9
\$\begingroup\$

I want to drive a strip of LEDs from a microcontroller using PWM to control the brightness. The strip I have takes about 1.5A at 12V. I'm only familiar with purely low power digital electronics so wanted to check if these assumptions are correct and get any advice :-

  • If I use an NPN transistor to drive this, the transistor when turned on will drop about 0.7v so will dissipate over 1Watt when turned on.
  • This would require a reasonably chunky transistor and a heat sink which I want to avoid if possible.
  • So I'd be better using a mosfet that has much lower resistance so I might be able to get away with a smaller one and perhaps no heatsink?

  • However looking at the spec of the various MOSFETs I can buy it looks like any that can pass this amount of current require considerably more than 3.3v I can get from my microcontroller to turn on fully.

  • So am I best to have a small NPN transistor switching 12v to the input of a mosfet to control the actual LED strip? (Sorry I can't draw a diagram on this computer but can add one later if needed)

Are my assumptions correct, and does anyone have any advice or a better way? I'd also be interested in recommendations for suitable parts although that's not my main question.

(Edit: I looked for other posts that answered this and didn't find anything that was quite what I wanted, if someone has a link to a duplicate then please post it and I'll happily close the question).

\$\endgroup\$
8
\$\begingroup\$

For 1.5 A at 12 Volts, switched by 3.3 Volts, here is a MOSFET solution that would work well. The MOSFET suggested here is an IRLML2502 available from eBay and other sites for as little as $2.35 for 10 with free shipping.

schematic

simulate this circuit – Schematic created using CircuitLab

The IRLML2502 has a maximum on-resistance of 0.08 Ohms at 2.5 Volts gate voltage, and less as the gate voltage gets closer to 3.3 Volts. It can withstand 20 Volts Drain to Source, so it will work well with a 12 Volt supply. Drain-Source current rating is greater than 3 Amperes, providing over 100% margin of safety.

At 0.08 Ohms and 1.5 Amperes, the MOSFET will dissipate 180 milliWatt when fully on. Even allowing for the switching edges of the PWM, dissipation will not exceed 250 mW or so, hence no heat sink is required for this application.

Regarding the assumptions:

  • NPN transistor drop and dissipation are correct, give or take a bit due to Vce of specific transistors
  • Chunky transistor (BJT), not really, but a TO-220 size would be typical, and yes, a heat sink would be required
  • Yes, see suggested MOSFET above
  • Not correct, there are several low-cost MOSFETs that turn on solidly well below 3.3 Volts, and can easily pass 1.5 Amperes
  • No, with an NPN BJT there is always a balancing act around base current etc. MOSFETs being voltage driven devices, work with less fuss

Some of your assumptions are correct. This answer provides one better way, and I am sure there are others.

\$\endgroup\$
  • \$\begingroup\$ Thank you, this is very helpful and I'll look up the spec of that device, I failed to find anything like that myself so this is very helpful. \$\endgroup\$ – John Burton May 1 '13 at 12:24
  • \$\begingroup\$ The device specifications are in the datasheet linked in the answer above, happy to have been of help. \$\endgroup\$ – Anindo Ghosh May 1 '13 at 12:55
  • \$\begingroup\$ The IRLML2502 is a good suggestion, but your circuit is not. You can drive that FET with 3.3 V on the gate, but you don't want to go lower. Your R2 and R1 form a voltage divider that severly reduces the gate drive. in this case, replace R2 with a short and lose R1 altogether, basically drive the gate directly from a digital CMOS output. Put a 10 kOhm pulldown on the gate if you want to make sure it wakes up off. That way it won't interfere with normal operation. \$\endgroup\$ – Olin Lathrop May 1 '13 at 12:57
  • \$\begingroup\$ Thank you @OlinLathrop. I guess I want the pulldown for safety as it looks like half turning on the device by accident would make it overheat very quickly... \$\endgroup\$ – John Burton May 1 '13 at 13:02
  • 1
    \$\begingroup\$ @hamsolo474 The gate to ground junction of the MOSFET is almost an infinite resistance, in that the DC current through R2 will be negligible. Perhaps you are modeling the gate junction as a short circuit. \$\endgroup\$ – Anindo Ghosh Nov 19 '17 at 13:24
3
\$\begingroup\$

First thought is this circuit: -

enter image description here

The MCU will turn on or off the BC547 (virtually any NPN will do) and this will apply (or remove) 12V to the gate of the P channel FET. You'll need a P channel fet with low on resistance. 0.1 Rds(on) will dissipate less than 0.2W so that's a good point to start hunting for the FET.

If you are switching in the 100's of hertz then 10k gate-to-source is OK for the FET but if you are in the several kHz region a 1k value would be better.

Possibly IRLML5203 is a decent choice - it has 0.098 ohms Rds(on), 30Vmax, 3Amax and is SOT23

\$\endgroup\$
  • \$\begingroup\$ This is pretty much what I was thinking of. Thank you for the advice and diagram :) \$\endgroup\$ – John Burton May 1 '13 at 12:25

protected by Kortuk May 1 '13 at 18:29

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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