0
\$\begingroup\$

I am wanting for safety a uC to determine a few environmental checks before a device is allowed to be activated, it is a relatively low voltage (2V) however high current (4A) load (powered by a beefy constant current regulator)

A relay requires some current to initiate and inductive kickback protection on the uC's side, so I do not wish to implement something such as that if there are other options. I had looked at some power nFETs to do this instead and do not understand some points.

Will 4A flow through the power nFET even though it may be located after the load? A simulation (i.e. falstad) shows current being the same before and after the load and through the nFET. If this is true, I will have to add heatsinking, however does not make much sense to me as the load will "use" the 4A mostly?

Can the nFET allow 2V to flow through it without much trouble when switched on by my uC, i.e. is without too much resistance?

nFET something simple like this: http://www.fairchildsemi.com/ds/FD%2FFDPF5N50T.pdf

\$\endgroup\$
  • 3
    \$\begingroup\$ This is NOT a complaint - just a hopefully useful comment: Your question is understandable but reflects some misunderstandings on your part of basic electronics in both terminology and practice. You are on the right path but will benefit from some reading up on MOSFET operation and general basic electronics switching. See my answer re actual question. \$\endgroup\$ – Russell McMahon Feb 10 '12 at 23:10
  • \$\begingroup\$ What is the drive voltage for the FET gate? What IS the load? How rapidly / often will it be switched? \$\endgroup\$ – Russell McMahon Feb 10 '12 at 23:40
  • 1
    \$\begingroup\$ Voltage DOES NOT flow. Current DOES. \$\endgroup\$ – Count Zero Feb 11 '12 at 11:31
  • \$\begingroup\$ @CountZero: some highly respected educators claim that there is no "flow of current", either. \$\endgroup\$ – davidcary Feb 23 '12 at 2:23
  • \$\begingroup\$ @davidcary: I see your point. What I wanted to make clear is the difference between voltage and current (apparently the OP confused them). I think we both agree that the expression 'flow of current' is deeply anchored in the technical language and a specialists in the field will know instantly that it refers in fact to a 'flow of electric charge'. This makes me think that the debate on the correctness of 'flow of current' is a matter of linguistics. \$\endgroup\$ – Count Zero Feb 23 '12 at 11:04
5
\$\begingroup\$

Placement

You are correct, the current before and after the load is always the same. Because the voltage that switches on the MOSFET is measured between gate and source, you will better put it between the load and ground so you get a larger voltage difference.

Cooling

To determine the need for a heatsink, first you need to calculate the power that is lost in the MOSFET. This is

$$P = R_{DS(on)} \cdot I^2$$

At Vgs = 10 V it has a resistance of 1.4 Ω, which at 4 A means a power of 22 W. As Russell mentions, you'd better go for a MOSFETs with a much lower Rds(on), there are MOSFETs available with an Rds(on) low enough to run without a heatsink, but let's do the calculation for this one anyway:
At an ambient temperature of 30°C and a thermal resistance of 62.5 K/W (given in the datasheet) the junction would reach over 1000°C which is far off scale in Fig. 8 and too hot even just estimated by common sense. So you definitely need a heatsink. Choose one, look up its thermal resistance in its datasheet, add it to the thermal resistance junction-to-case-to-sink of the MOSFET (1.9 K/W) and then redo the calculation.

Keep in mind that the heat must be dissipated out of the case (by a fan or by natural convection) to keep the ambient temperature.

On-Voltage

The MOSFET you have chosen has a typical Vgs(th) of 4 V. They don't give the Rds(on) for that voltage and neither is there a diagram of Rds(on) vs. Vgs, but from Fig. 2 you can see, that the Rds is still quite large at 4V. You didn't tell us the voltage your µC works at, but if it's 5 V or even 3.3 V you will need a MOSFET driver to drive this MOSFET. This is probably true for all MOSFETs of that range.

\$\endgroup\$
  • \$\begingroup\$ Very clear explanation, now if only I can accept both answers. It looks like I will want a very low Rds on such as what Russell has linked to, to save the need for 3 heatsinks in one project. They get a little expensive compared to everything else. Thank you. \$\endgroup\$ – Kenny Robinson Feb 11 '12 at 0:05
6
\$\begingroup\$

Basic operation:

Think of the FET as an on/off switch or tap.

Current flow is as water current flow.
Power supply or battery is like a pump or head of water.
Load is like a hydraulic motor or a weir that drops water head or a hole in a pipe that drops pressure as the water escapes.

Current flows FROM pump through load through switch back to pump inlet.

OR

Current flows FROM pump through switch then through load then back to pump inlet.

Either way current MUST flow through switch and load regardless of what order thy are in.


MOSFET CHOICE:

A properly specified MOSFET will easily meet your requirement.
The one you have indicated is a good one for high voltage applications but totally unsuited to this task due to its very high on resistance = Rdson .
This high resistance is a feature of the high voltage capability of the MOSFET and will not be a major issue in its intended role.
A MOSFET to meet your need can have a rated voltage Vds of say 10V or more but 20V or30V parts are safer and as cheap, and a rated current of say 5A, but 10A to 30A or even 50A parts are also often as cheap and have some advantages.

If a part with Rdson = 0.1 ohm is chosen then on dissipation at 2A = I^2R = 2^2 x 0.1 = 0.4 Watt. A TO220 package will run at this power level with no heatsinking and a DPak package surface device needs only modest copper heatsinking. If Rdson is say 0.02 ohm (easily achieved with modern devices) dissipation at 2A = 2^2 x 0.020 = 0.08W ~= 00.1 Wa and no heatsinking is needed with the above devices

**What is your system Vdd - ie what drive voltage is available?
2V / 3V3 / 5V / more ... Very important for FET selection.

Suitable MOSFET:

I did a Digikey search for 30v+, 10A+ NChannel MOSFTS with 2V max gate turn on voltage

Cheapest stocked item is a SMD package but hobbyist almost friendly
NXP (Philips) PSMN9R5-30YL - datasheet
63c/1 Digikey in stock 30V, 44A !, 9.8 milliohm !!!
Rson is not nice at 2.5V Vgs but is under 15 milliohm at 3V.
ie driving in a 3.3V or more system is good.

ST brand DPak STB85NF3L - datasheet

30v, 85A !!!,
Good at 3V drive
Superb at 5V drive
About Under 80c/1.

[TO220 STP55NF06 60V 55A $1.56/1 but TO220 may suit.]( STP55NF06)
Good at 2.5V gate drive
Superb at 34V gate drive
Works ~~ at 2V gate drive.


You do not say what the load is.
If it is inductive it MUST have a reverse diode placed across it. 1 x 1N400x will probably suffice.

If it is not inductive a reverse diode will not hurt, costs a few cents and will save you if the load is in fact more inductive than expected.
Murphy lurks everywhere.

\$\endgroup\$
  • \$\begingroup\$ Input will be 6V for constant current regulator, so I will likely use an LDO or something to be determined to power the microcontroller at 5V. The load is actually just light bulb/laser (maybe later) requiring a specific current at around 2V and I do not believe is that inductive. If 5V can turn those two on I am very happy, I just hadn't turned my brain on for a moment there. Thank you Russell! I am off to implement this in to my schematic and crunch some numbers (reading mosfet operations is on my to do list too. \$\endgroup\$ – Kenny Robinson Feb 10 '12 at 23:54
  • \$\begingroup\$ +1 for help choosing a better MOSFET. Regarding basic operation I think you should mention that/why low-side switching is preferred. \$\endgroup\$ – AndreKR Feb 11 '12 at 1:49
  • \$\begingroup\$ Use a reverse diode across the load. It's a lot cheaper than a microcontroller. \$\endgroup\$ – Russell McMahon Feb 11 '12 at 3:30
  • \$\begingroup\$ @Russell, I had meant for the uC require some input/warning and only after a certain period of time power the FET to allow the path from the high current load to flow. \$\endgroup\$ – Kenny Robinson Feb 11 '12 at 6:50
4
\$\begingroup\$

You seem to have a misconception of current. Current flows through a conductor in the same amount through all of it. It is like a freeway. When one car gets on the freeway another gets off and there are always cars moving in between. So no matter what device you put in series with your load it will have the same current going through it as the load.

Basically current cannot be "backed up" because you'll end up violating the conservation of energy/charge. When one electron enters some conductor another must leave(this even happens in capacitors where it seems like you are actually storing electrons). If this doesn't happen you have an excess of electrons and somewhere else a deficit. Even just a few electrons in this state can create huge forces so it can't ever really happen to any degree(lightning is a pretty extreme example but we are talking about basic electrons here ;).

Volts do not flow. The current through a fet(or any device) will create heat depending on the resistance of that device. A fet has a resistance associated with it. If you have the fet switched fully on then the power dissipated is P=I^2*R_ds(on). If P is too large(usually more than a W or two for power fets) then you must use heat sinks... which then require calculations using thermal resistances to determine the junction temperature of the fet(it can be quite involved).

The easiest way is to simply get a fet with reduced R_ds(in) or parallel them.

There are a ton of factors involved. The best I can say is experiment and have fun. When you burn up a few fets you'll quickly start to understand it all. It's not that complex but lots of little things to learn about. Fets are very cool devices and have a ton of uses but also have some issues.

Note that 4A for a 10mOhm power fet is only 16*0.01 = 160mW. Any power fet will do(but note it takes more to drive a power fet) without a heat sync. But a small signal mosfet with 10Ohms resistance will dissipate 16*10 = 160W. It will literally explode.

So the trade off is finding the right size mosfet that you can drive at the speed you need and can handle the current you want. There will always be mosfets that can be used without heat sinks but it's not always practical. You could parallel 1M mosfets if you wanted. This will reduce the resistance by a factor of 1M. The more mosfets you have in parallel the harder it becomes to turn them all on quick enough.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.