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I'm currently working on a project where I need to be able to switch 7m of 9W/m LED strip (connected from both sides) using an Arduino with PWM. So the power of the led strip is about 63W at 12V. I have a plenty of IRF540N mosfets lying around but I need to calculate if I need a heatsink, or not. The intended circuit looks like this:

schematic

Note that I don't want the high current to flow on the tiny PCB traces, that's the purpose of the screw terminal.

I know that a logic level MOSFET would be better, but if I understand the datasheet correctly, IRF540N should be able to switch even much higher currents with 5V at the gate than I need. Or would using e.g. IRL540N make any difference?

Do I need a heatsink? If so, how to choose one properly? Is there anything else I should be worried of considering my circuit? Thanks in advance!

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  • \$\begingroup\$ Welcome to EE.SE! What does the datasheet say about RdsON at 5 V gate voltage? \$\endgroup\$
    – winny
    Sep 1 '18 at 21:38
  • \$\begingroup\$ Thanks! It says 44mΩ at 10V. I don't see anything about 5V. \$\endgroup\$
    – MSKL
    Sep 1 '18 at 22:34
  • \$\begingroup\$ Just want to comment that you are making a mistake by sticking with that FET. You can find one that doesn't need a heatsink. Something with just a few mOhms at 5V. \$\endgroup\$
    – mkeith
    Sep 2 '18 at 6:33
  • \$\begingroup\$ @mkeith Sure thing! I just wanted to teach him how to read the Id-Vds graph to come to his own conclusion that it’s unsuitable. \$\endgroup\$
    – winny
    Sep 2 '18 at 7:39
  • \$\begingroup\$ @mkeith So is there any general usage FET with low resistance? I found the IRL8113, which is weirdly dirt cheap. Would that be suitable? Or is there any well known general purpose N channel FET with low RdsON? \$\endgroup\$
    – MSKL
    Sep 2 '18 at 13:55
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The IRF540 can have a gate threshold voltage of up to 4V according to the datasheet, so it would probably not be able to supply enough current for the LEDs to run close to 12V.

Actually, taking a closer look at the datasheet for the IRF540, figure 3, it seems that it would be more than adequate of driving a 5A load at 5V gate drive. I would try this if i were you. The heatsink is still necessary though!

The IRL540 on the other hand is more suitable as it is a logic level MOSFET. According to the datasheet, the RDS(on) is around 77mΩ.

63W at 12V equals 5.25A, lets just say 5A as there will be a small voltage drop across the MOSFET, and that will lower the current in the LEDs.

The dissipated power in the MOSFET will then be:

\$ P_{dissipated} =RDS_{on} *I^2\ = 77mΩ * 5^2 = \textbf{1.9W}\$

According to the datasheet, the junction to ambient thermal resistance is 62 °C/W

This means the MOSFETs temperature will increase with

\$1.9W * 62 °C/W = \textbf{117°C} \$ from ambient temperature.

Assuming 25°C degrees ambient, that will result in 142°C. That is still in spec assuming ideal condition.. But you know, the world is not ideal..

EDIT: I forgot to take into account that the on-resistance has an significant increase with temperature, so you should definitely put a heatsink on it!

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  • \$\begingroup\$ The answer is vague but implies that OP should get a heatsink. You still haven't answered it black on white, crystal clear, if OP should acquire one, and if so, how he should choose one. \$\endgroup\$
    – Harry Svensson
    Sep 1 '18 at 22:20
  • \$\begingroup\$ @HarrySvensson: I've added a quick note \$\endgroup\$
    – Linkyyy
    Sep 1 '18 at 22:24
  • \$\begingroup\$ Thanks for a clear answer, I'll probably go and stick with the logic level FETs. However, I don't understand the first paragraph. I'm using 5V logic, which is clearly above the threshold. Why is it not able to supply enough current then? \$\endgroup\$
    – MSKL
    Sep 1 '18 at 22:30
  • \$\begingroup\$ @MatyášSkalický. You must use the IRL540 which is rated for 100V and turns ON full at 5 volts. Also include a small heatsink. Check out Mouser, Digi-key, Newark, Allied. If it is too hot to touch double the size of the heatsink and/or add a cooling fan. \$\endgroup\$
    – user105652
    Sep 1 '18 at 22:50
  • \$\begingroup\$ @MatyášSkalický: Driving the FET close to its threshold voltage means that i may not fully turn on, and result in a lot of loss (high on-resistance). It may not be a problem when its 'only' 5A of current, but you need to prototype it and make some measurements(Voltage across it, and current). If you just want it to work the first time, you use the logic level FET which has known data available(and specified) in the datasheet at 5V :) \$\endgroup\$
    – Linkyyy
    Sep 1 '18 at 22:54
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The simple answer is that you really do need a heat sink.

Take a look at figures 1 and 2 of the data sheet. At a 4.5 volt gate voltage, a typical Vds at 5 amps is about 0.6 volts for a case temperature of 25 C. 0.6 volts times 5 amps is 3 watts. As Linkyyy pointed out, the nominal maximum thermal resistance is 63 degrees/watt, so this suggests a 190 degree rise, for a nominal temperature of 215 degrees C. If you now check figure 2, you'll see that at 175 degrees, for the same Vgs, Vds is now about 0.8 volts, and power is now 4 watts. This is not a good trend. For that matter, figure 9 establishes that the absolute maximum junction temperature is 175 C, so you know you're looking at trouble.

And all of this is predicated on using typical values. Trust me on this, if there is one thing you should learn it is never, ever, to design a circuit using typical values for critical functions. Always, always, always use worst-case. Murphy's Law reigns supreme here.

Granted, since this circuit isn't failing by an order of magnitude, or anything like that, you probably don't need all that much of a heat sink to stay safe. But you do need one.

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    \$\begingroup\$ Instead of a heatsink, OP should get a logic-level MOSFET or a proper gate driver for their IRF540. Even with a heatsink the current design is marginal since the Arduino output will be somewhat less than 5.0V and the IRF540 on-resistance is not guaranteed at such low gate voltages (only typical values are plotted, not max/min over part-to-part variation and over temperature). Thermal runaway or "hotspotting" is a possibility when operating so close to the linear region. Switching losses haven't been considered either. \$\endgroup\$
    – pericynthion
    Sep 2 '18 at 0:07
  • \$\begingroup\$ Or use two in parallel. Half the current would improve things. \$\endgroup\$
    – Passerby
    Sep 2 '18 at 3:31
  • \$\begingroup\$ @pericynthion - a) I'm with you on gate voltage - which is why I used the 4.5 volt curves rather than the 5 volt curves, and b) did you miss my insistence that typical is bad in this case? \$\endgroup\$
    – WhatRoughBeast
    Sep 2 '18 at 3:48
  • \$\begingroup\$ @whatroughbeast I didn't miss that - I'm agreeing with and reinforcing the bulk of what you've written, but I disagree with the conclusion that a heatsink is appropriate. A heatsink (even a large one) may not be sufficient to prevent part failure. \$\endgroup\$
    – pericynthion
    Sep 2 '18 at 10:20
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The IRF540 switches on well when the gate-source is 10V. Some of them switch with less voltage but others simply get very hot and cause dim LEDs. A Mosfet has a high gate capacitance and an Arduino cannot charge and discharge it quickly which causes heat in the Mosfet if you have it switching at a high frequency.

The datasheet of a IRL540 show it heating with a maximum of 2.2W when driving your LEDs slowly so a little heatsink will be fine.

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  • \$\begingroup\$ You've identified new problems for OP (Original poster = Matyáš Skalický). This answer would be complete if you answered them as well. - Just giving some feedback. \$\endgroup\$
    – Harry Svensson
    Sep 1 '18 at 22:17

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