I'm using a PT4115 LED driver to power a 3 W LED (700 mA) over a 1 meter cable. I've already had help here to use the same VIN for the LED driver and LED+ pin to power an MCU on the end of the cable, and it worked perfectly, as in the diagram below (I have only 4 wires inside the cable):

enter image description here

(Solution to my previous question )

Now I want to modify the circuit to PWM (dim) the LED not on the LED driver DIM PIN, which is 1 meter from the LED, (which will be set to 100% brightness), but with the MCU together with the LED at the end of the cable. So I chose an N-channel MOSFET with a low Rds (0.33 Ω at 4.5 V) for testing. It's the AO3400 MOSFET.

That extra wire from the cable will be used for other purposes, so I can't use it for dimming the LED with the driver DIM pin.

enter image description here

In my test circuit it works, but the MOSFET heats up a lot. Doing PWM, with whatever frequency I choose, or even using the MOSFET as an ON/OFF switch (to get full 700 mA current) it heats up a lot.

The negative pole of the LED is LED-, and not GND, so it goes to the LED driver and through another FET inside the driver. (set to allow 100% current, so no dimming in the FET inside the LED driver)

Is there a way to correct this excessive heat in the AO3400 mosfet? My space in the MCU/LED casing is limited.

Edit: I forgot to draw, but there´s a 10 kΩ resistor between the gate of the mosfet to GND, pulling it LOW. Also, this test circuit has the MCU pins at 3.3 V, not 5 V.

The MOSFET drawn in my schematics is an N-channel one, but I chose wrongly in TinyCAD and put a P-channel in the drawing. The correct MOSFET is an N-channel one, the AO3400.

  • \$\begingroup\$ On mobile these schematics are basically not intelligible. \$\endgroup\$
    – tobalt
    Commented Dec 18, 2022 at 15:55
  • \$\begingroup\$ There is a voltage drop between SW and GND across the internal FET of the PT4115 and I assume the AO3400N can not fully turn on with the provided gate voltage. \$\endgroup\$
    – Jens
    Commented Dec 18, 2022 at 16:33
  • 1
    \$\begingroup\$ Controlling a floating MOSFET looks adventurous. \$\endgroup\$
    – greybeard
    Commented Dec 18, 2022 at 17:06
  • \$\begingroup\$ @tobalt I used maybe an older version of Tinycad to draw the diagrams. The components when exported have their pins really .. thin... \$\endgroup\$
    – Rodrigo
    Commented Dec 18, 2022 at 17:10
  • 4
    \$\begingroup\$ This sounds very much like an XY problem. You would probably be better off using a completely different approach instead of trying to make this circuit work. What are you going to use the green wire for? With some luck, you could multiplex the dimming signal on that. \$\endgroup\$
    – TooTea
    Commented Dec 18, 2022 at 18:57

2 Answers 2


LEDs are current operated devices, therefore the LED driver is a current source. The FET is inserted as if the LED driver is a voltage source. BAD IDEA! By opening the circuit, the inductive kick causes the FET to breakover in turn causing the FET to get hot. The negative spike on the source may also cause a gate source break over as well. The chance of this circuit sustaining operation is very low. It may eventually take out the MCU and the LED driver.

To switch current an alternate, controlled, low resistance path must be provided around the LEDs. It must be designed to maintain the inductor current not interrupt it.

The P-MOSFET will divert the current around the LEDs turning them off when the FET is on and vice versa. The FET and R1 are both referred to CSN so that the FET will properly turn off.

The BJT Q1 is used to level shift the MCU signal to the FET gate. Choose R2 to provide adequate current for turn-on speed for the FET. Then choose R1 so that VGS does not exceed maximum rating.

PWM high turns the FET on, which means the LEDs off.


simulate this circuit – Schematic created using CircuitLab

Update: The problem with trying to insert a control point at the LED is that it is in the feedback loop for the constant current. The driver is expecting an LED load to be in place. The current is ramped up in the inductor based on the PWM duty cycle applied to DIM. I was hoping that the driver would function with a short to the rail as a true current source would within power limits. Based on comments from the OP, the driver cannot handle the modifications that I suggested. I now suggest to discontinue this suggestion and look for another. It is not clear to me why the PT4115 cannot be brought closer to the MCU.

Perhaps another will step up with a unique solution.

  • \$\begingroup\$ Hi! I tested the circuit using a BJT 2n2222 I had laying around a P Mosfet AO3401. For testing, I used for R1 and R2 10K resistors each. The circuit works, turn on and off normally, the P-FET doesn´t heat at all, but the Led driver heats more when the led is off than when the led is on. And when the LED is on, the power supply at 8.4V draws 350mA. When the LED is off, it still draws 100mA (and the inductor heats more than when it´s on) \$\endgroup\$
    – Rodrigo
    Commented Dec 18, 2022 at 23:17
  • \$\begingroup\$ I was afraid of that. The power supply current should be the same whether On or OFF. The voltage normally across the LEDs is being dropped by the LED driver heating it up and interfering with regulation. This complicates things. I don't have a solution. The driver was not intended to work this way. You may have to run extra wires for the added functionality that you need or as @graybeard says, "Put the driver with the Pro Micro". \$\endgroup\$
    – RussellH
    Commented Dec 18, 2022 at 23:46
  • \$\begingroup\$ Russel, in the led driver datasheet it says: "Switch Output. SW is the drain of the internal N-Ch MOSFET switch.". So, if inside the driver there´s an Nchannel Mosfet that turns on and off, what happens to this energy inside the inductor, when the led is off, and this mosfet is turned off? When using the driver the standard way, when the DIM pin is low, the current that flows is minimal. This is also in the datasheet: \$\endgroup\$
    – Rodrigo
    Commented Dec 19, 2022 at 1:10
  • 1
    \$\begingroup\$ "When a voltage applied to DIM falls below the threshold (0.3V nom.), the output switch is turned off. The internal regulator and voltage reference remain powered during shutdown to provide the reference for the shutdown circuit. Quiescent supply current during shutdown is nominally 95uA and switch leakage is below 5uA." \$\endgroup\$
    – Rodrigo
    Commented Dec 19, 2022 at 1:10
  • 1
    \$\begingroup\$ You can add some regular diodes in series with the MOSFET to keep the voltage above the minimum (but still low enough to turn off the LED). \$\endgroup\$
    – Oskar Skog
    Commented Dec 19, 2022 at 7:37

Use a high side switch. Or put the LED driver with the Pro Micro.

For all I can tell, you have an output pin close to 3.3 V to switch on a \$V_{TH}\$ 1.5 V MOSFET: \$V_S\$ will be below 1.8 V, \$V_D\$ around 5V with the PT4115's SW pulled low.
(I can't quite see what happens when the PT4115 turns off - pulling the gate to ground doesn't turn the MOSFET off because \$V_S\$ is driven below ground? Until the body diode inside the PT4115 conducts? Does that get hot, too?)

  • \$\begingroup\$ This end of the cable stays close to the body, it´s a headlamp. The led driver heats a little, so I want to keep it away with the battery. How would I wire a high side switch? The PT4115 and inductor heats a little, and normally (using the DIM pin for turning on/off/pwm) , when the led is off, it doesn´t heat at all \$\endgroup\$
    – Rodrigo
    Commented Dec 18, 2022 at 23:24
  • \$\begingroup\$ (Russel beat me in analysing what happens when a high side switch is turned off while there is energy in the inductor: Nothing good if it is a series switch. Manageable when shunting the LED current. I'd have said good to go until reading When the LED is off, […] the inductor heats more than when it´s on.) \$\endgroup\$
    – greybeard
    Commented Dec 18, 2022 at 23:37

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