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With 3.3v at the gate & 12v at the drain, how many milliAmpere can i let flow trough this mosfet keeping it at a resonable temperature?

and

What is a resonable temperature ?

40,60,80 or more °C


I just salvaged 3 IRLL3303 n-type mosfets. And a friend asked me if i could make him a rgb led lamp (for free). I'm trying to figrue out how many amperes i could let flow trough those mosfets but also keeping a low temperature.

  1. i'm using a 3.3v mcu.
  2. the volatage for the led strips is 12v.
  3. everything should kept at a resonable temperature even in summer (max 37°C room temp).

I tried different calculations from various sites but i always get different values and most of the time i can't find all the variables needed in the datasheet.

I linked the datasheet but here is the important stuff:

RDS(on) 0.045Ω (VGS = 4.5V, ID = 2.3A) but i use 3.3v on the gate.

Power Dissipation (PCB Mount) 1.0 W

Linear Derating Factor (PCB Mount) 8.3 mW/°C

Junction and Storage Temperature Range -55 to + 150 °C

Junction-to-Amb. (PCB Mount, steady state) normal=93°C/W max=120°C/W

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From the data sheet Fig. 2, you can see a curve at different VGS values. Using the lowest curve (3.0 V), you can see it crosses 2 A at 0.2 V (it's a log-log scale). Therefore the on-resistance is 0.2/2 = 0.1 ohm. This is at +150 C, which is also the maximum allowed junction temperature (this is quite hot and would burn your finger if you touched it). Note that this is actually a typical plot, not a guaranteed worst case value, but in practice FETs are within 20 % of this value. Therefore let's assume 0.12 ohm.

If you don't have a real heatsink, it might be reasonable to assume a thermal resistance of 100 deg/W (with it mounted on a large sheet of metal, it might be 30 deg/W or less). Assuming an ambient of 60 deg (also very warm), and allowing up to 150 C junction, you can have a temperature rise of 150-60 = 90 deg. At 100 deg/W, this allows 90/100 = 0.9 W dissipation.

Using P = I^2*R, and 0.12 ohm gives a current of sqrt(0.9/0.12), or just over 2 A.

The 12 V LED strip value doesn't directly matter (as long as the FET's max VDS rating exceeds this which it does).

In cases like this, it's always better to be on the safe side and round calculations to the 'safe' side.

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  • \$\begingroup\$ So the mosfet works at 3.3v?@Spehro Pefhany says it won't turn on.And if i can dissipate 0.9W, 0.9W/12V=75mA or "just over 2A"?@Spehro Pefhany says 350mA. now i'm confused \$\endgroup\$ – cocco Nov 14 '15 at 21:02
  • \$\begingroup\$ The plotted curves are typical rather than worst-case. So you might get this but it's not guaranteed : worst case will be closer to Spehro's answer. If you're making one, just measure Rds yourself at 3.3v, and decide... \$\endgroup\$ – Brian Drummond Nov 14 '15 at 21:14
  • \$\begingroup\$ @Brian Drummond measure the ohm resistance with my cheap multimeter between drain and source while i apply i 3.3v high on gate? then i get some milliohm as output. then sqrt(.9/output) ? that gives me the amperes. this would be at 150°C in the wors case? \$\endgroup\$ – cocco Nov 14 '15 at 21:24
  • \$\begingroup\$ what about "MOSFET is not guaranteed to switch 'on'" like @Spehro Pefhany said. true or not? btw thx for the math explaination (+1) even if not everything is clear.350mA vs 2000mA is a big difference. \$\endgroup\$ – cocco Nov 14 '15 at 21:39
  • \$\begingroup\$ Yes, it is not 'guaranteed' to switch on (in fact in the data sheet, I can only find a minimum VTH (1.0 V) -- no upper limit is specified. In practice, I expect it would work just fine in this application at 3.3 V. I wouldn't go into high volume production with a FET that has data sheets this sloppy. \$\endgroup\$ – jp314 Nov 14 '15 at 23:45
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This particular MOSFET is not guaranteed to switch 'on' very well with 3.3V on the gate. Suggest if your time is free put a bit less than 3.3V on the gate and measure Rds(on) for each part. Multiply that number by 1.5 and use that number for the 'hot' Rds(on), since it increases with temperature. Or just buy some better logic-level gate MOSFETs.

You will also need to estimate the worst-case temperature inside your enclosure and pick a maximum junction temperature to operate at, and decide whether you are using the minimum recommended footprint or something better.

For example if using the minimum recommended footprint and you want maximum junction temperature of 120 degrees C with a maximum ambient of 70 degrees C, then you can allow 50 degrees C rise. That's about 400mW allowable dissipation so if you are hoping to get 700mA you must have a cold Rds(on) with (say) 3.1V Vgs of about 0.54 ohm. It has a chance at that level, and a better chance at 350mA.

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  • \$\begingroup\$ "put a bit less than 3.3V on the gate" with a resistor voltage divider (3v 2v)?? sorry but i'm not an electronical engineer, all this math is new and so some parts of your answer are not very clear for me. the max room temperature is 37°Celsius. I will use standard electronics enclosure (small, not ventilated).So your saying that 120°C is a resonable temperature?Where do i see in the datasheet that it won't turn on at 3.3v? \$\endgroup\$ – cocco Nov 14 '15 at 20:35
  • \$\begingroup\$ also if you could maybe explain the calculation you did so i don't have to ask everytime. \$\endgroup\$ – cocco Nov 14 '15 at 20:37
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    \$\begingroup\$ oh ok thx. now everything is clear... yeah ... i made a led lamp with npn transistors. then as the mcu has it's own voltage regulator i just let in a little more voltage. but thats in the 1-2v range. mosfets voltage drop should be much lower. i can increase the voltage when the circuit is finished. Another thing is that those cheap led strips are already desgned bad and so they are not nearly as bright as they should. .. \$\endgroup\$ – cocco Nov 19 '15 at 18:33
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    \$\begingroup\$ Vgs high is not a problem (unless you reach the maximum which is usually 6-20V). Even the 2.5V drive ones turn on a tiny bit better with more voltage. They don't draw any gate current to speak of, so it's all good. The MOSFET might cost a bit more though. \$\endgroup\$ – Spehro Pefhany Nov 19 '15 at 18:38
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    \$\begingroup\$ i'm like you when it comes to programming , it needs to be perfect. Electronics is new for me.. so even if i would , without an oscilloscope i can't create perfect circuits. So if it turns on i'm already happy... i just don't wanna burn down the house. thank you for your patience! ;) \$\endgroup\$ – cocco Nov 19 '15 at 18:52

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