I'm trying to build a board capable of driving 12 V, single-channel, non-addressable strips with the WS2811 protocol for my project. In my case I want to be able to safely switch up to 5 A (but most times way less).

I've found two interesting boards online, this one with three channels and this one, like what I want, with just one channel.

They both share the same schematics, and as you can see use pretty similar designs. They use two MOSFETs to drive the output: the WS2811 IC drives a small RJU003N03 or NX7002AKVL, which then drives a beefier NTD5867NLT4G to switch the high-current flow for the LEDs. Both of those I can't find on AliExpress (which is where I usually look for cheap small batches of electronics for hobby projects).

On hand I have a bunch of IRF3708, which after tinkering with the datasheets I've come think could be driven directly from the IC skipping the smaller MOSFET, but I'm not sure. Do you think it's a feasible equivalent?

Schematic with IRF3708

Edit 1: looking at it more, shouldn't the 10 kΩ pull-up resistor be connected between OUTR and the 1 kΩ one? In the current configuration it acts as a voltage divider, which may not be ideal.

If it's not suitable, and I would need to keep the double MOSFET design, what components could I use? Since I will be soldering the components by hand, TO packages are preferred.

Also, what do you think about gate resistor values? One design uses 1 kΩ, the other 10 Ω. It's a big difference, what's more appropriate? 1 kΩ seems high, but 10 Ω too low to me. At 5 V, the current would be limited to 5 mA and 500 mA respectively. Given that the WS2811 is intended to be driving LEDs at a constant current of 18.5 mA, I'm a bit confused. 1 kΩ makes sense, if the current that it allows to flow is enough to make the gate switch fast enough for PWM. 10 Ω seems too low, but I'm guessing the small NX7002AKVL could be the limiting factor for appropriate values.

Edit 2: Updated schematic. Note: VCC is 5 V, drain can be 12 or 24 V (up to 30 V) final schematic

Edit 3: In the comments I mentioned the PWM frequency being 2 kHz, but after further research this may not be the case. From what I found online some sources state different values, but most seem to agree that for WS2811 chips the frequency is at least 400 Hz.

  • \$\begingroup\$ Revise your edit-2 – chip VDD goes to GND. Personally, if VCC is 12V and will be the only source to be chained then I'd place a local regulator for VDD (a resistor, zener and an MLCC will do) for each block, or for an alternative option/implementation check Fig.2 here (read its explanation as well). For edit-1, 1k or 10R doesn't matter as the gate will be shunted with a constant-current source. You don't have to place any resistor there. BTW, 10k can be decreased to 1k (if VCC is 12V) for better PWM signal as 10k will bring distorted PWM signal if the PWM frequency is 2 kHz. \$\endgroup\$ Jan 10 at 13:42
  • \$\begingroup\$ @RohatKılıç Fixed it, thanks for pointing it out. So I can safely remove the resistor between the output and the gate? Are there any advantages for having one? \$\endgroup\$
    – AleMonti
    Jan 10 at 13:53
  • \$\begingroup\$ A resistor basically limits the current but since the output is already current limited, there is no significant advantage for placing one. For the 12V-5V-VDD thing, I thought you'll chain the supply voltage for loads, sorry for the misunderstanding so just ignore what I said about it. \$\endgroup\$ Jan 10 at 13:57
  • \$\begingroup\$ No worries thanks a lot for your help! \$\endgroup\$
    – AleMonti
    Jan 10 at 14:05

1 Answer 1


There are two problems here:

  1. Supply: The max acceptable VDD for the chip is 7V and you plan to use the circuit with 12V. If you tie the VDD pin to input VCC, 12V, directly then it'll damage the chip. So you should consider modifying the circuit as shown in Fig.2 in the datasheet or use a separate, low-voltage regulator for VDD (5V is recommended for the sake of safety). – OP's edits reveal that VCC will be 5V i.e. 12V will not reach to the chips. So this item becomes invalid.

  2. Driving the MOSFET: I actually don't know if this is a problem. The thing is, the mark and space ratios will be swapped i.e. you'll get 70% duty-cycle if you send a code for 30%. Likewise, the output will be fully on when you send zero-brightness code. That's why there's an inverter (second MOSFET).

On hand I have a bunch of IRF3708, which after tinkering with the datasheets I've come think could be driven directly from the IC skipping the smaller MOSFET, but I'm not sure. Do you think it's a feasible equivalent?

Given that the WS2811 is intended to be driving LEDs at a constant current of 18.5 mA, I'm a bit confused.

See item 2 above for the main concern, inverted brightness code. And don't get confused about the "18.5mA constant drive" thing. The output configuration inside the chip should be something like this:


simulate this circuit – Schematic created using CircuitLab

So, as long as the current flowing through Q1 is lower than 18.5mA it should be fine.

When the INTERNAL_DRIVE_SIGNAL is high the output driver will be on and drawing current. When it's off the OUT-R will be floating. That's why there's a 10k resistor from VCC to the MOSFET's gate, to prevent it to be floating. When the OUT-R is on (drawing current) the MOSFET will be off because the gate will be shunted to the ground through the chip's internal driver. When the OUT-R is off (floating) the MOSFET will be turned on through 10k resistor from VCC.

From electrical point of view, there's no problem with using this MOSFET in the arrangement shown in your question as the voltages and current are suitable. But, as I stated above, the mark-space ratio will be inverted. If you don't want it to be inverted then go for two-MOSFET approach. The small MOSFET can be 2N7002, and the big one can be IRF3708.

NOTE: In the arrangement shown in the question, the 10k resistor and the MOSFET's input capacitance (around 2.4 nF) form a 6.6-kHz-low-pass filter but this could be a concern if the PWM frequency was high since the filter would chop off the sharp edges of the PWM signal and therefore bring some driving-related problems.

  • \$\begingroup\$ Regarding point 1, this is not an issue since I will be chaining these modules to other WS2812B strips from which I'll borrow the 5V supply and ground. I should have specified it in the original question so thanks for the heads up. As for point 2, the explanation for the inverted duty-cycle makes perfect sense. Now I see why two MOSFETs are needed. 2N7002 looks good. The PWM frequency of the outputs should be 2kHz if I'm not mistaken so those resistor values should be fine. \$\endgroup\$
    – AleMonti
    Jan 10 at 13:04

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