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I am trying to switch on/off the connection between a load and a variable power supply (going from -10 to 10 V) using a combination of NMOSFETs and PMOSFETs. I want to use MOSFETs because of limited power consumption in the overall design. Also, because I need the power supply and the voltage load to be of same value.

I want to turn off/off the connection between the load and the power supply using a digital output that goes from 0 to 5 V.

The problem is, I am able to design a circuit that can be completely shut off (when digital output is 0 V for instnace) , when the power supply goes from 0 to 10 V using a combination of PMOS and NMOS, but I am not able to do figure out how to do it when the power supply goes from -10 to 10 V.

I imagine the circuit to be something like this:

enter image description here

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  • \$\begingroup\$ Look in one of the most classical CMOS structures, the "transmission gate". Basically an electronic switch (think 4066). The trick is using one CMOS pair for gate, not only one \$\endgroup\$ Commented Jul 6, 2021 at 10:34
  • \$\begingroup\$ Thanks for the answer. So the trick is to use an IC? For instance 74HCT 4066 for instance? \$\endgroup\$ Commented Jul 6, 2021 at 10:54
  • \$\begingroup\$ Well, depends on the current… the 4066 can handle a few milliamps but you can do it in discretes for whatever current you need \$\endgroup\$ Commented Jul 7, 2021 at 11:21
  • \$\begingroup\$ If you are using discrete components, you almost certainly will not have access to the bulk (body) connection as a separate connection. Unless you have a clever solution on how to do this? \$\endgroup\$ Commented Jul 7, 2021 at 12:46
  • \$\begingroup\$ Well I cede to your argument, it can't be done discretely :( also the ESD protection would interfere, most probably. Ok, just use a fat load switch grounded on the negative rail and a suitably translated control \$\endgroup\$ Commented Jul 13, 2021 at 5:50

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The reason why your P-MOSFET doesn't seem to turn off when the supply voltage goes negative is because of the parasitic body diode in the PMOS. It becomes forward biased and conducts when the supply voltage goes negative.

enter image description here

You need two MOSFETs in order to prevent the body diode from conducting.

enter image description here

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  • \$\begingroup\$ Maybe I did not make myself clear. I want the voltage supply to conduct at -10 V, but I want it to turn off by changing the digital output from 0 to 5 V. \$\endgroup\$ Commented Jul 6, 2021 at 11:20
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    \$\begingroup\$ The circuit that I suggested will conduct even at -10V. You can't use a digital signal directly. You would have to feed it through a level shifter. \$\endgroup\$ Commented Jul 6, 2021 at 11:22
  • \$\begingroup\$ Okay, so I will have to use a digital output with a level shifter to control the gate voltage? But what about the load voltage? I need the load voltage to be equivalent to the voltage supply. \$\endgroup\$ Commented Jul 6, 2021 at 11:28
  • \$\begingroup\$ The load voltage will almost be equal the the supply voltage when the MOSFET arrangement is ON. And yes, you will need a level shifter. \$\endgroup\$ Commented Jul 6, 2021 at 11:35
  • \$\begingroup\$ Would you care to tell me if what type of MOSFETs you are referring to? I've tried different combinations, and none of them seem to work with simulations. Is the top one a NMOS or PMOS? And what about the bottom one? \$\endgroup\$ Commented Jul 6, 2021 at 11:49

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