-1
\$\begingroup\$

What the circuit does is very simple: once the PIR motion sensor detects a motion, the OUT pin is high (3.3v) and it will put the MOSFET in an open state. The LED strip will be therefore on.

It's working like a charm but I think I'm missing some components to make a more robust circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Some information about the voltage source: it's a 220v AC to 12v (adjustable) DC converter. It seems that the LED strip is slightly blinking from time to time. I don't know if it's because of the converter or the MOSFET.

Any help for a newbie will be appreciated.

\$\endgroup\$
  • 4
    \$\begingroup\$ A note on style: Your schematic is upside down! It's customary to put ground at the bottom. \$\endgroup\$ – duskwuff Oct 12 '17 at 20:12
  • 4
    \$\begingroup\$ @duskwuff: It's customary to put negative at the bottom. GND can be on the supply positive. \$\endgroup\$ – Transistor Oct 12 '17 at 20:14
  • \$\begingroup\$ Nothing is jumping out at me. You might put 100Ω between M1 gate and OUT for complicated but possibly unnecessary reasons. Check your connections re: blinking \$\endgroup\$ – Daniel Oct 12 '17 at 20:15
  • \$\begingroup\$ Generally, though, open-ended questions like this are somewhat frowned upon for SE. You might get some downvotes, just a heads up. \$\endgroup\$ – Daniel Oct 12 '17 at 20:17
  • 1
    \$\begingroup\$ No, the battery is upside down. The acid will leak out! \$\endgroup\$ – winny Oct 12 '17 at 20:47
6
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Circuit redrawn "right-way up".

The circuit looks fine although, as pointed out in the comments, you might add a resistor between the out pin and the gate. The gate has some capacitance and the resistor prevents it acting as an instantaneous short when OUT changes state.

Note the schematic convention is highest voltages at the top and lowest at the bottom. Signal flow is generally from left to right as with western writing. (This doesn't always work out and the power in this case flows from right to left to keep the schematic "clean".)

Some older transistor circuits in particular used a common positive. In those circuits GND appeared at the top of the schematic.

Note also how easy it is to identify the ground points, without having to trace wires, when the symbols are used.

\$\endgroup\$
  • 2
    \$\begingroup\$ +1 for redraw. Depending on how that is wired up, some decoupling / bulk caps on the sensor may be prudent too. \$\endgroup\$ – Trevor_G Oct 12 '17 at 20:28
1
\$\begingroup\$

You ought to be OK. But be aware that your gate drive is marginal. If you look at the data sheet you'll see that the gate threshold voltage is 1 to 2 volts. Since this provides a drain current of 250 uA, a 50% margin for a 1 amp current is not healthy. A good rule of thumb is 3 times Vgs(th). I'd recommend a level shifter like

schematic

simulate this circuit – Schematic created using CircuitLab

should give adequate margin.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.