0
\$\begingroup\$

I was looking for a MOSFET that I could use to control 3.7v 4A dc motors with for a quad copter. I'm very new to electrical stuff, but I read that I could use MOSFETs to control motors. I look online for MOSFETs meeting the specs I listed above, and noticed that the majority of MOSFETs have labels with extremely high currents. For instance https://www.amazon.com/INTERNATIONAL-RECTIFIER-IRLB3034PBF-MOSFET-220AB/dp/B00HKHW9UU/ref=sr_1_13?ie=UTF8&qid=1485926431&sr=8-13&keywords=mosfet says 195A on it, and it seems really small.

Can MOSFETs that small really handle currents that high? Wouldn't they just melt?

\$\endgroup\$
  • \$\begingroup\$ As stated by the datasheet, 195A is limited by the package. So, it can handle that current, while melting. \$\endgroup\$ – Wesley Lee Feb 1 '17 at 6:00
  • \$\begingroup\$ According to the datasheet, you can pass 195A IF you can keep the case at 25C while doing it. That will require more than just a heatsink. You will need some kind of liquid cooling, and the liquid will need to be below 25C. \$\endgroup\$ – mkeith Feb 1 '17 at 17:13
1
\$\begingroup\$

MOSFETs can handle currents, but there is always something to engineer around them. Normally you have to check power dissipation in your working conditions (caused by current and switching according to your system) and see if the MOSFET can dissipate it alone, or may require special layout or even heatsink. So when they state 195A, they may mean "with water cooling attached". Of course, this is not what they will write, they will just provide all required information so you will be able to check by yourself.

Take a look, this is your MOSFET's datasheet. They say Rdson is 1.5mR, so if you don't switch the resistor and current is 100A, the power dissipation is P = I^2*R = 100*100*0.0015 = 15W. You definitely would need serious heatsink for that. But for your 4A the power is 2.4mW, so it will stay cool.

One think you should know- your motors are probably brushless, which means you can't just switch them on and hope they rotate. You need a controller, either off-the-shelf or designed by you. And since moving a brushless motor is not trivial- go to ready made products first. Don't waste time.

\$\endgroup\$
1
\$\begingroup\$

Yes, it can pass the rated current, to use the full rated current you need heat sinks, MOSFETs device have a on resistance which defined the power dissipated by the device on ON condition at a particular current, based on this you need to size your heat sink

from the article Calculating power loss in switching MOSFETs

Every MOSFET has a resistive element, so it dissipates power as current is conducted through the device. The resistive parameter is described as on-resistance, or RDS(ON). These conduction losses are inversely proportional to the size of the MOSFET; the larger the switching transistor, the lower its RDS(ON) and, therefore, its conduction loss.

The other source of power loss is through switching losses. As the MOSFET switches on and off, its intrinsic parasitic capacitance stores and then dissipates energy during each switching transition. The losses are proportional to the switching frequency and the values of the parasitic capacitances. As the physical size of the MOSFET increases, its capacitance also increases; so, increasing MOSFET size also increases switching loss.

\$\endgroup\$
  • 1
    \$\begingroup\$ it would be nice, if the people who down vote also make a comment for the reason, so that the answer can be corrected fixing the error \$\endgroup\$ – Raj Feb 1 '17 at 6:14
  • \$\begingroup\$ I'm not the downvoter, but you should really attribute your quoted text so the origin is clear. You quoted from http://www.eetimes.com/document.asp?doc_id=1278970 I think. \$\endgroup\$ – Roger Rowland Feb 1 '17 at 7:12
  • \$\begingroup\$ Some people think that everybody must march to sounds of some anthem, so answer slightly different from their imaginary standards are downvoted. It's OK, don't pay attention. \$\endgroup\$ – Gregory Kornblum Feb 1 '17 at 8:23
  • \$\begingroup\$ I downvoted it before you edited it because it is misleading. Many mosfets can only carry the "full rated current" if the case is chilled to 25C which is practically impossible. The current rating is based on a junction temperature of something like 125C, while the case is 25C. \$\endgroup\$ – mkeith Feb 1 '17 at 17:08
0
\$\begingroup\$

Sometime long ago, it became the convention that power mosfet current ratings would be given with the case temperature at 25C. Maybe at one time this was reasonable. I don't know.

But it is practically impossible to keep a mosfet at 25C when it is dissipating, say, 100Watts.

I have noticed that some datasheets are now listing max current at an ambient temperature of 25C, which is more realistic. Bottom line is that you need to look in the datasheet and do a few calculations to determine whether a heatsink is needed and how large it should be.

In your application, most likely you can find a mosfet which can pass 4A without a heatsink. But do the calculations to confirm.

The failure mode would probably not involve much melting. Once the die gets too hot, it will not be a transistor anymore, and conduction may occur along undesired pathways, and eventually there will be smoke, and then that is the end of that.

\$\endgroup\$
0
\$\begingroup\$

Standard copper foil has thermal resistance of 70 degrees Centgrade per watt per square of foil.

Lets work with a 3cm by 3cm piece of 1-ounce-per-squarefoot of foil, and attach the MOSFET to the center. 1cm^2 is just about the right size footprint.

How many squares of foil are around that MOSFET? Eight, right? Thus the total thermal resistance is 70/8 = 9 degree Centigrade per watt.

Now grow the foil to 5cm by 5cm. How many squares surround the original 9? 5 +4 + 4 +3 = 16. The thermal resistance is 70/16 = 4.5 degree Centigrade per watt, for that extra ring of foil.

Our total Rth is 9 + 4.5 = 13.5 degree C per watt.

At some point, air flow will suffice.

Be cautious about getting the heat of the middle of the package and into the surrounding foil. Draw some tiny squares under the MOSFET footprint, and compute those thermal resistances also.

\$\endgroup\$
-1
\$\begingroup\$

Yes, they can handle the currents without melting. The trick is that they have extremely low "on" resistance, or "RDS(on)", in the low fractions of an Ohm. So, while the current is high, the voltage across the FET is very low. Heat is due to power dissipation, and power = current x voltage. Low voltage gets you low heat.

You need to transition these FETs between on and off very quickly, because they will heat up if they stay in an intermediate region (partially on) for more than a brief moment. Depending on the application, there will be other complications in designing a reliable circuit.

\$\endgroup\$

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