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I was originally planning to drive my 12V RGB LED strips (5m + 1m) from an Arduino Uno. It's not an addressable strip so I'll use a mosfet per channel. Since I am now looking to connect it to my Domoticz, I want to drive it from a NodeMCU (ESP8266) which only has 3.3V on the GPIOs. Will this influence the brightness of my strip?

I've got some IRLZ44N N-channel mosfets laying around that should work with the 3.3V GPIOs (Vgs(th) = 2V, if I recall correctly.)

EDIT: forgot to mention that I'll be drawing about 2.5A per channel (i.e. 2.5A per MOSFET) for the complete 6 meters.

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  • \$\begingroup\$ Vcc must be 3x Vgs(th) to achieve low RdsOn so U need 1V types, those are good for 4~5V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 19 '17 at 20:50
  • \$\begingroup\$ It would, but it may not be significant. Easiest thing would be to test it and see how it looks. \$\endgroup\$ – Passerby Nov 19 '17 at 21:46
  • \$\begingroup\$ See my added comments to your IRF520 question under @Spehro's answer. \$\endgroup\$ – Jim Fischer Nov 20 '17 at 22:21
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They probably will, in my opinion, but there is no guaranteed with such a MOSFET. The key number is the guarantee of Rds(on) at the lowest gate voltage shown in the "Specifications" section. Vgs(th) is the guarantee that the MOSFET will be almost entirely off (250uA). There is a region between Vgs(th) and your operating point and you want the voltage across the MOSFET to be small or it may get too hot and your LEDs may not be at full brightness.

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In this case, the Rds(on) is guaranteed to be less than 0.039 ohms at 25 degees C Tj with 4.0V Vgs. The transfer characteristics curve is only a typical characteristic and is not guaranteed so it is bad engineering to use it for your particular purpose. It may be appropriate for a hobby project to give it a try, of course.

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    \$\begingroup\$ FWIW, I concur with @Spehro's comments. Fig. 6 in the linked datasheet for the IRLZ44 part shows the typical gate charge vs. gate-to-source voltage graph. The flat (horizontal) part of the curve is called the "Miller plateau". The MOSFET is fully ON (triode operation) when the gate charge reaches the right-hand knee on that graph, which for the IRLZ44 occurs at around 4 Volts. With this in mind, when I need to drive a MOSFET with a logic signal, I start by selecting a MOSFET part whose Miller plateau occurs somewhere between VOL and VOH for the logic family I'm using. \$\endgroup\$ – Jim Fischer Nov 19 '17 at 20:26
  • \$\begingroup\$ So, with the 3.3V logic I won't be able to switch the MOSFET fully on (?) meaning that my LED strip won't be able to reach all colours. Or am I missing the point completely here? \$\endgroup\$ – user2693053 Nov 19 '17 at 21:25
  • \$\begingroup\$ 3.3V is not enough to be guaranteed to work well. It might work well enough or it might not, depending on the particular units you happen to have. So it's closer to tinkering than electrical engineering. \$\endgroup\$ – Spehro Pefhany Nov 20 '17 at 3:45
  • \$\begingroup\$ @JimFischer Thanks, that comment was really helpful. Although I am not exactly sure I understand. There are plenty of so called Mosfet driver modules available that work with an IRF520. If I look up that datasheet and try to find the 'Miller Plateau' they shouldn't be selling that for 3.3V logic, right? This module (more or less) hooks up to the 3.3V ports directly. \$\endgroup\$ – user2693053 Nov 20 '17 at 7:15
  • \$\begingroup\$ I won't go on record as saying that product shouldn't be used. That said, I personally wouldn't use the IRF520 in a switching circuit where the IRF520's gate is driven directly from a logic HIGH output voltage from a 3V3 logic system, where VOH=2.4V <= V_HIGH <= 3.3V. In fact, I'd contact the manufacturer of any "3V3 MOSFET driver" module based on the IRF520 and ask them for test data that proves the IRF520 is in fact driven fully into triode(linear) mode for gate-source voltages in the range 2.4V <= V_GS <= 3.3V, and for which values of drain current ID and drain-source voltage VDS. \$\endgroup\$ – Jim Fischer Nov 20 '17 at 22:00
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Try to find the datasheet for your IRLZ44N mosfet. From there you should look at the transfer characteristics shown in the Vgs-Id diagram. If you look at that diagram than you can see the Vgs required for the drain current needed. The drain current is the current needed by your RGB leds

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That would depend on how much current your LEDs take. Since you failed to mention that it's pretty hard to answer this question.

The graphs on the data sheet indicate the currents you can expect to drive at specific gate voltages. At 3.3V the current capability is about 20% of a five volt gate.

Whether that is enough, and whether your device will need extra heat-sinking in that state is what you need to calculate. I'd guess you will be ok if your current requirement is not huge. At2.5A it will be fine. Mosfet will be a little warmer than optimum, but ok.

However, since you appear to have 12V available, you might want to consider augmenting your circuit to have the GPIO switch that to drive the gate, while keeping the gate under the Vgs max (10V).

Or just buy a more appropriate MOSFET.

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    \$\begingroup\$ Driving with a higher voltage is appropriate ....but 12 V exceeds the maximum VGS voltage so may destroy the FET device. 3.3 V drive should work just fine out to at least 6-8 A. \$\endgroup\$ – Jack Creasey Nov 19 '17 at 19:25
  • \$\begingroup\$ @JackCreasey good point.. And yes, I think it's probably fine but without him telling us the load current.... anyhow, I edited the answer to mention the Vgs max, thanks. \$\endgroup\$ – Trevor_G Nov 19 '17 at 19:35
  • \$\begingroup\$ Excuse me for leaving out the current draw of my strips, I've edited my question and added the information. Since my LED strips will draw roughly 2.5A I don't see much of a problem there. Since I'm going to feed the NodeMCU 5V, I could use that to switch the MOSFETS. But it's not the ideal option. Once the LED strips get here I'll try the MOSFETS I currently have and buy some others if this doesn't work. \$\endgroup\$ – user2693053 Nov 19 '17 at 21:34

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