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I have an array of about 10x10 high-power LEDs driven by a ~200w LED driver.
I want to turn on/off the whole array using 3V3 logic with a single GPIO pin on some controller.
The 10x10 array consists of 10 parallel branches, each with 10 LEDs in series.

I connected a MOSFET in the following manner:

mosfet

where the MOSFET source goes to ground, the drain goes to the LED cathode in each of the 10 parallel branches, and the gate goes to a GPIO pin on a controller.

I can turn on the LED array by setting the gate to about 2.7 volts, but then it doesn't turn off when the gate goes back to 0 volts.
Further, the MOSFET gets so hot that the solder on the drain melts and the MOSFET gets disconnected (and then the LEDS turn off).
(Note: I'm currently using a bench power supply at the gate, for testing; not the GPIO pin yet.)

I'd now like to try a downscaled LED array, say 5x8, with a ~100W LED driver.

How can I accomplish the desired outcome (for the downscaled array)?
Is the answer to use a beefier MOSFET to control the entire array, or a more complicated circuit?

(The LED driver is something like this, but I'd like an answer that works independently of the driver brand. Ideally the answer would only depend on things like the driver voltage and current and LED array configuration.)

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  • \$\begingroup\$ Usually I'd say 2.7V isn't a high enough gate voltage, but according to this MOSFET's datasheet, it should be. Still try giving it a higher gate voltage anyway. Usually MOSFET gate voltage is "the higher, the better" - as long as it's under the limit - and typically a good voltage is more like 15 volts. This specific MOSFET could be subtly damaged if it got that hot - you bought a whole packet, right? \$\endgroup\$
    – user20574
    Commented Apr 15 at 15:42
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    \$\begingroup\$ Your LED driver may have an enable function or it might be worth looking into using one that does. At that power level, a relay might also be an attractive choice. \$\endgroup\$
    – vir
    Commented Apr 15 at 15:49
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    \$\begingroup\$ Note the rated RDSon for that FET is measured at VGS = 5 V. 15 V is absolute max, it would not be wise to attempt to run that as a normal gate voltage. 2.7 V VGS is inadequate, note the max VGS of 2.45 V is at a 1 mA drain current. \$\endgroup\$
    – Neil_UK
    Commented Apr 15 at 15:54
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    \$\begingroup\$ @Hari: MOSFETs work by creating a conductive channel between the source and drain. The channel is created by changing the voltage between the gate and (simplification) source terminal, Vgs. The larger the Vgs, the "stronger" the channel is. You have selected a logic level MOSFET, so good job there, but it's obviously still inadequate for the job. Also, please edit the configuration of the LED strings into the question. \$\endgroup\$
    – vir
    Commented Apr 15 at 16:35
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    \$\begingroup\$ it doesn't turn off when the gate goes back to 0 volts just disconnecting the bench supply from the gate won't turn off a LL power MOSFET any time soon: while operating without a proper driver, please add a resistor from base to source. I'd start with 10 kΩ. \$\endgroup\$
    – greybeard
    Commented Apr 15 at 17:03

1 Answer 1

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The chosen MOSFET is more than enough to accomplish the task, although it would be nice to know the driver voltage. There is no need for a "beefier" device.

The bench testing failure is due to mistreatment of the gate. A floating gate can easily be damaged by ESD. The gate voltage maximum is |15|V. Use a voltage dividing low pass filter to connect to the bench supply, possibly with transient suppression like a 10V Zener diode. At least a 10k to 100k resistor from gate to source will help. Use static dissipative techniques including wrist straps tp equalize charge differential between body-bench circuit.

To select a decent VGSon, use Figures 11 and 14 from the data sheet.

Figure 11 shows that at 15A drain current, 2.7V VGS should work but leaves little headroom. Any voltage above 3.0V is desirable. The higher VGS, the lower RDSon.

The Miller plateau in Figure 14 indicates a VGS of 3V will easily turn the device on.

This FET has a very high input capacitance (Ciss in the datasheet). Use a resistor in series with the gate to limit the drive current. Use diodes to protect the micro.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

enter image description here

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  • \$\begingroup\$ Thank you. If I wanted to control each branch with a separate MOSFET (so that, for instance, you end up with 10 MOSFETs, one for each branch), how then would you all control all 10 MOSFETs with a single extra MOSFET? Pointing out a source that talks about this would also help. \$\endgroup\$ Commented Apr 16 at 3:52
  • \$\begingroup\$ Also how many Ohms should R1 be? \$\endgroup\$ Commented Apr 16 at 17:31
  • \$\begingroup\$ Shouldn't the 1N4148 diode that protects the microcontroller be oriented in the opposite direction, to prevent current from flowing into the microcontroller GPIO pin? \$\endgroup\$ Commented Apr 17 at 5:39
  • \$\begingroup\$ The label that says "to micro", that does that mean? \$\endgroup\$ Commented Apr 17 at 5:40
  • \$\begingroup\$ To micro means connect to the microprocessor GPIO. \$\endgroup\$
    – user319836
    Commented Apr 17 at 6:32

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