# Power Dissipated by Reverse Polarity MOSFET Protection

#### brightnight1

Joined Jan 13, 2018
70
I want to implement a reverse polarity N FET or P FET to protect a low power PIR circuit. I expect the PIR circuit to use ~26uA, so I need a protection MOSFET that is very efficient when turned on, otherwise it will be wasting more power than the circuit is using.

From looking online, a low R(DS) ON resistance at 3.3 volts appears to be ~ .013 Ohms. Somehow I keep messing up my current calculation and therefore my power loss calculation. I think it’s because I am calculating the current for the FET and not using the 26UA for the power loss calculation?

Power losses are found by the R(DS) ON of the device and current load:

If I have a 3.3v supply and the R(DS) ON is 13 mOhms

I=3.3/.013 ohms = 253.8A? Clearly this is wrong

Power loss = (I)^2 *(R)= (253.8)^2*(.013) = 837 Watts

There’s no way that current calculation is correct… Mostly I’m confused why this gives me the wrong answer.

I think this calculation is correct using the actual current the circuit will be using but I’m not sure. It’s much closer to what I’d expect, but I’m still confused why the previous current calculation doesn’t work:

Power loss = (I)^2 *(R)= (.000026 A) ^2*(.013 Ohms) = 8.7X10^-12 Watts

Should I also be taking into account the current the MOSFET consumes just to stay on? Thanks in advance

#### ebp

Joined Feb 8, 2018
2,332
power = (current squared) times resistance

A one ohm FET will dissipate
(26 µA)^2 x 1 ohm = 6.76 x 10^-10 W = a really really tiny amount of power

1000 ohms would still be only 676 nanowatts
The load power requirement is 86 microwatts

Virtually any very small logic-level P-channel MOSFET in the high side (often preferable if the negative of the power supply must be directly connected to the negative of the powered circuit) or N-channel on the low side will do. The maximum gate-source threshold voltage needs to be in the range of 2.5 V or less.

The gate current will be nanoamps at 3.3 V. If you were using voltage higher than the maximum allowable gate-source voltage, then you would need to protect the gate and would need to be a little careful to keep the power in the protection circuit down.

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#### Xavier Pacheco Paulino

Joined Oct 21, 2015
728
The power losses of a MOSFET are determined by the RDS(on) of the device and the load current.

#### brightnight1

Joined Jan 13, 2018
70
Ah ok that makes much more sense, thank you!!!

Someone replied to a different post I made and suggested the SI7137DP-T1-GE3, looks like it makes a really good 3.3V logic level FET. Do you agree?
Drain to Source Voltage (Vdss) = 20V
Vgs(th) (Max) @ Id =1.4V @ 250µA
Rds On (Max) @ Id, Vgs =1.95 mOhm @ 25A, 10V

https://www.digikey.com/product-detail/en/vishay-siliconix/SI7137DP-T1-GE3/SI7137DP-T1-GE3TR-ND/2622441

Looking at a different FET, I was a little confused as to if the IPP024N06N3GXKSA1 is a logic level FET (info below). It’s an obsolete part but it on the info page on Digikey:
Vgs(th) (Max) @ Id4V @ 196µA

However, on the datasheet it says that the gate threshold voltage is 2V minimum, typical 3V and a max of 4. What the different values and I assume I want a max gate threshold voltage of lower then this (this isn’t logic level).

https://www.digikey.com/product-detail/en/infineon-technologies/IPP024N06N3GXKSA1/IPP024N06N3GXKSA1-ND/2081135

and most of the graphs range up to 5V, so perhaps this FET is for 5V logic?

#### ebp

Joined Feb 8, 2018
2,332
The Si part would be very good for voltages up to at least 15 V (you could go fully to 20, but you would need to be careful there would be no transients above that). It would be usable on copper no larger than the package outline up to 10 amps or so. At higher current you would want to have more copper area for cooling.

The second part would be a bit dubious for high current at anything less than about 6 volts G-S. The 4 volt maximum threshold voltage means it could be just barely starting to conduct at 4 volts. The typical data shows the "gate plateau voltage" (the horizontal bit in the gate charge plot) as 5 volts. I generally aim for at least a volt or two above that. It would probably be acceptable at 5 volts, but I wouldn't go lower.

For your application, something like a 2N7002 would be suitable for a low-side switch. It isn't strongly enhanced at only 3.3 volts G-S, but certainly adequately for very low loss at only 26 µA.

#### -live wire-

Joined Dec 22, 2017
912
One thing to be aware of, mosfets will be turned only partially on if the gate voltage is too low. That is generally pretty bad. While it will not be too big a deal for such low power, it is still something to be aware of. Look at the datasheet to find the minimum recommended gate voltage.

#### brightnight1

Joined Jan 13, 2018
70
For your application, something like a 2N7002 would be suitable for a low-side switch. It isn't strongly enhanced at only 3.3 volts G-S, but certainly adequately for very low loss at only 26 µA.

I really like the 2N7002 because they're only $.18, but Digikey says "Rds On (Max) @ Id, Vgs 7.5 Ohm @ 50mA, 5V" So wouldn't something like the SI7137DP-T1-GE3 which is "Rds On (Max) @ Id, Vgs 1.95 mOhm @ 25A, 10V" be much more efficient because it's Rds is so much lower? Only problem is the are SI7137DP-T1-GE3$2.73 each! I'll be using a lot of these on every board I make so cost is definitely something I'll have to balance in and take a deeper look at.

#### ebp

Joined Feb 8, 2018
2,332
Again, go back to the calculations. The current-squared is so close to zero that multiplying it by a few ohms still results in a tiny fraction of the power used by the circuit. In my calc for 1000 ohms, the power dissipated in that 1000 ohms is 680 nanowatts - about 0.8% of the power used by the circuit that is being powered, so the efficiency is over 99%. At ten ohms it would be 0.0008%.

Once you get up into higher currents, the 2N7002 would be less suitable. At 50 mA, for example, the power would be
(50e-3)^2 x 7.5 = 19 mW - assuming 3.3 V supply, the load power would be 165 mW, which is not a great ratio, but might be acceptable.​
At 100 mA the FET's dissipation would be about 75 mW vs. 330 mW for the load, so the ratio is starting to look decidedly unattractive. 100 mA at 7.5 ohms is 0.75 V, which you probably can't tolerate with a 3.3 V supply - many things that are spec'd to run at 3.3 V won't be happy with only 2.55 V. And you probably want gate-source voltage of more than 3.3 V to be sure the FET is fully enhanced.

Ignoring cost, typically you evaluate this sort of thing from multiple standpoints: voltage drop, the overall efficiency and the need for heatsinking for the FET. You might decide efficiency is OK with one type of FET, but if you spend just a little more you might get a FET that not only improves efficiency but saves either board area for heat dissipating copper (SMD) or a heatsink that not only costs a bit itself but requires labor to mount the device to it.

I recommend doing calculations like this with a spreadsheet, even though they are simple. It makes it very quick and easy to check all the numbers that interest you - voltage drop, power dissipation, efficiency, temperature rise. But when you use a spreadsheet, you always need to verify that you've gotten all the formulas right by doing "manual" calculations. Simple calc's like this are something you need to be thoroughly competent at in any case. You should be able to do more complicated calc's too. I have a nasty tendency to make some stupid error someplace early on in a long chain of calculations, so I like spreadsheets because I can go back and find where I forgot to multiply by the square root of 2 or some other dumb thing and can fix everything by fixing one formula instead of recalculating everything subsequent the hard way. I generally don't do very elaborate formulas in a spreadsheet, but rather use formulas that give intermediate values. Those values not only are often useful in themselves, but help with verifying that I've entered the formulas properly - do those values look reasonable?

#### crutschow

Joined Mar 14, 2008
26,321
So wouldn't something like the SI7137DP-T1-GE3 which is "Rds On (Max) @ Id, Vgs 1.95 mOhm @ 25A, 10V" be much more efficient because it's Rds is so much lower?
Yes it would, but how efficient is efficient enough?
Suppose you want to drop no more than 1mV across the Rds of the MOSFET when it is on.
That means an Rds of no more than 1mV/26μA = 38.5Ω with a power dissipation of 26e-6² * 38.5 = 26nW.

Is that low enough for you?
If so then a 2N7002 should work fine, but note that it must be placed in the negative line to work for reverse voltage protection.

Below is the LTspice simulation of the circuit using a 2N7002 for an input from +3.3V to -3.3V.
The is a 431μV drop across the MOSET at the start when the input voltage is +3.3V and the load current is 26μA.
The output voltage then drops to zero when the input goes negative, protecting against a reverse polarity input.
(Note the polarity of the MOSFET for proper operation of the circuit.)