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Hello I would like to add a high side switch for my circuit. Typically a physical switch can do this job but due to form factor restraints the switches that are rated for my application does not meet the cut. To mitigate this is to use a MOSFET along side a lower power rated switch that meets the size limitation.

Usually you use a P-Channel MOSFET for highside switching but im having trouble looking for a part that meets the specifications in my supplier (3.3v ON 10+V 5-10A PMOS are not that common it seems). N channel MOSFETS on the other hand has a lot more of variety, but typically they are used for low side switching.

Is it possible to use an NMOS as a highside switch ? what are the disadvantages?

In my circuit i have a Reverse voltag protection IC that controls an NMOS on the highside. This makes me think that it is possible. I will just use the same NMOS for the physical switch as the one i will use for the RVP in series.

You might say why not integrate the physical switch to the NMOS in RVP, that seems to not possible as discussed here. The circuit would now look something like this.

enter image description here

EDIT:

I have found a document from texas instrument which is very interesting contains parts which could have everything in my circuit be contained in a single ic. On page 4 contains a list of features that i would want. what cought my attention is this ic TPS25942 it is a high side switch with reverse voltage protection and just requires minimal passives. What is important is i has an enable (EN) pin wich can be used by a physical switch

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    \$\begingroup\$ Search High-side gate driver. But 100% duty cycle is required for reverse polarity applications which takes a lot of extra circuitry. Unless you get an purpose built IC with internal charge pump. Usually you use high side NMOS for motors and inverters which does not have this restriction. \$\endgroup\$ – DKNguyen Mar 24 '20 at 14:48
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    \$\begingroup\$ Not like that. Notice the capacitor on the protection IC? It's a charge pump to boost the gate drive voltage for Q1. I would simply connect Q2 gate to Q1 gate to turn ON, and to GND via R1 to turn off. (which would mean an SPDT swittch, or making R1 high enough that the IC can drive it. \$\endgroup\$ – Brian Drummond Mar 24 '20 at 14:48
  • \$\begingroup\$ Q2 won't do anything for you except act as a diode. If I were doing this, I would use a commercially available part. See, for example, analog.com/en/products/monitor-control-protection/… \$\endgroup\$ – Peter Smith Mar 24 '20 at 14:50
  • \$\begingroup\$ @BrianDrummond wont that be the same as putting a switch on the gate of Q1 ? which according to the other thread linked it wont work? \$\endgroup\$ – Jake quin Mar 24 '20 at 15:07
  • \$\begingroup\$ @PeterSmith i have taken a lookt at some of the IC and wow this seems to be a lot of components for just a switch \$\endgroup\$ – Jake quin Mar 24 '20 at 15:09
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Is it possible to use an NMOS as a highside switch ?

Yes it is possible, see this circuit:

enter image description here

The issue is that you need to apply a gate voltage to the High side NMOS that is higher than the supply voltage. If you just applied the supply voltage (\$V_{IN}\$ in the picture) then the NMOS would not be fully switched on, the NMOS would drop a significant voltage between drain and source. That voltage is somewhat larger than the threshold voltage (\$V_t\$) of that NMOS.

The reason for this is that the gate-source voltage \$V_{GS}\$ of the NMOS must be at least \$V_t\$ for the NMOS to even conduct at all.

For the lowest resistance between drain and source (and that's what we want as we want to use the NMOS as a switch) we must apply a large \$V_{GS}\$. As we want \$V_{IN}\$ = \$V_{OUT}\$ when the NMOS is on that means that at the gate of the NMOS we need a voltage that is higher than the supply voltage.

One way to make this voltage is to use a bootstrap circuit. The "Built-in diode" and the "Bootstrap capacitor" form the bootstrap circuit. When the low-side NMOS conducts, then the High-side NMOS must be off (Do you see why?) then the capacitor charges to \$V_{IN}\$.

When state changes and the low-side NMOS stops conducting and the high side NMOS is switched on voltage at the pin called "BOOT" raises above the supply voltage. That is then used to fully turn on the high side NMOS.

The disadvantage of this circuit is that the bootstrap capacitor needs to be charged before it is discharged (due to leakage currents etc.). So for an application where \$V_{OUT}\$ is switched on/off regularly, this will work. For an application where the high side NMOS is switched on and must remain switched on for a long time this circuit is not suitable. Then a charge pump or a boost converter could be used to make a voltage that is higher than the supply voltage.

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  • \$\begingroup\$ I see, why does the voltage on the gate have to be always greater than the input voltage if used as highside to turn it fully on, but when used properly (on the low side) you only have to a smaller voltage \$\endgroup\$ – Jake quin Mar 24 '20 at 16:27
  • \$\begingroup\$ On the compenent/material level why is that? \$\endgroup\$ – Jake quin Mar 24 '20 at 16:32
  • \$\begingroup\$ @Jakequin The turn-on condition for enhancement mode NFETs is that VGS > Vth so when the source voltage is the same as the input voltage you can see why the gate voltage must be higher than the input voltage... \$\endgroup\$ – Captainj2001 Mar 24 '20 at 18:00
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    \$\begingroup\$ On the compenent/material level why is that? Answering that here goes too far for an answer. Also, it is not needed as there are plenty of books and courses available that explain how MOSFETs work. You could start here: en.wikipedia.org/wiki/MOSFET do note that you might also need to study semiconductors and the PN junction as well to fully understand everything. But there's no need to learn all that if you just want to use a MOSFET in your circuit, Then just remember that a MOSFET will only turn on when \$V_{GS} > V_t\$, whatever the reason for that is. \$\endgroup\$ – Bimpelrekkie Mar 24 '20 at 19:44

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