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I've designed and built a circuit to PWM an LED strip from an ESP8266. I'm using the IRFZ44N for this as its \$V_{gs(th)}\$ is between 2V and 4V and has a low \$R_{DS(on)}\$. Everything seems to work, although the LED strip is very very faint; I would estimate somewhere around 5% of what I would expect.

Below is my schematic and screen captures from my oscilloscope. The reading at TP3 is where I believe the problem is visible, though I don't know why this is the case.

If I connect TP3 directly to ground I get the full brightness I'm expecting, so the strip works correctly.

One other possibly noteworthy point: The LED strip has a smattering of built in resistors (150\$\Omega\$, IIRC, but I don't know how that's all connected).

Researching other schematics show similar wiring, so I'm unsure where I've gone wrong. Any insight into what the issue may be would be greatly appreciated.

Schematic:

Schematic

TP1 (11.4V coming from the power supply; looks good to me):

TP1

TP4 (5.4V coming from the buck converter; looks good to me):

TP4

TP2 (3.68V PWM signal going to the gate of the MOSFET; looks good to me):

TP2

TP3 (I was expecting 11.4V max, 0V min here; doesn't look good to me):

TP3

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The max threshold voltage for your FET (the point where it passes 250 uA) is 4 V. It looks like your gate drive isn't even getting there. You need a low-threshold FET or a higher gate drive voltage.

enter image description here

The threshold voltage isn't when the FET is fully "on" it's the point where the FET just starts to conduct a small amount of current.

From the above spec you can see that the drain current will be 250 uA when the gate (to source) voltage is somewhere between 2 and 4 V (depending on the part and conditions) with the drain-source voltage equal to the gate-source voltage. You need to exceed the threshold voltage by a decent margin to get low RDSon.

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  • \$\begingroup\$ Thanks for the answer, John. I was under the impression that I only needed to exceed 2V to fully turn the MOSFET on and no more than 4V (as 4V is listed as the "Max")? I suppose I could test your theory by altering the output of the buck converter to around 4V and using that to drive the gate manually to see if that does indeed make a difference. Does that sound reasonable? \$\endgroup\$
    – OdinX
    Commented Dec 20, 2022 at 23:44
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    \$\begingroup\$ The threshold voltage is the point where the FET passes 250 uA of current (for this datasheet). 4V is the max voltage where you can expect that to happen. The RDSon of 0.024 ohm is specified for a Vgs of 10 V when you can consider it "fully on". This is not a logic-level part. \$\endgroup\$
    – John D
    Commented Dec 20, 2022 at 23:47
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    \$\begingroup\$ No, that's 250 uA of drain current (in the saturation region of operation, so more or less a current source). Some parts may require up to 4 V before the drain current reaches 250 uA, others may reach that well before 4 V, but you can't guarantee it. Answer edited for clarity. \$\endgroup\$
    – John D
    Commented Dec 20, 2022 at 23:59
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    \$\begingroup\$ Right ok, I've obviously got a lot more to learn! Thanks for the explanation and edits to your answer; makes sense. Looks like the IRLZ44N might have been more suitable. Thanks again for your help! \$\endgroup\$
    – OdinX
    Commented Dec 21, 2022 at 0:07
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    \$\begingroup\$ For future reference, you can read the graphs in the IRLZ44N datasheet you linked to see if it would do the job. Fig 1. shows that the transistor will happy conduct upwards of 10 A when you drive the gate with 3 V. How much it will actually conduct will depend on your load -- presumably, about 65 mA if the 150Ω series resistor is present. The IRLZ44N is a bit of overkill for this project -- you can probably get a BS107 or 2N7002 (or 2N2222 if you redesign around a BJT, or ZVN4525 if you're comfortable soldering SMD parts) to do the job for a lot cheaper. \$\endgroup\$
    – Matt S
    Commented Dec 21, 2022 at 1:35

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