# PIR sensor 3.3V output driving a mosfet loaded with 24V power LED

I'm trying to build a cheap motion activated LED light system.

Circuit is: 24V power --> LED --> LED --> resistor --> (drain) mosfet (source) --> GND

LED are 10V 10W on aluminium heat sink, resistor is 7.5ohm 5W

Flowing current is about 0.55A. I choosed to drive the LED at half its nominal current because I noticed that the light level does not increase proportionally to the current. Two LEDs at 0.5A emitt far more light than one LED at 1A, and heat less.

Due to the 3.3V output level of the PIR I choosed a FQP30N06L mosfet which should have a 2.5V max gate threshold voltage.

PIR sensor is powered with 5V, the GNDs of the 5V and 24V power suppliers are connected togheter. Mosfet gate is connected to the output of the PIR sensor which can be 0V or 3.3V. The system is working but the mosfet seems not to be fully on, the LED seems to be at 10%. When manually activated, switching mosfet gate to GND or +5V, the circuits works great, LED is fully ON or OFF, so I think the problem is related to the 3.3V output level of the sensor. I guess when powering with 24V, max gate threshold is not 2.5V anymore but raises, so 5V will work but 3.3V is not enough.

Whay would you do? Should I search for another mosfet which will work in this conditions, or I've better build or buy a gate driver circuit? TIA

• Figure 2 on the datasheet shows how much current you can source at various gate voltages. Although the FET is on at 2.5V, you're right it's not fully on at 3.3V. Aug 2, 2017 at 10:12
• Bear in mind that the threshold voltage is specified for a drain current of 250uA. Personally speaking I'd do this with a bipolar NPN rather than a MOSFET, I'm sure your PIR will have no problem supplying the base current. Aug 2, 2017 at 10:15
• Please use the schematic editor next time! Aug 2, 2017 at 14:15
• @RogerRowland you're right, I should have read more carefully the datasheet, it's clear that with 3.3V the mosfet isn't fully on. Aug 3, 2017 at 17:29
• @Finbarr I learn now that the treshold voltage is for only 250uA drain current. I'll keep it in mind next time I've to choose a component, I will look carefully at the Vgs-current diagrams. PIR won't have problems outputting the current needed by a BJT, but I prefer to use mosfets since I've already quite a lot of them. Aug 3, 2017 at 17:36

## 2 Answers

You can easily increase the gate voltage by 1.2 .. 1.4 V using a pull-up resistor and a couple of diodes:

This will toggle gate voltage between 1.2 and 4.5V instead of 0 and 3.3V. There may be a small current flowing into the PIR when it drives its output high, which is undesired but which the PIR will most probably cope with.

A more elaborate solution using an extra BJT would be:

simulate this circuit – Schematic created using CircuitLab

This will drive the gate between 0 and 5V, as you did manually, so it should work.

• Interesting. With your second solution, since I have to step the +5V down from the +24 I could instead go to +9V and have 0V or +9V at the gate, just to be sure the mosfet is fully on. Same for +12V. If I understand how it works, I just have to calculate R4 and R3 in order to have BJT base at about +3.3V, the PIR ON level. Using +9V and switching them will give the base a +3.0V. Or R3=12K and R4=22K should give 3.18V. PIR output is always 3.2V - 3.3V, even if powered at +9V or +12V. Aug 3, 2017 at 17:56

Thank you all for the suggestions, I learned some useful things. I solved it in a different way. I've found cheap PIR sensors working from 5V to 24V, the output is the same as the input and up to 5A. So I will do like this, using less components, less work, mosfet surely fully on:

simulate this circuit – Schematic created using CircuitLab

I don't use the power output of the PIR, I need only the signal because I want my system to be able to switch between: manual ON - automatic (PIR) - manual OFF. (sorry for the poor 3 way switch diagram, I didn't find one in the editor) Bye.

• If this is the right answer you can accept it - even though you wrote it yourself. Aug 11, 2017 at 12:38