.

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  • \$\begingroup\$ Thank you for your detailed answer. It is very insightful. Would the gate voltage required to trigger the devices be the same in both cases? \$\endgroup\$ – Egon Jan 22 '16 at 9:33
  • \$\begingroup\$ Yes, it will be the same. As long as Vgs is much larger than the Vt (threshold voltage) of the MOSFET it will switch on. In your circuits this is the case. \$\endgroup\$ – Bimpelrekkie Jan 22 '16 at 9:48

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