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I have following circuit. Really simple. But I keep missing why I cant power LED strip (5V 1A). Falstad simulator says overflowing current. I am not very good at simple hardware thing. I often solving it via microcontroller which is not now suitable for solution small as possible.

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Figure 1. Original circuit.

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Figure 2. Update to match @Transistor's answer.

What do you suggest to have minimal working circuit?

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  • \$\begingroup\$ Please be careful with your edits that you leave enough of the original question there so that my answer doesn't look stupid. I've added back the original schematic and captioned both. \$\endgroup\$
    – Transistor
    Nov 10, 2019 at 21:11

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Add a base resistor of 100 Ω or so. Your present design overloads the PIR because it has no current limiting and the base-emitter junction behaves like a diode.

Your question is missing the current requirements of the LED strip. If you edit and add that in we can see if the transistor is adequately rated.


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Figure 1. The transistor should be capable of 2 A. You may need heatsinking.

Ah okay. LED strip requires 1A. Added 100 ohm resistor which limits LED strip to 2.5V. If I need to have 5V then I need 1 ohm resistor right? But then base draws 2.4A ?

No, read it again. Add the resistor to the base. 100 Ω will limit the current from the microcontroller to the base to about \$ \frac {3.3 - 0.6} {100} = 27 \ \text{mA} \$. This should be enough to drive the transistor into saturation and provide a very low voltage drop between the collector and the emitter applying almost 5 V to your LED strip.

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  • \$\begingroup\$ Ah okay. LED strip requires 1A. Added 100 ohm resistor which limits LED strip to 2.5V. If I need to have 5V then I need 1 ohm resistor right? But then base draws 2.4A ? \$\endgroup\$ Nov 10, 2019 at 20:49
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Nov 10, 2019 at 20:59
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    \$\begingroup\$ The 100Ω resistor goes in series with the base of the transistor; not in series with the LED string. \$\endgroup\$ Nov 10, 2019 at 21:00
  • \$\begingroup\$ @Transistor so get only 100 ohm and its working solution ok? Thanks \$\endgroup\$ Nov 10, 2019 at 21:03

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