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Author Topic: How much current flows through a MOSFET when off?  (Read 1607 times)
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Given a circuit like this:



If the MOSFET is off (ie. the Arduino pin is LOW) then what current would you expect to flow between Drain and Source? I know "not much" but are we talking maybe 10 nA?

I am asking because of low-power applications, if we had (say) a motor connected to the MOSFET but did not want to drain the battery while it was off (and it might be off for a while) can we consider the current through the MOSFET to be negligible?

I was thinking of this project: http://arduiniana.org/projects/the-reverse-geo-cache-puzzle/

He used a Polulo switch which lets the circuit turn itself off, but I was wondering if simply going to sleep (and having the button connected to Reset) would achieve the same thing. Then you control external devices, like the GPS and the motor, through a MOSFET or two.



(edit)

Because the above graphic is linked to from a number of pages it has now been amended to have a 150 ohm gate resistor (R1) rather than a 10 ohm one originally shown. I mention this so that subsequent comments on the circuit make sense.
« Last Edit: June 27, 2013, 05:32:50 pm by Nick Gammon » Logged


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If the MOSFET is off (ie. the Arduino pin is LOW) then what current would you expect to flow between Drain and Source? I know "not much" but are we talking maybe 10 nA?

The datasheet will tell you but I think it will be more than that. I just looked at a couple and it's more like 10uA than 10nA.

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10uA is far less than the self-discharge rate of most batteries - you can ignore FET leakage
at room temperature.

In the diagram the gate current limiting resistor is shown as 10 ohms - no use for the Arduino,
150 ohms would be a more suitable value (keeps Arduino pin current within spec - less than
40mA)
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10uA is far less than the self-discharge rate of most batteries - you can ignore FET leakage
at room temperature.

But still far more than a sleeping AVR chip... smiley-sad

(which has tens of thousands of MOSFETS in it)

I just flicked through a few datasheets, the Intersil HUF76137P has a much lower 'off' current, probably less than a microamp...but still nowhere near the nanoamp range.

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10uA is far less than the self-discharge rate of most batteries - you can ignore FET leakage
at room temperature.

In the diagram the gate current limiting resistor is shown as 10 ohms - no use for the Arduino,
150 ohms would be a more suitable value (keeps Arduino pin current within spec - less than
40mA)

I did a whole lot of reading about gate resistors a while back. A lot of pages said "it depends", and then a lot of examples quoted in the 10R to 100R range. Just out of curiosity, what is the basis for suggesting 150R?
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Just out of curiosity, what is the basis for suggesting 150R?

It stops more than 40mA coming out of the Arduino pin.
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So the MOSFET would temporarily act as a complete short on the output pin? And thus we need to limit the current, is that it? So that would limit the current to 33.3 mA?
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Drain leakage of a MOSFET with a low threshold :

1 uA at 25 degrees C
20 uA at 150 degrees C

http://www.nxp.com/documents/data_sheet/PMV20XN.pdf

Summary of spec :
30v max drain voltage
4.9 amp
Threshold voltage .5 to 1.5 volts at 250uA and Vds=Vgs

For less drain leakage current, get a MOSFET with a higher threshold voltage.

Figure 7 gives details of drain current at sub-threshold gate voltages

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So the MOSFET would temporarily act as a complete short on the output pin?

Yep. You're charging/draining a capacitor every time you switch it on/off.
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"But still far more than a sleeping AVR chip...

(which has tens of thousands of MOSFETS in it)"

Yes, but the total *area* of all those MOSFETs are probably less then the area of the power device, which is what counts - to a first order - in leakage.
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So the MOSFET would temporarily act as a complete short on the output pin? And thus we need to limit the current, is that it? So that would limit the current to 33.3 mA?

A lot depends on the mosfet's specific gate capacitance as that capacitance value plus the output impedance of a arduino output pin determines the peak charge and discharge current value of the gate. The AVR output pins are pretty rugged so in some cases with smaller mosfets a series gate resistor may not even be required, for other larger mosfets they may be required to protect the output pin. The down side of using a series gate resistor is that the then larger resulting R/C charge/discharge time constant can slow the speed that the mosfet can switch from on to off and off to on, which can raise the mosfet's heat dissipation significantly depending on switching frequency and load current requirement. So there are tradeoffs to be considered depending of the specifications that are most important for the circuit under consideration and of course the specific mosfet being used.

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« Last Edit: June 26, 2013, 11:45:27 am by retrolefty » Logged

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Quote
Yes, but the total *area* of all those MOSFETs are probably less then the area of the power device, which is what counts - to a first order - in leakage.
The main reason is the CMOS technology (atmega) uses N-channel and P-channel transistors in series (wired between Vcc and Gnd). The single N-channel mosfet with a load wired in the drain is always "under power"..
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For a MOSFET I figure the cap will look like a short that you want to charge up quick.
So, 5V/150 = 33mA, should go pretty quick, especially if the gate capacitance is not too high, like this one:
http://www.digikey.com/product-detail/en/NTD5867NL-1G/NTD5867NL-1GOS-ND/2401422

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650pF is not the smallest I've seen, the input capacitance could be up to 2500pF.


* mosf.jpg (40.17 KB, 639x566 - viewed 16 times.)
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Drain leakage of a MOSFET with a low threshold :

1 uA at 25 degrees C
20 uA at 150 degrees C

http://www.nxp.com/documents/data_sheet/PMV20XN.pdf
If you check, you'll also see the leakage for higher-power MOSFETs is much
higher, 25 uA at room temp for IRL540, for instance.
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