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I am very confused about the way the below circuit functions. Below is an excerpt from a research paper which shows two NMOS transistors connected such that the body diodes face each other back to back. This is to form a bi-directional switch (low side), as it will be supplied with an AC voltage source.

My confusion arises from the way the bulk/substrate of the transistors are handled. The substrates of the MOSFETs are tied together. My understanding of the MOSFET's device physics is that, for forming the inversion layer of the MOSFET, the gate of the MOSFET has to be given a voltage higher than the substrate and also the body diode has to be reverse biased to avoid leakage. Taking note of these conditions, the bulk is connected to the lowest potential in case of an NMOS transistor and to the highest potential in case of a PMOS transistor. But generally speaking, the often told condition for turning on an NMOS device is Vgs>Vth, which still satisfies the bulk's requirement as it is most often connected to the source. Request your kind attention to my questions below:

  1. Will the MOSFET turn on if I keep my bulk floating, but still ensuring Vgs>Vth?
  2. If my understanding is correct and the bulk has to be connected to the lower potential, then in that case, how does the above circuit function when positive potential is applied at node SN.
  3. Also, please note that the above circuit is implemented in Bi-CMOS process and there is a flexibility with the bulk connection. Can the above circuit be implemented using discrete components?
  4. Isolated gate drive is recommended when driving a bi-directional switch http://goo.gl/0u5om2. Is there a way this can be done without isolated gate drive?

Thanks for reading.

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  • \$\begingroup\$ Note that the two transistors should be as Andy shows - or the switch is not bidirectional -UNLESS the no ref no real partt number provided 3600xx is a magic device which has body diode inverted as shown on two N Channel devices - and if do its still wrong as input needs to be to a common gate to source reference for both transistors. This is becoming a classic death by 1000 snippets question with power budget now being added. If you told us what you were actually doing and not ways that may work to do whatever it is a good solution would be easier to come by. You mention strings of ... \$\endgroup\$ – Russell McMahon Jun 18 '15 at 11:17
  • \$\begingroup\$ ... actually " ... the author in the paper uses a chain of inverters to drive the gate of the bi-directional mosfet without anything connected to the common point of the transistors ..." BUT you may be misunderstanding what he is doing and we do not know what are hidden constraints there are. \$\endgroup\$ – Russell McMahon Jun 18 '15 at 11:23
  • \$\begingroup\$ Here a few examples \$\endgroup\$ – Russell McMahon Jun 18 '15 at 11:23
  • \$\begingroup\$ @RussellMcMahon This circuit came up in the image search results you sent me: goo.gl/VEo1fq. I understand that I haven't divulged a lot of information for you to give me a good solution. That's because I am doing my PhD and this is part of my thesis. The 3600/2 is not a model number, but the W/L ratio of transistors in a Bi-CMOS process. I have mentioned the process information in my original post. \$\endgroup\$ – ashare Jun 18 '15 at 14:22
  • \$\begingroup\$ That circuit 'had me going' for a moment :-). It's MORE than meets the eye. That's a rather special driver IC. Those are two N Channel FETs and the gate drive is +ve ABOVE the battery rail - the LTC1154 contains a charge pump with internal to IC capacitors to provide gate drives of well above Vin. -data sheet here and LT AN53 app note here on use of it and its family as high side drivers, Dual FET used here nothing special. \$\endgroup\$ – Russell McMahon Jun 18 '15 at 15:21
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The two MOSFETs are drawn incorrectly in the question's picture. That is why you are confused. This is how they should be i.e. sources tied together: -

enter image description here

I'm not aware that there is a decent non-isolated gate-drive version.

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  • \$\begingroup\$ In CMOS process, the source and drain are symmetrical and is decided by the direction of current flow. Since this is an AC circuit, the source and drain actually changes positions in every cycle. I can understand your circuit as it satisfies the condition that the gate to source (bulk) is positive and can work pretty well too using an isolated gate drive. However, the author in the paper uses a chain of inverters to drive the gate of the bi-directional mosfet without anything connected to the common point of the transistors. Wondering how that works for him. \$\endgroup\$ – ashare Jun 17 '15 at 13:31
  • \$\begingroup\$ @ashare - It works for me :-) Or did - quite a while ago now. (Counts on fingers. Maybe 15 to 20 years ago ? :-) ) If the gate drive voltage will be correct when the FETS are on they tend to find their own way there quite well. You can pore over the circuit and wonder about leakage resistances and capacitive effects when on voltage transient is applied and more BUT it works quite well. Obviously a formal gate signal would ensure "rather better" switching waveforms. \$\endgroup\$ – Russell McMahon Jun 17 '15 at 14:09
  • \$\begingroup\$ @RussellMcMahon - Are you hinting that the turn on phenomenon is due to the leakage resistance and other capacitive effects which cannot be accounted for using the simplistic view of the MOSFET? \$\endgroup\$ – ashare Jun 18 '15 at 6:51
  • \$\begingroup\$ @ashare - the circuit I have shown works period. It is used in virtually all solid-state relays. For one half cycle of the AC one device is "normally biased" and for the other half cycle the other device is. It's as simple as that. Both FETs are turned on and off by the presence of a floating drive to the common gate and this is normally achieved using an optical photo-voltaic device hence it gives isolation. \$\endgroup\$ – Andy aka Jun 18 '15 at 7:08
  • \$\begingroup\$ @Andyaka I was commenting on his comment on what they did in 'the paper' and he was commenting on my comment. I'd not recommend the floating one sided drive system - just noting that it works (when it wants to) after a fashion. \$\endgroup\$ – Russell McMahon Jun 18 '15 at 7:47

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