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The maximum current my transistors are designed for is 1.5 A (from collector to emitter).

I am looking to use a Raspberry Pi GPIO pin to switch a white LED strip on/off, by placing a transistor in the place of the ground wire of the LED strip.

The LED strip has a maximum current of 2.4 A. Can I use two transistors in parallel to supply the required current?

Below I've attached a rough schematic of what I mean, here I've coloured the wires going into the transistors' collectors as green and emitters as orange to distinguish between the two.

Secondly, is this safe?

enter image description here

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    \$\begingroup\$ It depends. Not BJTs, without some method of making them share evenly. Why are you drawing your circuit upside down? \$\endgroup\$
    – Hearth
    Jul 4, 2022 at 17:41
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    \$\begingroup\$ ˙˙˙sʇɹnɥ pɐǝɥ ʎW \$\endgroup\$
    – TypeIA
    Jul 4, 2022 at 17:43
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    \$\begingroup\$ It depends on the transistor. Some transistors reduce their voltage drop as they get hotter while others do the opposite. This makes the difference between sharing the current versus one transistor hogging all the current. \$\endgroup\$
    – DKNguyen
    Jul 4, 2022 at 17:46
  • \$\begingroup\$ Apologies for drawing current like that, I'm relatively new to electrical engineering. Is there a way to determine whether it's possible with my transistor? \$\endgroup\$ Jul 4, 2022 at 18:06
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    \$\begingroup\$ You can also consider splitting the led strip in two, and drive the shorter strips with a single transistor each \$\endgroup\$
    – Pelle
    Jul 5, 2022 at 11:03

2 Answers 2

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Your schematic is hard to follow, normally it is top down and left to right. Your + should be on the top and your ground at the bottom. Yes you can parallel transistors but you need to add low ohm emitter resistors so they balance. You will lose between 0.7V to 1.4V depending on the transistor. Consider using a logic level N-Channel MOSFET, you will only need one and it will be much cooler. When you pick the MOSFET try to find something with a Vgs at least 2.5V or lower. With that I would put a 10K resistor from the GPIO pin to ground and something around 50 Ohm between the gate and GPIO pin. Be sure all grounds are connected. Source would be ground, Drain would be the output and the Gate is the input.

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  • \$\begingroup\$ Thanks for your response. Yeh I kinda realised this after posting sorry (my only experience with electronics is in physics experiments). After doing a bit of research, using a MOSFET does definitely seem like the sensible option! \$\endgroup\$ Jul 4, 2022 at 18:59
  • \$\begingroup\$ Why would I need a connection between the GPIO pin and ground (via the GPIO pin)? Surely you would only need a connection between the GPIO pin and the gate (via a resistor) \$\endgroup\$ Jul 5, 2022 at 12:51
  • \$\begingroup\$ @AdrienAmour: Perhaps as a pull-down to avoid unpredictable behaviour when the pin is configured as an input, with a floating gate. (Worst case being partially-on for an extended period due to a software bug leaving the pin high-impedance, resulting in much high power dissipation in the MOSFET than you specced it for, given the planned switch usage, always full-on or full-off with relatively quick transitions through partially conducting.) \$\endgroup\$ Jul 5, 2022 at 13:21
  • \$\begingroup\$ In my answer I stated " I would put a 10K resistor from the GPIO pin to ground" That does exactly what you are trying but does not become a voltage divider reducing the gate voltage. Putting the resistor on the gate in place of the GPIO forms the divider. \$\endgroup\$
    – Gil
    Jul 5, 2022 at 17:49
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Yes it can be made safe but you need to separate base resistors that will supply a total of 5% of the LEDs for a current ratio of 20 or less.

Because the current gain of the transistors reduces down to 20 or 10% of the maximum and also because the maximum current rating also implies the maximum temperature rise of 100° so consider using three of those cheap transistors in parallel with three resistors that adds up to 15 or 20% of the LED current.

The reason you cannot share the base resistor is because of the negative temperature coefficient of the base emitter and this results in current hogging when it gets hot and thermal runaway.

If it were an emitter followers then you would have a very low resistance on each emitter, but these are common emitters so the bases are isolated by Rb’s.

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